From 59332e34e4650b0b60a86024848dc8c69f9f0bdd Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Maximilian=20Ke=C3=9Fler?= Date: Wed, 16 Feb 2022 01:07:11 +0100 Subject: [PATCH] integrate into template (for sure broken for now) --- 2021_Algebra_I.tex | 2926 +++++++++++++++++++++++++++++++++++++++++ algebra.sty | 89 ++ summary/algebra.pdf | Bin 933934 -> 0 bytes summary/algebra.tex | 3019 ------------------------------------------- 4 files changed, 3015 insertions(+), 3019 deletions(-) delete mode 100644 summary/algebra.pdf delete mode 100644 summary/algebra.tex diff --git a/2021_Algebra_I.tex b/2021_Algebra_I.tex index d98ea36..74c619c 100644 --- a/2021_Algebra_I.tex +++ b/2021_Algebra_I.tex @@ -15,4 +15,2930 @@ \cleardoublepage +\begin{warning} + This is not an official script! + This document was written in preparation for the oral exam. It mostly follows the way \textsc{Prof. Franke} presented the material in his lecture rather closely. + There are probably errors. +\end{warning} + +\noindent The \LaTeX template by \textsc{Maximilian Kessler} is published under the MIT-License and can be obtained from \url{https://github.com/kesslermaximilian/LatexPackages}. % TODO +\newline + +\noindent $\fk$ is {\color{red} always} an algebraically closed field and $\fk^n$ is equipped with the Zariski-topology. +Fields which are not assumed to be algebraically closed have been renamed (usually to $\fl$). +\pagebreak + + +\section{Finiteness conditions} + +\subsection{Finitely generated and Noetherian modules} + +\begin{definition}[Generated submodule] + Let $R$ be a ring, $M$ an $R$-module, $S \se M$. + Then the following sets coincide + \begin{enumerate} + \item $\left\{ \sum_{s \in S'} r_{s} \cdot s ~ |~ S \se S' \text{finite}, r_s \in R, \right\}$ + \item $\bigcap_{\substack{S \se N \se M\\N \text{submodule}}} N$ + \item The $\se$-smallest submodule of $M$ containing $S$ + \end{enumerate} + + This subset of $N \se M$ is called the \vocab[Module!Submodule]{submodule of $M $ generated by $S$}. If $N= M$ we say that \vocab[Module!generated by subset $S$]{$ M$ is generated by $S$}. + $M$ is finitely generated $:\iff \E S \se M$ finite such that $M$ is generated by $S$. +\end{definition} + +\begin{definition}[Noetherian $R$-module] + $M$ is a \vocab{Noetherian} $R$-module if the following equivalent conditions hold: + \begin{enumerate} + \item Every submodule $N \se M$ is finitely generated. + \item Every sequence $N_0 \subset N_1 \subset \ldots$ of submodules terminates + \item Every set $\fM \neq \emptyset$ of submodules of $M$ has a $\se$-largest element. + \end{enumerate} +\end{definition} +\begin{proposition}[Hilbert's Basissatz]\label{basissatz} + If $R$ is a Noetherian ring, then the polynomial rings $R[X_1,\ldots, X_n]$ in finitely many variables are Noetherian. +\end{proposition} +\subsubsection{Properties of finite generation and Noetherianness} + +\begin{fact}[Properties of Noetherian modules] + \begin{enumerate} + \item Every Noetherian module over an arbitrary ring is finitely generated. + \item If $R$ is a Noetherian ring, then an $R$-module is Noetherian iff it is finitely generated. + \item Every submodule of a Noetherian module is Noetherian. + \end{enumerate} +\end{fact} +\begin{proof} + + \begin{enumerate} + \item By definition, $M$ is a submodule of itself. Thus it is finitely generated. + \item Since $M$ is finitely generated, there exists a surjective homomorphism $R^n \to M$. As $R$ is Noetherian, $R^n$ is Noethrian as well. + \item trivial + \end{enumerate} +\end{proof} + +\begin{fact} + Let $M, M', M''$ be $R$-modules. + \begin{enumerate} + \item Suppose $M \xrightarrow{p} M''$ is surjective. If $M$ is finitely generated (resp. Noetherian), then so is $M''$. + \item Let $M' \xrightarrow{f} M \xrightarrow{p} M'' \to 0$ be exact. If $M'$ and $M ''$ are finitely generated (reps. Noetherian), so is $M$. + \end{enumerate} +\end{fact} +\begin{proof} + \begin{enumerate} + \item Consider a sequence $M_0'' \subset M_1'' \subset \ldots \subset M''$. Then $p\inv M_i''$ yields a strictly ascending sequence. + If $M$ is generated by $S, |S| < \omega$, then $M''$ is generated by $p(S)$. + \item Because of 1. we can replace $M'$ by $f(M')$ and assume $0 \to M' \xrightarrow{f} M \xrightarrow{p} M'' \to 0$ to be exact. The fact about finite generation follows from \einfalg. + + If $M', M''$ are Noetherian, $N \se M$ a submodule, then $N' \coloneqq f\inv(N)$ and $N''\coloneqq p(N)$ are finitely generated. Since $0 \to N' \to N \to N'' \to 0$ is exact, $N$ is finitely generated. + + \end{enumerate} +\end{proof} +\subsection{Ring extensions of finite type} + +\begin{definition}[$R$-algebra] + Let $R$ be a ring. An $R$-algebra $(A, \alpha)$ is a ring $A$ with a ring homomorphism $R \xrightarrow{\alpha} A$. + $\alpha$ will usually be omitted. In general $\alpha$ is not assumed to be injective.\\ + \\ + An $R$-subalgebra is a subring $\alpha(R) \se A' \se A$.\\ +A morphism of $R$-algebras $A \xrightarrow{f} \tilde{A}$ is a ring homomorphism with $\tilde{\alpha} = f \alpha$. +\end{definition} + +\begin{definition}[Generated (sub)algebra, algebra of finite type] + Let $(A, \alpha)$ be an $R$-algebra. + \begin{align} + \alpha: R[X_1,\ldots,X_m] &\longrightarrow A[X_1,\ldots,X_m] \\ + P = \sum_{\beta \in \N^m} p_\beta X^{\beta} &\longmapsto \sum_{\beta \in \N^m} \alpha(p_\beta) X^{\beta} + \end{align} + is a ring homomorphism. We will sometimes write $P(a_1,\ldots,a_m)$ instead of $(\alpha(P))(a_1,\ldots,a_m)$. + + Fix $a_1,\ldots,a_m \in A^m$. Then we get a ring homomorphism $R[X_1,\ldots,X_m] \to A$. The image of this ring homomorphism is the $R$-subalgebra of $A$ \vocab[Algebra!generated subalgebra]{generated by the $a_i$}. + $A$ is \vocab[Algebra!of finite type]{of finite type} if it can be generated by finitely many $a_i \in I$. + + For arbitrary $S \se A$ the subalgebra generated by $S$ is the intersection of all subalgebras containing $S$ \\ + $=$ the union of subalgebras generated by finite $S' \se S$\\ + $= $ the image of $R[X_s | s \in S]$ under $P \mapsto (\alpha(P))(S)$. + +\end{definition} +\subsection{Finite ring extensions} % LECTURE 2 +\begin{definition}[Finite ring extension] + Let $R$ be a ring and $A$ an $R$-algebra. $A$ is a module over itself and the ringhomomorphism $R \to A$ allows us to derive an $R$-module structure on $A$. + $A$ \vocab[Algebra!finite over]{is finite over} $R$ / the $R$-algebra $A$ is finite / $A / R$ is finite if $A$ is finitely generated as an $R$-module. +\end{definition} +\begin{fact}[Basic properties of finiteness] + \begin{enumerate}[A] + \item Every ring is finite over itself. + \item A field extension is finite as a ring extension iff it is finite as a field extension. + \item $A$ finite $\implies$ $A$ of finite type. + \item $A / R$ and $B / A$ finite $\implies$ $B / R$ finite. + \end{enumerate} +\end{fact} +\begin{proof} + \begin{enumerate}[A] + \item $1$ generates $R$ as a module + \item trivial + \item Let $A $ be generated by $a_1,\ldots,a_n$ as an $R$-module. Then $A$ is generated by $a_1,\ldots,a_n$ as an $R$-algebra. + \item Let $A$ be generated by $a_1,\ldots,a_m$ as an $R$-module and $B$ by $b_1,\ldots,b_n$ as an $A$-module. + For every $b$ there exist $\alpha_j \in A$ such that $b = \sum_{j=1}^{n} \alpha_j b_j$. We have $\alpha_j = \sum_{i=1}^{m} \rho_{ij} a_i$ for some $\rho_{ij} \in R$ thus + $b = \sum_{i=1}^{m} \sum_{j=1}^{n} \rho_{ij} a_i b_j$ and the $a_ib_j$ generate $B$ as an $R$-module. +\end{enumerate} + +\end{proof} + +\subsection{Determinants and Caley-Hamilton} %LECTURE 2 TODO: move to int. elements? +This generalizes some facts about matrices to matrices with elements from commutative rings with $1$. +\footnote{Most of this even works in commutative rings without $ 1$, since $1$ simply can be adjoined.} +\begin{definition}[Determinant] + Let $A = (a_{ij}) \Mat(n,n,R)$. We define the determinant by the Leibniz formula \[ + \det(A) \coloneqq \sum_{\pi \in S_n} \sgn(\pi) \prod_{i=1}^{n} a_{i, \pi(i)} + \] + + Define $\text{Adj}(A)$ by $\text{Adj}(A)^{T}_{ij} \coloneqq (-1)^{i+j} \cdot M_{ij}$, where $M_{ij}$ is the determinant of the matrix resulting from $A$ after deleting the $i^{\text{th}}$ row and the $j^{\text{th}}$ column. +\end{definition} +\begin{fact} + \begin{enumerate} + \item $\det(AB) = \det(A)\det(B)$ + \item Development along a row or column works. + \item Cramer's rule: $A \cdot \text{Adj}(A) = \text{Adj}(A) \cdot A = \det(A) \cdot \mathbf{1}_n$. $A$ is invertible iff $\det(A)$ is a unit. + \item Caley-Hamilton: If $P_A = \det(T \cdot \mathbf{1}_n - A)$ \footnote{$T \cdot \mathbf{1}_n -A \in \Mat(n,n,A[T])$}, then $P_A(A) = 0$. + \end{enumerate} + +\end{fact} +\begin{proof} + All rules hold for the image of a matrix under a ring homomorphism if they hold for the original matrix. The converse holds in the case of injective ring homomorphisms. + Caley-Hamilton was shown for algebraically closed fields in LA2 using the Jordan normal form. + Fields can be embedded into their algebraic closure, thus Caley-Hamilton holds for fields. Every domain can be embedded in its field of quotients $\implies$ Caley-Hamilton holds for domains. + + In general, $A$ is the image of $(X_{i,j})_{i,j = 1}^{n} \in \Mat(n,n,S)$ where $S \coloneqq \Z[X_{i,j} | 1 \le i, j \le n]$ (this is a domain) under the morphism $S \to A$ of evaluation defined by $X_{i,j} \mapsto a_{i,j}$. Thus Caley-Hamilton holds in general. +\end{proof} %TODO: lernen + +\subsection{Integral elements and integral ring extensions} %LECTURE 2 +\begin{proposition}[on integral elements]\label{propinte} + Let $A$ be an $R$-algebra, $a \in A$. Then the following are equivalent: + \begin{enumerate}[A] + \item $\E n \in \N, (r_i)_{i=0}^{n-1}, r_i \in R: a^n = \sum_{i=0}^{n-1} r_i a^i$ + \item There exists a subalgebra $B \se A$ finite over $R$ and containing $a$. + \end{enumerate} + If $a_1, \ldots, a_k \in A$ satisfy these conditions, there is a subalgebra of $A$ finite over $R$ and containing all $a_i$. +\end{proposition} +\begin{definition}\label{intclosure} + Elements that satisfy the conditions from \ref{propinte} are called \vocab{integral over} $R$. + $A / R$ is \vocab[Algebra!integral]{integral}, if all $a \in A$ are integral over $R$. + The set of elements of $A$ integral over $R$ is called the \vocab{integral closure} of $R$ in $A$. +\end{definition} +\begin{proof} + \hskip 10pt + \begin{enumerate} + {\color{gray} \item[B $\implies$ A] Let $a \in A$ such that there is a subalgebra $B \se A$ containing $a$ and finite over $R$. + Let $(b_i)_{i=1}^{n}$ generate $B$ as an $R$-module. + \begin{align} + q: R^n &\longrightarrow B \\ + (r_1,\ldots,r_n) &\longmapsto \sum_{i=1}^{n} r_i b_i + \end{align} + is surjective. Thus there are $\rho_{i} = \left( r_{i,j} \right)_{j=1}^n \in R^n$ such that $a b_i = q(\rho_i)$. Let $\fA$ be the matrix with the $\rho_i$ as columns. + Then for all $v \in R^n: q(\fA \cdot v) = a \cdot q(v)$. By induction it follows that $q(P(\fA) \cdot v) = P(a)q(v)$ for all $P \in R[T]$. Applying this to $P(T) = \det(T\cdot \mathbf{1}_n - \fA)$ and using Caley-Hamilton, we obtain $P(a) \cdot q(v) = 0$. $P$ is monic. Since $q$ is surjective, we find $v \in R^{n} : q(v) = 1$. Thus $P(a) = 0$ and $a$ satisfies A. + } + \item[B $\implies$ A] if $R$ is Noetherian.\footnote{This suffices in the exam.} + Let $a \in A$ satisfy B. Let $B$ be a subalgebra of $A$ containing $b$ and finite over $R$. Let $M_n \se B$ be the $R$-submodule generated by the $a^i$ with $0 \le i < n$. As a finitely generated module over the Noetherian ring $R$, $B$ is a Noetherian $R$-module. Thus the ascending sequence $M_n$ stabilizes at some step $d$ and $a^d \in M_d$. Thus there are $(r_i)_{i=0}^{d-1} \in R^d$ such that $a^d = \sum_{i=0}^{d-1} r_ia^i$. + \item[A $\implies$ B] Let $a = (a_i)_{i=1}^n$ where all $a_i$ satisfy A, i.e. $a_i^{d_i} = \sum_{j=0}^{d_i - 1} r_{i,j}a_i^j$ with $r_{i,j} \in R$. Let $B \se A$ be the sub-$R$-module generated by $a^\alpha = \prod_{i=1}^n a_i^{\alpha_i}$ with $0 \le \alpha_i < d_i$. + $B$ is closed under $a_1 \cdot $ since \[a_1a^{\alpha} = \begin{cases} + a^{(\alpha_1 + 1, \alpha')} &\text{if } \alpha = (\alpha_1, \alpha'), 0 \le \alpha_1 < d_1 - 1\\ + \sum_{j=0}^{d_1 - 1} r_{i_1,j} a^{(j, \alpha')} &\text{if } \alpha_1 = d_1 - 1 + \end{cases}\] + By symmetry, this hold for all $a_i$. By induction on $|\alpha| = \sum_{i=1}^{n} \alpha_i$, $B$ is invariant under $a^{\alpha}\cdot $. Since these generate $B$ as an $R$-module, $B$ is multiplicatively closed. Thus A holds. Furthermore we have shown the final assertion of the proposition. + \end{enumerate} +\end{proof} +\begin{corollary}\label{cintclosure} + \begin{enumerate} + \item[Q] Every finite $R$-algebra $A$ is integral. + \item[R] The integral closure of $R$ in $A$ is an $R$-subalgebra of $A$ + \item[S] If $A$ is an $R$-algebra, $B$ an $A$-algebra and $b \in B$ integral over $R$, then it is integral over $A$. + \item[T] If $A$ is an integral $R$-algebra and $B$ any $A$-algebra, $b \in B$ integral over $A$, then $b$ is integral over $R$. + \end{enumerate} +\end{corollary} +\begin{proof} + \begin{enumerate} + \item[Q] Put $ B = A $ in B. + \item[R] For every $r \in R$ $\alpha(r)$ is a solution to $T - r = 0$, hence integral over $R$. + From B it follows, that the integral closure is closed under ring operations. + \item[S] trivial + \item[T] Let $b \in B$ such that $b^n = \sum_{i=0}^{n-1} a_ib^{i}$. Then there is a subalgebra $\tilde{A} \se A$ finite over $R$, such that all $a_i \in \tilde{A}$. + $b$ is integral over $\tilde{A} \implies \E \tilde{B} \se B$ finite over $\tilde{A}$ and $b \in \tilde{B}$. Since $\tilde{B} / \tilde{A} $ and $\tilde{A} / R$ are finite, $\tilde{B} / R$ is finite and $b$ satisfies B. + \end{enumerate} +\end{proof} + +\subsection{Finiteness, finite generation and integrality} %some more remarks on finiteness, finite generation and integrality + +\begin{fact}[Finite type and integral $\implies$ finite]\label{ftaiimplf} + If $A$ is an integral $R$-algebra of finite type, then it is a finite $R$-algebra. +\end{fact} +\begin{proof} + Let $A $ be generated by $\left( a_i \right) _{i=1}^{n}$ as an $R$- algebra. By the proposition on integral elements (\ref{propinte}), there is a finite $R$-algebra $B \se A$ such that all $a_i \in B$. + We have $B = A$, as $A$ is generated by the $a_i$ as an $R$-algebra. +\end{proof} +\begin{fact}[Finite type in tower] + If $A$ is an $R$-algebra of finite type and $B$ an $A$-algebra of finite type, then $B$ is an $R$-algebra of finite type. +\end{fact} +\begin{proof} + If $A / R$ is generated by $(a_i)_{i=1}^m$ and $B / A$ by $(b_j)_{j=1}^{n}$, then $B /R$ is generated by the $b_j$ and the images of the $a_i$ in $B$. +\end{proof} +{\color{red} + \begin{fact}[About integrality and fields] \label{fintaf} + Let $B$ be a domain integral over its subring $A$. Then $B$ is a field iff $A$ is a field. +\end{fact} +} +\begin{proof} + Let $B$ be a field and $a \in A \sm \{0\} $. Then $a\inv \in B$ is integral over $A$, hence $a^{-d} = \sum_{i=0}^{d-1} \alpha_i a^{-i}$ for some $\alpha_i \in A$. Multiplication by $a^{d-1}$ yields + $a\inv = \sum_{i=0}^{d-1} \alpha_i a^{d-1-i} \in A$. + + On the other hand, let $B$ be integral over the field $A$. Let $b \in B \sm \{0\}$. As $B$ is integral over $A$, there is a sub-$A$-algebra $\tilde{B} \se B, b \in \tilde{B}$ finitely generated as an $A$-module, i.e. a finite-dimensional $A$-vector space. Since $B$ is a domain, $\tilde{B} \xrightarrow{b\cdot } \tilde{B}$ is injective, hence surjective, thus $\E x \in \tilde{B} : b \cdot x \cdot 1$. +\end{proof} +\subsection{Noether normalization theorem} +\begin{lemma}\label{nntechlemma} + Let $S \se \N^n$ be finite. Then there exists $\vec k \in \N^n$ such that $k_1 =1$ and $w_{\vec k}(\alpha) \neq w_{\vec k}(\beta)$ for $\alpha \neq \beta \in S$, + where $w_{\vec k}(\alpha) = \sum_{i=1}^{n} k_i \alpha_i$. +\end{lemma} +\begin{proof} + Intuitive: + For $\alpha \neq \beta$ the equation $w_{(1, \vec \kappa)}(\alpha) = w_{(1, \vec \kappa)}(\beta)$ ($\kappa \in \R^{n-1}$) + defines a codimension $1$ affine hyperplane in $\R^{n-1}$. It is possible to choose $\kappa$ such that all $\kappa_i$ are $> \frac{1}{2}$ and with Euclidean distance $> \frac{\sqrt{n-1} }{2}$ from the union of these hyperplanes. By choosing the closest $\kappa'$ with integral coordinates, each coordinate will be disturbed by at most $\frac{1}{2}$, thus at Euclidean distance $\le \frac{\sqrt{n-1} }{2}$. + + More formally:\footnote{The intuitive version suffices in the exam.} + Define $M \coloneqq \max \{\alpha_i | \alpha \in S, 1 \le i \le n\} $. We can choose $k$ such that $k_i > (i-1) M k_{i-1}$. + Suppose $\alpha \neq \beta$. Let $i$ be the maximal index such that $\alpha_i \neq \beta_i$. Then the contributions of $\alpha_j$ (resp. $\beta_j$) with $1 \le j < i$ to $w_{\vec k}(\alpha)$ (resp. $w_{\vec k}(\beta)$) cannot undo the difference $k_i(\alpha_i - \beta_i)$. +\end{proof} + +\begin{theorem}[Noether normalization] \label{noenort} + Let $K$ be a field and $A$ a $K$-algebra of finite type. Then there are $a = (a_i)_{i=1}^{n} \in A$ which are algebraically independent over $K$, i.e. the ring homomorphism \begin{align} + \ev_a: K[X_1,\ldots,X_n] &\longrightarrow A \\ + P &\longmapsto P(a_1,\ldots,a_n) +\end{align} +is injective. $n$ and the $a_i$ can be chosen such that $A$ is finite over the image of $\ev_a$. +\end{theorem} +\begin{proof} + + Let $(a_i)_{i=1}^n$ be a minimal number of elements such that $A$ is integral over its $K$-subalgebra generated by $a_1, \ldots, a_n$. (Such $a_i$ exist, since $A$ is of finite type). + Let $\tilde{A}$ be the $K$-subalgebra generated by the $a_i$. + If suffices to show that the $a_i$ are algebraically independent. + Since $A$ is of finite type over $K$ and thus over $\tilde{A}$, by fact \ref{ftaiimplf} (integral and finite type $\implies$ finite) $A$ is finite over $\tilde{A}$. + Thus we only need to show that the $a_i$ are algebraically independent over $K$. + Assume there is $P \in K[X_1,\ldots,X_n] \sm \{0\} $ such that $P(a_1,\ldots,a_n) = 0$. Let $P = \sum_{\alpha \in \N^n} p_\alpha X^{\alpha}$ and $S = \{ \alpha \in \N^n | p_\alpha \neq 0\}$. For $\vec{k} = (k_i)_{i=1}^{n} \in \N^n$ and $\alpha \in \N^n$ we define $w_{\vec{k}}(\alpha) \coloneqq \sum_{i=1}^{n} k_i\alpha_i$. + + By \ref{nntechlemma} it is possible to choose $\vec{k} \in \N^n$ such that + $k_1 = 1$ and for $\alpha \neq \beta \in S$ we have $w_{\vec{k}}(\alpha) \neq w_{\vec{k}}(\beta)$. + + Define $b_i \coloneqq a_{i+1} - a^{k_{i+1}}_1$ for $1 \le i < n$. + \begin{claim} + $A$ is integral over the subalgebra $B$ generated by the $b_i$. + \end{claim} + \begin{subproof} + By the transitivity of integrality, it is sufficient to show that the $a_i$ are integral over $B$. + For $i > 1$ we have $a_i = b_{i-1} + a_1^{k_i}$. Thus it suffices to show this for $a_1$. + Define $Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots, b_{n-1} + T^{k_n}) \in B[T]$. + We have $0 = P(a_1,\ldots, a_n) = Q(a_1)$. Hence it suffices to show that the leading coefficient of $Q$ is a unit. + + We have + \[ + T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_i + 1})^{\alpha_{i+1}} = T^{w_{\vec k}(\alpha)} + \sum_{l = 0}^{w_{\vec k}(\alpha) - 1} \beta_{\alpha, l} T^l + \] + with suitable $\beta_{\alpha, l} \in B$. + + By the choice of $\vec k$, we have \[ + Q(T) = p_{\alpha} T^{w_{\vec k}(\alpha)} + \sum_{j=0}^{w_{\vec k}(\alpha) - 1} q_j T^j + \] + with $q_j \in B$ and $\alpha$ such that $w_{\vec k }(\alpha)$ is maximal subject to the condition $p_\alpha \neq 0$. + Thus the leading coefficient of $Q$ is a unit. + \end{subproof} + + This contradicts the minimality of $n$, as $B$ can be generated by $< n$ elements $b_i$. + +\end{proof} +\section{The Nullstellensatz and the Zariski topology} +\subsection{The Nullstellensatz} %LECTURE 1 +Let $\fk$ be a field, $R \coloneqq \fk[X_1,\ldots,X_n], I \se R$ an ideal. + +\begin{definition}[zero] + $x \in \fk^n$ is \vocab[Ideal!zero]{a zero of $I$} if $\A x \in I: P(x) = 0$. Let $\Va(I)$ denote the set of zeros if $I$ in $\fk^n$. + + The \vocab[Ideal!zero]{zero in a field extension $\mathfrak{i}$ of $\fk$} is defined similarly. +\end{definition} + +\begin{remark}[Set of zeros and generators] + Let $I$ be generated by $S$. Then $\{x \in R | \A s \in S: s(x) = 0\} = \Va(I)$. Thus zero sets of ideals correspond to solutions sets to systems of polynomial equations. + If $S, \tilde{S}$ generate the same ideal $I$ they have the same set of solutions. Therefore we only consider zero sets of ideals. +\end{remark} + +\begin{theorem}[Hilbert's Nullstellensatz (1)]\label{hns1} + If $\fk$ is algebraically closed and $I \subsetneq R$ a proper ideal, then $I$ has a zero in $\fk^n$. +\end{theorem} + +\begin{remark} + Will be shown later (see proof of \ref{hns1b}). + Trivial if $n = 1$: $R$ is a PID, thus $I = pR$ for some $p \in R$. Since $I \neq R$ $p = 0$ or $P$ is non-constant. $\fk$ algebraically closed $\leadsto$ there exists a zero of $p$.\\ + + If $\fk$ is not algebraically closed and $n > 0$, the theorem fails (consider $I = p(X_1) R$). +\end{remark} + +Equivalent\footnote{used in a vague sense here} formulation: + +\begin{theorem}[Hilbert's Nullstellensatz (2)] \label{hns2} + Let $L / K$ be an arbitrary field extension. Then $L / K$ is a finite field extension ($\dim_K L < \infty$) iff $L $ is a $K$-algebra of finite type. +\end{theorem} +\begin{proof} + \begin{itemize} + \item[$\implies$] If $(l_i)_{i=1}^{m}$ is a base of $L$ as a $K$-vector space, then $L$ is generated by the $l_i$ as a $K$-algebra. + \item[$\impliedby$ ] Apply the Noether normalization theorem (\ref{noenort}) to $A = L$. This yields an injective ring homomorphism $\ev_a: K[X_1,\ldots,X_n] \to A$ such that $A$ is finite over the image of $\ev_a$. + By the fact about integrality and fields (\ref{fintaf}), the isomorphic image of $\ev_a$ is a field. Thus $K[X_1,\ldots, X_n]$ is a field $\implies n = 0$. Thus $L / K$ is a finite ring extension, hence a finite field extension. + \end{itemize} +\end{proof} +\begin{remark} + We will see several additional proofs of this theorem. See \ref{hns2unc} and \ref{rfuncnft}. + All will be accepted in the exam. + + \ref{hns3} and \ref{hnsp} are closely related. +\end{remark} + +\begin{theorem}[Hilbert's Nullstellensatz (1b)] \label{hns1b} + Let $\fl$ be a field and $I \subset R = \fl[X_1,\ldots,X_m]$ a proper ideal. Then there are a finite field extension $\mathfrak{i}$ of $\fl$ and a zero of $I$ in $\mathfrak{i}^m$. +\end{theorem} + +\begin{proof} (HNS2 (\ref{hns2}) $\implies$ HNS1b (\ref{hns1b})) + $I \se \fm$ for some maximal ideal. $R / \fm$ is a field, since $\fm$ is maximal. + $R / \fm$ is of finite type, since the images of the $X_i$ generate it as a $\fl$-algebra. + There are thus a field extension $\fri / \fl$ and an isomorphism $R / \fm \xrightarrow{\iota} \fri$ of $\fl$-algebras. + By HNS2 (\ref{hns2}), $\fri / \fl$ is a finite field extension. + Let $x_i \coloneqq \iota (X_i \mod \fm)$. + \[ + P(x_1,\ldots,x_m) = \iota(P \mod \fm) + \] + Both sides are morphisms $R \to \fri$ of $\fl$-algebras. For for $P = X_i$ the equality is trivial. It follows in general, since the $X_i$ generate $R$ as a $\fl$-algebra. + + Thus $(x_1,\ldots,x_m)$ is a zero of $I$ (since $P \mod \fm = 0$ for $P \in I \se \fm$). + HNS1 (\ref{hns1}) can easily be derived from HNS1b. +\end{proof} + +\subsubsection{Nullstellensatz for uncountable fields} % from lecture 5 Yet another proof of the Nullstellensatz +The following proof of the Nullstellensatz only works for uncountable fields, but will be accepted in the exam. + + +\begin{lemma}\label{dimrfunc} + If $K$ is an uncountable field, then $\dim_K K(T)$ is uncountable. +\end{lemma} +\begin{proof} + We will show, that $S \coloneqq \left\{ \frac{1}{T - \kappa} | \kappa \in K\right\} $ is $K$-linearly independent. It follows that $\dim_K K(T) \ge \#S > \aleph_0$. + + Suppose $(x_{\kappa})_{\kappa \in K}$ is a selection of coefficients from $K$ such that $I \coloneqq \{\kappa \in K | x_{\kappa} \neq 0\} $ is finite and + \[ + g \coloneqq \sum_{\kappa \in K} \frac{x_\kappa}{T-\kappa} = 0 + \] + Let $d \coloneqq \prod_{\kappa \in I} (T - \kappa) $. Then for $\lambda \in I$ we have + \[ + 0 = (dg)(\lambda) = x_\lambda \prod_{\kappa \in I \sm \{\lambda\} } (\lambda - \kappa) + \] + This is a contradiction as $x_\lambda \neq 0$. +\end{proof} + +\begin{theorem}[Hilbert's Nullstellensatz for uncountable fields]\label{hns2unc} + If $K$ is an uncountable field and $L / K$ a field extension and $L$ of finite type as a $K$-algebra, then this field extension is finite. +\end{theorem} +\begin{proof} + If $(x_i)_{i=1}^{n}$ generate $L$ as an $K$-algebra, then the countably many monomials $x^{\alpha} = \prod_{i = 1}^{n} x_i^{\alpha_i} $ in the $x_i$ with $\alpha \in \N^n$ generate $L$ as a $K$-vector space. + Thus $\dim_K L \le \aleph_0$ and the same holds for any intermediate field $K \se M \se L$ . If $l \in L$ is transcendent over $K$ and $M = K(l)$, then $M \cong K(T)$ has uncountable dimension by \ref{dimrfunc}. Thus $L / K$ is algebraic, hence integral, hence finite (\ref{ftaiimplf}). +\end{proof} + +\subsection{The Zariski topology} +\subsubsection{Operations on ideals and \texorpdfstring{$\Va\left( I \right) $}{V(I)}} +Let $R$ be a ring and $I,J, I_\lambda \se R$ ideals, $\lambda \in \Lambda$. +\begin{definition}[Radical, product and sum of ideals] + \[ + \sqrt{I} \coloneqq \bigcap_{n=0} ^{\infty} \{ f \in R | f^n \in I\} + \] + \[ + I \cdot J \coloneqq \langle\{ i \cdot j | i \in I , j \in J\}\rangle_R + \] + + \[ + \sum_{\lambda \in \Lambda} I_\lambda \coloneqq \left\{\sum_{\lambda \in \Lambda'} i_\lambda | \Lambda' \se \Lambda \text{ finite}\right\} + \] +\end{definition} +\begin{fact} + The radical is an ideal in $R$ and $\sqrt{\sqrt{I} } = \sqrt{I}$.\\ + $I \cdot J$ is an ideal.\\ + $\sum_{\lambda \in \Lambda} I_\lambda$ coincides with the ideal generated by $\bigcap_{\lambda \in \Lambda} I_\lambda$ in $R$.\\ + $\bigcap_{\lambda \in \Lambda} I_\lambda$ is an ideal. +\end{fact} + +Let $R = \fk[X_1,\ldots,X_n]$ where $\fk$ is an algebraically closed field. + +\begin{fact} \label{fvop} + Let $I, J, (I_{\lambda})_{\lambda \in \Lambda}$ be ideals in $R$. $\Lambda$ may be infinite. + \begin{enumerate}[A] + \item $\Va(I) = \Va(\sqrt{I})$ + \item $\sqrt{J} \se \sqrt{I} \implies \Va(I) \se \Va(J)$ + \item $\Va(R) = \emptyset, \Va(\{0\} =\fk^n$ + \item $\Va(I \cap J) = \Va(I\cdot J) = \Va(I) \cup \Va(J)$ + \item $\Va(\sum_{\lambda \in \Lambda} I_\lambda) = \bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$ + \end{enumerate} +\end{fact} +\begin{proof} + \begin{enumerate} + \item[A-C] trivial + \item[D] $I \cdot J \se I \cap J \se I$. Thus $\Va(I) \se \Va(I \cap J) \se \Va(I \cdot J)$. By symmetry we have $\Va(I) \cup \Va(J) \se \Va(I \cap J) \se \Va(I \cdot J)$. + Let $x \not\in \Va(I) \cup \Va(J)$. Then there are $f \in I, g \in J$ such that $f(x) \neq 0, g(x) \neq 0$ thus $(f \cdot g)(x) \neq 0 \implies x \not\in \Va(I\cdot J)$. + Therefore \[ + \Va(I) \cup \Va(J) \se \Va(I \cap J) \se \Va(I \cdot J) \se \Va(I) \cup \Va(J) + \] + \item[E] $I_\lambda \se \sum_{\lambda \in \Lambda} I_\lambda \implies \Va(\sum_{\lambda \in \Lambda} I_\lambda) \se \Va(I_\lambda)$. + Thus $\Va(\sum_{\lambda \in \Lambda} I_\lambda) \se \bigcap_{\lambda \in \Lambda} \Va(I_\lambda)$. + On the other hand if $f \in \sum_{\lambda \in \Lambda} I_\lambda$ we have $f = \sum_{\lambda \in \Lambda} f_\lambda$. Thus $f$ vanishes on $\bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$ and we have $\bigcap_{\lambda \in \Lambda} \Va(I_\lambda) \se \Va(\sum_{\lambda \in \Lambda} I_\lambda)$. + \end{enumerate} +\end{proof} +\begin{remark} + There is no similar way to describe $\Va(\bigcap_{\lambda \in \Lambda} I_\lambda)$ in terms of the $\Va(I_{\lambda})$ when $\Lambda$ is infinite. + For instance if $n = 1, I_k \coloneqq X_1^k R$ then $\bigcap_{k=0}^\infty I_k = \{0\} $ but $\bigcup_{k=0}^{\infty} \Va(I_k) = \{0\}$. +\end{remark} +\subsubsection{Definition of the Zariski topology} +Let $\fk$ be algebraically closed, $R = \fk[X_1,\ldots,X_n]$. +\begin{corollary} (of \ref{fvop}) + There is a topology on $\fk^n$ for which the set of closed sets coincides with the set $\fA$ of subsets of the form $\Va\left(I \right) $ for ideals $I \se R$. + This topology is called the \vocab{Zariski-Topology} +\end{corollary} + +\begin{example}\label{zariskinothd} + Let $n = 1$. Then $R$ is a PID. Hence every ideal is a principal ideal and the Zariski-closed subsets of $\fk$ are the subsets of the form $\Va(P)$ for $P \in R$. +As $\Va(0) = \fk$ and $\Va(P)$ finite for $P \neq 0$ and $\{x_1,\ldots,x_n\} = \Va(\prod_{i=1}^{n}(T-x_i))$ the Zariski-closed subsets of $\fk$ are $\fk$ and the finite subsets. + Because $\fk$ is infinite, this topology is not Hausdorff. +\end{example} + +\subsubsection{Separation properties of topological spaces} +\begin{definition} + Let $X$ be a topological space. $X$ satisfies the separation properties $T_{0-2}$ if for any $x \neq y \in X$ + \begin{enumerate} + \item[$T_0$ ] $\E U \se X$ open such that $|U \cap \{x,y\}| = 1$ + \item[$T_1$ ] $\E U \se X$ open such that $x \in U, y \not\in U$. + \item[$T_2$ ] There are disjoined open sets $U, V \se X$ such that $x \in U, y \in V$. (Hausdorff) + \end{enumerate} +\end{definition} +\begin{remark} + Let $x \sim y :\iff$ the open subsets of $X$ containing $x$ are precisely the open subsets of $X$ containing $y$. Then $T_0$ holds iff $x \sim y \implies x =y$. +\end{remark} +\begin{fact} + $T_0 \iff$ every point is closed. +\end{fact} +\begin{fact} + The Zariski topology on $\fk^n$ is $T_1$ but for $n \ge 1$ not Hausdorff. For $n \ge 1$ the intersection of two non-empty open subsets of $\fk^n$ is always non-empty. +\end{fact} +\begin{proof} + $\{x\} $ is closed, as $\{x\} = V(\Span{X_1 - x_1, \ldots, X_n - x_n}_R)$. If $A = V(I), B = V(J)$ are two proper closed subsets of $\fk^n$ then $I \neq \{0\} , J \neq \{0\} $ and thus $IJ \neq \{0\} $. Therefore $A \cup B = V(IJ)$ is a proper closed subset of $\fk^n$. +\end{proof} + + +\subsubsection{Compactness properties of topological spaces} +Let $X$ be a topological space. +\begin{definition}[Compact, quasi-compact] + $X$ is called \vocab[Topological space!quasi-compact]{quasi-compact} if every open covering of $X$ has a finite subcovering. + It is called \vocab[Topological space!compact]{compact}, if it is quasi-compact and Hausdorff. +\end{definition} +\begin{definition}[Noetherian topological spaces] + $X$ is called \vocab{Noetherian}, if the following equivalent conditions hold: + \begin{enumerate}[A] + \item Every open subset of $X$ is quasi-compact. + \item Every descending sequence $A_0 \supseteq A_1 \supseteq \ldots$ of closed subsets of $X$ stabilizes. + \item Every non-empty set $\cM$ of closed subsets of $X$ has a $\se$-minimal element. + \end{enumerate} +\end{definition} +\begin{proof}\, + \begin{enumerate} + \item[A $\implies$ B] Let $A_j$ be a descending chain of closed subsets. Define $A \coloneqq \bigcap_{j = 0}^{\infty} A_j$. If A holds, the covering $X \sm A = \bigcup_{j = 0}^{\infty} (X \sm A_j)$ has a finite subcovering. + \item[B $\implies$ C] Suppose $\cM$ does not have a $\se$-minimal element. Using DC, one can construct a counterexample $A_1 \subsetneq A_2 \supsetneq \ldots$ to B. + \item[C $\implies$ A] Let $\bigcup_{i \in I} V_i$ be an open covering of an open subset $U \se X$. + By C, the set $\cM \coloneqq \{X \sm \bigcup_{i \in F} V_i | F \se I \text{ finite} \}$ has a $\se$-minimal element. + \end{enumerate} +\end{proof} + +\subsection{Another form of the Nullstellensatz and Noetherianness of \texorpdfstring{$\fk^n$}{kn}} +Let $\fk$ be algebraically closed, $R = \fk[X_1,\ldots,X_n]$. +For $f \in R$ let $V(f) = V(fR)$. +\begin{theorem}[Hilbert's Nullstellensatz (3)] \label{hns3} + Let $I \se R$ be an ideal. Then $V(I) \se V(f)$ iff $f \in \sqrt{I}$. +\end{theorem} +\begin{proof} + Suppose $f$ vanishes on all zeros of $I$. Let $R' \coloneqq \fk[X_1,\ldots,X_n,T]$, + $g(X_1,\ldots,X_n,T) \coloneqq 1 - T \cdot f(X_1,\ldots,X_n)$ + and $J \se R'$ the ideal generated by $g$ and the elements of $I$ (viewed as elements of $R'$ which are constant in the $T$-direction). + + If $f$ vanishes on all zeros of $I$, then $J$ has no zeros in $\fk^{n+1}$. + + Thus there exist $p_i \in I, i=1,\ldots,n, q_i \in \fk[X_1,\ldots,X_n,T], i = 1,\ldots,n$ and $q \in \fk[X_1,\ldots,X_n,T]$ such that + \[ + 1 = g \cdot q + \sum_{i=1}^{n} p_{i}q_i + \] + Formally substituting $\frac{1}{f(x_1,\ldots,x_n)}$ for $Y$, one obtains: + \[ + 1 = \sum_{i=1}^{n} p_{i}\left(x_1,\ldots,x_n\right) q_i\left( x_1,\ldots,x_n, \frac{1}{f(x_1,\ldots,x_n)} \right) + \] + Multiplying by a sufficient power of $f$, this yields an equation in $R$ : + \[ + f^d = \sum_{i=1}^{n} p_{i}(x_1,\ldots,_n) \cdot q_i'(x_1,\ldots,x_n) \in I + \] + Thus $f \in \sqrt{I}$. +\end{proof} + +\begin{corollary}\label{antimonbij} + \begin{align} + f: \{I \se R | I \text{ ideal}, I = \sqrt{I} \} &\longrightarrow \{A \se \fk^n | A \text{ Zariski-closed}\} \\ + I &\longmapsto V(I)\\ + \{f \in R | A \se V(f)\} &\longmapsfrom A + \end{align} + is a $\se$-antimonotonic bijection. +\end{corollary} +\begin{corollary} + The topological space $\fk^n$ is Noetherian. +\end{corollary} +\begin{proof} + Because the map from \ref{antimonbij} is antimonotonic, strictly decreasing chains of closed subsets of $\fk^n$ are mapped to strictly increasing chains of ideals in $R$. + By the Basissatz (\ref{basissatz}), $R$ is Noetherian. +\end{proof} + +% Lecture 04 + + +\subsection{Irreducible spaces} + +Let $X$ be a topological space. + +\begin{definition} + $X$ is called \vocab[Topological space!irreducible]{irreducible}, if $X \neq \emptyset$ and the following equivalent conditions hold: + \begin{enumerate}[A] + \item Every open $\emptyset \neq U \se X$ is dense. + \item The intersection of non-empty, open subsets $U, V \se X$ is non-empty. + \item If $A, B \se X$ are closed, $X = A \cup B$ then $X = A$ or $X = B$. + \item Every open subset of $X$ is connected. + \end{enumerate} +\end{definition} +\begin{proof}\, + \begin{itemize} + \item[$A \iff B$] by definition of denseness. + \item[B $\iff$ C] Let $U \coloneqq X \sm A, V \coloneqq X \sm B$. + \item[B $\implies$ D] Suppose $W$ is a non-connected open subset. Then there exists a decomposition $W = U \cup V$ into disjoint open subsets. + \item[D $\implies$ B] If $U,V \neq \emptyset$ are disjoint open subsets, then $U \cup V$ is non-connected. + \end{itemize} +\end{proof} +\begin{corollary} + Every irreducible topological space is connected. +\end{corollary} +\begin{example} + $\fk^n$ is irreducible as shown in \ref{zariskinothd}. +\end{example} + +\begin{fact} + \begin{enumerate}[A] + \item A single point is always irreducible. + \item If $X$ is Hausdorff then it is irreducible iff it has precisely one point. + \item $X$ is irreducible iff it cannot be written as a finite union of proper closed subsets. + \item $X$ is irreducible iff any finite intersection of non-empty open subsets is non-empty. ($\bigcap \emptyset \coloneqq X$) + \end{enumerate} +\end{fact} +\begin{proof} + \begin{enumerate} + \item[A,B] trivial + \item[C] $\implies$ : Induction on the cardinality of the union. $\impliedby $: $\bigcap \emptyset = X$ is non-empty and any intersection of two non-empty open subsets is non-empty. + \item[D] Follows from C. + \end{enumerate} +\end{proof} + +\subsubsection{Irreducible components} + +\begin{fact} + If $D \se X$ is dense, then $X$ is irreducible iff $D$ is irreducible with its induced topology. +\end{fact} +\begin{proof} + $X = \emptyset$ iff $D = \emptyset$. + Suppose $B$ is the union of its proper closed subsets $A,B$. Then $X = \overline{A} \cup \overline{B}$. These are proper closed subsets of $X$, as $\overline{A} \cap D = A \cap D$ (by closedness of $D$) and thus $\overline{A} \cap D \neq D$. + + On the other hand, if $U$ and $V$ are disjoint non-empty open subsets of $X$, then $U \cap D$ and $V \cap D$ are disjoint non-empty open subsets of $D$. +\end{proof} +\begin{definition}[Irreducible subsets] + A subset $Z \se X$ is called \vocab[Topological space!irreducible]{irreducible} if it is irreducible with its induced topology. + $Z$ is called an \vocab{irreducible component} of $X$, if it is irreducible and if every irreducible subset $Z \se Y \se X$ coincides with $Z$. +\end{definition} +\begin{corollary} + \begin{enumerate} + \item $Z \se X$ is irreducible iff $\overline{Z} \se X$ is irreducible. + \item Every irreducible component of $X$ is a closed subset of $X$. + \end{enumerate} +\end{corollary} +\begin{notation} + From now on, irreducible means irreducible and closed. +\end{notation} + +\subsubsection{Decomposition into irreducible subsets} +\begin{proposition} + Let $X$ be a Noetherian topological space. Then $X$ can be written as a finite union $X = \bigcup_{i = 1}^n Z_i$ of irreducible closed subsets of $X$. + One may additionally assume that $i \neq j \implies Z_i \not\subseteq Z_i$. With this minimality condition, $n$ and the $Z_i$ are unique (up to permutation) and $\{Z_1,\ldots,Z_n\}$ is the set of irreducible components of $X$. +\end{proposition} + +\begin{proof} + % i = ic + Let $\fM$ be the set of closed subsets of $X$ which cannot be decomposed as a union of finitely many irreducible subsets. + Suppose $\fM \neq \emptyset$. Then there exists a $\se$-minimal $Y \in \fM$. $Y$ cannot be empty or irreducible. Hence $Y = A \cup B$ where $A,B$ are proper closed subsets of $ Y$. By the minimality of $Y$, $A$ and $B$ can be written as a union of proper closed subsets $\lightning$. + + Let $X = \bigcup_{i = 1}^n Z_i$, where there are no inclusions between the $Z_i$. If $Y$ is an irreducible subsets of $X$, $Y = \bigcup_{i = 1}^n (Y \cap Z_i)$ and there exists $1 \le i \le n$ such that $Y = Y \cap Z_i$. + Hence $Y \se Z_i$. Thus the $Z_i$ are irreducible components. Conversely, if $Y$ is an irreducible component of $X$, $Y \se Z_i$ for some $i$ and $Y = Z_i$ by the definition of irreducible component. +\end{proof} +\begin{remark} + The proof of existence was an example of \vocab{Noetherian induction} : If $E$ is an assertion about closed subsets of a Noetherian topological space $X$ and $E$ holds for $A$ if it holds for all proper subsets of $A$, then $E(A)$ holds for every closed subset $A \se X$. +\end{remark} + +\begin{proposition}\label{bijiredprim} + By \ref{antimonbij} there exists a bijection + \begin{align} + f: \{I \se R | I \text{ ideal}, I = \sqrt{I} \} &\longrightarrow \{A \se \fk^n | A \text{ Zariski-closed}\} \\ + I &\longmapsto V(I)\\ + \{f \in R | A \se V(f)\} &\longmapsfrom A + \end{align} + + Under this correspondence $A \se \fk^n$ is irreducible iff $I \coloneqq f\inv(A)$ is a prime ideal. + Moreover, $\#A = 1$ iff $I$ is a maximal ideal. +\end{proposition} +\begin{proof} + By the Nullstellensatz (\ref{hns1}), $A = \emptyset \iff I = R$. Suppose $A = B \cup C$ is a decomposition into proper closed subsets $A = V(J), B = V(K)$ where $J = \sqrt{J}. K = \sqrt{K}$. + Since $A \neq B$ and $A \neq C$, there are $f \in J \sm I, g \in K \sm I$. $fg$ vanishes on $A = B \cup C$. By the Nullstellensatz (\ref{hns3}) $fg \in \sqrt{I} = I$ and $I$ fails to be prime. + + On the other hand suppose that $fg \in I, f \notin I, g \not\in I$. By the Nullstellensatz (\ref{hns3}) and $I = \sqrt{I} $ neither $f$ nor $g$ vanishes on all of $A$. Thus $(A \cap V(f)) \cup (A \cap V(g))$ is a decomposition and $A$ fails to be irreducible. + + The remaining assertion follows from the fact, that the bijection is $\se$-antimonotonic and thus maximal ideals correspond to minimal irreducible closed subsets, which are the one-point subsets as $\fk^n$ is T${}_1$. +\end{proof} +\subsection{Krull dimension} +\begin{definition} + Let $Z $ be an irreducible subset of the topological space $X$. Let $\codim(Z,X)$ be the maximum of the length $n$ of strictly increasing chains $Z \se Z_0 \subsetneq Z_1 \subsetneq \ldots \subsetneq Z_n$ of irreducible closed subsets of $X$ containing $Z$ or $\infty$ if such chains can be found for arbitrary $n$. + Let + \[ + \dim X \coloneqq \begin{cases} + - \infty &\text{if } X = \emptyset\\ + \sup_{\substack{Z \se X\\ Z \text{ irreducible}}} \codim(Z,X) & \text{otherwise} + \end{cases} + \] +\end{definition} +\begin{remark} + \begin{itemize} + \item In the situation of the definition $\overline{Z}$ is irreducible. Hence $\codim(Z,X)$ is well-defined and one may assume without losing much generality that $Z$ is closed. + \item Because a point is always irreducible, every non-empty topological space has an irreducible subset and for $X \neq \emptyset$, $\dim X$ is $\infty$ or $\max_{x \in X} \codim(\{x\}, X)$. + \item Even for Noetherian $X$, it may happen that $\codim(Z,X) = \infty$. + \item Even for if $X$ is Noetherian and $\codim(Z,X)$ is finite for all irreducible subsets $Z$ of $X$, $\dim X$ may be infinite. + \end{itemize} +\end{remark} +\begin{fact} + If $X = \{x\}$, then $\dim X = 0$. +\end{fact} +\begin{fact} + For every $x \in \fk$, $\codim( \{x\} ,\fk) = 1$. The only other irreducible closed subset of $\fk$ is $\fk$ itself, which has codimension zero. Thus $\dim \fk = 1$. +\end{fact} +\begin{fact} + Let $Y \se X$ be irreducible and $U \se X$ an open subset such that $U \cap Y \neq \emptyset$. Then we have a bijection + \begin{align} + f: \{A \se X | A \text{ irreducible, closed and } Y \se A\} &\longrightarrow \{B \se U | B \text{ irreducible, closed and } Y \cap U \se B\} \\ + A&\longmapsto A \cap U\\ + \overline{B}&\longmapsfrom B + \end{align} + where $\overline{B}$ denotes the closure in $X$. +\end{fact} +\begin{proof} + If $A$ is given and $B = A \cap U$, then $B \neq \emptyset$ and B is open hence (irreducibility of $A$) dense in $A$, hence $A = \overline{B}$. The fact that $B = \overline{B} \cap U$ is a general property of the closure operator. +\end{proof} +\begin{corollary}[Locality of Krull codimension] \label{lockrullcodim} + Let $Y \se X$ be irreducible and $U \se X$ an open subset such that $U \cap Y \neq \emptyset$. + Then $\codim(Y,X) = \codim(Y \cap U, U)$. +\end{corollary} +\begin{fact} + Let $Z \se Y \se X$ be irreducible closed subsets of the topological space $X$. Then + \[ + \codim(Z,Y) + \codim(Y,X) \le \codim(Z,X) \tag{CD+}\label{eq:cdp} + \] +\end{fact} +\begin{proof} + A chain of irreducible closed subsets between $Z$ and $Y$ and a chain of irreducible closed between $Y$ and $X$ can be spliced together. +\end{proof} +Taking the supremum over all $Z$ we obtain: +\begin{fact} + If $Y$ is an irreducible closed subset of the topological space $X$, then + \[ + \dim(Y) + \codim(Y,X) \le \dim(X) \tag{D+}\label{eq:dp} + \] +\end{fact} +In general, these inequalities may be strict. +\begin{definition}[Catenary topological spaces] + A topological space $T$ is called \vocab[Topological space!catenary]{catenary} if equality holds in \eqref{eq:cdp} whenever $X$ is an irreducible closed subset of $T$. +\end{definition} + +\subsubsection{Krull dimension of \texorpdfstring{$\fk^n$}{kn}} % from lecture 04 +\begin{theorem}\label{kdimkn} + $\dim \fk^n = n$ and $\fk^n$ is catenary. Moreover, if $X$ is an irreducible closed subset of $\fk^n$, then equality occurs in \eqref{eq:dp}. +\end{theorem} +\begin{proof} + Considering + \[ + \{0\} \subsetneq \fk \times \{0\} \subsetneq \fk^2 \times \{0\} \subsetneq \ldots \subsetneq \fk^n + \] + it is clear that $\codim(\{0\}, \fk^n) \ge n$.Translation by $x \in \fk^n$ gives us $\codim(\{x\} , \fk^n) \ge n$. + + The opposite inequality follows from \ref{upperbounddim} ($Z = \fk^n$ $\dim \fk^n \le \trdeg(\fK(Z) / \fk) = \trdeg(Q(\fk[X_1,\ldots,X_n]) / \fk) = n$). + + The theorem is a special case of \ref{htandtrdeg}. + % DIMT +\end{proof} + +\begin{lemma}\label{ufdprimeideal} + Every non-zero prime ideal $\fp$ of a UFD $R$ contains a prime element. +\end{lemma} +\begin{proof} + Let $p \in \fp \sm \{0\} $ with the minimal number of prime factors, counted by multiplicity. + If $p $ was a unit, then $\fp \supseteq pR = R$. If $p = ab$ with non-units $a,b$, it follows that $a \in \fp$ or $b \in \fp$ contradicting the minimality assumption. + Thus $p$ is a prime element of $R$. +\end{proof} + +\begin{proposition}[Irreducible subsets of codimension one]\label{irredcodimone} + Let $p \in R = \fk[X_1,\ldots, X_n]$ be a prime element. Then the irreducible subset $X = V(p) \se \fk^n$ has codimension one, and every codimension one subset of $\fk^n$ has this form. +\end{proposition} +\begin{proof} + Since $pR$ is a prime ideal, $X = V(p)$ is irreducible. Since $p \neq 0$, $X$ is a proper subset of $\fk^n$. + If $X \se Y \se \fk^n$ is irreducible and closed, then $Y = V(\fq)$ for some prime ideal $\fp \se pR$. + If $Y \neq \fk^n$, then $\fp \neq \{0\}$. By \ref{ufdprimeideal} there exists a prime element $q \in \fq$. As $\fq \se pR$ we have $p \divides q$. + By the irreducibility of $p$ and $q$ it follows that $p \sim q$. Hence $\fq = pR$ and $X = Y$. + + Suppose $X = V(\fp) \se \fk^n$ is closed, irreducible and of codimension one. + Then $\fp \neq \{0\}$, hence $X \neq \fk^n$. By \ref{ufdprimeideal} there is a prime element $p \in \fp$. If $\fp \neq pR$, then + $X \subsetneq V(p) \subsetneq \fk^n$ contradicts $\codim(X, \fk^n) = 1$. +\end{proof} + +% Lecture 05 +\subsection{Transcendence degree} +\subsubsection{Matroids} +\begin{definition}[Hull operator] + \npr + Let $X$ be a set, $\cP(X)$ the power set of $X$. A \vocab{Hull operator} on $X$ is a map $\cP(X) \xrightarrow{\cH} \cP(X)$ such that + \begin{enumerate} + \item[H1] $\A A \in \cP(X) ~ A \se \cH(A)$. + \item[H2] $A \se B \se X \implies \cH(A) \se \cH(B)$. + \item[H3] $\cH(\cH(X)) = \cH(X)$. + \end{enumerate} + + We call $\cH$ \vocab{matroidal} if in addition the following conditions hold: + \begin{enumerate} + \item[M] If $m,n \in X$ and $A \se X$ then $m \in \cH( \{n\} \cup A) \sm \cH(A) \iff n \in \cH(\{m\} \cup A) \sm \cH(A).$ + \item[F] $\cH(A) = \bigcup_{F \se A \text{ finite}} \cH(F)$. + \end{enumerate} + In this case, $S \se X$ is called \vocab{Independent subset}, if $s \not\in \cH(S \sm \{s\})$ for all $s \in S$ and + \vocab[Generating subset]{generating} if $X = \cH(S)$. + $S$ is called a \vocab{base}, if it is both generating and independent. +\end{definition} + +\begin{theorem} + If $\cH$ is a matroidal hull operator on $X$, then a basis exists, every independent set is contained in a base and two arbitrary bases have the same cardinality. +\end{theorem} + + +\begin{example} + Let $K$ be a field, $V$ a $K$-vector space and $\cL(T)$ the $K$-linear hull of $T$ for $T \se V$. + Then $\cL$ is a matroidal hull operator on $V$. +\end{example} + +\subsubsection{Transcendence degree} +\begin{lemma} + Let $L / K$ be a field extension and let $\cH(T)$ be the algebraic closure in $L$ of the subfield of $L$ generated by $K$ and $T$.\footnote{This is the intersection of all subfields of $L$ containing $K \cup T$, or the field of quotients of the sub-$K$-algebra of $L$ generated by $T$.} + Then $\cH$ is a matroidal hull operator. +\end{lemma} +\begin{proof}\npr + H1, H2 and F are trivial. For an algebraically closed subfield $K \se M \se L$ we have $\cH(M) = M$. Thus $\cH(\cH(T)) = \cH(T)$ (H3). + + Let $x,y \in L$, $T \se L$ and $x \in \cH(T \cup \{y\}) \sm \cH(T)$. We have to show that $y \in \cH(T \cup \{x\}) \sm \cH(T)$. + If $y \in \cH(T)$ we have $\cH(T \cup \{y\}) \se \cH(\cH(T)) = \cH(T) \implies x \in \cH(T) \sm \cH(T) \lightning$. + Hence it is sufficient to show $y \in \cH(T \cup \{x\})$. \Wlog $T = \emptyset$ (replace $K$ be the subfield generated by $K \cup T$). + Then $x$ is algebraic over the subfield $M$ of $L$ generated by $K \cup \{y\}$. Thus there exists $0 \neq P \in M[T]$ with $P(x) = 0$. + The coefficients $p_i$ of $P$ belong to the field of quotients of the $K$-subalgebra of $L$ generated by $y$. There are thus polynomials $Q_i, R \in K[Y]$ such that $p_i = \frac{Q_i(y)}{R(y)}$, $R(y) \neq 0$. + Let + \[ + Q(X,Y) \coloneqq \sum_{i=0}^{\infty} X^i Q_i(Y) = \sum_{i,j=0}^{\infty} q_{i,j}X^i Y^j = \sum_{j=0}^{\infty} Y^j \hat{Q_j}(X) \in K[X,Y] + \]. + Then $Q(x,y) = 0$. + Let $\hat{p_j} \coloneqq \hat{Q_j}(x)$. Then $\hat{P}(y) = 0$. As $Q \neq 0$ there is $(i,j) \in \N^2$ such that $q_{i,j} \neq 0$ and then $\hat{p_j} \neq 0$ as $x \not\in \cH(\emptyset)$. Thus $\hat{P} \in \hat{M}[X] \sm \{0\} $, where $\hat{M}$ is the subfield of $L$ generated by $K$ and $x$. Thus $y$ is algebraic over $\hat{M}$ and $y \in \cH(\{x\})$, +\end{proof} +\begin{definition}[Transcendence Base] + Let $L / K$ be a field extension and $\cH(T)$ the algebraic closure in $L$ of the subfield generated by $K$ and $T$. A base for $(L, \cH)$ is called a \vocab{transcendence base} and the \vocab{transcendence degree} $\trdeg(L / K)$ is defined as the cardinality of any transcendence base of $L / K$. +\end{definition} +\begin{remark} + $L / K$ is algebraic iff $\trdeg(L / K) = 0$. +\end{remark} + +\subsection{Inheritance of Noetherianness and of finite type by subrings and subalgebras / Artin-Tate} +The following will lead to another proof of the Nullstellensatz, which uses the transcendence degree. +\begin{remark} + There exist non-Noetherian domains, which are subrings of Noetherian domains (namely the field of quotients is Noetherian). +\end{remark} + +\begin{theorem}[Eakin-Nagata] + Let $A$ be a subring of the Noetherian ring $B$. If the ring extension $B / A$ is finite (i.e. $B$ finitely generated as an $A$-module) then $A$ is Noetherian. +\end{theorem} +\begin{dfact}\label{noethersubalg} + Let $R$ be Noetherian and let $B$ be a finite $R$-algebra. Then every $R$-subalgebra $A \se B$ is finite over $R$. +\end{dfact} +\begin{proof} + Since $B$ a finitely generated $R$-module and $R$ a Noetherian ring, $B$ is a Noetherian $R$-module (this is a stronger assertion than Noetherian algebra). + Thus the sub- $R$-module $A$ is finitely generated. +\end{proof} + +\begin{proposition}[Artin-Tate] + \label{artintate} + Let $A$ be a subalgebra of the $R$-algebra $B$, where $R$ is Noetherian. If $ B / R$ is of finite type and $B / A$ is finite, then $A / R$ is also of finite type. + + \begin{figure}[H] + \centering + \begin{tikzcd} + A \arrow[hookrightarrow]{rr}{\se}& & B \\ + &R \arrow{ul}{\alpha} \arrow{ur}{\alpha} \text{~(Noeth.)} + \end{tikzcd} + \end{figure} + +\end{proposition} +\begin{proof} + Let $(b_i)_{i=1}^{m}$ generate $B$ as an $A$-module and $(\beta_j)_{j=1}^m$ as an $R$-algebra. + There are $a_{ijk} \in A$ such that $b_i b_j = \sum_{k=1}^{m} a_{ijk}b_k$. And $\alpha_{ij} \in A$ such that $\beta_i = \sum_{j=1}^{m} \alpha_{ij}b_j$. Let $\tilde{A}$ be the sub- $R$-algebra of $A$ generated by the $a_{ijk}$ and $\alpha_{ij}$. $\tilde{A}$ is of finite type over $ R$, hence Noetherian. The $\tilde{A}$-submodule generated by $1$ and the $b_i$ is a sub-$R$-algebra containing the $\beta_i$ and thus coincides with $B$. + Hence $B / \tilde{A}$ is finite. Since $A \se B, A / \tilde{A}$ is finite (\ref{noethersubalg}). + Hence $A / \tilde{A}$ is of finite type. By the transitivity of ``of finite type'', it follows that $A / R$ is of finite type. + \begin{figure}[H] + \centering + \begin{tikzcd} + \tilde A \arrow[hookrightarrow]{r}{\se}& A \arrow[hookrightarrow]{r}{\se} & B \\ + &R \arrow[bend left, swap]{ul}{\alpha} \arrow{u}{\alpha} \arrow[bend right]{ur}{\alpha} + \end{tikzcd} + \end{figure} + +\end{proof} +\subsubsection{Artin-Tate proof of the Nullstellensatz} +Let $K$ be a field and $R = K[X_1,\ldots,X_n]$. +\begin{definition}[Rational functions] + Let $K(X_1,\ldots,X_n) \coloneqq Q(R)$ be the field of quotients of $R$. + + $K(X_1,\ldots,X_n)$ is called the \vocab{field of rational functions} in $n$ variables over $K$. +\end{definition} +\begin{lemma}[Infinitely many prime elements] + There are infinitely many multiplicative equivalence classes of prime elements in $R$. +\end{lemma} +\begin{proof} + Suppose $(P_i)_{i =1}^m$ is a complete (up to multiplicative equvialence) lsit of prime elements of $R$. + $m > 0$, as $X_1$ is prime. The polynomial $f \coloneqq 1 + \prod_{i=1}^{m} P_i $ is non-constant, hence not a unit in $R$. Hence there exists a prime divisor $P \in R$. As no $P_i$ divides $f$, $P$ cannot be multiplicatively equivalent to any $P_i \lightning$. +\end{proof} +\begin{lemma}[Ring of rational functions not of finite type]\label{rfuncnft} + If $n > 0$, then $K(X_1,\ldots,X_n) / K$ is not of finite type. +\end{lemma} +\begin{proof} + Suppose $(f_i)_{i=1}^m$ generate $K(X_1,\ldots,X_n)$ as a $K$-algebra. Let $f_i = \frac{a_i}{b}, a_i \in R, b \in R \sm \{0\}$. + Then $bf_i \in R$, and as the $f_i$ generate $K(X_1,\ldots,X_n)$ as a $K$-algebra, for every $g \in K(X_1,\ldots,X_n)$ there is $N \in \N$ with + \[ + b^Ng \in R \tag{+} \label{bNginR} + \] + However, if $b = \eps \prod_{i=1}^{l} P_i$ is a decomposition of $b$ into prime factors $P_i$ and a unit $\eps$ in $R$ and $g = \frac{1}{P}$, wehere $P \in R$ is a prime element not multiplicatively equvalent to any $P_i$, + then \eqref{bNginR} fails for any $N \in \N$. +\end{proof} + +The Nullstellensatz (\ref{hns2}) can be reduced to the case of \ref{rfuncnft}: + +\begin{proof}(Artin-Tate proof of HNS) + Let $(l_i)_{i=1}^n$ be a transcendence base of $L / K$. If $n = 0$ then $L / K$ is algebraic, hence an integral ring extension, hence a finite ring extension (\ref{ftaiimplf}). + + Suppose $n > 0$. Let $\tilde R \se L$ be the $K$-subalgebra generated by the $l_i$. $\tilde R \cong R \coloneqq K[X_1,\ldots,X_n]$, as the $l_i$ are algebraically independent. + As they are a transcendence base, $L$ is algebraic over the field of quotients $Q(\tilde R)$, hence integral over $Q(\tilde R)$. + + As $L / K$ is of finite type, so is $L / Q(\tilde R)$ and it follows that $L / Q(\tilde R)$ is a finite ring extension. + By Artin-Tate (\ref{artintate}), $Q(\tilde K)$ is of finite type over $K$. This contradicts \ref{rfuncnft}, as $R \cong \tilde R \implies K(X_1,\ldots,X_n) \cong Q(\tilde R)$. +\end{proof} + + +\subsection{Transcendence degree and Krull dimension} +Let $R = \fk[X_1,\ldots,X_n]$. +%i = ic +\begin{notation} + Let $X \se \fk^n$ be an irreducible closed subset. Then $X = V(\fp)$ for a unique prime ideal $\fp \se R$. + Let $\fK(X) \coloneqq Q(R / \fp)$ denote the field of quotients of $R / \fp$. +\end{notation} +\begin{remark} + As the elements of $\fp$ vanish on $X$, $R / \fp$ may be viewed as the ring of polynomials and $\fK(X)$ as the field of rational functions on $X$. +\end{remark} +\begin{theorem}\label{trdegandkdim} + If $X \se \fk^n$ is irreducible, then $\dim X = \trdeg (\fk(X) / \fk)$ and $\codim(X, \fk^n) = n - \trdeg(\fK(X) / \fk)$. + More generally if $Y \se \fk^n$ is irreducible and $X \se Y$, then $\codim(X,Y) = \trdeg(\fK(Y) / \fk) - \trdeg(\fK(X) / \fk)$. +\end{theorem} +\begin{proof} + % DIMT + One part will be shown in "A first result on dimension theory" (\ref{upperboundcodim}) + and other one in "Aplication to dimension theory: Proof of $\dim Y = \trdeg(\fK(Y) / \fk)$" (\ref{lowerbounddimy}). + The theorem is a special case of \ref{htandtrdeg}. +\end{proof} +\begin{remark} + Loosely speaking, the Krull dimension of $X$ is equal to the maximal number of $\fk$-algebraically independent rational functions on $X$. + This is yet another indication that the notion of dimension is the ``correct'' one. +\end{remark} +\begin{remark} + \ref{kdimkn} follows. +\end{remark} + + + + + +% Lecture 06 + +\subsection{The spectrum of a ring} +\begin{definition}[Spectrum] + Let $R$ be a commutative ring. + \begin{itemize} + \item Let $\Spec R$ denote the set of prime ideals and $\mSpec R \se \Spec R$ the set of maximal ideals of $R$. + \item For an ideal $I \se R$ let $V(I) \coloneqq \{\fp \in \Spec R | I \se \fp\}$ + \item We equip $\Spec R$ with the \vocab{Zariski-Topology} for which the closed subsets are the subsets of the form $V(I)$, where $I$ runs over the set of ideals in $R$. + \end{itemize} +\end{definition} + +\begin{remark} + When $R = \fk[X_1,\ldots,X_n]$, the notation $V(I)$ clashes with the previous notation. When several types of $V(I)$ will be in use, they will be distinguished using indices. +\end{remark} +\begin{remark} + Let $(I_{\lambda})_{\lambda \in \Lambda}$ and $(l_j)_{j=1}^n$ be ideals in $R$, where $\Lambda$ may be infinite. We have $V(\sum_{\lambda \in \Lambda} I_\lambda ) = \bigcap_{\lambda \in \Lambda} V(I_\lambda)$ and $V(\bigcap_{j=1}^n I_j) = V(\prod_{j=1}^{n} I_j) = \bigcup_{j = 1}^n V(I_j)$. + Thus, the Zariski topology on $\Spec R$ is a topology. +\end{remark} +\begin{remark} + Let $R = \fk[X_1,\ldots,X_n]$. Then there exists a bijection (\ref{antimonbij}, \ref{bijiredprim}) between $\Spec R$ and the set of irreducible closed subsets of $\fk^n$ sending $\fp \in \Spec R$ to $V_{\fk^n}(\fp)$ and identifying the one-point subsets with $\mSpec R$. + This defines a bijection $\fk^n \cong \mSpec R$ which is a homeomorphism if $\mSpec R$ is equipped with the induced topology from the Zariski topology on $\Spec R$. +\end{remark} + +\subsection{Localization of rings} +\begin{definition}[Multiplicative subset] + A \vocab{multiplicative subset} of a ring $R$ is a subset $S \se R$ such that $\prod_{i=1}^{n} f_i \in S $ when $n \in \N$ and all $f_i \in S$. +\end{definition} +\begin{proposition} + Let $S \se R$ be a multiplicative subset. Then there is a ring homomorphism $R \xrightarrow{i} R_S$ such that $i(S) \se R_S^{\times }$ and $i$ has the \vocab{universal property} for such ring homomorphisms: + If $R \xrightarrow{j} T$ is a ring homomorphism with $j(S) \se T^{\times }$, then there is a unique ring homomorphism $R_S \xrightarrow{\iota} T$ with $j = \iota i$. + + \begin{figure}[H] + \centering + \begin{tikzcd} + R \arrow{r}{i}\arrow{d}{j}& R_S \arrow[dotted]{ld}{\Eone \iota}\\ + T + \end{tikzcd} + \end{figure} + +\end{proposition} +\begin{proof} + The construction is similar to the construction of the field of quotients: + + Let $R_S \coloneqq (R \times S) / \sim $, where $(r,s) \sim (\rho, \sigma) : \iff \E t \in S ~ t \sigma r = ts\rho$.\footnote{$t$ does not appear in the construction of the field of quotients, but is important if $S$ contains zero divisors.} + $[r,s] + [\rho, \sigma] \coloneqq [r\sigma + \rho s, s \sigma]$, $[r,s] \cdot [\rho, \sigma] \coloneqq [r \cdot \rho, s \cdot \sigma]$. + + In order proof the universal property define $\iota([r,s]) \coloneqq \frac{j(r)}{j(s)}$. + The universal property characterizes $R_S$ up to unique isomorphism. + +\end{proof} +\begin{remark} + $i$ is often not injective and $\Ker(i) = \{r \in R | \E s \in S ~ s \cdot r = 0\} $. + In particular $(r = 1)$, $R_S$ is the null ring iff $0 \in S$. +\end{remark} +\begin{notation} + Let $S \se R$ be a multiplicative subset of $R$. We write $\frac{r}{s}$ for $[r,s]$. + The ring homomorphism $R \xrightarrow{i} R_S$ i given by $i(r) = \frac{r}{1}$. + For $X \se R_S$ let $X \sqcap R$ denote $i\inv(X)$. +\end{notation} +\begin{definition}[$S$-saturated ideal] + An ideal $I \se R$ is called \vocab[Ideal!S-saturated]{$S$-saturated} if for all $s \in S, r \in R$ + $rs \in I \implies r \in I$. +\end{definition} +\begin{fact}\label{primeidealssat} + A prime ideal $\fp \se \Spec R$ is $S$-saturated iff $\fp \cap S = \emptyset$. +\end{fact} +Because the elements of $S$ become units in $R_S$, $J \sqcap R$ is an $S$-saturated ideal in $R$ when $J$ is an ideal in $R_S$. + +\begin{fact}\label{ssatiis} + Let $I \se R$ be an $S$-saturated ideal and let $I_S$ denote the ideal $\{\frac{r}{s} | r \in R, s \in S\} \se R_S$. + Then for all $r \in R, s \in S$ + we have $\frac{r}{s} \in I_S \iff r \in I$. +\end{fact} +\begin{proof} + Clearly $i \in I \implies \frac{i}{s} \in I_S$. If $\frac{i}{s} \in J$ there are $\iota \in I$, $\sigma \in S$ such that $\frac{i}{s} = \frac{\iota}{\sigma}$ in $R_S$. + This equation holds iff there exists $t \in S$ such that $ts\iota = t \sigma i$. But $ts \iota \in I$ hence $i \in I$, as $I$ is $S $-saturated. +\end{proof} +\begin{fact}\label{invimgprimeideal} + The inverse image of a prime ideal under any ring homomorphism is a prime ideal. +\end{fact} + +\begin{proposition}\label{idealslocbij} + \begin{align} + f: \{I \se R | I \text{ $S$-saturated ideal}\} &\longrightarrow \left\{J \se R_S | J \text{ ideal}\right\} \\ + I &\longmapsto I_S \coloneqq \left\{\frac{i}{s} | i \in I, s \in S\right\}\\ + J \sqcap R &\longmapsfrom J\\ + \end{align} + is a bijection. Under this bijection $I$ is a prime ideal iff $f(I)$ is. +\end{proposition} + +\begin{proof} + Applying \ref{ssatiis} to $s = 1$ gives $I_S \sqcap R = I$, when $I$ is $S$-saturated. + + Conversely, if $J$ is given and $I = J \sqcap R, \frac{r}{s} \in R_S$, then by \ref{ssatiis} $\frac{r}{s} \in IR_S \iff r \in I$. + But as $\frac{r}{1} = s \cdot \frac{r}{s}$ and $s \in R_S^{\times }$, we have $r \in I \iff \frac{r}{1} \in J \iff \frac{r}{s} \in J$ . + We have thus shown that the two maps between sets of ideals are well-defined and inverse to each other. + + By \ref{invimgprimeideal}, $J \in \Spec R_S \implies f\inv(J) = J \cap R \in \Spec R_S$. + Suppose $I \in \Spec R$, $\frac{a}{s} \cdot \frac{b}{t} \in I_S$ for some $a,b \in R, s,t \in S$. + By \ref{ssatiis} $ab \in I$. Thus $a \in I \lor b \in I$, hence $\frac{a}{s} \in I_S \lor \frac{b}{t} \in I_S$ and we have $I_S \in \Spec R_S$. + + + +\end{proof} + +% Some more remarks on localization + +\begin{remark}\label{locandquot} + Let $R$ be a domain. If $S = R \sm \{0\}$, then $R_S$ is the field of quotients $Q(R)$. + If $S \se R \sm \{0\} $, then + \[ + R_S \cong \left\{ \frac{a}{s} \in K | a \in R, s \in S\right\} + \] + In particular $Q(R) \cong Q(R_S)$. +\end{remark} + +\begin{definition}[$S$-saturation]\label{ssaturation} + Let $R$ be any ring, $I \se R$ an ideal. Even if $I$ is not $S$-saturated, $J = I_S \coloneqq \{\frac{i}{s} | i \in I, s \in S\}$ is an ideal in $R_S$, and $I_S \sqcap R = \{r \in R | s\cdot r \in I, s \in S\}$ is called the \vocab[Ideal!$S$-saturation]{$S$-saturation of $I$ } which is the smallest $S$-saturated ideal containing $I$. + + +\end{definition} +\begin{lemma}\label{locandfactor} + In the situation of \ref{ssaturation}, if $\overline{S}$ denotes the image of $S$ in $R / I$, there is a canonical isomorphism $R_S / I_S \cong (R / I)_{\overline{S}}$. +\end{lemma} +\begin{proof} + We show that both rings have the universal property for ring homomorphisms $R \xrightarrow{\tau} T$ with $\tau(I) = \{0\} $ and $\tau(S) \se T^{\times }$. + For such $\tau$, by the fundamental theorem on homomorphisms (Homomorphiesatz) there is a unique $R/I \xrightarrow{\tau_1} T$ such that $\tau = \tau_1 \pi_{R,I}$. + We have $\tau_1(\overline{S}) = \tau(S) \se T^{\times }$, hence there is a unique $(R / I)_{\overline{S}} \xrightarrow{\tau_2} T$ such that the composition $R / I \to (R / I)_{\overline{S}} \xrightarrow{\tau_2} T $ equals $\tau_1$. It is easy to see that this is the only one for which $R \to R / I \to (R / I)_{\overline{S}} \xrightarrow{\tau_2} T$ equals $\tau$. + + + Similarly, by the universal property of $R_S$ there is a unique $R_S \xrightarrow{\tau_3} T$ whose composition with $R \to R_S$ equals $\tau$. + $\tau_3(I_{S}) = 0$, hence a unique $R_S / I_S \xrightarrow{\tau_4} T$ whose composition with $\pi_{R_S, I_S}$ equals $\tau_3$ exists. + This is the only one for which the composition $R \to R_S \to R_S / I_S \xrightarrow{\tau_4} T$ equals $\tau$. + +\begin{figure}[H] + \centering + \begin{tikzcd} + R \arrow{r}{\tau}\arrow[swap]{d}{\pi_{R,I}}& T & R\arrow[swap]{l}{\tau}\arrow{d}{}\\ + R / I \arrow[dotted]{ru}{\Eone \tau_1}\arrow{d}{} & & R_S \arrow[dotted, swap]{lu}{\Eone \tau_3}\arrow{d}{\pi_{R_S, I_S}}\\ + (R / I)_{\overline{S}} \arrow[dotted,bend right]{ruu}{\Eone \tau_2} & & R_S / I_S \arrow[dotted, bend left, swap]{luu}{\Eone \tau_4}\\ + \end{tikzcd} +\end{figure} + + +\end{proof} + + + +\subsection{A first result of dimension theory} + +\begin{notation} + Let $R$ be a ring, $\fp \in \Spec R$. Let $\fk(\fp)$ denote the field of quotients of the domain $R / \fp$. This is called the \vocab{residue field} of $\fp$. +\end{notation} + +% i = ic +\begin{proposition}\label{trdegresfield} + Let $\fl$ be a %% ?? +field, $A$ a $\fl$-algebra of finite type and $\fp, \fq \in \Spec A$ with $\fp \subsetneq \fq$. +Then \[ + \trdeg(\fk(\fp) / \fl) > \trdeg(\fk(\fq) / \fl) +\] +\end{proposition} +\begin{proof} + Replacing $A$ by $A / \fp$, we may assume $\fp = \{0\} $ and $A$ to be a domain. Then $\fk(\fp) = Q(A / \fp) = Q(A)$. + + If $\fq$ is a maximal ideal, $\fk(\fq) = A / \fq$ is of finite type over $\fl$, hence a finite field extension of $\fl$ by the Nullstellensatz (\ref{hns2}). + Thus, $\trdeg(\fk(\fq) / \fl) = 0$. + If $\trdeg(Q(A) / \fl) = 0$, $A$ would be integral over $\fl$, hence a field (fact about integrality and fields, \ref{fintaf}). But if $A$ is a field, $\fp = \{0\}$ is a maximal ideal of $A$, hence $\fq = \fp \lightning$. + This finishes the proof for $\fq \in \mSpec A$. + We will use the following lemma to reduce the general case to this case: + \begin{lemma}\label{ltrdegresfieldtrbase} + There are algebraically independent $a_1,\ldots,a_n \in A$ whose images in $A / \fq$ form a transcendence base for $\fk(\fq) / \fl$. + \end{lemma} + \begin{subproof} + There exist $a_1,\ldots,a_n \in A$ such that $\fk(\fq)$ is algebraic over the subfield generated by $\fl$ and their images $\overline{a_i}$ (for instance generators of $A$ as a $\fl$-algebra). + We may assume that $n$ is minimal. If the $a_i$ are $\fl$-algebraically dependent, then w.l.o.g. $\overline{a_n}$ can be assumed to be algebraic over the subfield generated by $\fl$ and the $\overline{a_i}, 1\le i 0$, then $L^{\Aut(L / K)} = \{l \in L | \E n \in \N ~ l^{p^n} \in K\}$. +\end{proposition} +\begin{proof} + In both cases $L^G \supseteq$ is easy to see. + + If $K \se M \se L$ is an intermediate field, then $L$ is normal over $M$. If $\sigma \in \Aut(M /K)$, an application of Zorn's lemma to the set of all $(N, \vartheta)$ where $N$ is an intermediate field $M \se N \se L$ and $N \xrightarrow{\vartheta} L$ a ring homomorphism such that $\vartheta\defon{M} = \sigma$ shows that $\sigma$ has an extension to an element of $\Aut(L / K)$. % TODO make this rigorous + If $M$ is normal over $K$, it is easily seen to be $\Aut(L / K)$ invariant. + Thus $L^G$ is the union of $M^{\Aut(M / K)}$ over all intermediate fields which are finite and normal over $K$, and it is sufficient to show the proposition for finite normal extensions $L / K$. + + \begin{itemize} + \item Characteristic $0$: The extension is normal, hence Galois, and the assertion follows from Galois theory. + \item Characteristic $p > 0$: Let $l \in L^G$ and $P \in K[T]$ be the minimal polynomial of $l$ over $K$. + We show that $l^{p^n} \in K$ for some $n \in \N$ by induction on $\deg(l / K) \coloneqq \deg(P)$. + + If $\deg(l / K) = 1$, we have $l \in K$. + Otherwise, assume that the assertion has been shown for elements of $L^G$ whose degree over $K$ is smaller than $\deg( l / K)$. + Let $\overline{L}$ be an algebraic closure of $L$ and $\lambda$ a zero of $P$ in $\overline{L}$. + If $M = K(l) \se L$, then there is a ring homomorphism $M - \overline{L}$ sending $l$ to $\lambda$. This can be extended to a ring homomorphism $L \xrightarrow{\sigma} \overline{L}$. We have $\sigma \in G$ because $L / K$ is normal. Hence $\lambda = \sigma(l) = l$, as $l \in L^G$. Thus $l$ is the only zero of $P$ in $\overline{L}$ and because $\deg P >1$ it is a multiple zero. + It is shown in the Galois theory lecture % TODO: link to EinfAlg + that this is possible only when $P(T) = Q(T^p)$ for some $Q \in K[T]$. Then $Q(l^p) = 0$ and the induction assumption can be applied to $x = l^p$ showing $x^{p^m} \in K$ hence $l^{p^{m+1}} \in K$ for some $m \in \N$. + \end{itemize} +\end{proof} +\subsubsection{Integral closure and normal domains} + +\begin{definition}[Integral closure, normal domains] + Let $A$ be a domain with field of quotients $Q(A)$ and let $L$ be a field extension of $Q(A)$. + By \ref{intclosure} the set of elements of $L$ integral over $A$ is a subring of $L$, the \vocab{integral closure} of $A$ in $L$. + $A$ is \vocab{Domain!integrally closed} in $L$ if the integral closure of $A$ in $L$ equals $A$. + $A$ is \vocab{Domain!normal} if it is integrally closed in $Q(A)$. +\end{definition} + +\begin{proposition}\label{ufdnormal} + Any factorial domain (UFD) is normal. +\end{proposition} +\begin{proof} + Let $x \in Q(A)$ be integral over $A$. Then there is a normed polynomial $P \in A[T]$ with $P(x) = 0$. + In \einfalg it was shown that $A[T]$ is a UFD and that the prime elements of $A[T]$ are the elements which are irreducible in $Q(A)[T]$ and for which the $\gcd$ of the coefficients is $\sim 1$. % TODO reference + The prime factors of a normed polynomial are all normed up to multiplicative equivalence. We may thus assume $P$ to be irreducible in $Q(A)[T]$. + But then $\deg P = 1$ as $x$ is a zero of $P$ in $Q(A)$, hence $P(T) = T - x$ and $x \in A$ as $P \in A[T]$. + + + Alternative proof\footnote{\url{http://www.math.lsa.umich.edu/~tfylam/Math221/2.pdf}}: + Let $x = \frac{a}{b} \in Q(A)$ be integral over $A$. \Wlog $\gcd(a,b) = 1$. Then $x^n + c_{n-1} x^{n-1} + \ldots + c_0 = 0$ for some $c_i \in A$. + Multiplication with $b^n$ yields $a^n + c_{n-1} b a^{n-1} + \ldots +c_0 b^n = 0$. Thus $b | a^n$. Since $\gcd(a,b) = 1$ it follows that $b$ is a unit, hence $x \in A$. +\end{proof} + +\begin{remark} + It follows from \ref{cintclosure} and \ref{locandquot} that the integral closure of $A$ in some field extension $L$ of $Q(A)$ is always normal. +\end{remark} +\begin{remark} + A finite field extension of $\Q$ is called an \vocab{algebraic number field} (ANF). If $K$ is an ANF, let $\cO_K$ (the \vocab[Ring of integers in an ANF]{ring of integers in $K$}) be the integral closure of $\Z$ in $K$. + One can show that this is a finitely generated (hence free, by results of \einfalg % EINFALG + ) abelian group. + We have $\cO_{\Q} = \Z$ by the proposiiton. +\end{remark} + +\subsubsection{Action of \texorpdfstring{$\Aut(L / K)$}{Aut(L / K)} on prime ideals of a normal ring extension} + +\begin{theorem}\label{autonprime} + Let $A$ be a normal domain, $L$ a normal field extension of $K \coloneqq Q(A)$, $B$ the integral closure of $A$ in $L$ and $\fp \in \Spec A$. + Then $G \coloneqq \Aut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq \cap A = \fp\}$. +\end{theorem} + + +\begin{proof} + Let $\fq, \fr$ be prime ideals of $B$ above the given $\fp \in \Spec A$. + We must show that there exists $\sigma \in G$ such that $\fq = \sigma(\fr)$. + This is equivalent to $\fq \se \sigma(\fr)$, since the Krull going-up theorem (\ref{cohenseidenberg}) applies to the integral ring extension $B / A$, showing that there are no inclusions between different elements of $\Spec B$ lying above $\fp \in \Spec A$. + + If $L / K$ is finite and there is no such $\sigma$, then by prime avoidance (\ref{primeavoidance}) there is $ x \in \fq \sm \bigcup_{\sigma \in G} \sigma(\fr)$. + As $\fr$ is a prime ideal, $y = \prod_{\sigma \in G} \sigma(x) \in \fq \sm \fr$.\footnote{$\prod_{\sigma \in G} \sigma(x) = \prod_{\sigma \in G} \sigma\inv(x)$} + By the characterization of $L^G$ for normal field extensions (\ref{characfixnormalfe}), there is a positive integer $k$ with $y^k \in K$. + As $A$ is normal, we have $y^k \in K \cap B = A$. + Thus $y^k \in (A \cap \fq) \sm (A \cap \fr) = \fp \sm \fp = \emptyset \lightning$. + + If $L / K$ is not finite, one applies Zorn's lemma to the poset of pairs $(M, \sigma)$ where $M$ is an intermediate field and $\sigma \in \Aut(M / K)$ such that $\sigma(\fr \cap M) = \fq \cap M$. + +\end{proof} +\begin{remark} + The theorem is very important for its own sake. For instance, if $K$ is an ANF which is a Galois extension of $\Q$ it shows that $\Gal(K / \Q)$ transitively acts on the set of prime ideals of $\cO_K$ over a given prime number $p$. More generally, if $L / K$ is a Galois extension of ANF then $\Gal(L / K)$ transitively acts on the set of $\fq \in \Spec \cO_L$ for which $\fq \cap K$ is a given $\fp \in \Spec \cO_K$. +\end{remark} + +\subsubsection{A going-down theorem} +\begin{theorem}[Going-down for integral extensions of normal domains (Krull)]\label{gdkrull} + Let $B$ be a domain which is integral over its subring $A$. If $A$ is a normal domain, then going-down holds for $B / A$. +\end{theorem} + +\begin{proof} + It follows from the assumptions that the field of quotients $Q(B)$ is an algebraic field extension of $Q(A)$. + There is an algebraic extension $L$ of $Q(B)$ such that $L / Q(A)$ is normal (for instance an algebraic closure of $Q(B)$). + Let $C$ be the integral closure of $A$ in $L$. Then $B \se C$ and $C / B$ is integral. +\begin{figure}[H] + \centering + \begin{tikzcd} + Q(A) \arrow[hookrightarrow]{r}{} & Q(B) \arrow[hookrightarrow]{r}{} & L \coloneqq \overline{Q(B)} \\ + A \arrow[hookrightarrow]{u}{}\arrow[hookrightarrow]{r}{} & B \arrow[hookrightarrow]{r}{}\arrow[hookrightarrow]{u}{}& C \arrow[hookrightarrow]{u}{}\\ + \end{tikzcd} +\end{figure} + + \begin{claim} + Going-down holds for $C / A$. + \end{claim} + \begin{subproof} + Let $\fp \se \tilde \fp$ be an inclusion of prime ideals of $A$ and $\tilde \fr \in \Spec C$ with $\tilde \fr \cap A = \tilde \fp$. + By going-up for integral ring extensions (\ref{cohenseidenberg}), $\Spec C \xrightarrow{\cdot \cap A} \Spec A$ is surjectiv. Thus there is $\fr' \in \Spec C$ such that $\fr' \cap A = \fp$. By going up for $C / A$ there is $\tilde \fr' \in \Spec C$ with $\tilde \fr' \cap A = \tilde \fp, \fr' \se \tilde \fr'$. + By the theorem about the action of the automorphism group on prime ideals of a normal ring extension (\ref{autonprime}) there exists a $\sigma \in \Aut(L / Q(A))$ with $\sigma(\tilde \fr') = \tilde \fr$. Then $\fr \coloneqq \sigma(\fr')$ satisfies $\fr \se \tilde \fr$ and $\fr \cap A = \fp$. + \end{subproof} + If $\fp \se \tilde \fp$ is an inclusion of elements of $\Spec A$ and $\tilde \fq \in \Spec B$ with $\tilde \fp \cap A = \tilde \fp$, by the surjectivity of $\Spec C \xrightarrow{\cdot \cap B} \Spec B$ (\ref{cohenseidenberg}) there is $\tilde \fr \in \Spec C$ with $\tilde \fr \cap B = \fq$. + By going-down for $C / A$, there is $\fr \in \Spec C$ with $\fr \se \tilde \fr$ and $\fr \cap A = \fp$. + Then $\fq \coloneqq \fr \cap B \in \Spec B, \fq \se \tilde \fq$ and $\fq \cap A = \fp$. Thus going-down holds for $B / A$. +\end{proof} + +\begin{remark}[Universally Japanese rings] + A Noetherian ring $A$ is called universally Japanese if for every $\fp \in \Spec A$ and every finite field extension $L$ of $\fk(\fp)$, the integral closure of $A / \fp$ in $L$ is a finitely generated $A$-module. This notion was coined by Grothendieck because the condition was extensively studied by the Japanese mathematician Nataga Masayoshji. + By a hard result of Nagata, algebras of finite type over a universally Japanese ring are universally Japanese. + Every field is universally Japanese, as is every PID of characteristic $0$. + There are, however, examples of Noetherian rings which fail to be universally Japanese. +\end{remark} + +\begin{dexample}[Counterexample to going down] + Let $R = \fk[X,Y]$ and $A = \fk[X,Y, \frac{X}{Y}]$. Then going down does not hold for $A / R$: + + For any ideal $Y \in \fq \se A$ we have $X = \frac{X}{Y} \cdot Y \in \fq$. + Consider $(Y)_R \subsetneq (X,Y)_R \se \fq \cap R$. As $(X,Y)_R$ is maximal and the preimage of a prime ideal is prime and thus proper, we have $(X,Y)_R = \fq \cap R$. + The prime ideal $(\frac{X}{Y},Y)_A = (\frac{X}{Y}, X,Y)_A$ is lying over $(X,Y)_R$, so going down is violated. +\end{dexample} + + +\subsubsection{Proof of \texorpdfstring{$\codim(\{y\},Y) = \trdeg(\fK(Y) / \fk)$}{codim(\{y\},Y) = trdeg(K(Y) /k)}} +\label{proofcodimletrdeg} +This is part of the proof of \ref{trdegandkdim}. %TODO: reorder + +\begin{proof} + % DIMT + Let $B = \fk[X_1,\ldots,X_n]$ and $X \se Y = V(\fp) \se \fk^n$ irreducible closed subsets of $\fk^n$. + We want to show that $\codim(X,Y) = \trdeg(\fK(Y) / \fk) - \trdeg(\fK(X) / \fk)$. + $\le $ was shown in \ref{upperboundcodim}. + $\dim Y \ge \trdeg(\fK(Y) / \fk)$ was shown in \ref{lowerbounddimy} by + + Applying Noether normalization to $A \coloneqq B / \fp$, giving us $(f_i)_{i=1}^d \in A^d$ such that the $f_i$ are algebraically independent and $A$ finite over the subalgebra generated by them. + We then used going-up to lift a chain of prime ideals corresponding to $\fk^d \supsetneq \{0\} \times \fk^{n-1} \supsetneq \ldots \supsetneq \{0\}$ under $Y \xrightarrow{F = (f_1,\ldots,f_d)} \fk^d$ to a chain of prime ideals in $A$. + This was done left-to-right as going-up was used to make prime ideals larger. In particular, when $\{0\} \in \fk^d$ has several preimages under $F$ we cannot control to which of them the maximal ideal terminating the lifted chain belongs. Thus, we can show that in the inequality + \[ + \codim(\{y\}, Y) \le d = \trdeg(\fK(Y) \sm \fk) + \] + (see \ref{upperboundcodim}) + equality holds for at least one pint $y \in F\inv(\{0\})$ but cannot rule out that there are other $y \in F\inv(\{0\})$ for which the inequality becomes strict. + However using going-down (\ref{gdkrull}) for $F$, we can use a similar argument, but start lifting of the chain at the right end for the point $y \in Y$ for which we would like to show equality. + From this $\codim(X,Y) \ge \trdeg(\fK(Y) / \fk) - \trdeg(\fK(X) / \fk)$ can be derived similarly to \ref{upperboundcodim}. + Thus + \[ + \codim(X,Y) = \trdeg(\fK(Y) / \fk) - \trdeg(\fK(X) / \fk) + \] + follows (see \ref{htandcodim} and \ref{htandtrdeg}). +\end{proof} +\begin{remark} + The going-down theorem used to prove this is somewhat more general, as it does not depend on $\fk$ being algebraically closed. +\end{remark} + + + + +% Lecture 09 +% i = ic + +\subsection{The height of a prime ideal} +In order to complete the proof of \ref{proofcodimletrdeg} and show $\codim(X,Y) = \trdeg(\fK(Y) / \fk) - \trdeg(\fK(X) / \fk)$, +we need to localize the $\fk$-algebra with respect to a multiplicative subset and replace the ground field by a larger subfield of that localization which is no longer algebraically closed. +To formulate a result which still applies in this context, we need the following: +\begin{definition}[Height of a prime ideal] + Let $A$ be a ring, $\fp \in \Spec A$. We define the \vocab[Height of a prime ideal]{height of the prime ideal $\fp$}, $\hght(\fp)$, to be the largest $k \in \N$ such that there is a strictly decreasing sequence $\fp = \fp_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_k$ of prime ideals of $A$, or $\infty$ if there is no finite upper bound on the length of such sequences. +\end{definition} +\begin{example} + Let $A = \fk[X_1,\ldots,X_n]$, $X = V(\fp)$ for a prime ideal $\fp$. + By the correspondence between irreducible subsets of $\fk^n$ and prime ideals in $A$ (\ref{bijiredprim}), + the $\fp_i$ correspond to irreducible subsets $X_i \se \fk^n$ containing $X$. Thus $\hght(\fp) = \codim(X, \fk^n)$. +\end{example} + +\begin{example}\label{htandcodim} + Let $B = \fk[X_1,\ldots,X_n], \fq \in \Spec B$ and let $A \coloneqq B / \fp$. + Let $Y \coloneqq V(\fq) \se \fk^n$, $\tilde \fp \coloneqq \pi_{B, \fq}\inv(\fp)$, where $B \xrightarrow{\pi_{B, \fp}} A $ is the projection to the ring of residue classes, and let $X = V(\tilde \fp)$. + By \ref{idealslocbij} we have a bijection between the prime ideals $\fr \se \fp$ of $A$ contained in $\fp$ and the prime ideals and the prime ideals $\tilde \fr \in \Spec B$ with $\fq \se \tilde \fr \se \tilde \fp$: + \begin{align} + f: \{\fr \in \Spec A | \fr \se \fp \} &\longrightarrow \{\tilde \fr \in \Spec B | \fq \se \tilde \fr \se \tilde \fp\} \\ + \fr &\longmapsto \pi_{B, \fq}\inv(\fr)\\ + \tilde \fr / \fq &\longmapsfrom \tilde \fr + \end{align} + By \ref{bijiredprim}, the $\tilde \fr$ are in canonical bijection with the irreducible subsets $Z$ of $Y$ containing $X$. + Thus, the chains $\fp = \fp_0 \supsetneq \ldots \supsetneq \fp_k$ are in canonical bijection with the chains $X = X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_k \se Y$ of irreducible subsets and + $\hght(\fp) = \codim(X,Y)$. +\end{example} + + +\begin{remark} + Let $A$ be an arbitrary ring. One can show that there is a bijection between $\Spec A$ and the set of irreducible subsets $Y \se \Spec A$: + \begin{align} + f: \Spec A &\longrightarrow \{Y \se \Spec A | Y\text{irreducible}\} \\ + \fp &\longmapsto \Vs(\fp)\\ + \bigcup_{\fp \in Y} \fp &\longmapsfrom Y + \end{align} + Thus, the chains $\fp = \fp_0 \supsetneq \ldots \supsetneq \fp_k$ are in canonical bijection with the chains $V(\fp) = X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_k \se \Spec A$ of irreducible subsets, and $\hght(\fp) = \codim(V(\fp), \Spec A)$. +\end{remark} + + +\subsubsection{The relation between \texorpdfstring{$\hght(\fp)$}{ht(p)} and \texorpdfstring{$\trdeg$}{trdeg}} +We will use the following +\begin{lemma}\label{extendtotrbase} + Let $\fl$ be an arbitrary field, $A$ a $\fl$-algebra of finite type which is a domain, $K \coloneqq Q(A)$ the field of quotients and let $(a_i)_{i=1}^n$ be $\fl$-algebraically independent elements of $A$. Then there exist a natural number $m \ge n$ and a transcendence base $(a_i)_{i = 1}^m$ for $K / \fl$ with $a_i \in A$ for $1 \le i \le m$. +\end{lemma} +\begin{proof} + The proof is similar to the proof of \ref{ltrdegresfieldtrbase}. + There are a natural number $m \ge n$ and elements $(a_i)_{i = n+1}^m \in A^{m-n}$ which generate $K$ in the sense of a matroid used in the definition of $\trdeg$. + For instance, one can use generators of the $\fl$-algebra $A$. We assume $m$ to be minimal and claim that $(a_i)_{i=1}^m$ are $\fl$-algebraically independent. + Otherwise there is $j \in \N$, $1 \le j \le m$ such that $a_j$ is algebraic over the subfield of $K$ generated by $\fl$ and the $(a_i)_{i=1}^{j-1}$. We have $j > n$ by the algebraic independence of $(a_i)_{i=1}^n$. + Exchanging $x_j$ and $x_m$, we may assume $j = m$. But then $K$ is algebraic over its subfield generated by $\fl$ and the $(a_i)_{i=1}^{m-1} $, contradicting the minimality of $m$. +\end{proof} + +\begin{theorem}\label{htandtrdeg} + Let $\fl$ be an arbitrary field, $A$ a $\fl$-algebra of finite type which is a domain, and $\fp \in \Spec A$. + Let $K \coloneqq Q(A)$ be the field of quotients of $A$. Then + \[ + \hght(\fp) = \trdeg(K /\fl) - \trdeg(\fk(\fp) / \fl) + \] +\end{theorem} +\begin{remark} + By example \ref{htandcodim}, theorem \ref{trdegandkdim} is a special case of this theorem. %(\ref{htandtrdeg}). +\end{remark} +\begin{proof} + If $\fp = \fp_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_k$ is a chain of prime ideals in $A$, we have $\trdeg(\fk(\fp_i) / \fl) < \trdeg(\fk(\fp_{i+1}) / \fl)$ by \ref{trdegresfield} (``A first result of dimension theory''). +Thus +\[ + k \le \trdeg(\fk(\fp_k) / \fl) - \trdeg(\fk(\fp) / \fl) \le \trdeg(K / \fl) - \trdeg(\fk(\fp) / \fl) +\] +where the last inequality is another application of \ref{trdegresfield} (using $K = Q(A) = Q(A / \{0\}) = \fk(\{0\})$ and the fact that $\{0\} \se \fp_k$ is a prime ideal). +Hence \[ + \hght(\fp) \le \trdeg( K / \fl) - \trdeg(\fk(\fp) / \fl) +\] +and it remains to show the opposite inequality. + +\begin{claim} + For any maximal ideal $\fp \in \mSpec A$ \[ + \hght(\fm) \ge \trdeg(K / \fl) + \] +\end{claim} +\begin{subproof} + By the Noether normalization theorem (\ref{noenort}), there are $(x_i)_{i=1}^d \in A^d$ which are algebraically independent over $\fl$ such that $A$ is finite over the subalgebra $S$ generated by the $x_i$. We have $d = \trdeg(K / \fl)$ as the $x_i$ form a transcendence base of $K / \fl$. +\begin{claim} + We can choose $x_i \in \fm$ +\end{claim} +\begin{subproof} + By the Nullstellensatz (\ref{hns2}), $\fk(\fm) = A / \fm$ is a finite field extension of $\fl$. Hence there exists a normed polynomial $P_i \in \fl[T]$ with $P_i(x_i \mod \fm) = 0$ in $\fk(\fm)$. + Let $\tilde x_i \coloneqq P_i(x_i) \in \fm$ and $\tilde S$ the subalgebra generated by the $\tilde x_i$. As $P_i(x_i) - \tilde x_i = 0$, $x_i$ is integral over $\tilde S$ and so is $S / \tilde S$. It follows that $A / \tilde S$ is integral, hence finite by \ref{ftaiimplf}. Replacing $x_i$ by $\tilde x_i$, we may thus assume that $x_i \in \fm$. + + +\end{subproof} +% TODO: fix names A_1 = A_S, k_1 = R_S + The ring homomorphism $\ev_x : R = \fl[X_1,\ldots,X_d] \xrightarrow{P \mapsto P(x_1,\ldots,x_d)} A$ is injective. Because $R$ is a UFD, $R$ is normal (\ref{ufdnormal}). Thus the going-down theorem (\ref{gdkrull}) applies to the integral $R$-algebra $A$. + For $0 \le i \le d$, let $\fp_i \se R$ be the ideal generated by $(X_j)_{j=i+1}^d$. We have $\fm \sqcap R = \fp_0$ as all $X_i \in \fm$, hence $X_i \in \fm \sqcap R$ and $\fp_0$ is a maximal ideal. + By applying going-down and induction on $i$, there is a chain $\fm = \fq_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_d$ of elements of $\Spec A$ such that $\fq_i \sqcap R = \fp_i$. + It follows that $\hght(\fm) \ge d$. +\end{subproof} +This finishes the proof in the case of $\fp \in \mSpec A$. + +To reduce the general case to that special case, we proceed as in \ref{trdegresfield}: +By lemma \ref{ltrdegresfieldtrbase} there are $a_1,\ldots,a_n \in A$ whose images in $A / \fp$ form a transcendence base for $\fk(\fp) / \fl$. +As these images are $\fl$-algebraically independent, the same holds for the $a_i$ themselves. + +By lemma \ref{extendtotrbase} we can extend $(a_{i})_{i=1}^n$ to a transcendence base $(a_i)_{i=1}^m \in A^m$ of $K / \fl$. +Let $R \se A$ denote the $\fl$-subalgebra generated by $a_1,\ldots,a_n$ and let $S \coloneqq R \sm \{0\}$. +Let $A_1 \coloneqq A_S$ and $\fp_S$ the prime ideal corresponding to $\fp$ under $\Spec(A_1) \cong \{\fr \in \Spec A | \fr \cap S = \emptyset\}$ (\ref{idealslocbij}). +As in \ref{locandquot}, $A_1$ is a domain with $Q(A_1) \cong K = Q(A)$ and by \ref{locandfactor} $A_1 / \fp_S \cong (A / \fp)_{\overline{S}}$, where $\overline{S}$ denotes the image of $S$ in $A / \fp$. +As in \ref{trdegresfield}, $\fk(\fp_S) \cong \fk(\fp)$ is integral over $A_1 / \fp_S$. +From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1 / \fp_S$ is a field. Hence $\fp_S \in \mSpec(A_1)$ and the special case can be applied to $\fp_S$ and $A_1 / \fl_1$, showing that $\hght(\fp_S) \ge e = \trdeg(K / \fl_1)$. We have $\trdeg(K / \fl_1) = m - n$, as $(a_i)_{i = n+1}^m$ is a transcendence base for $K / \fl_1$. By the description of $\Spec A_S$ (\ref{idealslocbij}), a chain $\fp_S = \fq_0 \supsetneq \ldots \supsetneq \fp_e$ of prime ideals in $A_S$ defines a similar chain $\fp_i \coloneqq \fq_i \sqcap A$ in $A$ with $\fp_0 = \fp$. Thus $\hght(\fp) \ge e$. +\end{proof} + +\begin{remark} + As a consequence of his principal ideal theorem, Krull has shown the finiteness of $\hght(\fp)$ for $\fp \in \Spec A$ when $A$ is a Noetherian ring. But $\dim A = \sup_{\fp \in \Spec A} \hght(\fp) = \sup_{\fm \in \mSpec A} \hght(\fm)$, the Krull dimension of the Noetherian topological space $\Spec A$ may nevertheless be infinite. +\end{remark} +\begin{dexample}[Noetherian ring with infinite dimension]\footnote{\url{https://math.stackexchange.com/questions/1109732/noetherian-ring-with-infinite-krull-dimension-nagatas-example}} + Let $A = \fk[X_i | i \in \N]$ and $m_1, m_2, \ldots \in \N$ an increasing sequence such that $m_{i+1}-m_i > m_i - m_{i-1}$. + Let $\fp_i \coloneqq (X_{m_{i}+1},\ldots,X_{m_{i+1}})$ and $S \coloneqq A \sm \bigcup_{i \in \N} \fp_i$. + $S$ is multiplicatively closed. + $A_S$ is Noetherian but $\hght((\fp_i)_S) = m_{i+1}- m_{i}$ hence $\dim(A_S) = \infty$. +\end{dexample} + +% Lecture 10 + + +\subsection{Dimension of products} + + +\begin{proposition}\label{dimprod} + Let $X \se \fk^n$ and $Y \se \fk^n$ be irreducible and closed. Then $X \times Y$ is also an irreducible closed subset of $\fk^{m+n}$. + Moreover, $\dim(X \times Y) = \dim(X) + \dim(Y)$ and $\codim(X \times Y, \fk^{m+n}) = \codim(X, \fk^m) + \codim(Y, \fk^n)$. +\end{proposition} +\begin{proof} + Let $X = V(\fp)$ and $Y = V(\fq)$ where $\fp \in \Spec \fk[X_1,\ldots,X_m]$ and $\fq \in \Spec \fk[X_1,\ldots,X_n]$. + We denote points of $\fk^{m+n}$ as $x = (x',x'')$ with $x' \in \fk^m, x''\in\fk^n$. Then $X \times Y$ is the set of zeroes of the ideal in $\fk[X_1,\ldots,X_{m+n}]$ generated by the polynomials $f(x) = \phi(x')$, with $\phi$ running over $\fp$ and $g(x) = \gamma(x'')$ with $\gamma$ running over $\fq$. + Thus $X \times Y$ is closed in $\fk^{m+n}$. + We must also show irreducibility. $X \times Y \neq \emptyset$ is obvious. + + Assume that $X \times Y = A_1 \cup A_2$, where the $A_i \se \fk^{m+n}$ are closed. + For $x' \in \fk^m, x' \times Y$ is homeomorphic to the irreducible $Y$. Thus $X = X_1 \cup X_2$ where $X_i = \{x \in X | \{x\} \times Y \se A_i\} $. + Because $X_i = \bigcap_{y \in Y} \{x \in X | (x,y) \in A_i\}$, this is closed. As $X$ is irreducible, there is $i \in \{1;2\} $ which $X_i = X$. Then $X \times Y = A_i$ confirming the irreducibility of $X \times Y$. + + Let $a = \dim X$ and $b = \dim Y$ and $X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_a = X$,$Y_0 \subsetneq Y_1 \subsetneq \ldots \subsetneq Y_b = Y$ be chains of irreducible subsets. By the previous result, + $X_0 \times Y_0 \subsetneq X_1 \times Y_0 \subsetneq \ldots \subsetneq X_a \times Y_0 \subsetneq X_a \times Y_1 \subsetneq \ldots \subsetneq X_a \times Y_a = X \times Y$ is a chain of irreducible subsets. + Thus $\dim(X \times Y) \ge a + b = \dim X + \dim Y$. + Similarly one derives $\codim(X \times Y, \fk^{m+n}) \ge \codim(X, \fk^m) + \codim(Y, \fk^n)$. + By \ref{trdegandkdim} we have $\dim(A) + \codim(A, \fk^l) = l$ for irreducible subsets of $\fk^l$. Thus equality must hold in the previous two inequalities. + +\end{proof} +\subsection{The nil radical} +\begin{notation} + Let $\Vspec(I)$ denote the set of $\fp \in \Spec A$ containing $I$. +\end{notation} + +\begin{proposition}[Nil radical] + For a ring $A$, $\bigcap_{\fp \in \Spec A} \fp = \sqrt{\{0\} } = \{a \in A | \E k \in \N ~ a^k = 0\} \text{\reflectbox{$\coloneqq$}} \nil(A)$, the set of nilpotent elements of $A$. + This is called the \vocab{nil radical} of $A$. +\end{proposition} +\begin{proof} + It is clear that elements of $\sqrt{\{0\} } $ must belong to all prime ideals. Conversely, let $a \in A \sm \sqrt{\{0\} }$. Then $S = a^{\N}$ is a multiplicative subset of $A$ not containing $0$. + The localisation $A_S$ of $A$ is thus not the null ring. Hence $\Spec A_S \neq \emptyset$. If $\fq \in \Spec A_S$, then by the description of $\Spec A_S$ (\ref{idealslocbij}), $\fp \coloneqq \fq \sqcap A$ is a prime ideal of $A$ disjoint from $S$, hence $a \not\in \fp$. +\end{proof} + +\begin{corollary}\label{sqandvspec} + For an ideal $I$ of $R$, $\sqrt{I} = \bigcap_{\fp \in \Vspec(I)} \fp$. +\end{corollary} +\begin{proof} + This is obtained by applying the proposition to $A = R / I $ and using the bijection $\Spec( R / I) \cong V(I)$ sending $\fp \in V(I)$ to $\fp \coloneqq \fp / I$ and $\fq \in \Spec(R / I)$ to its inverse image $\fp$ in $R$. +\end{proof} +\subsubsection{Closed subsets of \texorpdfstring{$\Spec R$}{Spec R}} +\begin{proposition}\label{bijspecideal} + There is a bijection + \begin{align} + f: \{A \se \Spec R | A\text{ closed}\} &\longrightarrow \{I \se R | I \text{ ideal and } I = \sqrt{I} \} \\ + A &\longmapsto \bigcap_{\fp \in A} \fp\\ + \Vspec(I) &\longmapsfrom I + \end{align} + Under this bijection, the irreducible subsets correspond to the prime ideals and the closed points $\{\fm\}, \fm \in \Spec A$ to the maximal ideals. +\end{proposition} +\begin{proof} + If $A = \Vspec(I)$, then by \ref{sqandvspec} $\sqrt{I} = \bigcap_{\fp \in A} \fp$. Thus, an ideal with $\sqrt{I} = I$ can be recovered from $\Vspec( I)$. Since $\Vspec(J) = \Vspec(\sqrt{J})$, the map from ideals with $\sqrt{I} = I$ to closed subsets is surjective. + + Sine $R$ corresponds to $\emptyset$, the proper ideals correspond to non-empty subsets of $\Spec R$. Assume that $\Vspec(I) = \Vspec(J_1) \cup \Vspec(J_2)$, where the decomposition is proper and the ideals coincide with their radicals. + Let $g = f_1f_2$ with $f_k \in J_k \sm I$. Since $\Vspec(g) \supseteq \Vspec(f_k) \supseteq \Vspec(I_k), \Vspec(I) \se \Vspec(g)$. Hence $g \in \sqrt{I} = I$. + As $f_k \not\in I$, $I$ fails to be a prime ideal. + Conversely, assume that $f_1f_2 \in I$ while the factors are not in $I$. Since $I = \sqrt{I}, \Vspec(f_k) \not\supseteq \Vspec(I)$. But $\Vspec(f_1) \cup \Vspec(f_2) = \Vspec(f_1f_2) \supseteq \Vspec(I)$. + The proper decomposition $\Vspec(I) = \left( \Vspec(I) \cap \Vspec(f_1) \right) \cup \left( \Vspec(I) \cap \Vspec(f_2) \right) $ now shows that $\Vspec(I)$ fails to be irreducible. + The final assertion is trivial. +\end{proof} + +\begin{corollary} + If $R$ is a Noetherian ring, then $\Spec R$ is a Noetherian topological space. +\end{corollary} +\begin{remark} + It is not particularly hard to come up with examples which show that the converse implication does not hold. +\end{remark} +\begin{dexample} + Let $A = \fk[X_n | n \in \N] / I$ where $I$ denotes the ideal generated by $\{X_i^2 | i \in \N\}$. + $A$ is not Noetherian, since the ideal $J$ generated by $\{X_i | i \in \N\} $ is not finitely generated. + $A / J \cong \fk$, hence $J$ is maximal. As every prime ideal must contain $\nil(A) \supseteq J$, $J$ is the only prime ideal. + Thus $\Spec A$ contains only one element and is hence Noetherian. +\end{dexample} + +\begin{corollary}[About the smallest prime ideals containing $I$ ]\label{smallestprimesvi} + If $R$ is Noetherian and $I \se R$ an ideal, then the set $\Vspec(I) = \{\fp \in \Spec R | I \se \fp\}$ has finitely many $\se$-minimal elements $(\fp_i)_{i=1}^k$ and every element of $V(I)$ contains at least one $\fp_i$. + The $\Vspec(\fp_i)$ are precisely the irreducible components of $V(I)$. Moreover $\bigcap_{i=1}^k \fp_i = \sqrt{I}$ and $k > 0$ if $I$ is a proper ideal. +\end{corollary} +\begin{proof} + If $\Vspec(I) = \bigcup_{i=1}^k \Vspec(\fp_i)$ is the decomposition into irreducible components then every $\fq \in \Vspec(I)$ must belong to at least one $\Vspec(\fp_i)$, hence $\fp_i \se \fq$. Also $\fp_i \in \Vspec(\fp_i) \se \Vspec(I)$. + It follows that the sets of $\se$-minimal elements of $\Vspec(I)$ and of $\{\fp_1,\ldots,\fp_k\} $ coincide. + As there are no non-trivial inclusions between the $\Vspec(\fp_i)$, there are no non-trivial inclusions between the $\fp_i$ and the assertion follows. + The final remark is trivial. +\end{proof} +\begin{corollary} + If $R$ is any ring, $\hght(\fp) = \codim(\Vspec(\fp), \Spec R)$. +\end{corollary} + + +\subsection{The principal ideal theorem} +Krull was able to show: +\begin{theorem}[Principal ideal theorem / Hauptidealsatz]\label{pitheorem} + Let $A$ be a Noetherian ring, $a \in A$ and $\fp \in \Spec A$ a $\se$-minimal element of $\Vspec(a)$. Then $\hght(\fp) \le 1$. +\end{theorem} +\begin{proof} + Probably not relevant for the exam. +\end{proof} +\begin{remark} + Intuitively, the theorem says that by imposing a single equation one ends up in codimension at most $1$. This would not be true in real analysis (or real algebraic geometry) as the equation $\sum_{i=1}^{n} X_i^2 = 0$ shows. By \ref{smallestprimesvi}, if $a$ is a non-unit then a $\fp \in \Spec A$ to which the theorem applies can always be found. + Using induction on $k$, Krull was able to derive: +\end{remark} + +\begin{theorem}[Generalized principal ideal theorem] + Let $A$ be a Noetherian ring, $(a_i)_{i=1}^k \in A$ and $\fp \in \Spec A$ a $\se$-minimal element of $\bigcap_{i=1}^k V(a_i)$, the set of prime ideals containing all $a_i$. + Then $\hght(\fp) \le k$. +\end{theorem} +Modern approaches to the principal ideal theorem usually give a direct proof of this more general theorem. + +\begin{corollary} + If $R$ is a Noetherian ring and $\fp \in \Spec R$, then $\hght(\fp) < \infty$. +\end{corollary} +\begin{proof} + If $\fp$ is generated by $(f_i)_{i=1}^k$, then $\hght(\fp) \le k$. +\end{proof} +\subsubsection{Application to the dimension of intersections} + +\begin{remark}\label{smallestprimeandirredcomp} +Let $R = \fk[X_1,\ldots,X_n]$ and $I \se R$ an ideal. + + If $(\fp_i)_{i=1}^k$ are the smallest prime ideals of $R$ containing $I$, then $(\Va(\fp_i))_{i=1}^k$ are the irreducible components of $\Va(I)$. +\end{remark} +\begin{proof} + The $\Va(\fp_i)$ are irreducible, there are no non-trivial inclusions between them and $ \Va(I) = \Va(\sqrt{I}) = \Va(\bigcap_{i=1}^k \fp_i) = \bigcup_{i=1}^k \Va(\fp_i)$. +\end{proof} + +\begin{corollary}[of the principal ideal theorem] + \label{corpithm} + Let $X \se \fk^n$ be irreducible, $(f_i)_{i=1}^k$ elements of $R = \fk[X_1,\ldots,X_n]$ and $Y$ an irreducible component of $A = X \cap \bigcap_{i=1}^k V(f_i)$. + Then $\codim(Y,X) \le k$. +\end{corollary} +\begin{remark} + This confirms the naive geometric intuition that by imposing $k$ equations one ends up in codimension at most $k$. +\end{remark} +\begin{proof} + If $X = v(\fp), X \cap \bigcap_{i=1}^k V(f_i) = V(I)$ where $I \se R$ is the ideal generated by $\fp$ and the $f_i$. + By \ref{smallestprimeandirredcomp}, $Y = V(\fq)$ where $\fq$ is the smallest prime ideal containing $I$. + Then $\fq / \fp$ is a smallest prime ideal of $R / \fp$ containing all $(f_i \mod \fp)_{i=1}^k$. + By the principal ideal theorem (\ref{pitheorem}), $\hght(\fq / \fp) \le k$ and the assertion follows from example \ref{htandcodim}. +\end{proof} +\begin{remark}\label{affineproblem} + Note that the intersection $X \cap \bigcap_{i=1}^k V(f_i)$ can easily be empty, even when $k$ is much smaller than $\dim X$. +\end{remark} + +\begin{corollary}\label{codimintersection} + Let $A$ and $B$ be irreducible subsets of $\fk^n$. If $C$ is an irreducible component of $A \cap B$, then $\codim(C, \fk^n) \le \codim(A, \fk^n) + \codim(B, \fk^n)$. +\end{corollary} +\begin{dremark} + Equivalently, $\dim(C) \ge \dim(A) + \dim(B)-n$. +\end{dremark} +\begin{proof} + Let $X = A \times B \se \fk^{2n}$, where we use $(X_1,\ldots,X_n,Y_1,\ldots,Y_n)$ as coordinates of $\fk^{2n}$. + Let $\Delta \coloneqq \{(x_1,\ldots,x_n,x_1,\ldots,x_n) | x \in \fk^n\} $ be the diagonal in $\fk^n \times \fk^n$. + The projection $\fk^{2n}\to \fk^n$ to the $X$-coordinates defines a homeomorphism between $(A \times B) \cap \Delta$ and $A \cap B$. + Thus, $C$ is homeomorphic to an irreducible component $C'$ of $(A \times B) \cap \Delta$ and + \begin{align} + \codim(C, \fk^n) = n - \dim(C) = n - \dim(C') = n - \dim(A \times B) + \codim(C', A \times B)\\ + \overset{\text{\ref{corpithm}}}{\le }2n - \dim(A \times B) \overset{\text{\ref{dimprod}}}{=} 2n - \dim(A) - \dim(B) = \codim(A,\fk^n) + \codim(B, \fk^n) + \end{align} + by the general properties of dimension and codimension, \ref{corpithm} applied to $(X_i - Y_i)_{i=1}^n$, + the result about the dimension of products (\ref{dimprod}) and again the general properties of dimension and codimension. + +\end{proof} +\begin{remark} + As in \ref{affineproblem}, $A \cap B$ can easily be empty, even when $A$ and $B$ have codimension $1$ and $n$ is very large. +\end{remark} + +\subsubsection{Application to the property of being a UFD} +\begin{proposition}\limrel + Let $R$ be a Noetherian domain. Then $R$ is a UFD iff every $\fp \in \Spec R$ with $\hght(\fp)= 1$\footnote{In other words, every $\se$-minimal element of the set of non-zero prime ideals of $R$ } is a principal ideal. +\end{proposition} +\begin{proof} + Every element of every Noetherian domain can be written as a product of irreducible elements.\footnote{Consider the set of principal ideals $rR$ where $r$ is not a product of irreducible elements.} + Thus, $R$ is a UFD iff every irreducible element of $R$ is prime. + + + Assume that this is the case. Let $\fp \in \Spec R, \hght(\fp) = 1$. + Let $p \in \fp \sm \{0\}$. Replacing $p$ by a prime factor of $p$, we may assume $p$ to be prime. Thus $\{0\} \subsetneq pR \se \fp$ is a chain of prime ideals and since $\hght(\fp) = 1$ it follows that $\fp = pR$. + + Conversely, assume that every $\fp \in \Spec R$ with $\hght(\fp)=1$ is a principal ideal. Let $f \in R$ be irreducible. + Let $\fp \in \Spec R$ be a $\se$-minimal element of $V(f)$. By the principal ideal theorem (\ref{pitheorem}), $\hght(\fp)=1$. + Thus $\fp = pR$ for some prime element $p$. We have $p | f$ since $f \in \fp$. As $f$ is irreducible, $p$ and $f$ are multiplicatively equivalent. Thus $f$ is a prime element. +\end{proof} + +\subsection{The Jacobson radical}\limrel +\begin{proposition} + For a ring $A, \bigcap_{\fm \in \mSpec A} \fm = \{a \in A | \A x \in A ~ 1 - ax \in A^{\times }\} \text{\reflectbox{$\coloneqq$}} \rad(A)$, the \vocab{Jacobson radical} of $A$. +\end{proposition} +\begin{proof} + Suppose $\fm \in \mSpec A$ and $a \in A \sm \fm$. Then $a \mod \fm \neq 0$ and $A / \fm$ is a field. Hence $a \mod \fm$ has an inverse $x \mod \fm$. + $1 - ax \in \fm$, hence $1 - ax \not\in A^{\times}$ and $a $ is not al element of the RHS. + + Conversely, let $a \in A$ belong to all $\fm \in \mSpec A$. If there exists $x \in A$ such that $1 - ax \not\in A^{\times }$ then $(1-ax) A$ was a proper ideal in $A$, hence contained in a maximal ideal $\fm$. As $a \in \fm, 1 = (1-ax) + ax \in \fm$, a contradiction. + Hence every element of $\bigcap_{\fm \in \mSpec A} \fm$ belongs to the right hand side. +\end{proof} + +\begin{example} + If $A$ is a local ring, then $\rad(A) = \fm_A$. +\end{example} +\begin{example} + If $A$ is a PID with infinitely many multiplicative equivalence classes of prime elements (e.g. $\Z$ of $\fk[X]$), then $\rad(A) = \{0\}$: + Prime ideals of a PID are maximal. Thus if $x \in \rad(A)$, every prime element divides $x$. If $x \neq 0$, it follows that $x$ has infinitely many prime divisors. + However every PID is a UFD. +\end{example} +\begin{example} +If $A$ is a PID for which $p_1,\ldots,p_n$ is a list of representatives of the multiplicative equivalence classes of prime elements, then +$\rad(A) = f A$ where $f = \prod_{i=1}^{n} p_i$. +\end{example} + +% proof of the pitheorem probably won't be relevant in the exam +% last 2 slides are of "limited relevance" (3 option questions), and may improve grade, but 1.0 can be obtained without it + + + +% Lecture 11 + +\section{Projective spaces} +Let $\fl$ be any field. +\begin{definition} + For a $\fl$-vector space $V$, let $\bP(V)$ be the set of one-dimensional subspaces of $V$. + Let $\bP^n(\fl) \coloneqq \bP(\fl^{n+1})$, the \vocab[Projective space]{$n$-dimensional projective space over $\fl$}. + + If $\fl$ is kept fixed, we will often write $\bP^n$ for $\bP^n(\fl)$. + + When dealing with $\bP^n$, the usual convention is to use $0$ as the index of the first coordinate. + + We denote the one-dimensional subspace generated by $(x_0,\ldots,x_n) \in \fk^{n+1} \sm \{0\}$ by $[x_0,\ldots,x_n] \in \bP^n$. + If $x = [x_0,\ldots,x_n] \in \bP^n$, the $(x_{i})_{i=0}^n$ are called \vocab{homogeneous coordinates} of $x$. + At least one of the $x_{i}$ must be $\neq 0$. +\end{definition} +\begin{remark} + There are points $[1,0], [0,1] \in \bP^1$ but there is no point $[0,0] \in \bP^1$. +\end{remark} +\begin{definition}[Infinite hyperplane] + For $0 \le i \le n$ let $U_i \se \bP^n$ denote the set of $[x_0,\ldots,x_{n}]$ with $x_{i}\neq 0$. + This is a correct definition since two different sets $[x_0,\ldots,x_{n}]$ and $[\xi_0,\ldots,\xi_n]$ of homogeneous coordinates for the same point $x \in \bP^n$ differ by scaling with a $\lambda \in \fl^{\times}$, $x_i = \lambda \xi_i$. Since not all $x_i$ may be $0$, $\bP^n = \bigcup_{i=0}^n U_i$. We identify $\bA^n = \bA^n(\fl) = \fl^n$ with $U_0$ by identifying $(x_1,\ldots,x_n) \in \bA^n$ with $[1,x_1,\ldots,x_n] \in \bP^n$. + Then $\bP^1 = \bA^1 \cup \{\infty\} $ where $\infty=[0,1]$. More generally, when $n > 0$ $\bP^n \sm \bA^n$ can be identified with $\bP^{n-1}$ identifying $[0,x_1,\ldots,x_n] \in \bP^n \sm \bA^n$ with $[x_1,\ldots,x_n] \in \bP^{n-1}$. + + Thus $\bP^n$ is $\bA^n \cong \fl^n$ with a copy of $\bP^{n-1}$ added as an \vocab{infinite hyperplane} . +\end{definition} + +\subsubsection{Graded rings and homogeneous ideals} +\begin{notation} + Let $\bI = \N$ or $\bI = \Z$. +\end{notation} +\begin{definition} + By an \vocab[Graded ring]{$\bI$-graded ring} $A_\bullet$ we understand a ring $A$ with a collection $(A_d)_{d \in \bI}$ of subgroups of the additive group $(A, +)$ such that $A_a \cdot A_b \se A_{a + b}$ for $a,b \in \bI$ and such that $A = \bigoplus_{d \in \bI} A_d$ in the sense that every $r \in A$ has a unique decomposition $r = \sum_{d \in \bI} r_d$ with $r_d \in A_d$ and but finitely many $r_d \neq 0$. + + We call the $r_d$ the \vocab{homogeneous components} of $r$. + + An ideal $I \se A$ is called \vocab{homogeneous} if $r \in I \implies \A d \in \bI ~ r_d \in I_d$ where $I_d \coloneqq I \cap A_d$. + + By a \vocab{graded ring} we understand an $\N$-graded ring. Tin this case, $A_{+} \coloneqq \bigoplus_{d=1}^{\infty} A_d = \{r \in A | r_0 = 0\} $ is called the \vocab{augmentation ideal} of $A$. +\end{definition} +\begin{remark}[Decomposition of $1$] + If $1 = \sum_{d \in \bI} \eps_d$ is the decomposition into homogeneous components, then $\eps_a = 1 \cdot \eps_a = \sum_{b \in \bI} \eps_a\eps_b$ with $\eps_a\eps_b \in A_{a+b}$. + By the uniqueness of the decomposition into homogeneous components, $\eps_a \eps_0 = \eps_a$ and $b \neq 0 \implies \eps_a \eps_b = 0$. + Applying the last equation with $a = 0$ gives $b\neq 0 \implies \eps_b = \eps_0 \eps _b = 0$. + Thus $1 = \eps_0 \in A_0$. +\end{remark} +\begin{remark} + The augmentation ideal of a graded ring is a homogeneous ideal. +\end{remark} + +% Graded rings and homogeneous ideals (2) + +\begin{proposition}\footnote{This holds for both $\Z$-graded and $\N$-graded rings.} + \begin{itemize} + \item A principal ideal generated by a homogeneous element is homogeneous. + \item The operations $\sum, \bigcap, \sqrt{}$ preserve homogeneity. + \item An ideal is homogeneous iff it can be generated by a family of homogeneous elements. + \end{itemize} +\end{proposition} +\begin{proof} + Most assertions are trivial. We only show that $J$ homogeneous $\implies \sqrt{J} $ homogeneous. + Let $A$ be $\bI$-graded, $f \in \sqrt{J} $ and $f = \sum_{d \in \bI} f_d$ the decomposition. + To show that all $f_d \in \sqrt{J} $, we use induction on $N_f \coloneqq \# \{d \in \bI | f_d \neq 0\}$. + $N_f = 0$ is trivial. Suppose $N_f > 0$ and $e \in \bI$ is maximal with $f_e \neq 0$. + For $l \in \N$, the $le$-th homogeneous component of $f^l$ is $f_e^l$. Choosing $l$ large enough such that $f^l \in J$ and using the homogeneity of $J$, we find $f_e \in \sqrt{J}$. + As $\sqrt{J} $ is an ideal, $\tilde f \coloneqq f - f_e \in \sqrt{J} $. As $N_{\tilde f} = N_f -1$, the induction assumption may be applied to $\tilde f$ and shows $f_d \in \sqrt{J} $ for $d \neq e$. +\end{proof} +\begin{fact} + A homogeneous ideal is finitely generated iff it can be generated by finitely many of its homogeneous elements. + In particular, this is always the case when $A$ is a Noetherian ring. +\end{fact} + + +\subsubsection{The Zariski topology on $\bP^n$} +\begin{notation} + Recall that for $\alpha \in \N^{n+1}$ $|\alpha| = \sum_{i=0}^{n} \alpha_i$ and $x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}$. +\end{notation} +\begin{definition}[Homogeneous polynomials] + Let $R$ be any ring and $f = \sum_{\alpha \in \N^{n+1}} f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n]$. + We say that $f$ is \vocab{homogeneous of degree $d$} if $|\alpha| \neq d \implies f_\alpha = 0$ . + We denote the subset of homogeneous polynomials of degree $d$ by $R[X_0,\ldots,X_n]_d \se R[X_0,\ldots,X_n]$. +\end{definition} +\begin{remark} + This definition gives $R$ the structure of a graded ring. +\end{remark} +\begin{definition}[Zariski topology on $\bP^n(\fk)$]\label{ztoppn} + Let $A = \fk[X_0,\ldots,X_n]$.\footnote{As always, $\fk$ is algebraically closed} + For $f \in A_d = \fk[X_0,\ldots,X_n]_d$, the validity of the equation $f(x_0,\ldots,x_{n}) = 0$ does not depend on the choice of homogeneous coordinates, as + \[ + f(\lambda x_0,\ldots, \lambda x_n) 0 \lambda^d f(x_0,\ldots,x_n) + \] + Let $\Vp(f) \coloneqq \{x \in \bP^n | f(x) = 0\}$. + + We call a subset $X \se \bP^n$ Zariski-closed if it can be represented as + \[ + X = \bigcap_{i=1}^k \Vp(f_i) + \] + where the $f_i \in A_{d_i}$ are homogeneous polynomials. +\end{definition} +\pagebreak +\begin{fact} + If $X = \bigcap_{i = 1}^k \Vp(f_i) \se \bP^n$ is closed, then $Y = X \cap \bA^n$ can be identified with the closed subset + \[ + \{(x_1,\ldots,x_n) \in \fk^n | f_i(1,x_1,\ldots,x_n) = 0, 1 \le i \le k\} \se \fk^n + \] + Conversely, if $Y \se \fk^n$ is closed it has the form + \[ + \{(x_1,\ldots,x_n) \in \fk^n | g_i(x_1,\ldots,x_n) = 0, 1 \le i \le k\} + \] + and can thus be identified with $X \cap \bA^n$ where $X \coloneqq \bigcap_{i=1}^k \Vp(f_i)$ is given by \[f_i(X_0,\ldots,X_n) \coloneqq X_0^{d_i} g_i(X_1 / X_0,\ldots, X_n / X_0), d_i \ge \deg(g_i)\] + Thus, the Zariski topology on $\fk^n$ can be identified with the topology induced by the Zariski topology on $\bA^n = U_0$, and the same holds for $U_i$ with $0 \le i \le n$. + + In this sense, the Zariski topology on $\bP^n$ can be thought of as gluing the Zariski topologies on the $U_i \cong \fk^n$. +\end{fact} + +% The Zariski topology on P^n (2) + +\begin{definition} + Let $I \se A = \fk[X_0,\ldots,X_n]$ be a homogeneous ideal. + Let $\Vp(I) \coloneqq \{[x_0,\ldots,_n] \in \bP^n | \A f \in I ~ f(x_0,\ldots,x_n) = 0\}$ + As $I$ is homogeneous, it is sufficient to impose this condition for the homogeneous elements $f \in I$. + Because $A$ is Noetherian, $I$ can finitely generated by homogeneous elements $(f_i)_{i=1}^k$ and $\Vp(I)=\bigcap_{i=1}^k \Vp(f_i)$ as in \ref{ztoppn}. + Conversely, if the homogeneous $f_i$ are given, then $I = \langle f_1,\ldots,f_k \rangle_A$ is homogeneous. +\end{definition} +\begin{remark} + Note that $V(A) = V(A_+) = \emptyset$. +\end{remark} +\begin{fact} + For homogeneous ideals in $A$ and $m \in \N$, we have: + \begin{itemize} + \item $\Vp(\sum_{\lambda \in \Lambda} I_\lambda) = \bigcap_{\lambda \in \Lambda} \Vp(I_\lambda)$ + \item $\Vp(\bigcap_{k=1}^m I_k) = \Vp(\prod_{k=1}^{m} I_k) = \bigcup_{k=1}^m \Vp(I_k)$ + \item $\Vp(\sqrt{I}) = \Vp(I)$ + \end{itemize} +\end{fact} +\begin{fact} + If $X = \bigcup_{\lambda \in \Lambda} U_\lambda$ is an open covering of a topological space then $X$ is Noetherian iff there is a finite subcovering and all $U_\lambda$ are Noetherian. +\end{fact} +\begin{proof} + By definition, a topological space is Noetherian $\iff$ all open subsets are quasi-compact. +\end{proof} +\begin{corollary} + The Zariski topology on $\bP^n$ is indeed a topology. + The induced topology on the open set $\bA^n = \bP^n \sm \Vp(X_0) \cong \fk^n$ is the Zariski topology on $\fk^n$. + The same holds for all $U_i = \bP^n \sm \Vp(X_i) \cong \fk^n$. + Moreover, the topological space $\bP^n$ is Noetherian. +\end{corollary} + +\subsection{Noetherianness of graded rings} +\begin{proposition} + For a graded ring $R_{\bullet}$, the following conditions are equivalent: + \begin{enumerate}[A] + \item $R$ is Noetherian. + \item Every homogeneous ideal of $R_{\bullet}$ is finitely generated. + \item Every chain $I_0\se I_1 \se \ldots$ of homogeneous ideals terminates. + \item Every set $\fM \neq \emptyset$ of homogeneous ideals has a $\se$-maximal element. + \item $R_0$ is Noetherian and the ideal $R_+$ is finitely generated. + \item $R_0$ is Noetherian and $R / R_0$ is of finite type. + \end{enumerate} +\end{proposition} +\begin{proof} + \noindent\textbf{A $\implies$ B,C,D} trivial. + + \noindent\textbf{B $\iff$ C $\iff$ D} similar to the proof about Noetherianness. + + \noindent\textbf{B $\land$ C $\implies $E} B implies that $R_+$ is finitely generated. Since $I \oplus R_+$ is homogeneous for any homogeneous ideal $I \se R_0$, C implies the Noetherianness of $R_0$. + +\noindent\textbf{E $\implies$ F} Let $R_+$ be generated by $f_i \in R_{d_i}, d_i > 0$ as an ideal. + \begin{claim} + The $R_0$-subalgebra $\tilde R$ of $R$ generated by the $f_i$ equals $R$. + \end{claim} + \begin{subproof} + It is sufficient to show that every homogeneous $f \in R_d$ belongs to $\tilde R$. We use induction on $d$. The case of $d = 0$ is trivial. + Let $d > 0$ and $R_e \se \tilde R$ for all $e < d$. + as $f \in R_+$, $f = \sum_{i=1}^{k} g_if_i$. Let $f_a = \sum_{i=1}^{k} g_{i, a-d_i} f_i$, where $g_i = \sum_{b=0}^{\infty} g_{i,b}$ is the decomposition into homogeneous components. + Then $f = \sum_{a=0}^{\infty} f_a$ is the decomposition of $f$ into homogeneous components, hence $a \neq d \implies f_a = 0 $. Thus we may assume $g_i \in R_{d-d_i}$. + As $d_i > 0$, the induction assumption may now be applied to $g_i$, hence $g_i \in \tilde R$, hence $f \in \tilde R$. + \end{subproof} + + \noindent\textbf{F $\implies$ A} Hilbert's Basissatz (\ref{basissatz}) + +\end{proof} + + + +% Lecture 12 + +\subsection{The projective form of the Nullstellensatz and the closed subsets of $\bP^n$} +Let $A = \fk[X_0,\ldots,X_n]$. +\begin{proposition}[Projective form of the Nullstellensatz]\label{hnsp} + If $I \se A$ is a homogeneous ideal and $f \in A_d$ with $d>0$, then $\Vp(I) \se \Vp(f) \iff f \in \sqrt{I}$. +\end{proposition} +\begin{proof} + $\impliedby$ is clear. Let $\Vp(I) \se \Vp(f)$. If $x = (x_0,\ldots,x_n) \in \Va(I)$, then either $x = 0$ in which case $f(x) = 0$ since $d > 0$ + or the point $[x_0,\ldots,x_n] \in \bP^n$ is well-defined and belongs to $\Vp(I) \se \Vp(f)$, hence $f(x) = 0$. + Thus $\Va(I) \se \Va(f)$ and $f \in \sqrt{I}$ be the Nullstellensatz (\ref{hns3}). +\end{proof} + +\begin{definition}\footnote{This definition is not too important, the characterization in the following remark suffices.}. + For a graded ring $R_\bullet$, let $\Proj(R_\bullet)$ be the set of $\fp \in \Spec R$ such that $\fp$ is a homogeneous ideal and $\fp \not\supseteq R_+$. +\end{definition} +\begin{remark}\label{proja} + As the elements of $A_0 \sm \{0\}$ are units in $A$ it follows that for every homogeneous ideal $I$ we have $I \se A_+$ or $I = A$. + In particular, $\Proj(A_\bullet) = \{\fp \in \Spec A \sm A_+ | \fp \text{ is homogeneous}\} $. +\end{remark} +\begin{proposition}\label{bijproj} + There is a bijection + \begin{align} + f: \{I \se A_+ | I \text{ homogeneous ideal}, I = \sqrt{I}\} &\longrightarrow \{X \se \bP^n | X \text{ closed}\} \\ + I &\longmapsto \Vp(I)\\ + \langle \{f \in A_d | d > 0, X \se \Vp(f)\} \rangle & \longmapsfrom X + \end{align} + Under this bijection, the irreducible subsets correspond to the elements of $\Proj(A_\bullet)$. +\end{proposition} +\begin{proof} + From the projective form of the Nullstellensatz it follows that $f$ is injective and that $f\inv(\Vp\left( I \right)) = \sqrt{I} = I$. + If $X \se \bP^n$ is closed, then $X = \Vp(J)$ for some homogeneous ideal $J \se A$. \Wlog $J = \sqrt{J}$. If $J \not\se A_+$, then $J = A$ (\ref{proja}), hence $X = \Vp(J) = \emptyset = \Vp(A_+)$. + Thus we may assume $J \se A_+$, and $f$ is surjective. + + + Suppose $\fp \in \Proj(A_\bullet)$. Then $\fp \neq A_+$ hence $X = \Vp(\fp) \neq \emptyset$ by the proven part of the proposition. + Assume $X = X_1 \cup X_2$ is a decomposition into proper closed subsets, where $X_k = \Vp(I_k)$ for some $I_k \se A_+, I_k = \sqrt{I_k}$. Since $X_k$ is a proper subset of $X$, there is $f_k \in I_k \sm \fp$. + We have $\Vp(f_1f_2) \supseteq \Vp(f_k) \supseteq \Vp(I_k)$ hence $\Vp(f_1f_2) \supseteq \Vp(I_1) \cup \Vp(I_2) = X = \Vp(\fp)$ and it follows that $f_1f_2\in \sqrt{\fp} = \fp \lightning$. + + Assume $X = \Vp(\fp)$ is irreducible, where $\fp = \sqrt{\fp} \in A_+$ is homogeneous. The $\fp \neq A_+$ as $X = \emptyset$ otherwise. Assume that $f_1f_2 \in \fp$ but $f_i \not\in A_{d_i} \sm \fp$. + Then $X \not \se \Vp(f_i)$ by the projective Nullstellensatz when $d_i > 0$ and because $\Vp(1) = \emptyset$ when $d_i = 0$. + Thus $X = (X \cap \Vp\left( f_1 \right)) \cup (X \cap \Vp(f_2))$ is a proper decomposition $\lightning$. + By lemma \ref{homprime}, $\fp$ is a prime ideal. + +\end{proof} +\begin{remark} + It is important that $I \se A_{\color{red} +}$, since $\Vp(A) = \Vp(A_+) = \emptyset$ would be a counterexample. +\end{remark} +\begin{corollary} + $\bP^n$ is irreducible. +\end{corollary} +\begin{proof} + Apply \ref{bijproj} to $\{0\} \in \Proj(A_\bullet)$. +\end{proof} + +\subsection{Some remarks on homogeneous prime ideals} +\begin{lemma}\label{homprime} + Let $R_\bullet$ be an $\bI$ graded ring ($\bI = \N$ or $\bI = \Z$). + A homogeneous ideal $I \se R$ is a prime ideal iff $1 \not\in I$ and for homogeneous elements $f, g \in R , fg \in I \implies f \in I \lor g \in I$. +\end{lemma} +\begin{proof} + $\implies$ is trivial. + It suffices to show that for arbitrary $f,g \in R fg \in I \implies f \in I \lor g \in I$. + Let $f = \sum_{d \in \bI} f_d, g = \sum_{d \in \bI} g_d $ be the decompositions into homogeneous components. + If $f \not\in I$ and $g \not\in I$ there are $d,e \in I$ with $f_d \in I, g_e \in I$, and they may assumed to be maximal with this property. + As $I$ is homogeneous and $fg \in I$, we have $(fg)_{d+e} \in I$ but + \[ + (fg)_{d+e} = f_dg_e + \sum_{\delta = 1}^{\infty} (f_{d + \delta} g_{e - \delta} + f_{d - \delta} g_{e + \delta}) + \] + where $f_dg_e \not\in I$ by our assumption on $I$ and all other summands on the right hand side are $\in I$ (as $f_{d+ \delta} \in I$ and $g_{e + \delta} \in I$ by the maximality of $d$ and $e$), a contradiction. +\end{proof} + +\begin{remark} + If $R_\bullet$ is $\N$-graded and $\fp \in \Spec R_0$, then $\fp \oplus R_+ = \{r \in R | r_0 \in \fp\} $ is a homogeneous prime ideal of $R$. + \[\{\fp \in \Spec R | \fp \text{ is a homogeneous ideal of } R_\bullet\} = \Proj(R_\bullet) \sqcup \{\fp \oplus R_+ | \fp \in \Spec R_0\}\] +\end{remark} + +\subsection{Dimension of $\bP^n$} +\begin{proposition} + \begin{itemize} + \item $\bP^n$ is catenary. + \item $\dim(\bP^n) = n$. Moreover, $\codim(\{x\} ,\bP^n) = n$ for every $x \in \bP^n$. + \item If $X \se \bP^n$ is irreducible and $x \in X$, then $\codim(\{x\}, X) = \dim(X) = n - \codim(X, \bP^n)$. + \item If $X \se Y \se \bP^n$ are irreducible subsets, then $\codim(X,Y) = \dim(Y) - \dim(X)$. + \end{itemize} +\end{proposition} +\begin{proof} + Let $X \se \bP^n$ be irreducible. If $x \in X$, there is an integer $0 \le i \le n$ and $X \in U_i = \bP^n \sm \Vp(X_i)$. + \Wlog $i = 0$. Then $\codim(X, \bP^n) = \codim(X \cap \bA^n, \bA^n)$ by the locality of Krull codimension (\ref{lockrullcodim}). + Applying this with $X = \{x\}$ and our results about the affine case gives the second assertion. + If $Y$ and $Z$ are also irreducible with $X \se Y \se Z$, then $\codim(X,Y) = \codim(X \cap \bA^n, Y \cap \bA^n)$, $\codim(X,Z) = \codim(X \cap \bA^n, Z \cap \bA^n)$ and $\codim(Y,Z) = \codim(Y \cap \bA^n, Z \cap \bA^n)$. + Thus + \begin{align} + \codim(X,Y) + \codim(Y,Z) &= \codim(X \cap \bA^n, Y \cap \bA^n) + \codim(Y \cap \bA^n, Z \cap \bA^n)\\ + &= \codim(X \cap \bA^n, Z \cap \bA^n)\\ + &= \codim(X, Z) + \end{align} + because $\fk^n$ is catenary and the first point follows. + The remaining assertions can easily be derived from the first two. +\end{proof} + +\subsection{The cone $C(X)$} +\begin{definition} + If $X \se \bP^n$ is closed, we define the \vocab{affine cone over $X$} + \[ + C(X) = \{0\} \cup \{(x_0,\ldots,x_n) \in \fk^{n+1} \sm \{0\} | [x_0,\ldots,x_n] \in X\} + \] + If $X = \Vp(I)$ where $I \se A_+ = \fk[X_0,\ldots,X_n]_+$ is homogeneous, then $C(X) = \Va(I)$. +\end{definition} +\begin{proposition}\label{conedim} + \begin{itemize} + \item $C(X)$ is irreducible iff $X$ is irreducible or $X = \emptyset$. + \item If $X$ is irreducible, then + + $\dim(C(X)) = \dim(X) + 1$ and + + $\codim(C(X), \fk^{n+1}) = \codim(X, \bP^n)$ + \end{itemize} +\end{proposition} +\begin{proof} + The first assertion follows from \ref{bijproj} and \ref{bijiredprim} (bijection of irreducible subsets and prime ideals in the projective and affine case). + + Let $d = \dim(X)$ and + \[ + X_0 \subsetneq \ldots \subsetneq X_d = X \subsetneq X_{d+1} \subsetneq \ldots \subsetneq X_n = \bP^n + \] + be a chain of irreducible subsets of $\bP^n$. Then + \[ + \{0\} \subsetneq C(X_0) \subsetneq \ldots \subsetneq C(X_d) = C(X) \subsetneq \ldots \subsetneq C(X_n) = \fk^{n+1} + \] + is a chain of irreducible subsets of $\fk^{n+1}$. Hence $\dim(C(X)) \ge 1 + d$ and $\codim(C(X), \fk^{n+1}) \ge n-d$. Since $\dim(C(X)) + \codim(C(X), \fk^{n+1}) = \dim(\fk^{n+1}) = n+1$, the two inequalities must be equalities. +\end{proof} +\subsubsection{Application to hypersurfaces in $\bP^n$} +\begin{definition}[Hypersurface] + Let $n > 0$. + By a \vocab{hypersurface} in $\bP^n$ or $\bA^n$ we understand an irreducible closed subset of codimension $1$. +\end{definition} + +\begin{corollary} + If $P \in A_d$ is a prime element, then $H = \Vp(P)$ is a hypersurface in $\bP^n$ and every hypersurface $H$ in $\bP^n$ can be obtained in this way. +\end{corollary} +\begin{proof} + If $H = \Vp(P)$ then $C(H) = \Va(P)$ is a hypersurface in $\fk^{n+1}$ by \ref{irredcodimone}. By \ref{conedim}, $H$ is irreducible and of codimension $1$. + + Conversely, let $H$ be a hypersurface in $\bP^n$. By \ref{conedim}, $C(H)$ is a hypersurface in $\fk^{n+1}$, hence $C(H) = \Vp(P)$ for some prime element $P \in A$ (again by \ref{irredcodimone}). + We have $H = \Vp(\fp)$ for some $\fp \in \Proj(A)$ and $C(H) = \Va(\fp)$. By the bijection between closed subsets of $\fk^{n+1}$ and ideals $I = \sqrt{I} \se A$ (\ref{antimonbij}), $\fp = P \cdot A$. +Let $P = \sum_{k=0}^{d}P_k$ with $P_d \neq 0$ be the decomposition into homogeneous components. +If $P_e $ with $e < d$ was $\neq 0$, it could not be a multiple of $P$ contradicting the homogeneity of $\fp = P \cdot A$. Thus, $P$ is homogeneous of degree $d$. +\end{proof} +\begin{definition} + A hypersurface $H \se \bP^n$ has \vocab{degree $d$} if $H = \Vp(P)$ where $P \in A_d$ is an irreducible polynomial. +\end{definition} + +\subsubsection{Application to intersections in $\bP^n$ and Bezout's theorem} +\begin{corollary} + Let $A \se \bP^n$ and $B \se \bP^n$ be irreducible subsets of dimensions $a$ and $b$. If $a+ b \ge n$, then $A \cap B \neq \emptyset$ and every irreducible component of $A \cap B$ as dimension $\ge a + b - n$. +\end{corollary} + +\begin{remark} + This shows that $\bP^n$ indeed fulfilled the goal of allowing for nicer results of algebraic geometry because ``solutions at infinity'' to systems of algebraic equations are present in $\bP^n$ + (see \ref{affineproblem}). +\end{remark} + +\begin{proof} + The lower bound on the dimension of irreducible components of $A \cap B$ is easily derived from the similar affine result (corollary of the principal ideal theorem, \ref{codimintersection}). + From the definition of the affine cone it follows that $C(A \cap B) = C(A) \cap C(B)$. + We have $\dim(C(A)) = a+1$ and $\dim(C(B)) = b + 1$ by \ref{conedim}. + If $A \cap B = \emptyset$, then $C(A) \cap C(B) = \{0\}$ with $\{0\} $ as an irreducible component, contradicting the lower bound $a + b + 1 - n > 0$ for the dimension of irreducible components of $C(A) \cap C(B)$ (again \ref{codimintersection}). +\end{proof} +\begin{remark}[Bezout's theorem] + If $A \neq B$ are hypersurfaces of degree $a$ and $b$ in $\bP^2$, then $A \cap B$ has $ab$ points counted by (suitably defined) multiplicity. +\end{remark} + + +%TODO Proof of "Dimension of P^n" +% SLIDE APPLICATION TO HYPERSURFACES IN $\P^n$ +%ERROR: C(H) = V_A(P) +%If n = 0, P = 0, V_P(P) = \emptyset is a problem! + + + +% Lecture 13 +\section{Varieties} + +\subsection{Sheaves} + +\begin{definition}[Sheaf] + Let $X$ be any topological space. + + A \vocab{presheaf} $\cG$ of sets (or rings, (abelian) groups) on $X$ associates a set (or rings, or (abelian) group) $\cG(U)$ to every open subset $U$ of $X$, and a map (or ring or group homomorphism) $\cG(U) \xrightarrow{r_{U,V}} \cG(V)$ to every inclusion $V \se U$ of open subsets of $X$ such that $r_{U,W} = r_{V,W} r_{U,V}$ for inclusions $U \se V \se W$ of open subsets. + + Elements of $\cG(U)$ are often called \vocab{sections} of $\cG$ on $U$ or \vocab{global sections} when $U = X$. + + Let $U \se X$ be open and $U = \bigcup_{i \in I} U_i$ an open covering. + A family $(f_i)_{i \in I} \in \prod_{i \in I} \cG(U_i)$ is called \vocab[Sections!compatible]{compatible} if $r_{U_i, U_i \cap U_j}(f_i) = r_{U_j, U_i \cap U_j}(f_j)$ for all $i,j \in I$. + + Consider the map + \begin{align} + \phi_{U, (U_i)_{i \in I}}: \cG(U) &\longrightarrow \{(f_i)_{i \in I} \in \prod_{i \in I} \cG(U_i) | r_{U_i, U_i \cap U_j}(f_i) = r_{U_j, U_i \cap U_j}(f_j) \text{ for } i,j \in I \} \\ + f &\longmapsto (r_{U, U_i}( f))_{i \in I} + \end{align} + + A presheaf is called \vocab[Presheaf!separated]{separated} if $\phi_{U, (U_i)_{i \in I}}$ is injective for all such $U$ and $(U_i)_{i \in I}$.\footnote{This also called ``locality''.} + It satisfies \vocab{gluing} if $\phi_{U, (U_i)_{i \in I}}$ is surjective. + + A presheaf is called a \vocab{sheaf} if it is separated and satisfies gluing. + + The bijectivity of the $\phi_{U, (U_i)_{i \in I}}$ is called the \vocab{sheaf axiom}. +\end{definition} +\begin{dtrivial} + A presheaf is a contravariant functor $\cG : \cO(X) \to C$ where $\cO(X)$ denotes the category of open subsets of $X$ with inclusions as morphisms and $C$ is the category of sets, rings or (abelian) groups. +\end{dtrivial} +\begin{definition} + A subsheaf $\cG'$ is defined by subsets (resp. subrings or subgroups) $\cG'(U) \se \cG(U)$ for all open $U \se X$ such that the sheaf axioms still hold. +\end{definition} +\begin{remark} + If $\cG$ is a sheaf on $X$ and $\Omega \se X$ open, then $\cG\defon{\Omega}(U) \coloneqq \cG(U)$ for open $U \se \Omega$ and $r_{U,V}^{(\cG\defon{\Omega})}(f) \coloneqq r_{U,V}^{(\cG)}(f)$ is a sheaf of the same kind as $\cG$ on $\Omega$. +\end{remark} +\begin{remark} + The notion of restriction of a sheaf to a closed subset, or of preimages under general continuous maps, can be defined but this is a bit harder. +\end{remark} +\begin{notation} + It is often convenient to write $f \defon{V}$ instead of $r_{U,V}(f)$. +\end{notation} +\begin{remark} + Applying the \vocab{sheaf axiom} to the empty covering of $U = \emptyset$, one finds that $\cG(\emptyset) = \{0\} $. +\end{remark} + + + + +\subsubsection{Examples of sheaves} +\begin{example} + Let $G$ be a set and let $\fG(U)$ be the set of arbitrary maps $U \xrightarrow{f} G$. We put $r_{U,V}(f) = f\defon{V}$. + It is easy to see that this defines a sheaf. + If $\cdot $ is a group operation on $G$, then $(f\cdot g)(x) \coloneqq f(x)\cdot g(x)$ defines the structure of a sheaf of group on $\fG$. + Similarly, a ring structure on $G$ can be used to define the structure of a sheaf of rings on $\fG$. +\end{example} +\begin{example} + If in the previous example $G$ carries a topology and $\cG(U) \se \fG(U)$ is the subset (subring, subgroup) of continuous functions $U \xrightarrow{f} G$, then $\cG$ is a subsheaf of $\fG$, called the sheaf of continuous $G$-valued functions on (open subsets of) $X$. +\end{example} + +\begin{example} + If $X = \R^n$, $\bK \in \{\R, \C\}$ and $\cO(U)$ is the sheaf of $\bK$-valued $C^{\infty}$-functions on $U$, then $\cO$ is a subsheaf of the sheaf (of rings) of $\bK$-valued continuous functions on $X$. +\end{example} +\begin{example} + If $X = \C^n$ and $\cO(U)$ the set of holomorphic functions on $X$, then $\cO$ is a subsheaf of the sheaf of $\C$-valued $C^{\infty}$-functions on $X$. +\end{example} +\subsubsection{The structure sheaf on a closed subset of $\fk^n$} + +Let $X \se \fk^n$ be open. Let $R = \fk[X_1,\ldots,X_n]$. +\begin{definition}\label{structuresheafkn} + For open subsets $U \se X$, let $\cO_X(U)$ be the set of functions $U \xrightarrow{\phi} \fk$ such that every $x \in U$ has a neighbourhood $V$ such that there are $f,g \in R$ such that for $y \in V$ we have $g(y) \neq 0$ and $\phi(y) = \frac{f(y)}{g(y)}$. +\end{definition} + +\begin{remark}\label{structuresheafcontinuous} + $\cO_X$ is a subsheaf (of rings) of the sheaf of $\fk$-valued functions on $X$. + The elements of $\cO_X(U)$ are continuous: + Let $M \se \fk$ be closed. We must show the closedness of $N \coloneqq \phi\inv(M)$ in $U$. For $M = \fk$ this is trivial. Otherwise $M$ is finite and we may assume $M = \{t\} $ for some $t \in \fk$. For $x \in U$, there are open $V_x \se U$ and $f_x, g_x \in R$ such that $\phi = \frac{f_x}{g_x}$ on $V_x$. + Then $N \cap V_x = V(f_x - t\cdot g_x) \cap V_x)$ is closed in $V_x$. As the $V_x$ cover $U$ and $U$ is quasi-compact, $N$ is closed in $U$. +\end{remark} + +\begin{proposition}\label{structuresheafri} + Let $X = V(I)$ where $I = \sqrt{I} \se R$ is an ideal. Let $A = R / I$. Then + \begin{align} + \phi: A &\longrightarrow \cO_X(X) \\ + f \mod I &\longmapsto f\defon{X} + \end{align} +is an isomorphism. +\end{proposition} + +\begin{proof} + It is easy to see that the map $A \to \cO_X(X)$ is well-defined and a ring homomorphism. + Its injectivity follows from the Nullstellensatz and $I = \sqrt{I}$ (\ref{hns3}). + + + Let $\phi \in \cO_X(X)$. for $x \in X$, there are an open subset $U_x \se X$ and $f_x, g_x \in R$ such that $\phi = \frac{f_x}{g_x}$ on $U_x$. + \begin{claim} + \Wlog we can assume $U_x = X \sm V(g_x)$. + \end{claim} + \begin{subproof} + The closed subsets $(X \sm U_x) \se \fk^n$ has the form $X\sm U_x = V(J_x)$ for some ideal $J_x \se R$. + As $x \not\in X \sm V_x$ there is $h_x \in J_x$ with $h_x(x) \neq 0$. + Replacing $U_x$ by $X \sm V(h_x)$, $f_x$ by $f_xh_x$ and $g_x$ by $g_xh_x$, we may assume $U_x = X \sm V(g_x)$. + \end{subproof} + \begin{claim} + \Wlog we can assume $V(g_x) \se V(f_x)$. + \end{claim} + \begin{subproof} + Replace $f_x$ by $f_xg_x$ and $g_x$ by $g_x^2$. + \end{subproof} + As $X$ is quasi-compact, there are finitely many points $(x_i)_{i=1}^m$ such that the $U_{x_i}$ cover $X$. + Let $U_i \coloneqq U_{x_i}, f_i \coloneqq f_{x_i}, g_i \coloneqq g_{x_i}$. + + As the $U_i = X \sm V(g_i)$ cover $X$, $V(I) \cap \bigcap_{i=1}^m V(g_i) = X \cap \bigcap_{i=1}^m V(g_i) = \emptyset$. + By the Nullstellensatz (\ref{hns1}) the ideal of $R$ generated by $I$ and the $a_i$ equals $R$. + There are thus $n \ge m \in \N$ and elements $(g_i)_{i = m+1}^n$ of $I$ and $(a_i)_{i=1}^n \in R^n$ such that $1 = \sum_{i=1}^{n} a_ig_i$. + Let for $i > m$ $f_i \coloneqq 0$, $F = \sum_{i=1}^{n} a_if_i = \sum_{i=1}^{m} a_if_i \in R$. + + \begin{claim} + For all $x \in X $ ~ $f_i(x) = \phi(x) g_i(x)$. + \end{claim} + \begin{subproof} + If $x \in V_i$ this follows by our choice of $f_i$ and $g_i$. + If $x \in X \sm V_i$ or $i > m$ both sides are zero. + \end{subproof} + It follows that + \[ + \phi(x) = \phi(x) \cdot 1 = \phi(x) \cdot \sum_{i=1}^{n} a_i(x) g_i(x) = \sum_{i=1}^{n} a_i(x) f_i(x) = F(x) + \] + Hence $\phi = F\defon{X}$. +\end{proof} +\subsubsection{The structure sheaf on closed subsets of $\bP^n$} +Let $X \se \bP^n$ be closed and $R_\bullet = \fk[X_0,\ldots,X_n]$ with its usual grading. + +\begin{definition}\label{structuresheafpn} + For open $U \se X$, let $\cO_X(U)$ be the set of functions $U \xrightarrow{\phi} \fk$ such that for every $x \in U$, there are an open subset $W \se U$, a natural number $d$ and $f,g \in R_d$ such that $W \cap \Vp(g) = \emptyset$ and $\phi(y) = \frac{f(y_0,\ldots,y_n)}{g(y_0,\ldots,y_n)}$ for $y = [y_0,\ldots,y_n] \in W$. +\end{definition} + +\begin{remark} + This is a subsheaf of rings of the sheaf of $\fk$-valued functions on $X$. +Under the identification $\bA^n =\fk^n$ with $\bP^n \sm \Vp(X_0)$, one has $\cO_X \defon{X \sm \Vp(X_0)} = \cO_{X \cap \bA^n}$ as subsheaves of the sheaf of $\fk$-valued functions, where the second sheaf is a sheaf on a closed subset of $\fk^n$: + +Indeed, if $W$ is as in the definition then $\phi([1,y_1,\ldots,y_n]) = \frac{f(1,y_1,\ldots,y_n)}{g(1,y_1,\ldots,y_n)}$ for $[1,y_1,\ldots,y_n] \in W$. +Conversely if $\phi([1,y_1,\ldots,y_n]) = \frac{f(y_1,\ldots,y_n)}{g(y_1,\ldots,y_n)}$ on an open subset $W $ of $X \cap \bA^n$ then +$\phi([y_0,\ldots,y_n]) = \frac{F(y_0,\ldots,y_n)}{G(y_0,\ldots,y_n)}$ on $W$ where $F(X_0,\ldots,X_n) \coloneqq X_0^d f(\frac{X_1}{X_0}, \ldots, \frac{X_n}{X_0})$ and $G(X_0,\ldots,X_n) = X_0^d g(\frac{X_1}{X_0},\ldots, \frac{X_n}{X_0})$ with a sufficiently large $d \in \N$. +\end{remark} +\begin{remark} + It follows from the previous remark and the similar result in the affine case that the elements of $\cO_X(U)$ are continuous on $U \sm V(X_0)$. + Since the situation is symmetric in the homogeneous coordinates, they are continuous on all of $U$. +\end{remark} +The following is somewhat harder than in the affine case: +\begin{proposition} + If $X$ is connected (e.g. irreducible), then the elements of $\cO_X\left( X \right) $ are constant functions on $X$. +\end{proposition} + + + +% Lecture 14 + +\subsection{The notion of a category} +\begin{definition} + A \vocab{category} $\cA$ consists of: + \begin{itemize} + \item A class $\Ob \cA$ of \vocab[Objects]{objects of $\cA$}. + \item For two arbitrary objects $A, B \in \Ob \cA$, a \textbf{set} $\Hom_\cA(A,B)$ of \vocab[Morphism]{morphisms for $A$ to $B$ in $\cA$}. + \item A map $\Hom_\cA(B,C) \times \Hom_\cA(A,B) \xrightarrow{\circ} \Hom_\cA(A,C)$, the composition of morphisms, for arbitrary triples $(A,B,C)$ of objects of $\cA$. + \end{itemize} + The following conditions must be satisfied: + \begin{enumerate}[A] + \item For morphisms $A \xrightarrow{f} B\xrightarrow{g} C \xrightarrow{h} D$, we have $h \circ (g \circ f) = (h \circ g) \circ f$. + \item For every $A \in \Ob(\cA)$, there is an $\Id_A \in \Hom_{\cA}(A,A)$ such that $\Id_A \circ f = f$ (reps. $g \circ \Id_A = g$) for arbitrary morphisms $B \xrightarrow{f} A$ (reps. $A \xrightarrow{g} C).$ + \end{enumerate} + + A morphism $X \xrightarrow{f} Y$ is called an \vocab[Isomorphism]{isomorphism (in $\cA $)} if there is a morphism $Y \xrightarrow{g} X$ (called the \vocab[Inverse morphism]{inverse $f\inv$ of $f$)} such that $g \circ f = \Id_X$ and $f \circ g = \Id_Y$. +\end{definition} +\begin{remark} + \begin{itemize} + \item The distinction between classes and sets is important here. + \item We will usually omit the composition sign $\circ$. + \item It is easy to see that $\Id_A$ is uniquely determined by the above condition $B$, and that the inverse $f\inv$ of an isomorphism $f$ is uniquely determined. + \end{itemize} +\end{remark} +\subsubsection{Examples of categories} +\begin{example} + \begin{itemize} + \item The category of sets. + \item The category of groups. + \item The category of rings. + \item If $R$ is a ring, the category of $R$-modules and the category $\Alg_R$ of $R$-algebras + \item The category of topological spaces + \item The category $\Var_\fk$ of varieties over $\fk$ (see \ref{defvariety}) + \item If $\cA$ is a category, then the \vocab{opposite category} or \vocab{dual category} is defined by $\Ob(\cA\op) = \Ob(\cA)$ and $\Hom_{\cA\op}(X,Y) = \Hom_\cA(Y,X)$. + \end{itemize} + In most of these cases, isomorphisms in the category were just called `isomorphism'. The isomorphisms in the category of topological spaces are the homeomophisms. +\end{example} +\subsubsection{Subcategories} +\begin{definition}[Subcategories] + A \vocab{subcategory} of $\cA$ is a category $\cB$ such that $\Ob(\cB) \se \Ob(\cA)$, such that $\Hom_\cB(X,Y) \se \Hom_\cA(X,Y)$ for objects $X$ and $Y$ of $\cB$, such that for every object $X \in \Ob(\cB)$, the identity $\Id_X$ of $X$ is the same in $\cB$ as in $\cA$, and such that for composable morphisms in $\cB$, their compositions in $\cA$ and $\cB$ coincide. + We call $\cB$ a \vocab{full subcategory} of $\cA$ if in addition $\Hom_\cB(X,Y) = \Hom_\cA(X,Y)$ for arbitrary $X,Y \in \Ob(\cB)$. +\end{definition} +\begin{example} + \begin{itemize} + \item The category of abelian groups is a full subcategory of the category of groups. + It can be identified with the category of $\Z$-modules. + \item The category of finitely generated $R$-modules as a full subcategory of the category of $R$-modules. + \item The category of $R$-algebras of finite type as a full subcategory of $\Alg_R$. + \item The category of affine varieties over $\fk$ as a full subcategory of the category of varieties over $\fk$. + \end{itemize} +\end{example} + +\subsubsection{Functors and equivalences of categories} +\begin{definition} + A \vocab[Functor!covariant]{(covariant) functor} (resp. \vocab[Functor!contravariant]{contravariant functor}) between categories $\cA \xrightarrow{F} \cB$ is a map $\Ob(\cA) \xrightarrow{F} \Ob(\cB)$ with a family of maps $\Hom_\cA(X,Y) \xrightarrow{F} \Hom_\cB(F(X),F(Y))$ (resp. $\Hom_\cA(X,Y) \xrightarrow{F} \Hom_\cB(F(Y),F(X))$ in the case of contravariant functors), where $X$ and $Y$ are arbitrary objects of $\cA$, such that the following conditions hold: + \begin{itemize} + \item $F(\Id_X) = \Id_{F(X)}$ + \item For morphisms $X \xrightarrow{f} Y \xrightarrow{g} Z$ in $\cA$, we have $F(gf) = F(g)F(f)$ ( resp. $F(gf) = F(f)F(g)$) + \end{itemize} + A functor is called \vocab[Functor!essentially surjective]{essentially surjective} if every object of $\cB$ is isomorphic to an element of the image of $\Ob(\cA) \xrightarrow{F} \Ob(\cB)$. + A functor is called \vocab[Functor!full]{full} (resp. \vocab[Functor!faithful]{faithful}) if it induces surjective (resp. injective) maps between sets of morphisms. + It is called an \vocab{equivalence of categories} if it is full, faithful and essentially surjective. +\end{definition} +\begin{example} + \begin{itemize} + \item There are \vocab[Functor!forgetful]{forgetful functors} from rings to abelian groups or from abelian groups to sets which drop the multiplicative structure of a ring or the group structure of a group. + \item If $\fk$ is any vector space there is a contravariant functor from $\fk$-vector spaces to itself sending $V$ to its dual vector space $V\se$ and $V \xrightarrow{f} W$ to the dual linear map $W\st \xrightarrow{f\st} V\st$. + When restricted to the full subcategory of finite-dimensional vector spaces it becomes a contravariant self-equivalence of that category. + \item The embedding of a subcategory is a faithful functor. In the case of a full subcategory it is also full. + \end{itemize} +\end{example} + + + +\subsection{The category of varieties} + +\begin{definition}[Algebraic variety]\label{defvariety} + An \vocab{algebraic variety} or \vocab{prevariety} over $\fk$ is a pair $(X, \cO_X)$, where $X$ is a topological space and $\cO_X$ a subsheaf of the sheaf of $\fk$-valued functions on $X$ such that for every $x \in X$, there are a neighbourhood $U_x$ of $x$ in $X$, an open subset $V_x$ of a closed subset $Y_x$ of $\fk^{n_x}$\footnote{By the result of \ref{affopensubtopbase} it can be assumed that $V_x = Y_x$ without altering the definition.} and a homeomorphism $V_x \xrightarrow{\iota_x} U_x$ such that for every open subset $V \se U_x$ and every function $V\xrightarrow{f} \fk$, we have $f \in \cO_X(V) \iff \iota\st_x(f) \in \cO_{Y_x}(\iota_x\inv(V))$, + + In this, the \vocab{pull-back} $\iota_x\st(f)$ of $f$ is defined by $(\iota_x\st(f))(\xi) \coloneqq f(\iota_x(\xi))$. + + + A morphism $(X, \cO_X) \to (Y, \cO_Y)$ of varieties is a continuous map $X \xrightarrow{\phi} Y$ such that for all open $U \se Y$ and $f \in \cO_Y(U)$, $\phi\st(f) \in \cO_X(\phi\inv(U))$. + An isomorphism is a morphism such that $\phi$ is bijective and $\phi\inv$ also is a morphism of varieties. +\end{definition} +\begin{example} + \begin{itemize} + \item If $(X, \cO_X)$ is a variety and $U \se X$ open, then $(U, \cO_X\defon{U})$ is a variety (called an \vocab{open subvariety} of $X$), and the embedding $U \to X$ is a morphism of varieties. + \item If $X$ is a closed subset of $\fk^n$ or $\bP^n$, then $(X, \cO_X)$ is a variety, where $\cO_X$ is the structure sheaf on $X$ (\ref{structuresheafkn}, reps. \ref{structuresheafpn}). + A variety is called \vocab[Variety!affine]{affine} (resp. \vocab[Variety!projective]{projective}) if it is isomorphic to a variety of this form, with $X $ closed in $\fk^n$ (resp. $\bP^n$). + A variety which is isomorphic to and open subvariety of $X$ is called \vocab[Variety!quasi-affine]{quasi-affine} (resp. \vocab[Variety!quasi-projective]{quasi-projective}). + \item If $X = V(X^2 - Y^3) \se \fk^2$ then $\fk \xrightarrow{t \mapsto (t^3,t^2)} X$ is a morphism which is a homeomorphism of topological spaces but not an isomorphism of varieties. + % TODO + + \item The composition of two morphisms $X \to Y \to Z$ of varieties is a morphism of varieties. + \item $X\xrightarrow{\Id_X} X$ is a morphism of varieties. + \end{itemize} +\end{example} + +\subsubsection{The category of affine varieties} +\begin{lemma}\label{localinverse} + Let $X$ be any $\fk$-variety and $U \se X$ open. + \begin{enumerate}[i)] + \item All elements of $\cO_X(U)$ are continuous. + \item If $U \se X$ is open, $U \xrightarrow{\lambda} \fk$ any function and every $x \in U$ has a neighbourhood $V_x \se U$ such that $\lambda \defon{V_x} \in \cO_X(V_x)$, then $\lambda \in \cO_X(U)$. + \item If $\vartheta \in \cO_X(U)$ and $\vartheta(x) \neq 0$ for all $x \in U$, then $\vartheta \in \cO_X(U)^{\times }$. + \end{enumerate} +\end{lemma} +\begin{proof} + \begin{enumerate}[i)] + \item The property is local on $U$, hence it is sufficient to show it in the quasi-affine case. This was done in \ref{structuresheafcontinuous}. + \item For the second part, let $\lambda_x \coloneqq \lambda \defon{V_x} $. + We have $\lambda_x\defon{V_x \cap V_y} = \lambda \defon{V_x \cap V_y} = \lambda_y \defon{V_x \cap V_y} $. + The $V_x$ cover $U$. By the sheaf axiom for $\cO_X$ there is $\ell \in \cO_X(U)$ with $\ell\defon{V_x} =\lambda_x$. It follows that $\ell=\lambda$. + \item By the definition of variety, every $x \in U$ has a quasi-affine neighbourhood $V \se U$. We can assume $U$ to be quasi-affine and $X = V(I) \se \fk^n$, as the general assertion follows by an application of ii). + If $x \in U$ there are a neighbourhood $x \in W \se U$ and $a,b \in R = \fk[X_1,\ldots,X_n]$ such that $\vartheta(y) = \frac{a(y)}{b(y)}$ for $y \in W$, with $b(y) \neq 0$. + Then $a(x) \neq 0$ as $\vartheta(x) \neq 0$. Replacing $W$ by $W \sm V(a)$, we may assume that $a$ has no zeroes on $W$. + Then $\lambda(y) = \frac{b(y)}{a(y)}$ for $y \in W$ has a non-vanishing denominator and $\lambda \in \cO_X(U)$. + We have $\lambda \cdot \vartheta = 1$, thus $\vartheta \in \cO_X(U)^{\times}$. + \end{enumerate} + + +\end{proof} +\begin{proposition}[About affine varieties] + \label{propaffvar} + \begin{itemize} + \item Let $X,Y$ be varieties over $\fk$. Then the map + \begin{align} + \phi: \Hom_{\Var_\fk}(X,Y) &\longrightarrow \Hom_{\Alg_\fk}(\cO_Y(Y), \cO_X(X)) \\ + (X \xrightarrow{f} Y) &\longmapsto (\cO_Y(Y) \xrightarrow{f\st} \cO_X(X)) + \end{align} + is injective when $Y$ is quasi-affine and bijective when $Y$ is affine. + \item The contravariant functor + \begin{align} + F: \Var_\fk &\longrightarrow \Alg_\fk \\ + X &\longmapsto \cO_X(X)\\ + (X\xrightarrow{f} Y) &\longmapsto (\cO_X(X) \xrightarrow{f\st} \cO_Y(Y)) + \end{align} + restricts to an equivalence of categories between the category of affine varieties over $\fk$ and the full subcategory $\cA$ of $\Alg_\fk$, + having the $\fk$-algebras $A$ of finite type with $\nil A = \{0\} $ as objects. + \end{itemize} +\end{proposition} + +\begin{remark} + It is clear that $\nil(\cO_X(X)) = \{0\}$ for arbitrary varieties. For general varieties it is however not true that $\cO_X(X)$ is a $\fk$-algebra of finite type. + There are counterexamples even for quasi-affine $X$. %TODO + + If, however, $X$ is affine, we may assume w.l.o.g. that $X = V(I)$ where $I = \sqrt{I} \se R$ is an ideal with $R = \fk[X_1,\ldots,X_n]$. + Then $\cO_X(X) \cong R / I$ (see \ref{structuresheafri}) is a $\fk$-algebra of finite type. +\end{remark} + +\begin{proof} + + + + + It suffices to investigate $\phi$ when $Y$ is an open subset of $V(I) \se \fk^n$, where $I = \sqrt{I} \se R$ is an ideal and $Y = V(I)$ when $Y$ is affine. + Let $(f_1,\ldots,f_n)$ be the components of $X \xrightarrow{f} Y \se \fk^n$. Let $Y \xrightarrow{\xi_i} \fk$ be the $i$-th coordinate. + By definition $f_i = f\st(\xi_i) $. Thus $f$ is uniquely determined by $\cO_Y(Y) \xrightarrow{f\st} \cO_X(X)$. + Conversely, let $Y = V(I)$ and $\cO_Y(Y) \xrightarrow{\phi} \cO_X(X)$ be a morphism of $\fk$-algebras. Define $f_i \coloneqq \phi(\xi_i)$ and consider $X \xrightarrow{f = (f_1,\ldots,f_n)} Y\se \fk^n$. + \begin{claim} + $f$ has image contained in $Y$. + \end{claim} + \begin{subproof} + For $x \in X, \lambda \in I$ we have $\lambda(f(x)) = (\phi(\lambda \mod I))(x) = 0$ as $\phi$ is a morphism of $\fk$-algebras. + Thus $f(x) \in V(I) = Y$. + \end{subproof} + \begin{claim} + $f$ is a morphism in $\Var_\fk$ + \end{claim} + \begin{subproof} + For open $\Omega \se Y, U = f\inv(\Omega) = \{x \in X | \A \lambda \in J ~ (\phi(\lambda))(x) \neq 0\}$ is open in $X$, where $Y \sm \Omega = V(J)$. + If $\lambda \in \cO_Y(\Omega)$ and $x \in U$, then $f(x)$ has a neighbourhood $V$ such that there are $a,b \in R$ with $\lambda(v) = \frac{a(v)}{b(v)}$ and $b(v) \neq 0$ for all $v \in V$. + Let $W \coloneqq f\inv(V)$. Then $\alpha \coloneqq \phi(a)\defon{W} \in \cO_X(W)$, $\beta \coloneqq \phi(b)\defon{W} \in \cO_X(W)$. + By the second part of \ref{localinverse} $\beta \in \cO_X(W)^{\times}$ and $f\st(\lambda)\defon{W} = \frac{\alpha}{\beta} \in \cO_X(W)$. + The first part of \ref{localinverse} shows that $f\st(\lambda) \in \cO_X(U)$. + \end{subproof} + By definition of $f$, we have $f\st = \phi$. This finished the proof of the first point. + + + + \begin{claim} + The functor in the second part maps affine varieties to objects of $\cA$ and is essentially surjective. + \end{claim} + \begin{subproof} + It follows from the remark that the functor maps affine varieties to objects of $\cA$. + + If $A \in \Ob(\cA)$ then $ A /\fk$ is of finite type, thus $A \cong R / I$ for some $n$. + Since $\nil(A) = \{0\}$ we have $I = \sqrt{I}$, as for $x \in \sqrt{I}$, $x \mod I \in \nil(R / I) \cong \nil(A) = \{0\}$. + Thus $A \cong\cO_X(X)$ where $X = V(I)$. + \end{subproof} + Fullness and faithfulness of the functor follow from the first point. +\end{proof} + +\begin{remark} + Note that giving a contravariant functor $\cC \to \cD$ is equivalent to giving a functor $\cC \to \cD\op$. We have thus shown that the category of affine varieties is equivalent to $\cA\op$, where $\cA \subsetneq \Alg_\fk$ is the full subcategory of $\fk$-algebras $A$ of finite type with $\nil(A) = \{0\}$. +\end{remark} +\subsubsection{Affine open subsets are a topology base} + +\begin{definition} + A set $\cB$ of open subsets of a topological space $X$ is called a \vocab{topology base} for $X$ if every open subset of $X$ can be written as a (possibly empty) union of elements of $\cB$. +\end{definition} +\begin{fact} +If $X$ is a set, then $\cB \se \cP(X)$ is a base for some topology on $X$ iff $X = \bigcup_{U \in \cB} U$ and for arbitrary $U, V \in \cB, U \cap V$ is a union of elements of $\cB$. +\end{fact} +\begin{definition} + Let $X$ be a variety. + An \vocab{affine open subset} of $X$ is a subset which is an affine variety. + +\end{definition} +\begin{proposition}\label{oxulocaf} + Let $X$ be an affine variety over $\fk$, $\lambda \in \cO_X(X)$ and $U = X \sm V(\lambda)$. + Then $U$ is an affine variety and the morphism $\phi: \cO_X(X)_\lambda \to \cO_X(U)$ defined by the restriction $\cO_X(X) \xrightarrow{\cdot |_U } \cO_X(U)$ and the universal property of the localization is an isomorphism. +\end{proposition} +\begin{proof} + Let $X$ be an affine variety over $\fk, \lambda \in \cO_X(X)$ and $U = X \sm V(\lambda)$. The fact that $\lambda\defon{U} \in \cO_x(U)^{\times}$ follows from \ref{localinverse}. + Thus the universal property of the localization $\cO_X(X)_\lambda$ can be applied to $\cO_X(X) \xrightarrow{\cdot |_U} \cO_X(U)$. + \begin{figure}[H] + \centering + \begin{tikzcd} + \cO_X(X) \arrow{d}{\cdot |_U}\arrow{r}{x \mapsto \frac{x}{1}} & \cO_X(X)_\lambda \arrow[dotted, bend left]{dl}{\Eone \phi} \\ + \cO_X(U) & + \end{tikzcd} + \hspace{50pt} + \begin{tikzcd} + &Y \arrow[bend right, swap]{ld}{\pi_0} \arrow[bend right, swap]{d}{\pi}&\cO_Y(Y) \cong A_\lambda \arrow{d}{\fs}& \\ + X \arrow[hookrightarrow]{r}{}& U \arrow[swap]{u}{\sigma} & \cO_X(U) + \end{tikzcd} + \end{figure} + For the rest of the proof, we may assume $X = V(I) \se \fk^n$ where $I = \sqrt{I} \se R \coloneqq\fk[X_1,\ldots,X_n]$ is an ideal. + Then $A \coloneqq \cO_X(X) \cong R / I$ and there is $\ell \in R$ such that $\ell\defon{X} = \lambda$. + Let $Y = V(J) \se \fk^{n+1}$ where $J \se \fk[Z,X_1,\ldots,X_n]$ is generated by the elements of $I$ and $1 - Z\ell(X_1,\ldots,X_n)$. + + Then $\cO_Y(Y) \cong \fk[Z,X_1,\ldots,X_n] / J \cong A[Z] / (1 -\lambda Z) \cong A_\lambda$. + By the proposition about affine varieties (\ref{propaffvar}), the morphism $\fs: \cO_Y(Y) \cong A_\lambda \to \cO_X(U)$ corresponds to a morphism $U \xrightarrow{\sigma} Y$. + We have $\fs(Z \mod J) = \lambda\inv$ and $\fs(X_i \mod J) = X_i \mod I$. + Thus $\sigma(x) = (\lambda(x)\inv, x)$ for $x \in U$. + Moreover, the projection $Y \xrightarrow{\pi_0} X$ dropping the $Z$-coordinate has image contained in $U$, as for $(z,x) \in Y$ the equation + \[ + 1 = z\lambda(x) + \] + implies $\lambda(x) \neq 0$. It thus defines a morphism $Y \xrightarrow{\pi} U$ and by the description of $\sigma$ it follows that $\sigma \pi = \Id_U$. + Similarly it follows that $\sigma \pi = \Id_Y$. Thus, $\sigma$ and $\pi$ are inverse to each other. +\end{proof} +\begin{corollary}\label{affopensubtopbase} + The affine open subsets of a variety $X$ are a topology base on $X$. +\end{corollary} +\begin{proof} + Let $X = V(I) \se \fk^n$ with $I = \sqrt{I}$. If $U \se X$ is open then $X \sm U = V(J)$ with $J \supseteq I$ and $U = \bigcup_{f \in J} (X \sm V(f))$. + Thus $U$ is a union of affine open subsets. The same then holds for arbitrary quasi-affine varieties. + + Let $X$ be any variety, $U \se X$ open and $x \in U$. + By the definition of variety, $x$ has a neighbourhood $V_x$ which is quasi-affine, and replacing $V_x$ by $U \cap V_x$ which is also quasi-affine we may assume $V_x \se U$. + $V_x$ is a union of its affine open subsets. Because $U$ is the union of the $V_x$, $U$ as well is a union of affine open subsets. +\end{proof} + + +% Lecture 14A TODO? + +% Lecture 15 + +% CRTPROG + +\subsection{Stalks of sheaves} + +\begin{definition}[Stalk] + Let $\cG$ be a presheaf of sets on the topological space $X$, and let $x \in X$. + The \vocab{stalk} (\vocab[Stalk]{Halm}) of $\cG$ at $x$ is the set of equivalence classes of pairs $(U, \gamma)$, where $U$ is an open neighbourhood of $x$ and $\gamma \in \cG(U)$ + and the equivalence relation $\sim $ is defined as follows: + $( U , \gamma) \sim (V, \delta)$ iff there exists an open neighbourhood $W \se U \cap V$ of $x$ such that $\gamma \defon{W} = \delta \defon{W}$. + + + If $\cG$ is a presheaf of groups, one can define a groups structure on $\cG_x$ by + \[ + ((U, \gamma) / \sim ) \cdot \left( (V,\delta) / \sim \right) = (U \cap V, \gamma \defon{U \cap V} \cdot \delta\defon{U \cap V}) / \sim + \] + + If $\cG$ is a presheaf of rings, one can similarly define a ring structure on $\cG_x$. + + + If $U$ is an open neighbourhood of $x \in X$, then we have a map (resp. homomorphism) + \begin{align} + \cdot_x : \cG(U) &\longrightarrow \cG_x \\ + \gamma &\longmapsto \gamma_x \coloneqq (U, \gamma) / \sim + \end{align} + +\end{definition} +\begin{fact} + Let $\gamma,\delta \in \cG(U)$. If $\cG$ is a sheaf\footnote{or, more generally, a separated presheaf} and if for all $x \in U$, we have $\gamma_x = \delta_x$, then $\gamma = \delta$. + + In the case of a sheaf, the image of the injective map $\cG(U) \xrightarrow{\gamma \mapsto (\gamma_x)_{x \in U}} \prod_{x \in U} \cG_x$ + is the set of all $(g_x)_{x \in U} \in \prod_{x \in U} \cG_x $ satisfying the following \vocab{coherence condition}: + For every $x \in U$, there are an open neighbourhood $W_x \se U$ of $x$ and $g^{(x)} \in \cG(W_x)$ with $g_y^{(x)} = g_y$ for all $y \in W_x$. +\end{fact} +\begin{proof} + Because of $\gamma_x = \delta_x$, there is $x \in W_x \se U$ open such that $\gamma\defon{W_x} = \delta\defon{W_x}$. As the $W_x$ cover $U$, $\gamma = \delta$ by the sheaf axiom. +\end{proof} +\begin{definition} + Let $\cG$ be a sheaf of functions. + Then $\gamma_x$ is called the \vocab{germ} of the function $\gamma$ at $x$. + The \vocab[Germ!value at $x$]{value at $x$ } of $g = (U, \gamma) / \sim \in \cG_x$ defined as $g(x) \coloneqq \gamma(x)$, which is independent of the choice of the representative $\gamma$. +\end{definition} +\begin{remark} + If $\cG$ is a sheaf of $C^{\infty}$-functions (resp. holomorphic functions), then $\cG_x$ is called the ring of germs of $C^\infty$-functions (resp. of holomorphic functions) at $x$. + +\end{remark} +\subsubsection{The local ring of an affine variety} +\begin{definition} + If $X$ is a variety, the stalk $\cO_{X,x}$ of the structure sheaf at $x$ is called the \vocab{local ring} of $X$ at $x$. + This is indeed a local ring, with maximal ideal $\fm_x = \{f \in \cO_{X,x} | f(x) = 0\}$. +\end{definition} +\begin{proof} + By \ref{localring} it suffices to show that $\fm_x$ is a proper ideal, which is trivial, and that the elements of $\cO_{X,x} \sm \fm_x$ are units in $\cO_{X,x}$. + Let $g = (U, \gamma)/\sim \in \cO_{X,x}$ and $g(x) \neq 0$. + $\gamma$ is Zariski continuous (first point of \ref{localinverse}). Thus $V(\gamma)$ is closed. By replacing $U$ by $U \sm V(\gamma)$ we may assume that $\gamma$ vanishes nowhere on $U$. + By the third point of \ref{localinverse} we have $\gamma \in \cO_X(U)^{\times}$. + $(\gamma\inv)_x$ is an inverse to $g$. +\end{proof} + +\begin{proposition}\label{proplocalring} + Let $X = \Va(I) \se \fk^n$ be equipped with its usual structure sheaf, where $I = \sqrt{I} \se R = \fk[X_1,\ldots,X_n]$ . Let $x \in X$ and $A = \cO_X(X) \cong R / I$. + $\{P \in R | P(x) = 0\} \text{\reflectbox{$\coloneqq$}} \fn_x \se R$ is maximal, $I \se \fn_x$ and $\fm_x \coloneqq \fn_x / I$ is the maximal ideal of elements of $A$ vanishing at $x$. + If $\lambda \in A \sm \fm_x$, we have $\lambda_x \in \cO_{X,x}^{\times}$, where $\lambda_x$ denotes the image under $A \cong \cO_X(X) \to \cO_{X,x}$. + By the universal property of the localization, there exists a unique ring homomorphism $A_{\fm_x} \xrightarrow{\iota} \cO_{X,x}$ + such that + \begin{figure}[H] + \centering + \begin{tikzcd} + A \arrow{r}{} \arrow{d}{\lambda \mapsto \lambda_x} & A_{\fm_x} \arrow[dotted, bend left]{ld}{\Eone \iota} \\ + \cO_{X,x} + \end{tikzcd} + \end{figure} + commutes. + + The morphism $A_{\fm_x}\xrightarrow{\iota} \cO_{X,x}$ is an isomorphism. + +\end{proposition} +\begin{proof} + To show surjectivity, let $\ell = (U, \lambda) / \sim \in \cO_{X,x}$, where $U$ is an open neighbourhood of $x$ in $X$. + We have $X \sm U = V(J)$ where $J \se A$ is an ideal. As $x \in U$ there is $f \in J$ with $f(x) \neq 0$. Replacing $U $ by $X \sm V(f)$ we may assume $U = X \sm V(f)$. + By \ref{oxulocaf}, $\cO_X(U) \cong A_f$, and $\lambda = f^{-n}\vartheta$ for some $n \in \N$ and $\vartheta \in A$. + Then $\ell = \iota(f^{-n} \vartheta)$ where the last fraction is taken in $A_{\fm_x}$. + + + Let $\lambda = \frac{\vartheta}{g} \in A_{\fm_x}$ with $\iota(\lambda) = 0$. + It is easy to see that $\iota(\lambda) = (X \sm V(g), \frac{\vartheta}{g}) / \sim $. + Thus there is an open neighbourhood $U$ of $x$ in $X \sm V(g)$ such that $\vartheta$ vanishes on $U$. + Similar as before there is $h \in A$ with $h(x) \neq 0$ and $W = X \sm V(h) \se U$. + By the isomorphism $\cO_X(W) \cong A_h$, there is $n \in \N$ with $h^{n}\vartheta = 0$ in $A$. Since $h \not\in \fm_x$, $h$ is a unit and the image of $\vartheta$ in $A_{\fm_x}$ vanishes, implying $\lambda = 0$. +\end{proof} +\subsubsection{Intersection multiplicities and Bezout's theorem} +\begin{definition} + Let $R = \fk[X_0,X_1,X_2]$ equipped with its usual grading and let $x \in \bP^{2}$. +Let $G \in R_g, H \in R_h$ be homogeneous polynomials with $x \in V(G) \cap V(h)$. +Let $\ell\in R_1$ such that $\ell(x) \neq 0$. Then $x \in U = \bP^2 \sm V(\ell)$ and the rational functions $\gamma = \ell^{-g}G, \eta = \ell^{-h}H$ are elements of $\cO_{\bP^2}(U)$. +Let $I_x(G,H) \se \cO_{\bP^2,x}$ denote the ideal generated by $\gamma_x$ and $\eta_x$. + + +\noindent The dimension $\dim_{\fk}(\cO_{X,x} / I_x(G,H)) \text{\reflectbox{$\coloneqq$}} i_x(G,H)$ is called the \vocab{intersection multiplicity} of $G$ and $H$ at $x$. +\end{definition} +\begin{remark} + If $\tilde \ell \in R_1$ also satisfies $\tilde \ell(x) \neq 0$, then the image of $\tilde \ell / \ell$ under $\cO_{\bP^2}(U) \to \cO_{\bP^2,x}$ is a unit, showing that the image of $\tilde \gamma = \tilde \ell^{-g} G$ in $\cO_{\bP^2,x}$ is multiplicatively equivalent to $\gamma_x$, and similarly for $\eta_x$. + Thus $I_x(G,H)$ does not depend on the choice of $\ell \in R_1$ with $\ell(x) \neq 0$. +\end{remark} +\begin{theorem}[Bezout's theorem] + In the above situation, assume that $V(H)$ and $V(G)$ intersect properly in the sense that $V(G) \cap V(H) \se \bP^2$ has no irreducible component of dimension $\ge 1$. + Then + \[ + \sum_{x \in V(G) \cap V(H)} i_x(G,H) = gh + \] + Thus, $V(G) \cap V(H)$ has $gh$ elements counted by multiplicity. +\end{theorem} +\printvocabindex +\end{document} +\if\false + +% TODO REMARK ABOUT ZORNS LEMMA (LECTURE 1) + +% TODO REMARK ABOUT FIN PRESENTED MODULES (LECTURE 2) + + +% TODO: LECTURE 9 LEMMA + + + + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +% ÜBERSICHT % +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + + +% List of forms of HNS + + +\begin{itemize} + \item[HNS2 $\implies$ HNS1b] Let $I \se \fl[X_1,\ldots,X_n]$. $I \se \fm$ maximal. $R / \fm$ is isomorphic to a field extension of $\fl$. Finite by HNS2. + \item[NNT $\implies$ HNS2] Apply NNT to $L / K$ $\leadsto$ alg. independent $a_i$ such that $L$ is finite over the image of $K[X_1,\ldots,X_n] \xrightarrow{\ev_a} L$ ($\ev_a : P \to P(a_1,\ldots,a_n)$). + $\ev_a(K[X_1,\ldots,X_n])$ is a field by fact about integrality and fields \ref{fintaf}. Hence $n = 0$ and $L / K$ is finite. + \item[UNCHNS2] $K$ uncountable, $L / K$ fin. type. Then $\dim_K L$ is countable. Suppose $l \in L$ is not integral. Then $K(l) \cong K(T)$ and $\dim_K L \ge \dim_K K(T) \ge \aleph_1$. + Thus $L / K$ algebraic $\implies$ integral $\implies$ finite. + \item[HNS3] ($V(I) \se V(f) \iff f \in \sqrt{I} $). Suppose $V(I) \se V(f)$. $R' \coloneqq \fk[X_1,\ldots,X_n, T], J \se R'$ the ideal generated by $I$ and $g(X_1,\ldots,X_n,T) \coloneqq 1 - Tf(X_1,\ldots,X_n)$. +\end{itemize} + + +% Proofs +Def of integrality (<=>) + + +Fact about integrality and field: + % TODO + + +Technical lemma for Noether normalization: For $S \se \N^n$ finite, there exists $k \in \N^n$ such that $k_1 = 1$ and $s_1 \neq s_2 \in S \implies \langle k, s_1 \rangle \neq \langle k, s_2 \rangle$: +For $s_1 \neq s_2$, % TODO + +Noether normalization: +$a_i \in A$ minimal such that $A$ is integral over the subalgebra genereted by the $a_i$. +Suppose $\E P \in K[X_1,\ldots,X_n] \sm \{ 0\} ~ P(a_1,\ldots,a_n) = 0$. $P = \sum_{\alpha \in \N^n} p_\alpha X^\alpha, S \coloneqq \{ \alpha \in \N^n | p_\alpha \neq 0\}$. +Choose $k$ as in the lemma. +$b_i \coloneqq a_{i+1} - a_1^{k_{i+1}}, 1 \le i \trdeg(\fk(\fq) / \fl)$: +\Wlog $\fp = \{0\}$ and $A$ a domain ($A' \coloneqq A / \fp$). +For $\fq \in \mSpec A$, $\fk(\fq) = A / \fq$ finite type, hence finite (HNS) $\implies \trdeg(\fk(\fq) / \fl) = 0$. +$\trdeg(Q(A) / \fl) = 0 \implies A$ integral over $\fk$ $\implies$ $A$ a field $\implies \fp = \fq \lightning$. + +If $\fq \not\in \mSpec A$, let $a_1,\ldots,a_n \in A$ alg. independent such that the $\overline{a_i}$ are a transcendence base for $\fk(\fq) / \fk$ +Let $R$ be the ring generated by $\fl$ and the $a_i$. Localize with respect to $S \coloneqq R \sm \{0\}$. +%TODO +% TODO: LERNEN + + +% Dim k^n +$\dim(\fk^n)$ +$ \ge n$ build chian +$\le n$ a first result in dim T ($\fp \subsetneq \fq \implies \trdeg(\fk(\fq) / \fl) < \trdeg(\fk(\fp) / \fl)$. Thus $\codim(X,Y) \le \trdeg(\fK(Y) / \fl) - \trdeg(\fK(X) / \fl)$. + +TODO +% List of proofs of HNS + + +% Going up + + +% TODO proof of dim Y = trdeg(K(Y) / k) +$\dim Y \ge \trdeg(\fk(Y) / \fk)$: Noether normalization. Subalgebra $\cong \fk[X_1,\ldots,X_d]$. Lift chain of prime ideals using going up. + +% TODO prime avoidance + + +Action of $\Aut(L/K)$ on prime ideals of a normal ring extension. $A$ normal domain, $L / Q(A)$ normal field extension, $B$ int closure of $A$ in $L$, $\fp \in \Spec A$. +Then $\Aut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq \cap A = \fp\}$ : + +\begin{itemize} + \item $\fq, \fr \in \Spec B$ lying over $\fp$. + \item only need to show $\fq \se \sigma(\fr)$ for some $\sigma \in G$ (Krull going-up, no inclusions) + \item Suppose not. Then $x \in \fq \sm \bigcup_{\sigma \in G} \sigma(\fr)$ (prime aviodance) + \item $y = \prod_{\sigma \in G} \sigma(x) \in \fq \sm \fr$ ($\fr$ prime ideal) + \item $\E k \in \N$ s.t. $y^k \in K$ ($y \in L^G$) + \item $y^k \in K \cap B = A $ ($A$ normal). Thus $y^k \in (A \cap \fq) \sm (A \cap \fr) = \fp \sm \fp$. + \item $L / K$ infinite: Apply Zorn to pairs $(M, \sigma)$ where $K \se M \se L$ and $\sigma \in \Aut(M /K)$ s.t. $\sigma(\fr \cap M) = \fq \cap M$. +\end{itemize} + + +Going down Krull %TODO + +The ht p and trdeg +================== +% TODO % TODO % TODO % + +% Definitions +Zariski-Topology, Spec, $\fk^n$ +Residue field $\fk(\fp) \coloneqq Q(A / \fp), \fK(V(\fp)) \coloneqq \fk(\fp)$. TODO? +% Counterexamples + no going-up +% list of definitions of codim, dim, trdeg, ht +Original (Noether normalization) +Artin-Tate +Uncountable fields +\begin{landscape} +\section{Übersicht} +{\rowcolors{2}{gray!10}{white} + \begin{longtable}{lll} + \end{longtable} +} + +\end{landscape} + + \end{document} diff --git a/algebra.sty b/algebra.sty index b34deb2..5ea351c 100644 --- a/algebra.sty +++ b/algebra.sty @@ -1 +1,90 @@ \ProvidesPackage{algebra}[2022/02/10 - Style file for notes of Algebra I] + +\RequirePackage[english, numberall]{mkessler-fancythm} +\RequirePackage{hyperref} +\RequirePackage[english, index]{mkessler-vocab} +\RequirePackage{mkessler-hypersetup} +\input{/home/jrpie/templates/latex/math.tex} + +\RequirePackage[utf8x]{inputenc} +\RequirePackage{babel} + + +\RequirePackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry} + +% Kopf- und Fußzeilen +\RequirePackage{scrlayer-scrpage, lastpage} +\setkomafont{pageheadfoot}{\large\textrm} +\lohead{\head} +\rohead{\Namen} +\cfoot*{\thepage{}/\pageref{LastPage}} + +% Position des Titels +\RequirePackage{titling} +\setlength{\droptitle}{-1.0cm} + + +\RequirePackage[normalem]{ulem} +\RequirePackage{pdflscape} +\RequirePackage{longtable} +\RequirePackage{xcolor} +\RequirePackage{dsfont} +\RequirePackage{wrapfig} +\RequirePackage[shortlabels]{enumitem} + +\RequirePackage{tikz-cd} +\newcommand{\fp}{\ensuremath \mathfrak{p}} +\newcommand{\fq}{\ensuremath \mathfrak{q}} +\newcommand{\fr}{\ensuremath \mathfrak{r}} +\newcommand{\fn}{\ensuremath \mathfrak{n}} +\DeclareMathOperator{\codim}{codim} +\DeclareMathOperator{\trdeg}{trdeg} +\DeclareMathOperator{\hght}{ht} +\DeclareMathOperator{\Spec}{Spec} +\DeclareMathOperator{\mSpec}{mSpec} +\DeclareMathOperator{\Proj}{Proj} +\DeclareMathOperator{\Ob}{Ob} +\DeclareMathOperator{\Hom}{Hom} +\DeclareMathOperator{\Alg}{\mathfrak{Alg}} +\DeclareMathOperator{\Var}{\mathfrak{Var}} +\DeclareMathOperator{\op}{{}^{\text{op}}} +\newcommand{\Wlog}{W.l.o.g. } +%\newcommand{\wlog}{w.l.o.g. } +%\RequirePackage{ebgaramond} +%\RequirePackage{ebgaramond-maths} +\title{\textbf{Algebra 1}} +\newcommand{\Namen}{} +\author{Lecturer: \textsc{Prof. Dr. Jens Franke}\\\small{Notes: \textsc{Josia Pietsch}}} +\newcommand{\head}{Algebra 1} +\subtitle{Summer semester 2021, University Bonn} +\date{\today} + + +\newcommand{\einfalg}{Einführung in die Algebra} +\newcommand{\fk}{\ensuremath\mathfrak{k}} +\newcommand{\fl}{\ensuremath\mathfrak{l}} +\newcommand{\fs}{\ensuremath\mathfrak{s}} +\newcommand{\fri}{\ensuremath\mathfrak{i}} +\newcommand{\fm}{\ensuremath\mathfrak{m}} +\newcommand{\Vspec}{\ensuremath V_{\mathbb{S}}}%\Spec}} +\newcommand{\Vs}{\ensuremath V_{\mathbb{S}}}%\Spec}} +\newcommand{\Va}{\ensuremath V_{\mathbb{A}}}%\Spec}} +\newcommand{\Vp}{\ensuremath V_{\mathbb{P}}}%\Spec}} +\newcommand{\Pn}{\bP^n}%\Spec}} +\newcommand{\Span}[1]{\langle#1\rangle} +\newcommand{\npr}{\footnote{Not relevant for the exam.}} +\newcommand{\limrel}{\footnote{Limited relevance for the exam.}} % may appear in 3x questions +\DeclareMathOperator{\Mat}{Mat} +\DeclareMathOperator{\ev}{ev} +\DeclareMathOperator{\Ker}{Ker} +\DeclareMathOperator{\Aut}{Aut} +\DeclareMathOperator{\Gal}{Gal} +\DeclareMathOperator{\nil}{\mathfrak{nil}} +\DeclareMathOperator{\rad}{\mathfrak{rad}} +\RequirePackage{stackengine} +\stackMath +\usetikzlibrary{arrows.meta, + quotes, babel} + +\newcommand{\iiff}{\item[$\iff$]} +\hypersetup{colorlinks, linkcolor=red} diff --git a/summary/algebra.pdf b/summary/algebra.pdf deleted file mode 100644 index 827bfa3ca50d776d752ab14ebf33f951e4113aa9..0000000000000000000000000000000000000000 GIT binary patch literal 0 HcmV?d00001 literal 933934 zcma&NQ;;YyyQST>ZQHhO+qP}nwr$%sciZmXZCkU?%vAk2=i|ZLBPb!&Q9>Z2Nb=SrHzZJ69K)L zjiHOFh^eu?i76BxAC$9;lc}LCl*i_^j#cI?At`xzhty%b6*2OSo=Vo_LZ=1*C7|IN7}N=*>O+l}ZGdjHS& z#pozD84wNPU=FA{JKzCQWdwMVG0D-s6{Zgi{Tj8eE=e=z3DrpRuTC$v1P~+B00!k; zBqH+}kfKl!XDLP6@}y@y`byoxXuH_RD0bmye!!XIkYEPxhcP3uW5!;=Mvf$6AKeu_ 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-\usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry} - -% Kopf- und Fußzeilen -\usepackage{scrlayer-scrpage, lastpage} -\setkomafont{pageheadfoot}{\large\textrm} -\lohead{\head} -\rohead{\Namen} -\cfoot*{\thepage{}/\pageref{LastPage}} - -% Position des Titels -\usepackage{titling} -\setlength{\droptitle}{-1.0cm} - - -\usepackage[normalem]{ulem} -\usepackage{pdflscape} -\usepackage{longtable} -\usepackage{xcolor} -\usepackage{dsfont} -\usepackage{wrapfig} -\usepackage[shortlabels]{enumitem} - -\usepackage{tikz-cd} -\newcommand{\fp}{\ensuremath \mathfrak{p}} -\newcommand{\fq}{\ensuremath \mathfrak{q}} -\newcommand{\fr}{\ensuremath \mathfrak{r}} -\newcommand{\fn}{\ensuremath \mathfrak{n}} -\DeclareMathOperator{\codim}{codim} -\DeclareMathOperator{\trdeg}{trdeg} -\DeclareMathOperator{\hght}{ht} -\DeclareMathOperator{\Spec}{Spec} -\DeclareMathOperator{\mSpec}{mSpec} -\DeclareMathOperator{\Proj}{Proj} -\DeclareMathOperator{\Ob}{Ob} -\DeclareMathOperator{\Hom}{Hom} -\DeclareMathOperator{\Alg}{\mathfrak{Alg}} -\DeclareMathOperator{\Var}{\mathfrak{Var}} -\DeclareMathOperator{\op}{{}^{\text{op}}} -\newcommand{\Wlog}{W.l.o.g. } -%\newcommand{\wlog}{w.l.o.g. } -%\usepackage{ebgaramond} -%\usepackage{ebgaramond-maths} -\title{\textbf{Algebra 1}} -\newcommand{\Namen}{} -\author{Lecturer: \textsc{Prof. Dr. Jens Franke}\\\small{Notes: \textsc{Josia Pietsch}}} -\newcommand{\head}{Algebra 1} -\subtitle{Summer semester 2021, University Bonn} -\date{\today} - - -\newcommand{\einfalg}{Einführung in die Algebra} -\newcommand{\fk}{\ensuremath\mathfrak{k}} -\newcommand{\fl}{\ensuremath\mathfrak{l}} -\newcommand{\fs}{\ensuremath\mathfrak{s}} -\newcommand{\fri}{\ensuremath\mathfrak{i}} -\newcommand{\fm}{\ensuremath\mathfrak{m}} -\newcommand{\Vspec}{\ensuremath V_{\mathbb{S}}}%\Spec}} -\newcommand{\Vs}{\ensuremath V_{\mathbb{S}}}%\Spec}} -\newcommand{\Va}{\ensuremath V_{\mathbb{A}}}%\Spec}} -\newcommand{\Vp}{\ensuremath V_{\mathbb{P}}}%\Spec}} -\newcommand{\Pn}{\bP^n}%\Spec}} -\newcommand{\Span}[1]{\langle#1\rangle} -\newcommand{\npr}{\footnote{Not relevant for the exam.}} -\newcommand{\limrel}{\footnote{Limited relevance for the exam.}} % may appear in 3x questions -\DeclareMathOperator{\Mat}{Mat} -\DeclareMathOperator{\ev}{ev} -\DeclareMathOperator{\Ker}{Ker} -\DeclareMathOperator{\Aut}{Aut} -\DeclareMathOperator{\Gal}{Gal} -\DeclareMathOperator{\nil}{\mathfrak{nil}} -\DeclareMathOperator{\rad}{\mathfrak{rad}} -\usepackage{stackengine} -\stackMath -\usetikzlibrary{arrows.meta, - quotes, babel} - -\newcommand{\iiff}{\item[$\iff$]} -\hypersetup{colorlinks, linkcolor=red} - -\begin{document} - -\maketitle -\tableofcontents - -\begin{warning} - This is not an official script! - This document was written in preparation for the oral exam. It mostly follows the way \textsc{Prof. Franke} presented the material in his lecture rather closely. - There are probably errors. -\end{warning} - -\noindent The \LaTeX template by \textsc{Maximilian Kessler} is published under the MIT-License and can be obtained from \url{https://github.com/kesslermaximilian/LatexPackages}. % TODO -\newline - -\noindent $\fk$ is {\color{red} always} an algebraically closed field and $\fk^n$ is equipped with the Zariski-topology. -Fields which are not assumed to be algebraically closed have been renamed (usually to $\fl$). -\pagebreak -\section{Finiteness conditions} - -\subsection{Finitely generated and Noetherian modules} - -\begin{definition}[Generated submodule] - Let $R$ be a ring, $M$ an $R$-module, $S \se M$. - Then the following sets coincide - \begin{enumerate} - \item $\left\{ \sum_{s \in S'} r_{s} \cdot s ~ |~ S \se S' \text{finite}, r_s \in R, \right\}$ - \item $\bigcap_{\substack{S \se N \se M\\N \text{submodule}}} N$ - \item The $\se$-smallest submodule of $M$ containing $S$ - \end{enumerate} - - This subset of $N \se M$ is called the \vocab[Module!Submodule]{submodule of $M $ generated by $S$}. If $N= M$ we say that \vocab[Module!generated by subset $S$]{$ M$ is generated by $S$}. - $M$ is finitely generated $:\iff \E S \se M$ finite such that $M$ is generated by $S$. -\end{definition} - -\begin{definition}[Noetherian $R$-module] - $M$ is a \vocab{Noetherian} $R$-module if the following equivalent conditions hold: - \begin{enumerate} - \item Every submodule $N \se M$ is finitely generated. - \item Every sequence $N_0 \subset N_1 \subset \ldots$ of submodules terminates - \item Every set $\fM \neq \emptyset$ of submodules of $M$ has a $\se$-largest element. - \end{enumerate} -\end{definition} -\begin{proposition}[Hilbert's Basissatz]\label{basissatz} - If $R$ is a Noetherian ring, then the polynomial rings $R[X_1,\ldots, X_n]$ in finitely many variables are Noetherian. -\end{proposition} -\subsubsection{Properties of finite generation and Noetherianness} - -\begin{fact}[Properties of Noetherian modules] - \begin{enumerate} - \item Every Noetherian module over an arbitrary ring is finitely generated. - \item If $R$ is a Noetherian ring, then an $R$-module is Noetherian iff it is finitely generated. - \item Every submodule of a Noetherian module is Noetherian. - \end{enumerate} -\end{fact} -\begin{proof} - - \begin{enumerate} - \item By definition, $M$ is a submodule of itself. Thus it is finitely generated. - \item Since $M$ is finitely generated, there exists a surjective homomorphism $R^n \to M$. As $R$ is Noetherian, $R^n$ is Noethrian as well. - \item trivial - \end{enumerate} -\end{proof} - -\begin{fact} - Let $M, M', M''$ be $R$-modules. - \begin{enumerate} - \item Suppose $M \xrightarrow{p} M''$ is surjective. If $M$ is finitely generated (resp. Noetherian), then so is $M''$. - \item Let $M' \xrightarrow{f} M \xrightarrow{p} M'' \to 0$ be exact. If $M'$ and $M ''$ are finitely generated (reps. Noetherian), so is $M$. - \end{enumerate} -\end{fact} -\begin{proof} - \begin{enumerate} - \item Consider a sequence $M_0'' \subset M_1'' \subset \ldots \subset M''$. Then $p\inv M_i''$ yields a strictly ascending sequence. - If $M$ is generated by $S, |S| < \omega$, then $M''$ is generated by $p(S)$. - \item Because of 1. we can replace $M'$ by $f(M')$ and assume $0 \to M' \xrightarrow{f} M \xrightarrow{p} M'' \to 0$ to be exact. The fact about finite generation follows from \einfalg. - - If $M', M''$ are Noetherian, $N \se M$ a submodule, then $N' \coloneqq f\inv(N)$ and $N''\coloneqq p(N)$ are finitely generated. Since $0 \to N' \to N \to N'' \to 0$ is exact, $N$ is finitely generated. - - \end{enumerate} -\end{proof} -\subsection{Ring extensions of finite type} - -\begin{definition}[$R$-algebra] - Let $R$ be a ring. An $R$-algebra $(A, \alpha)$ is a ring $A$ with a ring homomorphism $R \xrightarrow{\alpha} A$. - $\alpha$ will usually be omitted. In general $\alpha$ is not assumed to be injective.\\ - \\ - An $R$-subalgebra is a subring $\alpha(R) \se A' \se A$.\\ -A morphism of $R$-algebras $A \xrightarrow{f} \tilde{A}$ is a ring homomorphism with $\tilde{\alpha} = f \alpha$. -\end{definition} - -\begin{definition}[Generated (sub)algebra, algebra of finite type] - Let $(A, \alpha)$ be an $R$-algebra. - \begin{align} - \alpha: R[X_1,\ldots,X_m] &\longrightarrow A[X_1,\ldots,X_m] \\ - P = \sum_{\beta \in \N^m} p_\beta X^{\beta} &\longmapsto \sum_{\beta \in \N^m} \alpha(p_\beta) X^{\beta} - \end{align} - is a ring homomorphism. We will sometimes write $P(a_1,\ldots,a_m)$ instead of $(\alpha(P))(a_1,\ldots,a_m)$. - - Fix $a_1,\ldots,a_m \in A^m$. Then we get a ring homomorphism $R[X_1,\ldots,X_m] \to A$. The image of this ring homomorphism is the $R$-subalgebra of $A$ \vocab[Algebra!generated subalgebra]{generated by the $a_i$}. - $A$ is \vocab[Algebra!of finite type]{of finite type} if it can be generated by finitely many $a_i \in I$. - - For arbitrary $S \se A$ the subalgebra generated by $S$ is the intersection of all subalgebras containing $S$ \\ - $=$ the union of subalgebras generated by finite $S' \se S$\\ - $= $ the image of $R[X_s | s \in S]$ under $P \mapsto (\alpha(P))(S)$. - -\end{definition} -\subsection{Finite ring extensions} % LECTURE 2 -\begin{definition}[Finite ring extension] - Let $R$ be a ring and $A$ an $R$-algebra. $A$ is a module over itself and the ringhomomorphism $R \to A$ allows us to derive an $R$-module structure on $A$. - $A$ \vocab[Algebra!finite over]{is finite over} $R$ / the $R$-algebra $A$ is finite / $A / R$ is finite if $A$ is finitely generated as an $R$-module. -\end{definition} -\begin{fact}[Basic properties of finiteness] - \begin{enumerate}[A] - \item Every ring is finite over itself. - \item A field extension is finite as a ring extension iff it is finite as a field extension. - \item $A$ finite $\implies$ $A$ of finite type. - \item $A / R$ and $B / A$ finite $\implies$ $B / R$ finite. - \end{enumerate} -\end{fact} -\begin{proof} - \begin{enumerate}[A] - \item $1$ generates $R$ as a module - \item trivial - \item Let $A $ be generated by $a_1,\ldots,a_n$ as an $R$-module. Then $A$ is generated by $a_1,\ldots,a_n$ as an $R$-algebra. - \item Let $A$ be generated by $a_1,\ldots,a_m$ as an $R$-module and $B$ by $b_1,\ldots,b_n$ as an $A$-module. - For every $b$ there exist $\alpha_j \in A$ such that $b = \sum_{j=1}^{n} \alpha_j b_j$. We have $\alpha_j = \sum_{i=1}^{m} \rho_{ij} a_i$ for some $\rho_{ij} \in R$ thus - $b = \sum_{i=1}^{m} \sum_{j=1}^{n} \rho_{ij} a_i b_j$ and the $a_ib_j$ generate $B$ as an $R$-module. -\end{enumerate} - -\end{proof} - -\subsection{Determinants and Caley-Hamilton} %LECTURE 2 TODO: move to int. elements? -This generalizes some facts about matrices to matrices with elements from commutative rings with $1$. -\footnote{Most of this even works in commutative rings without $ 1$, since $1$ simply can be adjoined.} -\begin{definition}[Determinant] - Let $A = (a_{ij}) \Mat(n,n,R)$. We define the determinant by the Leibniz formula \[ - \det(A) \coloneqq \sum_{\pi \in S_n} \sgn(\pi) \prod_{i=1}^{n} a_{i, \pi(i)} - \] - - Define $\text{Adj}(A)$ by $\text{Adj}(A)^{T}_{ij} \coloneqq (-1)^{i+j} \cdot M_{ij}$, where $M_{ij}$ is the determinant of the matrix resulting from $A$ after deleting the $i^{\text{th}}$ row and the $j^{\text{th}}$ column. -\end{definition} -\begin{fact} - \begin{enumerate} - \item $\det(AB) = \det(A)\det(B)$ - \item Development along a row or column works. - \item Cramer's rule: $A \cdot \text{Adj}(A) = \text{Adj}(A) \cdot A = \det(A) \cdot \mathbf{1}_n$. $A$ is invertible iff $\det(A)$ is a unit. - \item Caley-Hamilton: If $P_A = \det(T \cdot \mathbf{1}_n - A)$ \footnote{$T \cdot \mathbf{1}_n -A \in \Mat(n,n,A[T])$}, then $P_A(A) = 0$. - \end{enumerate} - -\end{fact} -\begin{proof} - All rules hold for the image of a matrix under a ring homomorphism if they hold for the original matrix. The converse holds in the case of injective ring homomorphisms. - Caley-Hamilton was shown for algebraically closed fields in LA2 using the Jordan normal form. - Fields can be embedded into their algebraic closure, thus Caley-Hamilton holds for fields. Every domain can be embedded in its field of quotients $\implies$ Caley-Hamilton holds for domains. - - In general, $A$ is the image of $(X_{i,j})_{i,j = 1}^{n} \in \Mat(n,n,S)$ where $S \coloneqq \Z[X_{i,j} | 1 \le i, j \le n]$ (this is a domain) under the morphism $S \to A$ of evaluation defined by $X_{i,j} \mapsto a_{i,j}$. Thus Caley-Hamilton holds in general. -\end{proof} %TODO: lernen - -\subsection{Integral elements and integral ring extensions} %LECTURE 2 -\begin{proposition}[on integral elements]\label{propinte} - Let $A$ be an $R$-algebra, $a \in A$. Then the following are equivalent: - \begin{enumerate}[A] - \item $\E n \in \N, (r_i)_{i=0}^{n-1}, r_i \in R: a^n = \sum_{i=0}^{n-1} r_i a^i$ - \item There exists a subalgebra $B \se A$ finite over $R$ and containing $a$. - \end{enumerate} - If $a_1, \ldots, a_k \in A$ satisfy these conditions, there is a subalgebra of $A$ finite over $R$ and containing all $a_i$. -\end{proposition} -\begin{definition}\label{intclosure} - Elements that satisfy the conditions from \ref{propinte} are called \vocab{integral over} $R$. - $A / R$ is \vocab[Algebra!integral]{integral}, if all $a \in A$ are integral over $R$. - The set of elements of $A$ integral over $R$ is called the \vocab{integral closure} of $R$ in $A$. -\end{definition} -\begin{proof} - \hskip 10pt - \begin{enumerate} - {\color{gray} \item[B $\implies$ A] Let $a \in A$ such that there is a subalgebra $B \se A$ containing $a$ and finite over $R$. - Let $(b_i)_{i=1}^{n}$ generate $B$ as an $R$-module. - \begin{align} - q: R^n &\longrightarrow B \\ - (r_1,\ldots,r_n) &\longmapsto \sum_{i=1}^{n} r_i b_i - \end{align} - is surjective. Thus there are $\rho_{i} = \left( r_{i,j} \right)_{j=1}^n \in R^n$ such that $a b_i = q(\rho_i)$. Let $\fA$ be the matrix with the $\rho_i$ as columns. - Then for all $v \in R^n: q(\fA \cdot v) = a \cdot q(v)$. By induction it follows that $q(P(\fA) \cdot v) = P(a)q(v)$ for all $P \in R[T]$. Applying this to $P(T) = \det(T\cdot \mathbf{1}_n - \fA)$ and using Caley-Hamilton, we obtain $P(a) \cdot q(v) = 0$. $P$ is monic. Since $q$ is surjective, we find $v \in R^{n} : q(v) = 1$. Thus $P(a) = 0$ and $a$ satisfies A. - } - \item[B $\implies$ A] if $R$ is Noetherian.\footnote{This suffices in the exam.} - Let $a \in A$ satisfy B. Let $B$ be a subalgebra of $A$ containing $b$ and finite over $R$. Let $M_n \se B$ be the $R$-submodule generated by the $a^i$ with $0 \le i < n$. As a finitely generated module over the Noetherian ring $R$, $B$ is a Noetherian $R$-module. Thus the ascending sequence $M_n$ stabilizes at some step $d$ and $a^d \in M_d$. Thus there are $(r_i)_{i=0}^{d-1} \in R^d$ such that $a^d = \sum_{i=0}^{d-1} r_ia^i$. - \item[A $\implies$ B] Let $a = (a_i)_{i=1}^n$ where all $a_i$ satisfy A, i.e. $a_i^{d_i} = \sum_{j=0}^{d_i - 1} r_{i,j}a_i^j$ with $r_{i,j} \in R$. Let $B \se A$ be the sub-$R$-module generated by $a^\alpha = \prod_{i=1}^n a_i^{\alpha_i}$ with $0 \le \alpha_i < d_i$. - $B$ is closed under $a_1 \cdot $ since \[a_1a^{\alpha} = \begin{cases} - a^{(\alpha_1 + 1, \alpha')} &\text{if } \alpha = (\alpha_1, \alpha'), 0 \le \alpha_1 < d_1 - 1\\ - \sum_{j=0}^{d_1 - 1} r_{i_1,j} a^{(j, \alpha')} &\text{if } \alpha_1 = d_1 - 1 - \end{cases}\] - By symmetry, this hold for all $a_i$. By induction on $|\alpha| = \sum_{i=1}^{n} \alpha_i$, $B$ is invariant under $a^{\alpha}\cdot $. Since these generate $B$ as an $R$-module, $B$ is multiplicatively closed. Thus A holds. Furthermore we have shown the final assertion of the proposition. - \end{enumerate} -\end{proof} -\begin{corollary}\label{cintclosure} - \begin{enumerate} - \item[Q] Every finite $R$-algebra $A$ is integral. - \item[R] The integral closure of $R$ in $A$ is an $R$-subalgebra of $A$ - \item[S] If $A$ is an $R$-algebra, $B$ an $A$-algebra and $b \in B$ integral over $R$, then it is integral over $A$. - \item[T] If $A$ is an integral $R$-algebra and $B$ any $A$-algebra, $b \in B$ integral over $A$, then $b$ is integral over $R$. - \end{enumerate} -\end{corollary} -\begin{proof} - \begin{enumerate} - \item[Q] Put $ B = A $ in B. - \item[R] For every $r \in R$ $\alpha(r)$ is a solution to $T - r = 0$, hence integral over $R$. - From B it follows, that the integral closure is closed under ring operations. - \item[S] trivial - \item[T] Let $b \in B$ such that $b^n = \sum_{i=0}^{n-1} a_ib^{i}$. Then there is a subalgebra $\tilde{A} \se A$ finite over $R$, such that all $a_i \in \tilde{A}$. - $b$ is integral over $\tilde{A} \implies \E \tilde{B} \se B$ finite over $\tilde{A}$ and $b \in \tilde{B}$. Since $\tilde{B} / \tilde{A} $ and $\tilde{A} / R$ are finite, $\tilde{B} / R$ is finite and $b$ satisfies B. - \end{enumerate} -\end{proof} - -\subsection{Finiteness, finite generation and integrality} %some more remarks on finiteness, finite generation and integrality - -\begin{fact}[Finite type and integral $\implies$ finite]\label{ftaiimplf} - If $A$ is an integral $R$-algebra of finite type, then it is a finite $R$-algebra. -\end{fact} -\begin{proof} - Let $A $ be generated by $\left( a_i \right) _{i=1}^{n}$ as an $R$- algebra. By the proposition on integral elements (\ref{propinte}), there is a finite $R$-algebra $B \se A$ such that all $a_i \in B$. - We have $B = A$, as $A$ is generated by the $a_i$ as an $R$-algebra. -\end{proof} -\begin{fact}[Finite type in tower] - If $A$ is an $R$-algebra of finite type and $B$ an $A$-algebra of finite type, then $B$ is an $R$-algebra of finite type. -\end{fact} -\begin{proof} - If $A / R$ is generated by $(a_i)_{i=1}^m$ and $B / A$ by $(b_j)_{j=1}^{n}$, then $B /R$ is generated by the $b_j$ and the images of the $a_i$ in $B$. -\end{proof} -{\color{red} - \begin{fact}[About integrality and fields] \label{fintaf} - Let $B$ be a domain integral over its subring $A$. Then $B$ is a field iff $A$ is a field. -\end{fact} -} -\begin{proof} - Let $B$ be a field and $a \in A \sm \{0\} $. Then $a\inv \in B$ is integral over $A$, hence $a^{-d} = \sum_{i=0}^{d-1} \alpha_i a^{-i}$ for some $\alpha_i \in A$. Multiplication by $a^{d-1}$ yields - $a\inv = \sum_{i=0}^{d-1} \alpha_i a^{d-1-i} \in A$. - - On the other hand, let $B$ be integral over the field $A$. Let $b \in B \sm \{0\}$. As $B$ is integral over $A$, there is a sub-$A$-algebra $\tilde{B} \se B, b \in \tilde{B}$ finitely generated as an $A$-module, i.e. a finite-dimensional $A$-vector space. Since $B$ is a domain, $\tilde{B} \xrightarrow{b\cdot } \tilde{B}$ is injective, hence surjective, thus $\E x \in \tilde{B} : b \cdot x \cdot 1$. -\end{proof} -\subsection{Noether normalization theorem} -\begin{lemma}\label{nntechlemma} - Let $S \se \N^n$ be finite. Then there exists $\vec k \in \N^n$ such that $k_1 =1$ and $w_{\vec k}(\alpha) \neq w_{\vec k}(\beta)$ for $\alpha \neq \beta \in S$, - where $w_{\vec k}(\alpha) = \sum_{i=1}^{n} k_i \alpha_i$. -\end{lemma} -\begin{proof} - Intuitive: - For $\alpha \neq \beta$ the equation $w_{(1, \vec \kappa)}(\alpha) = w_{(1, \vec \kappa)}(\beta)$ ($\kappa \in \R^{n-1}$) - defines a codimension $1$ affine hyperplane in $\R^{n-1}$. It is possible to choose $\kappa$ such that all $\kappa_i$ are $> \frac{1}{2}$ and with Euclidean distance $> \frac{\sqrt{n-1} }{2}$ from the union of these hyperplanes. By choosing the closest $\kappa'$ with integral coordinates, each coordinate will be disturbed by at most $\frac{1}{2}$, thus at Euclidean distance $\le \frac{\sqrt{n-1} }{2}$. - - More formally:\footnote{The intuitive version suffices in the exam.} - Define $M \coloneqq \max \{\alpha_i | \alpha \in S, 1 \le i \le n\} $. We can choose $k$ such that $k_i > (i-1) M k_{i-1}$. - Suppose $\alpha \neq \beta$. Let $i$ be the maximal index such that $\alpha_i \neq \beta_i$. Then the contributions of $\alpha_j$ (resp. $\beta_j$) with $1 \le j < i$ to $w_{\vec k}(\alpha)$ (resp. $w_{\vec k}(\beta)$) cannot undo the difference $k_i(\alpha_i - \beta_i)$. -\end{proof} - -\begin{theorem}[Noether normalization] \label{noenort} - Let $K$ be a field and $A$ a $K$-algebra of finite type. Then there are $a = (a_i)_{i=1}^{n} \in A$ which are algebraically independent over $K$, i.e. the ring homomorphism \begin{align} - \ev_a: K[X_1,\ldots,X_n] &\longrightarrow A \\ - P &\longmapsto P(a_1,\ldots,a_n) -\end{align} -is injective. $n$ and the $a_i$ can be chosen such that $A$ is finite over the image of $\ev_a$. -\end{theorem} -\begin{proof} - - Let $(a_i)_{i=1}^n$ be a minimal number of elements such that $A$ is integral over its $K$-subalgebra generated by $a_1, \ldots, a_n$. (Such $a_i$ exist, since $A$ is of finite type). - Let $\tilde{A}$ be the $K$-subalgebra generated by the $a_i$. - If suffices to show that the $a_i$ are algebraically independent. - Since $A$ is of finite type over $K$ and thus over $\tilde{A}$, by fact \ref{ftaiimplf} (integral and finite type $\implies$ finite) $A$ is finite over $\tilde{A}$. - Thus we only need to show that the $a_i$ are algebraically independent over $K$. - Assume there is $P \in K[X_1,\ldots,X_n] \sm \{0\} $ such that $P(a_1,\ldots,a_n) = 0$. Let $P = \sum_{\alpha \in \N^n} p_\alpha X^{\alpha}$ and $S = \{ \alpha \in \N^n | p_\alpha \neq 0\}$. For $\vec{k} = (k_i)_{i=1}^{n} \in \N^n$ and $\alpha \in \N^n$ we define $w_{\vec{k}}(\alpha) \coloneqq \sum_{i=1}^{n} k_i\alpha_i$. - - By \ref{nntechlemma} it is possible to choose $\vec{k} \in \N^n$ such that - $k_1 = 1$ and for $\alpha \neq \beta \in S$ we have $w_{\vec{k}}(\alpha) \neq w_{\vec{k}}(\beta)$. - - Define $b_i \coloneqq a_{i+1} - a^{k_{i+1}}_1$ for $1 \le i < n$. - \begin{claim} - $A$ is integral over the subalgebra $B$ generated by the $b_i$. - \end{claim} - \begin{subproof} - By the transitivity of integrality, it is sufficient to show that the $a_i$ are integral over $B$. - For $i > 1$ we have $a_i = b_{i-1} + a_1^{k_i}$. Thus it suffices to show this for $a_1$. - Define $Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots, b_{n-1} + T^{k_n}) \in B[T]$. - We have $0 = P(a_1,\ldots, a_n) = Q(a_1)$. Hence it suffices to show that the leading coefficient of $Q$ is a unit. - - We have - \[ - T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_i + 1})^{\alpha_{i+1}} = T^{w_{\vec k}(\alpha)} + \sum_{l = 0}^{w_{\vec k}(\alpha) - 1} \beta_{\alpha, l} T^l - \] - with suitable $\beta_{\alpha, l} \in B$. - - By the choice of $\vec k$, we have \[ - Q(T) = p_{\alpha} T^{w_{\vec k}(\alpha)} + \sum_{j=0}^{w_{\vec k}(\alpha) - 1} q_j T^j - \] - with $q_j \in B$ and $\alpha$ such that $w_{\vec k }(\alpha)$ is maximal subject to the condition $p_\alpha \neq 0$. - Thus the leading coefficient of $Q$ is a unit. - \end{subproof} - - This contradicts the minimality of $n$, as $B$ can be generated by $< n$ elements $b_i$. - -\end{proof} -\section{The Nullstellensatz and the Zariski topology} -\subsection{The Nullstellensatz} %LECTURE 1 -Let $\fk$ be a field, $R \coloneqq \fk[X_1,\ldots,X_n], I \se R$ an ideal. - -\begin{definition}[zero] - $x \in \fk^n$ is \vocab[Ideal!zero]{a zero of $I$} if $\A x \in I: P(x) = 0$. Let $\Va(I)$ denote the set of zeros if $I$ in $\fk^n$. - - The \vocab[Ideal!zero]{zero in a field extension $\mathfrak{i}$ of $\fk$} is defined similarly. -\end{definition} - -\begin{remark}[Set of zeros and generators] - Let $I$ be generated by $S$. Then $\{x \in R | \A s \in S: s(x) = 0\} = \Va(I)$. Thus zero sets of ideals correspond to solutions sets to systems of polynomial equations. - If $S, \tilde{S}$ generate the same ideal $I$ they have the same set of solutions. Therefore we only consider zero sets of ideals. -\end{remark} - -\begin{theorem}[Hilbert's Nullstellensatz (1)]\label{hns1} - If $\fk$ is algebraically closed and $I \subsetneq R$ a proper ideal, then $I$ has a zero in $\fk^n$. -\end{theorem} - -\begin{remark} - Will be shown later (see proof of \ref{hns1b}). - Trivial if $n = 1$: $R$ is a PID, thus $I = pR$ for some $p \in R$. Since $I \neq R$ $p = 0$ or $P$ is non-constant. $\fk$ algebraically closed $\leadsto$ there exists a zero of $p$.\\ - - If $\fk$ is not algebraically closed and $n > 0$, the theorem fails (consider $I = p(X_1) R$). -\end{remark} - -Equivalent\footnote{used in a vague sense here} formulation: - -\begin{theorem}[Hilbert's Nullstellensatz (2)] \label{hns2} - Let $L / K$ be an arbitrary field extension. Then $L / K$ is a finite field extension ($\dim_K L < \infty$) iff $L $ is a $K$-algebra of finite type. -\end{theorem} -\begin{proof} - \begin{itemize} - \item[$\implies$] If $(l_i)_{i=1}^{m}$ is a base of $L$ as a $K$-vector space, then $L$ is generated by the $l_i$ as a $K$-algebra. - \item[$\impliedby$ ] Apply the Noether normalization theorem (\ref{noenort}) to $A = L$. This yields an injective ring homomorphism $\ev_a: K[X_1,\ldots,X_n] \to A$ such that $A$ is finite over the image of $\ev_a$. - By the fact about integrality and fields (\ref{fintaf}), the isomorphic image of $\ev_a$ is a field. Thus $K[X_1,\ldots, X_n]$ is a field $\implies n = 0$. Thus $L / K$ is a finite ring extension, hence a finite field extension. - \end{itemize} -\end{proof} -\begin{remark} - We will see several additional proofs of this theorem. See \ref{hns2unc} and \ref{rfuncnft}. - All will be accepted in the exam. - - \ref{hns3} and \ref{hnsp} are closely related. -\end{remark} - -\begin{theorem}[Hilbert's Nullstellensatz (1b)] \label{hns1b} - Let $\fl$ be a field and $I \subset R = \fl[X_1,\ldots,X_m]$ a proper ideal. Then there are a finite field extension $\mathfrak{i}$ of $\fl$ and a zero of $I$ in $\mathfrak{i}^m$. -\end{theorem} - -\begin{proof} (HNS2 (\ref{hns2}) $\implies$ HNS1b (\ref{hns1b})) - $I \se \fm$ for some maximal ideal. $R / \fm$ is a field, since $\fm$ is maximal. - $R / \fm$ is of finite type, since the images of the $X_i$ generate it as a $\fl$-algebra. - There are thus a field extension $\fri / \fl$ and an isomorphism $R / \fm \xrightarrow{\iota} \fri$ of $\fl$-algebras. - By HNS2 (\ref{hns2}), $\fri / \fl$ is a finite field extension. - Let $x_i \coloneqq \iota (X_i \mod \fm)$. - \[ - P(x_1,\ldots,x_m) = \iota(P \mod \fm) - \] - Both sides are morphisms $R \to \fri$ of $\fl$-algebras. For for $P = X_i$ the equality is trivial. It follows in general, since the $X_i$ generate $R$ as a $\fl$-algebra. - - Thus $(x_1,\ldots,x_m)$ is a zero of $I$ (since $P \mod \fm = 0$ for $P \in I \se \fm$). - HNS1 (\ref{hns1}) can easily be derived from HNS1b. -\end{proof} - -\subsubsection{Nullstellensatz for uncountable fields} % from lecture 5 Yet another proof of the Nullstellensatz -The following proof of the Nullstellensatz only works for uncountable fields, but will be accepted in the exam. - - -\begin{lemma}\label{dimrfunc} - If $K$ is an uncountable field, then $\dim_K K(T)$ is uncountable. -\end{lemma} -\begin{proof} - We will show, that $S \coloneqq \left\{ \frac{1}{T - \kappa} | \kappa \in K\right\} $ is $K$-linearly independent. It follows that $\dim_K K(T) \ge \#S > \aleph_0$. - - Suppose $(x_{\kappa})_{\kappa \in K}$ is a selection of coefficients from $K$ such that $I \coloneqq \{\kappa \in K | x_{\kappa} \neq 0\} $ is finite and - \[ - g \coloneqq \sum_{\kappa \in K} \frac{x_\kappa}{T-\kappa} = 0 - \] - Let $d \coloneqq \prod_{\kappa \in I} (T - \kappa) $. Then for $\lambda \in I$ we have - \[ - 0 = (dg)(\lambda) = x_\lambda \prod_{\kappa \in I \sm \{\lambda\} } (\lambda - \kappa) - \] - This is a contradiction as $x_\lambda \neq 0$. -\end{proof} - -\begin{theorem}[Hilbert's Nullstellensatz for uncountable fields]\label{hns2unc} - If $K$ is an uncountable field and $L / K$ a field extension and $L$ of finite type as a $K$-algebra, then this field extension is finite. -\end{theorem} -\begin{proof} - If $(x_i)_{i=1}^{n}$ generate $L$ as an $K$-algebra, then the countably many monomials $x^{\alpha} = \prod_{i = 1}^{n} x_i^{\alpha_i} $ in the $x_i$ with $\alpha \in \N^n$ generate $L$ as a $K$-vector space. - Thus $\dim_K L \le \aleph_0$ and the same holds for any intermediate field $K \se M \se L$ . If $l \in L$ is transcendent over $K$ and $M = K(l)$, then $M \cong K(T)$ has uncountable dimension by \ref{dimrfunc}. Thus $L / K$ is algebraic, hence integral, hence finite (\ref{ftaiimplf}). -\end{proof} - -\subsection{The Zariski topology} -\subsubsection{Operations on ideals and \texorpdfstring{$\Va\left( I \right) $}{V(I)}} -Let $R$ be a ring and $I,J, I_\lambda \se R$ ideals, $\lambda \in \Lambda$. -\begin{definition}[Radical, product and sum of ideals] - \[ - \sqrt{I} \coloneqq \bigcap_{n=0} ^{\infty} \{ f \in R | f^n \in I\} - \] - \[ - I \cdot J \coloneqq \langle\{ i \cdot j | i \in I , j \in J\}\rangle_R - \] - - \[ - \sum_{\lambda \in \Lambda} I_\lambda \coloneqq \left\{\sum_{\lambda \in \Lambda'} i_\lambda | \Lambda' \se \Lambda \text{ finite}\right\} - \] -\end{definition} -\begin{fact} - The radical is an ideal in $R$ and $\sqrt{\sqrt{I} } = \sqrt{I}$.\\ - $I \cdot J$ is an ideal.\\ - $\sum_{\lambda \in \Lambda} I_\lambda$ coincides with the ideal generated by $\bigcap_{\lambda \in \Lambda} I_\lambda$ in $R$.\\ - $\bigcap_{\lambda \in \Lambda} I_\lambda$ is an ideal. -\end{fact} - -Let $R = \fk[X_1,\ldots,X_n]$ where $\fk$ is an algebraically closed field. - -\begin{fact} \label{fvop} - Let $I, J, (I_{\lambda})_{\lambda \in \Lambda}$ be ideals in $R$. $\Lambda$ may be infinite. - \begin{enumerate}[A] - \item $\Va(I) = \Va(\sqrt{I})$ - \item $\sqrt{J} \se \sqrt{I} \implies \Va(I) \se \Va(J)$ - \item $\Va(R) = \emptyset, \Va(\{0\} =\fk^n$ - \item $\Va(I \cap J) = \Va(I\cdot J) = \Va(I) \cup \Va(J)$ - \item $\Va(\sum_{\lambda \in \Lambda} I_\lambda) = \bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$ - \end{enumerate} -\end{fact} -\begin{proof} - \begin{enumerate} - \item[A-C] trivial - \item[D] $I \cdot J \se I \cap J \se I$. Thus $\Va(I) \se \Va(I \cap J) \se \Va(I \cdot J)$. By symmetry we have $\Va(I) \cup \Va(J) \se \Va(I \cap J) \se \Va(I \cdot J)$. - Let $x \not\in \Va(I) \cup \Va(J)$. Then there are $f \in I, g \in J$ such that $f(x) \neq 0, g(x) \neq 0$ thus $(f \cdot g)(x) \neq 0 \implies x \not\in \Va(I\cdot J)$. - Therefore \[ - \Va(I) \cup \Va(J) \se \Va(I \cap J) \se \Va(I \cdot J) \se \Va(I) \cup \Va(J) - \] - \item[E] $I_\lambda \se \sum_{\lambda \in \Lambda} I_\lambda \implies \Va(\sum_{\lambda \in \Lambda} I_\lambda) \se \Va(I_\lambda)$. - Thus $\Va(\sum_{\lambda \in \Lambda} I_\lambda) \se \bigcap_{\lambda \in \Lambda} \Va(I_\lambda)$. - On the other hand if $f \in \sum_{\lambda \in \Lambda} I_\lambda$ we have $f = \sum_{\lambda \in \Lambda} f_\lambda$. Thus $f$ vanishes on $\bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$ and we have $\bigcap_{\lambda \in \Lambda} \Va(I_\lambda) \se \Va(\sum_{\lambda \in \Lambda} I_\lambda)$. - \end{enumerate} -\end{proof} -\begin{remark} - There is no similar way to describe $\Va(\bigcap_{\lambda \in \Lambda} I_\lambda)$ in terms of the $\Va(I_{\lambda})$ when $\Lambda$ is infinite. - For instance if $n = 1, I_k \coloneqq X_1^k R$ then $\bigcap_{k=0}^\infty I_k = \{0\} $ but $\bigcup_{k=0}^{\infty} \Va(I_k) = \{0\}$. -\end{remark} -\subsubsection{Definition of the Zariski topology} -Let $\fk$ be algebraically closed, $R = \fk[X_1,\ldots,X_n]$. -\begin{corollary} (of \ref{fvop}) - There is a topology on $\fk^n$ for which the set of closed sets coincides with the set $\fA$ of subsets of the form $\Va\left(I \right) $ for ideals $I \se R$. - This topology is called the \vocab{Zariski-Topology} -\end{corollary} - -\begin{example}\label{zariskinothd} - Let $n = 1$. Then $R$ is a PID. Hence every ideal is a principal ideal and the Zariski-closed subsets of $\fk$ are the subsets of the form $\Va(P)$ for $P \in R$. -As $\Va(0) = \fk$ and $\Va(P)$ finite for $P \neq 0$ and $\{x_1,\ldots,x_n\} = \Va(\prod_{i=1}^{n}(T-x_i))$ the Zariski-closed subsets of $\fk$ are $\fk$ and the finite subsets. - Because $\fk$ is infinite, this topology is not Hausdorff. -\end{example} - -\subsubsection{Separation properties of topological spaces} -\begin{definition} - Let $X$ be a topological space. $X$ satisfies the separation properties $T_{0-2}$ if for any $x \neq y \in X$ - \begin{enumerate} - \item[$T_0$ ] $\E U \se X$ open such that $|U \cap \{x,y\}| = 1$ - \item[$T_1$ ] $\E U \se X$ open such that $x \in U, y \not\in U$. - \item[$T_2$ ] There are disjoined open sets $U, V \se X$ such that $x \in U, y \in V$. (Hausdorff) - \end{enumerate} -\end{definition} -\begin{remark} - Let $x \sim y :\iff$ the open subsets of $X$ containing $x$ are precisely the open subsets of $X$ containing $y$. Then $T_0$ holds iff $x \sim y \implies x =y$. -\end{remark} -\begin{fact} - $T_0 \iff$ every point is closed. -\end{fact} -\begin{fact} - The Zariski topology on $\fk^n$ is $T_1$ but for $n \ge 1$ not Hausdorff. For $n \ge 1$ the intersection of two non-empty open subsets of $\fk^n$ is always non-empty. -\end{fact} -\begin{proof} - $\{x\} $ is closed, as $\{x\} = V(\Span{X_1 - x_1, \ldots, X_n - x_n}_R)$. If $A = V(I), B = V(J)$ are two proper closed subsets of $\fk^n$ then $I \neq \{0\} , J \neq \{0\} $ and thus $IJ \neq \{0\} $. Therefore $A \cup B = V(IJ)$ is a proper closed subset of $\fk^n$. -\end{proof} - - -\subsubsection{Compactness properties of topological spaces} -Let $X$ be a topological space. -\begin{definition}[Compact, quasi-compact] - $X$ is called \vocab[Topological space!quasi-compact]{quasi-compact} if every open covering of $X$ has a finite subcovering. - It is called \vocab[Topological space!compact]{compact}, if it is quasi-compact and Hausdorff. -\end{definition} -\begin{definition}[Noetherian topological spaces] - $X$ is called \vocab{Noetherian}, if the following equivalent conditions hold: - \begin{enumerate}[A] - \item Every open subset of $X$ is quasi-compact. - \item Every descending sequence $A_0 \supseteq A_1 \supseteq \ldots$ of closed subsets of $X$ stabilizes. - \item Every non-empty set $\cM$ of closed subsets of $X$ has a $\se$-minimal element. - \end{enumerate} -\end{definition} -\begin{proof}\, - \begin{enumerate} - \item[A $\implies$ B] Let $A_j$ be a descending chain of closed subsets. Define $A \coloneqq \bigcap_{j = 0}^{\infty} A_j$. If A holds, the covering $X \sm A = \bigcup_{j = 0}^{\infty} (X \sm A_j)$ has a finite subcovering. - \item[B $\implies$ C] Suppose $\cM$ does not have a $\se$-minimal element. Using DC, one can construct a counterexample $A_1 \subsetneq A_2 \supsetneq \ldots$ to B. - \item[C $\implies$ A] Let $\bigcup_{i \in I} V_i$ be an open covering of an open subset $U \se X$. - By C, the set $\cM \coloneqq \{X \sm \bigcup_{i \in F} V_i | F \se I \text{ finite} \}$ has a $\se$-minimal element. - \end{enumerate} -\end{proof} - -\subsection{Another form of the Nullstellensatz and Noetherianness of \texorpdfstring{$\fk^n$}{kn}} -Let $\fk$ be algebraically closed, $R = \fk[X_1,\ldots,X_n]$. -For $f \in R$ let $V(f) = V(fR)$. -\begin{theorem}[Hilbert's Nullstellensatz (3)] \label{hns3} - Let $I \se R$ be an ideal. Then $V(I) \se V(f)$ iff $f \in \sqrt{I}$. -\end{theorem} -\begin{proof} - Suppose $f$ vanishes on all zeros of $I$. Let $R' \coloneqq \fk[X_1,\ldots,X_n,T]$, - $g(X_1,\ldots,X_n,T) \coloneqq 1 - T \cdot f(X_1,\ldots,X_n)$ - and $J \se R'$ the ideal generated by $g$ and the elements of $I$ (viewed as elements of $R'$ which are constant in the $T$-direction). - - If $f$ vanishes on all zeros of $I$, then $J$ has no zeros in $\fk^{n+1}$. - - Thus there exist $p_i \in I, i=1,\ldots,n, q_i \in \fk[X_1,\ldots,X_n,T], i = 1,\ldots,n$ and $q \in \fk[X_1,\ldots,X_n,T]$ such that - \[ - 1 = g \cdot q + \sum_{i=1}^{n} p_{i}q_i - \] - Formally substituting $\frac{1}{f(x_1,\ldots,x_n)}$ for $Y$, one obtains: - \[ - 1 = \sum_{i=1}^{n} p_{i}\left(x_1,\ldots,x_n\right) q_i\left( x_1,\ldots,x_n, \frac{1}{f(x_1,\ldots,x_n)} \right) - \] - Multiplying by a sufficient power of $f$, this yields an equation in $R$ : - \[ - f^d = \sum_{i=1}^{n} p_{i}(x_1,\ldots,_n) \cdot q_i'(x_1,\ldots,x_n) \in I - \] - Thus $f \in \sqrt{I}$. -\end{proof} - -\begin{corollary}\label{antimonbij} - \begin{align} - f: \{I \se R | I \text{ ideal}, I = \sqrt{I} \} &\longrightarrow \{A \se \fk^n | A \text{ Zariski-closed}\} \\ - I &\longmapsto V(I)\\ - \{f \in R | A \se V(f)\} &\longmapsfrom A - \end{align} - is a $\se$-antimonotonic bijection. -\end{corollary} -\begin{corollary} - The topological space $\fk^n$ is Noetherian. -\end{corollary} -\begin{proof} - Because the map from \ref{antimonbij} is antimonotonic, strictly decreasing chains of closed subsets of $\fk^n$ are mapped to strictly increasing chains of ideals in $R$. - By the Basissatz (\ref{basissatz}), $R$ is Noetherian. -\end{proof} - -% Lecture 04 - - -\subsection{Irreducible spaces} - -Let $X$ be a topological space. - -\begin{definition} - $X$ is called \vocab[Topological space!irreducible]{irreducible}, if $X \neq \emptyset$ and the following equivalent conditions hold: - \begin{enumerate}[A] - \item Every open $\emptyset \neq U \se X$ is dense. - \item The intersection of non-empty, open subsets $U, V \se X$ is non-empty. - \item If $A, B \se X$ are closed, $X = A \cup B$ then $X = A$ or $X = B$. - \item Every open subset of $X$ is connected. - \end{enumerate} -\end{definition} -\begin{proof}\, - \begin{itemize} - \item[$A \iff B$] by definition of denseness. - \item[B $\iff$ C] Let $U \coloneqq X \sm A, V \coloneqq X \sm B$. - \item[B $\implies$ D] Suppose $W$ is a non-connected open subset. Then there exists a decomposition $W = U \cup V$ into disjoint open subsets. - \item[D $\implies$ B] If $U,V \neq \emptyset$ are disjoint open subsets, then $U \cup V$ is non-connected. - \end{itemize} -\end{proof} -\begin{corollary} - Every irreducible topological space is connected. -\end{corollary} -\begin{example} - $\fk^n$ is irreducible as shown in \ref{zariskinothd}. -\end{example} - -\begin{fact} - \begin{enumerate}[A] - \item A single point is always irreducible. - \item If $X$ is Hausdorff then it is irreducible iff it has precisely one point. - \item $X$ is irreducible iff it cannot be written as a finite union of proper closed subsets. - \item $X$ is irreducible iff any finite intersection of non-empty open subsets is non-empty. ($\bigcap \emptyset \coloneqq X$) - \end{enumerate} -\end{fact} -\begin{proof} - \begin{enumerate} - \item[A,B] trivial - \item[C] $\implies$ : Induction on the cardinality of the union. $\impliedby $: $\bigcap \emptyset = X$ is non-empty and any intersection of two non-empty open subsets is non-empty. - \item[D] Follows from C. - \end{enumerate} -\end{proof} - -\subsubsection{Irreducible components} - -\begin{fact} - If $D \se X$ is dense, then $X$ is irreducible iff $D$ is irreducible with its induced topology. -\end{fact} -\begin{proof} - $X = \emptyset$ iff $D = \emptyset$. - Suppose $B$ is the union of its proper closed subsets $A,B$. Then $X = \overline{A} \cup \overline{B}$. These are proper closed subsets of $X$, as $\overline{A} \cap D = A \cap D$ (by closedness of $D$) and thus $\overline{A} \cap D \neq D$. - - On the other hand, if $U$ and $V$ are disjoint non-empty open subsets of $X$, then $U \cap D$ and $V \cap D$ are disjoint non-empty open subsets of $D$. -\end{proof} -\begin{definition}[Irreducible subsets] - A subset $Z \se X$ is called \vocab[Topological space!irreducible]{irreducible} if it is irreducible with its induced topology. - $Z$ is called an \vocab{irreducible component} of $X$, if it is irreducible and if every irreducible subset $Z \se Y \se X$ coincides with $Z$. -\end{definition} -\begin{corollary} - \begin{enumerate} - \item $Z \se X$ is irreducible iff $\overline{Z} \se X$ is irreducible. - \item Every irreducible component of $X$ is a closed subset of $X$. - \end{enumerate} -\end{corollary} -\begin{notation} - From now on, irreducible means irreducible and closed. -\end{notation} - -\subsubsection{Decomposition into irreducible subsets} -\begin{proposition} - Let $X$ be a Noetherian topological space. Then $X$ can be written as a finite union $X = \bigcup_{i = 1}^n Z_i$ of irreducible closed subsets of $X$. - One may additionally assume that $i \neq j \implies Z_i \not\subseteq Z_i$. With this minimality condition, $n$ and the $Z_i$ are unique (up to permutation) and $\{Z_1,\ldots,Z_n\}$ is the set of irreducible components of $X$. -\end{proposition} - -\begin{proof} - % i = ic - Let $\fM$ be the set of closed subsets of $X$ which cannot be decomposed as a union of finitely many irreducible subsets. - Suppose $\fM \neq \emptyset$. Then there exists a $\se$-minimal $Y \in \fM$. $Y$ cannot be empty or irreducible. Hence $Y = A \cup B$ where $A,B$ are proper closed subsets of $ Y$. By the minimality of $Y$, $A$ and $B$ can be written as a union of proper closed subsets $\lightning$. - - Let $X = \bigcup_{i = 1}^n Z_i$, where there are no inclusions between the $Z_i$. If $Y$ is an irreducible subsets of $X$, $Y = \bigcup_{i = 1}^n (Y \cap Z_i)$ and there exists $1 \le i \le n$ such that $Y = Y \cap Z_i$. - Hence $Y \se Z_i$. Thus the $Z_i$ are irreducible components. Conversely, if $Y$ is an irreducible component of $X$, $Y \se Z_i$ for some $i$ and $Y = Z_i$ by the definition of irreducible component. -\end{proof} -\begin{remark} - The proof of existence was an example of \vocab{Noetherian induction} : If $E$ is an assertion about closed subsets of a Noetherian topological space $X$ and $E$ holds for $A$ if it holds for all proper subsets of $A$, then $E(A)$ holds for every closed subset $A \se X$. -\end{remark} - -\begin{proposition}\label{bijiredprim} - By \ref{antimonbij} there exists a bijection - \begin{align} - f: \{I \se R | I \text{ ideal}, I = \sqrt{I} \} &\longrightarrow \{A \se \fk^n | A \text{ Zariski-closed}\} \\ - I &\longmapsto V(I)\\ - \{f \in R | A \se V(f)\} &\longmapsfrom A - \end{align} - - Under this correspondence $A \se \fk^n$ is irreducible iff $I \coloneqq f\inv(A)$ is a prime ideal. - Moreover, $\#A = 1$ iff $I$ is a maximal ideal. -\end{proposition} -\begin{proof} - By the Nullstellensatz (\ref{hns1}), $A = \emptyset \iff I = R$. Suppose $A = B \cup C$ is a decomposition into proper closed subsets $A = V(J), B = V(K)$ where $J = \sqrt{J}. K = \sqrt{K}$. - Since $A \neq B$ and $A \neq C$, there are $f \in J \sm I, g \in K \sm I$. $fg$ vanishes on $A = B \cup C$. By the Nullstellensatz (\ref{hns3}) $fg \in \sqrt{I} = I$ and $I$ fails to be prime. - - On the other hand suppose that $fg \in I, f \notin I, g \not\in I$. By the Nullstellensatz (\ref{hns3}) and $I = \sqrt{I} $ neither $f$ nor $g$ vanishes on all of $A$. Thus $(A \cap V(f)) \cup (A \cap V(g))$ is a decomposition and $A$ fails to be irreducible. - - The remaining assertion follows from the fact, that the bijection is $\se$-antimonotonic and thus maximal ideals correspond to minimal irreducible closed subsets, which are the one-point subsets as $\fk^n$ is T${}_1$. -\end{proof} -\subsection{Krull dimension} -\begin{definition} - Let $Z $ be an irreducible subset of the topological space $X$. Let $\codim(Z,X)$ be the maximum of the length $n$ of strictly increasing chains $Z \se Z_0 \subsetneq Z_1 \subsetneq \ldots \subsetneq Z_n$ of irreducible closed subsets of $X$ containing $Z$ or $\infty$ if such chains can be found for arbitrary $n$. - Let - \[ - \dim X \coloneqq \begin{cases} - - \infty &\text{if } X = \emptyset\\ - \sup_{\substack{Z \se X\\ Z \text{ irreducible}}} \codim(Z,X) & \text{otherwise} - \end{cases} - \] -\end{definition} -\begin{remark} - \begin{itemize} - \item In the situation of the definition $\overline{Z}$ is irreducible. Hence $\codim(Z,X)$ is well-defined and one may assume without losing much generality that $Z$ is closed. - \item Because a point is always irreducible, every non-empty topological space has an irreducible subset and for $X \neq \emptyset$, $\dim X$ is $\infty$ or $\max_{x \in X} \codim(\{x\}, X)$. - \item Even for Noetherian $X$, it may happen that $\codim(Z,X) = \infty$. - \item Even for if $X$ is Noetherian and $\codim(Z,X)$ is finite for all irreducible subsets $Z$ of $X$, $\dim X$ may be infinite. - \end{itemize} -\end{remark} -\begin{fact} - If $X = \{x\}$, then $\dim X = 0$. -\end{fact} -\begin{fact} - For every $x \in \fk$, $\codim( \{x\} ,\fk) = 1$. The only other irreducible closed subset of $\fk$ is $\fk$ itself, which has codimension zero. Thus $\dim \fk = 1$. -\end{fact} -\begin{fact} - Let $Y \se X$ be irreducible and $U \se X$ an open subset such that $U \cap Y \neq \emptyset$. Then we have a bijection - \begin{align} - f: \{A \se X | A \text{ irreducible, closed and } Y \se A\} &\longrightarrow \{B \se U | B \text{ irreducible, closed and } Y \cap U \se B\} \\ - A&\longmapsto A \cap U\\ - \overline{B}&\longmapsfrom B - \end{align} - where $\overline{B}$ denotes the closure in $X$. -\end{fact} -\begin{proof} - If $A$ is given and $B = A \cap U$, then $B \neq \emptyset$ and B is open hence (irreducibility of $A$) dense in $A$, hence $A = \overline{B}$. The fact that $B = \overline{B} \cap U$ is a general property of the closure operator. -\end{proof} -\begin{corollary}[Locality of Krull codimension] \label{lockrullcodim} - Let $Y \se X$ be irreducible and $U \se X$ an open subset such that $U \cap Y \neq \emptyset$. - Then $\codim(Y,X) = \codim(Y \cap U, U)$. -\end{corollary} -\begin{fact} - Let $Z \se Y \se X$ be irreducible closed subsets of the topological space $X$. Then - \[ - \codim(Z,Y) + \codim(Y,X) \le \codim(Z,X) \tag{CD+}\label{eq:cdp} - \] -\end{fact} -\begin{proof} - A chain of irreducible closed subsets between $Z$ and $Y$ and a chain of irreducible closed between $Y$ and $X$ can be spliced together. -\end{proof} -Taking the supremum over all $Z$ we obtain: -\begin{fact} - If $Y$ is an irreducible closed subset of the topological space $X$, then - \[ - \dim(Y) + \codim(Y,X) \le \dim(X) \tag{D+}\label{eq:dp} - \] -\end{fact} -In general, these inequalities may be strict. -\begin{definition}[Catenary topological spaces] - A topological space $T$ is called \vocab[Topological space!catenary]{catenary} if equality holds in \eqref{eq:cdp} whenever $X$ is an irreducible closed subset of $T$. -\end{definition} - -\subsubsection{Krull dimension of \texorpdfstring{$\fk^n$}{kn}} % from lecture 04 -\begin{theorem}\label{kdimkn} - $\dim \fk^n = n$ and $\fk^n$ is catenary. Moreover, if $X$ is an irreducible closed subset of $\fk^n$, then equality occurs in \eqref{eq:dp}. -\end{theorem} -\begin{proof} - Considering - \[ - \{0\} \subsetneq \fk \times \{0\} \subsetneq \fk^2 \times \{0\} \subsetneq \ldots \subsetneq \fk^n - \] - it is clear that $\codim(\{0\}, \fk^n) \ge n$.Translation by $x \in \fk^n$ gives us $\codim(\{x\} , \fk^n) \ge n$. - - The opposite inequality follows from \ref{upperbounddim} ($Z = \fk^n$ $\dim \fk^n \le \trdeg(\fK(Z) / \fk) = \trdeg(Q(\fk[X_1,\ldots,X_n]) / \fk) = n$). - - The theorem is a special case of \ref{htandtrdeg}. - % DIMT -\end{proof} - -\begin{lemma}\label{ufdprimeideal} - Every non-zero prime ideal $\fp$ of a UFD $R$ contains a prime element. -\end{lemma} -\begin{proof} - Let $p \in \fp \sm \{0\} $ with the minimal number of prime factors, counted by multiplicity. - If $p $ was a unit, then $\fp \supseteq pR = R$. If $p = ab$ with non-units $a,b$, it follows that $a \in \fp$ or $b \in \fp$ contradicting the minimality assumption. - Thus $p$ is a prime element of $R$. -\end{proof} - -\begin{proposition}[Irreducible subsets of codimension one]\label{irredcodimone} - Let $p \in R = \fk[X_1,\ldots, X_n]$ be a prime element. Then the irreducible subset $X = V(p) \se \fk^n$ has codimension one, and every codimension one subset of $\fk^n$ has this form. -\end{proposition} -\begin{proof} - Since $pR$ is a prime ideal, $X = V(p)$ is irreducible. Since $p \neq 0$, $X$ is a proper subset of $\fk^n$. - If $X \se Y \se \fk^n$ is irreducible and closed, then $Y = V(\fq)$ for some prime ideal $\fp \se pR$. - If $Y \neq \fk^n$, then $\fp \neq \{0\}$. By \ref{ufdprimeideal} there exists a prime element $q \in \fq$. As $\fq \se pR$ we have $p \divides q$. - By the irreducibility of $p$ and $q$ it follows that $p \sim q$. Hence $\fq = pR$ and $X = Y$. - - Suppose $X = V(\fp) \se \fk^n$ is closed, irreducible and of codimension one. - Then $\fp \neq \{0\}$, hence $X \neq \fk^n$. By \ref{ufdprimeideal} there is a prime element $p \in \fp$. If $\fp \neq pR$, then - $X \subsetneq V(p) \subsetneq \fk^n$ contradicts $\codim(X, \fk^n) = 1$. -\end{proof} - -% Lecture 05 -\subsection{Transcendence degree} -\subsubsection{Matroids} -\begin{definition}[Hull operator] - \npr - Let $X$ be a set, $\cP(X)$ the power set of $X$. A \vocab{Hull operator} on $X$ is a map $\cP(X) \xrightarrow{\cH} \cP(X)$ such that - \begin{enumerate} - \item[H1] $\A A \in \cP(X) ~ A \se \cH(A)$. - \item[H2] $A \se B \se X \implies \cH(A) \se \cH(B)$. - \item[H3] $\cH(\cH(X)) = \cH(X)$. - \end{enumerate} - - We call $\cH$ \vocab{matroidal} if in addition the following conditions hold: - \begin{enumerate} - \item[M] If $m,n \in X$ and $A \se X$ then $m \in \cH( \{n\} \cup A) \sm \cH(A) \iff n \in \cH(\{m\} \cup A) \sm \cH(A).$ - \item[F] $\cH(A) = \bigcup_{F \se A \text{ finite}} \cH(F)$. - \end{enumerate} - In this case, $S \se X$ is called \vocab{Independent subset}, if $s \not\in \cH(S \sm \{s\})$ for all $s \in S$ and - \vocab[Generating subset]{generating} if $X = \cH(S)$. - $S$ is called a \vocab{base}, if it is both generating and independent. -\end{definition} - -\begin{theorem} - If $\cH$ is a matroidal hull operator on $X$, then a basis exists, every independent set is contained in a base and two arbitrary bases have the same cardinality. -\end{theorem} - - -\begin{example} - Let $K$ be a field, $V$ a $K$-vector space and $\cL(T)$ the $K$-linear hull of $T$ for $T \se V$. - Then $\cL$ is a matroidal hull operator on $V$. -\end{example} - -\subsubsection{Transcendence degree} -\begin{lemma} - Let $L / K$ be a field extension and let $\cH(T)$ be the algebraic closure in $L$ of the subfield of $L$ generated by $K$ and $T$.\footnote{This is the intersection of all subfields of $L$ containing $K \cup T$, or the field of quotients of the sub-$K$-algebra of $L$ generated by $T$.} - Then $\cH$ is a matroidal hull operator. -\end{lemma} -\begin{proof}\npr - H1, H2 and F are trivial. For an algebraically closed subfield $K \se M \se L$ we have $\cH(M) = M$. Thus $\cH(\cH(T)) = \cH(T)$ (H3). - - Let $x,y \in L$, $T \se L$ and $x \in \cH(T \cup \{y\}) \sm \cH(T)$. We have to show that $y \in \cH(T \cup \{x\}) \sm \cH(T)$. - If $y \in \cH(T)$ we have $\cH(T \cup \{y\}) \se \cH(\cH(T)) = \cH(T) \implies x \in \cH(T) \sm \cH(T) \lightning$. - Hence it is sufficient to show $y \in \cH(T \cup \{x\})$. \Wlog $T = \emptyset$ (replace $K$ be the subfield generated by $K \cup T$). - Then $x$ is algebraic over the subfield $M$ of $L$ generated by $K \cup \{y\}$. Thus there exists $0 \neq P \in M[T]$ with $P(x) = 0$. - The coefficients $p_i$ of $P$ belong to the field of quotients of the $K$-subalgebra of $L$ generated by $y$. There are thus polynomials $Q_i, R \in K[Y]$ such that $p_i = \frac{Q_i(y)}{R(y)}$, $R(y) \neq 0$. - Let - \[ - Q(X,Y) \coloneqq \sum_{i=0}^{\infty} X^i Q_i(Y) = \sum_{i,j=0}^{\infty} q_{i,j}X^i Y^j = \sum_{j=0}^{\infty} Y^j \hat{Q_j}(X) \in K[X,Y] - \]. - Then $Q(x,y) = 0$. - Let $\hat{p_j} \coloneqq \hat{Q_j}(x)$. Then $\hat{P}(y) = 0$. As $Q \neq 0$ there is $(i,j) \in \N^2$ such that $q_{i,j} \neq 0$ and then $\hat{p_j} \neq 0$ as $x \not\in \cH(\emptyset)$. Thus $\hat{P} \in \hat{M}[X] \sm \{0\} $, where $\hat{M}$ is the subfield of $L$ generated by $K$ and $x$. Thus $y$ is algebraic over $\hat{M}$ and $y \in \cH(\{x\})$, -\end{proof} -\begin{definition}[Transcendence Base] - Let $L / K$ be a field extension and $\cH(T)$ the algebraic closure in $L$ of the subfield generated by $K$ and $T$. A base for $(L, \cH)$ is called a \vocab{transcendence base} and the \vocab{transcendence degree} $\trdeg(L / K)$ is defined as the cardinality of any transcendence base of $L / K$. -\end{definition} -\begin{remark} - $L / K$ is algebraic iff $\trdeg(L / K) = 0$. -\end{remark} - -\subsection{Inheritance of Noetherianness and of finite type by subrings and subalgebras / Artin-Tate} -The following will lead to another proof of the Nullstellensatz, which uses the transcendence degree. -\begin{remark} - There exist non-Noetherian domains, which are subrings of Noetherian domains (namely the field of quotients is Noetherian). -\end{remark} - -\begin{theorem}[Eakin-Nagata] - Let $A$ be a subring of the Noetherian ring $B$. If the ring extension $B / A$ is finite (i.e. $B$ finitely generated as an $A$-module) then $A$ is Noetherian. -\end{theorem} -\begin{dfact}\label{noethersubalg} - Let $R$ be Noetherian and let $B$ be a finite $R$-algebra. Then every $R$-subalgebra $A \se B$ is finite over $R$. -\end{dfact} -\begin{proof} - Since $B$ a finitely generated $R$-module and $R$ a Noetherian ring, $B$ is a Noetherian $R$-module (this is a stronger assertion than Noetherian algebra). - Thus the sub- $R$-module $A$ is finitely generated. -\end{proof} - -\begin{proposition}[Artin-Tate] - \label{artintate} - Let $A$ be a subalgebra of the $R$-algebra $B$, where $R$ is Noetherian. If $ B / R$ is of finite type and $B / A$ is finite, then $A / R$ is also of finite type. - - \begin{figure}[H] - \centering - \begin{tikzcd} - A \arrow[hookrightarrow]{rr}{\se}& & B \\ - &R \arrow{ul}{\alpha} \arrow{ur}{\alpha} \text{~(Noeth.)} - \end{tikzcd} - \end{figure} - -\end{proposition} -\begin{proof} - Let $(b_i)_{i=1}^{m}$ generate $B$ as an $A$-module and $(\beta_j)_{j=1}^m$ as an $R$-algebra. - There are $a_{ijk} \in A$ such that $b_i b_j = \sum_{k=1}^{m} a_{ijk}b_k$. And $\alpha_{ij} \in A$ such that $\beta_i = \sum_{j=1}^{m} \alpha_{ij}b_j$. Let $\tilde{A}$ be the sub- $R$-algebra of $A$ generated by the $a_{ijk}$ and $\alpha_{ij}$. $\tilde{A}$ is of finite type over $ R$, hence Noetherian. The $\tilde{A}$-submodule generated by $1$ and the $b_i$ is a sub-$R$-algebra containing the $\beta_i$ and thus coincides with $B$. - Hence $B / \tilde{A}$ is finite. Since $A \se B, A / \tilde{A}$ is finite (\ref{noethersubalg}). - Hence $A / \tilde{A}$ is of finite type. By the transitivity of ``of finite type'', it follows that $A / R$ is of finite type. - \begin{figure}[H] - \centering - \begin{tikzcd} - \tilde A \arrow[hookrightarrow]{r}{\se}& A \arrow[hookrightarrow]{r}{\se} & B \\ - &R \arrow[bend left, swap]{ul}{\alpha} \arrow{u}{\alpha} \arrow[bend right]{ur}{\alpha} - \end{tikzcd} - \end{figure} - -\end{proof} -\subsubsection{Artin-Tate proof of the Nullstellensatz} -Let $K$ be a field and $R = K[X_1,\ldots,X_n]$. -\begin{definition}[Rational functions] - Let $K(X_1,\ldots,X_n) \coloneqq Q(R)$ be the field of quotients of $R$. - - $K(X_1,\ldots,X_n)$ is called the \vocab{field of rational functions} in $n$ variables over $K$. -\end{definition} -\begin{lemma}[Infinitely many prime elements] - There are infinitely many multiplicative equivalence classes of prime elements in $R$. -\end{lemma} -\begin{proof} - Suppose $(P_i)_{i =1}^m$ is a complete (up to multiplicative equvialence) lsit of prime elements of $R$. - $m > 0$, as $X_1$ is prime. The polynomial $f \coloneqq 1 + \prod_{i=1}^{m} P_i $ is non-constant, hence not a unit in $R$. Hence there exists a prime divisor $P \in R$. As no $P_i$ divides $f$, $P$ cannot be multiplicatively equivalent to any $P_i \lightning$. -\end{proof} -\begin{lemma}[Ring of rational functions not of finite type]\label{rfuncnft} - If $n > 0$, then $K(X_1,\ldots,X_n) / K$ is not of finite type. -\end{lemma} -\begin{proof} - Suppose $(f_i)_{i=1}^m$ generate $K(X_1,\ldots,X_n)$ as a $K$-algebra. Let $f_i = \frac{a_i}{b}, a_i \in R, b \in R \sm \{0\}$. - Then $bf_i \in R$, and as the $f_i$ generate $K(X_1,\ldots,X_n)$ as a $K$-algebra, for every $g \in K(X_1,\ldots,X_n)$ there is $N \in \N$ with - \[ - b^Ng \in R \tag{+} \label{bNginR} - \] - However, if $b = \eps \prod_{i=1}^{l} P_i$ is a decomposition of $b$ into prime factors $P_i$ and a unit $\eps$ in $R$ and $g = \frac{1}{P}$, wehere $P \in R$ is a prime element not multiplicatively equvalent to any $P_i$, - then \eqref{bNginR} fails for any $N \in \N$. -\end{proof} - -The Nullstellensatz (\ref{hns2}) can be reduced to the case of \ref{rfuncnft}: - -\begin{proof}(Artin-Tate proof of HNS) - Let $(l_i)_{i=1}^n$ be a transcendence base of $L / K$. If $n = 0$ then $L / K$ is algebraic, hence an integral ring extension, hence a finite ring extension (\ref{ftaiimplf}). - - Suppose $n > 0$. Let $\tilde R \se L$ be the $K$-subalgebra generated by the $l_i$. $\tilde R \cong R \coloneqq K[X_1,\ldots,X_n]$, as the $l_i$ are algebraically independent. - As they are a transcendence base, $L$ is algebraic over the field of quotients $Q(\tilde R)$, hence integral over $Q(\tilde R)$. - - As $L / K$ is of finite type, so is $L / Q(\tilde R)$ and it follows that $L / Q(\tilde R)$ is a finite ring extension. - By Artin-Tate (\ref{artintate}), $Q(\tilde K)$ is of finite type over $K$. This contradicts \ref{rfuncnft}, as $R \cong \tilde R \implies K(X_1,\ldots,X_n) \cong Q(\tilde R)$. -\end{proof} - - -\subsection{Transcendence degree and Krull dimension} -Let $R = \fk[X_1,\ldots,X_n]$. -%i = ic -\begin{notation} - Let $X \se \fk^n$ be an irreducible closed subset. Then $X = V(\fp)$ for a unique prime ideal $\fp \se R$. - Let $\fK(X) \coloneqq Q(R / \fp)$ denote the field of quotients of $R / \fp$. -\end{notation} -\begin{remark} - As the elements of $\fp$ vanish on $X$, $R / \fp$ may be viewed as the ring of polynomials and $\fK(X)$ as the field of rational functions on $X$. -\end{remark} -\begin{theorem}\label{trdegandkdim} - If $X \se \fk^n$ is irreducible, then $\dim X = \trdeg (\fk(X) / \fk)$ and $\codim(X, \fk^n) = n - \trdeg(\fK(X) / \fk)$. - More generally if $Y \se \fk^n$ is irreducible and $X \se Y$, then $\codim(X,Y) = \trdeg(\fK(Y) / \fk) - \trdeg(\fK(X) / \fk)$. -\end{theorem} -\begin{proof} - % DIMT - One part will be shown in "A first result on dimension theory" (\ref{upperboundcodim}) - and other one in "Aplication to dimension theory: Proof of $\dim Y = \trdeg(\fK(Y) / \fk)$" (\ref{lowerbounddimy}). - The theorem is a special case of \ref{htandtrdeg}. -\end{proof} -\begin{remark} - Loosely speaking, the Krull dimension of $X$ is equal to the maximal number of $\fk$-algebraically independent rational functions on $X$. - This is yet another indication that the notion of dimension is the ``correct'' one. -\end{remark} -\begin{remark} - \ref{kdimkn} follows. -\end{remark} - - - - - -% Lecture 06 - -\subsection{The spectrum of a ring} -\begin{definition}[Spectrum] - Let $R$ be a commutative ring. - \begin{itemize} - \item Let $\Spec R$ denote the set of prime ideals and $\mSpec R \se \Spec R$ the set of maximal ideals of $R$. - \item For an ideal $I \se R$ let $V(I) \coloneqq \{\fp \in \Spec R | I \se \fp\}$ - \item We equip $\Spec R$ with the \vocab{Zariski-Topology} for which the closed subsets are the subsets of the form $V(I)$, where $I$ runs over the set of ideals in $R$. - \end{itemize} -\end{definition} - -\begin{remark} - When $R = \fk[X_1,\ldots,X_n]$, the notation $V(I)$ clashes with the previous notation. When several types of $V(I)$ will be in use, they will be distinguished using indices. -\end{remark} -\begin{remark} - Let $(I_{\lambda})_{\lambda \in \Lambda}$ and $(l_j)_{j=1}^n$ be ideals in $R$, where $\Lambda$ may be infinite. We have $V(\sum_{\lambda \in \Lambda} I_\lambda ) = \bigcap_{\lambda \in \Lambda} V(I_\lambda)$ and $V(\bigcap_{j=1}^n I_j) = V(\prod_{j=1}^{n} I_j) = \bigcup_{j = 1}^n V(I_j)$. - Thus, the Zariski topology on $\Spec R$ is a topology. -\end{remark} -\begin{remark} - Let $R = \fk[X_1,\ldots,X_n]$. Then there exists a bijection (\ref{antimonbij}, \ref{bijiredprim}) between $\Spec R$ and the set of irreducible closed subsets of $\fk^n$ sending $\fp \in \Spec R$ to $V_{\fk^n}(\fp)$ and identifying the one-point subsets with $\mSpec R$. - This defines a bijection $\fk^n \cong \mSpec R$ which is a homeomorphism if $\mSpec R$ is equipped with the induced topology from the Zariski topology on $\Spec R$. -\end{remark} - -\subsection{Localization of rings} -\begin{definition}[Multiplicative subset] - A \vocab{multiplicative subset} of a ring $R$ is a subset $S \se R$ such that $\prod_{i=1}^{n} f_i \in S $ when $n \in \N$ and all $f_i \in S$. -\end{definition} -\begin{proposition} - Let $S \se R$ be a multiplicative subset. Then there is a ring homomorphism $R \xrightarrow{i} R_S$ such that $i(S) \se R_S^{\times }$ and $i$ has the \vocab{universal property} for such ring homomorphisms: - If $R \xrightarrow{j} T$ is a ring homomorphism with $j(S) \se T^{\times }$, then there is a unique ring homomorphism $R_S \xrightarrow{\iota} T$ with $j = \iota i$. - - \begin{figure}[H] - \centering - \begin{tikzcd} - R \arrow{r}{i}\arrow{d}{j}& R_S \arrow[dotted]{ld}{\Eone \iota}\\ - T - \end{tikzcd} - \end{figure} - -\end{proposition} -\begin{proof} - The construction is similar to the construction of the field of quotients: - - Let $R_S \coloneqq (R \times S) / \sim $, where $(r,s) \sim (\rho, \sigma) : \iff \E t \in S ~ t \sigma r = ts\rho$.\footnote{$t$ does not appear in the construction of the field of quotients, but is important if $S$ contains zero divisors.} - $[r,s] + [\rho, \sigma] \coloneqq [r\sigma + \rho s, s \sigma]$, $[r,s] \cdot [\rho, \sigma] \coloneqq [r \cdot \rho, s \cdot \sigma]$. - - In order proof the universal property define $\iota([r,s]) \coloneqq \frac{j(r)}{j(s)}$. - The universal property characterizes $R_S$ up to unique isomorphism. - -\end{proof} -\begin{remark} - $i$ is often not injective and $\Ker(i) = \{r \in R | \E s \in S ~ s \cdot r = 0\} $. - In particular $(r = 1)$, $R_S$ is the null ring iff $0 \in S$. -\end{remark} -\begin{notation} - Let $S \se R$ be a multiplicative subset of $R$. We write $\frac{r}{s}$ for $[r,s]$. - The ring homomorphism $R \xrightarrow{i} R_S$ i given by $i(r) = \frac{r}{1}$. - For $X \se R_S$ let $X \sqcap R$ denote $i\inv(X)$. -\end{notation} -\begin{definition}[$S$-saturated ideal] - An ideal $I \se R$ is called \vocab[Ideal!S-saturated]{$S$-saturated} if for all $s \in S, r \in R$ - $rs \in I \implies r \in I$. -\end{definition} -\begin{fact}\label{primeidealssat} - A prime ideal $\fp \se \Spec R$ is $S$-saturated iff $\fp \cap S = \emptyset$. -\end{fact} -Because the elements of $S$ become units in $R_S$, $J \sqcap R$ is an $S$-saturated ideal in $R$ when $J$ is an ideal in $R_S$. - -\begin{fact}\label{ssatiis} - Let $I \se R$ be an $S$-saturated ideal and let $I_S$ denote the ideal $\{\frac{r}{s} | r \in R, s \in S\} \se R_S$. - Then for all $r \in R, s \in S$ - we have $\frac{r}{s} \in I_S \iff r \in I$. -\end{fact} -\begin{proof} - Clearly $i \in I \implies \frac{i}{s} \in I_S$. If $\frac{i}{s} \in J$ there are $\iota \in I$, $\sigma \in S$ such that $\frac{i}{s} = \frac{\iota}{\sigma}$ in $R_S$. - This equation holds iff there exists $t \in S$ such that $ts\iota = t \sigma i$. But $ts \iota \in I$ hence $i \in I$, as $I$ is $S $-saturated. -\end{proof} -\begin{fact}\label{invimgprimeideal} - The inverse image of a prime ideal under any ring homomorphism is a prime ideal. -\end{fact} - -\begin{proposition}\label{idealslocbij} - \begin{align} - f: \{I \se R | I \text{ $S$-saturated ideal}\} &\longrightarrow \left\{J \se R_S | J \text{ ideal}\right\} \\ - I &\longmapsto I_S \coloneqq \left\{\frac{i}{s} | i \in I, s \in S\right\}\\ - J \sqcap R &\longmapsfrom J\\ - \end{align} - is a bijection. Under this bijection $I$ is a prime ideal iff $f(I)$ is. -\end{proposition} - -\begin{proof} - Applying \ref{ssatiis} to $s = 1$ gives $I_S \sqcap R = I$, when $I$ is $S$-saturated. - - Conversely, if $J$ is given and $I = J \sqcap R, \frac{r}{s} \in R_S$, then by \ref{ssatiis} $\frac{r}{s} \in IR_S \iff r \in I$. - But as $\frac{r}{1} = s \cdot \frac{r}{s}$ and $s \in R_S^{\times }$, we have $r \in I \iff \frac{r}{1} \in J \iff \frac{r}{s} \in J$ . - We have thus shown that the two maps between sets of ideals are well-defined and inverse to each other. - - By \ref{invimgprimeideal}, $J \in \Spec R_S \implies f\inv(J) = J \cap R \in \Spec R_S$. - Suppose $I \in \Spec R$, $\frac{a}{s} \cdot \frac{b}{t} \in I_S$ for some $a,b \in R, s,t \in S$. - By \ref{ssatiis} $ab \in I$. Thus $a \in I \lor b \in I$, hence $\frac{a}{s} \in I_S \lor \frac{b}{t} \in I_S$ and we have $I_S \in \Spec R_S$. - - - -\end{proof} - -% Some more remarks on localization - -\begin{remark}\label{locandquot} - Let $R$ be a domain. If $S = R \sm \{0\}$, then $R_S$ is the field of quotients $Q(R)$. - If $S \se R \sm \{0\} $, then - \[ - R_S \cong \left\{ \frac{a}{s} \in K | a \in R, s \in S\right\} - \] - In particular $Q(R) \cong Q(R_S)$. -\end{remark} - -\begin{definition}[$S$-saturation]\label{ssaturation} - Let $R$ be any ring, $I \se R$ an ideal. Even if $I$ is not $S$-saturated, $J = I_S \coloneqq \{\frac{i}{s} | i \in I, s \in S\}$ is an ideal in $R_S$, and $I_S \sqcap R = \{r \in R | s\cdot r \in I, s \in S\}$ is called the \vocab[Ideal!$S$-saturation]{$S$-saturation of $I$ } which is the smallest $S$-saturated ideal containing $I$. - - -\end{definition} -\begin{lemma}\label{locandfactor} - In the situation of \ref{ssaturation}, if $\overline{S}$ denotes the image of $S$ in $R / I$, there is a canonical isomorphism $R_S / I_S \cong (R / I)_{\overline{S}}$. -\end{lemma} -\begin{proof} - We show that both rings have the universal property for ring homomorphisms $R \xrightarrow{\tau} T$ with $\tau(I) = \{0\} $ and $\tau(S) \se T^{\times }$. - For such $\tau$, by the fundamental theorem on homomorphisms (Homomorphiesatz) there is a unique $R/I \xrightarrow{\tau_1} T$ such that $\tau = \tau_1 \pi_{R,I}$. - We have $\tau_1(\overline{S}) = \tau(S) \se T^{\times }$, hence there is a unique $(R / I)_{\overline{S}} \xrightarrow{\tau_2} T$ such that the composition $R / I \to (R / I)_{\overline{S}} \xrightarrow{\tau_2} T $ equals $\tau_1$. It is easy to see that this is the only one for which $R \to R / I \to (R / I)_{\overline{S}} \xrightarrow{\tau_2} T$ equals $\tau$. - - - Similarly, by the universal property of $R_S$ there is a unique $R_S \xrightarrow{\tau_3} T$ whose composition with $R \to R_S$ equals $\tau$. - $\tau_3(I_{S}) = 0$, hence a unique $R_S / I_S \xrightarrow{\tau_4} T$ whose composition with $\pi_{R_S, I_S}$ equals $\tau_3$ exists. - This is the only one for which the composition $R \to R_S \to R_S / I_S \xrightarrow{\tau_4} T$ equals $\tau$. - -\begin{figure}[H] - \centering - \begin{tikzcd} - R \arrow{r}{\tau}\arrow[swap]{d}{\pi_{R,I}}& T & R\arrow[swap]{l}{\tau}\arrow{d}{}\\ - R / I \arrow[dotted]{ru}{\Eone \tau_1}\arrow{d}{} & & R_S \arrow[dotted, swap]{lu}{\Eone \tau_3}\arrow{d}{\pi_{R_S, I_S}}\\ - (R / I)_{\overline{S}} \arrow[dotted,bend right]{ruu}{\Eone \tau_2} & & R_S / I_S \arrow[dotted, bend left, swap]{luu}{\Eone \tau_4}\\ - \end{tikzcd} -\end{figure} - - -\end{proof} - - - -\subsection{A first result of dimension theory} - -\begin{notation} - Let $R$ be a ring, $\fp \in \Spec R$. Let $\fk(\fp)$ denote the field of quotients of the domain $R / \fp$. This is called the \vocab{residue field} of $\fp$. -\end{notation} - -% i = ic -\begin{proposition}\label{trdegresfield} - Let $\fl$ be a %% ?? -field, $A$ a $\fl$-algebra of finite type and $\fp, \fq \in \Spec A$ with $\fp \subsetneq \fq$. -Then \[ - \trdeg(\fk(\fp) / \fl) > \trdeg(\fk(\fq) / \fl) -\] -\end{proposition} -\begin{proof} - Replacing $A$ by $A / \fp$, we may assume $\fp = \{0\} $ and $A$ to be a domain. Then $\fk(\fp) = Q(A / \fp) = Q(A)$. - - If $\fq$ is a maximal ideal, $\fk(\fq) = A / \fq$ is of finite type over $\fl$, hence a finite field extension of $\fl$ by the Nullstellensatz (\ref{hns2}). - Thus, $\trdeg(\fk(\fq) / \fl) = 0$. - If $\trdeg(Q(A) / \fl) = 0$, $A$ would be integral over $\fl$, hence a field (fact about integrality and fields, \ref{fintaf}). But if $A$ is a field, $\fp = \{0\}$ is a maximal ideal of $A$, hence $\fq = \fp \lightning$. - This finishes the proof for $\fq \in \mSpec A$. - We will use the following lemma to reduce the general case to this case: - \begin{lemma}\label{ltrdegresfieldtrbase} - There are algebraically independent $a_1,\ldots,a_n \in A$ whose images in $A / \fq$ form a transcendence base for $\fk(\fq) / \fl$. - \end{lemma} - \begin{subproof} - There exist $a_1,\ldots,a_n \in A$ such that $\fk(\fq)$ is algebraic over the subfield generated by $\fl$ and their images $\overline{a_i}$ (for instance generators of $A$ as a $\fl$-algebra). - We may assume that $n$ is minimal. If the $a_i$ are $\fl$-algebraically dependent, then w.l.o.g. $\overline{a_n}$ can be assumed to be algebraic over the subfield generated by $\fl$ and the $\overline{a_i}, 1\le i 0$, then $L^{\Aut(L / K)} = \{l \in L | \E n \in \N ~ l^{p^n} \in K\}$. -\end{proposition} -\begin{proof} - In both cases $L^G \supseteq$ is easy to see. - - If $K \se M \se L$ is an intermediate field, then $L$ is normal over $M$. If $\sigma \in \Aut(M /K)$, an application of Zorn's lemma to the set of all $(N, \vartheta)$ where $N$ is an intermediate field $M \se N \se L$ and $N \xrightarrow{\vartheta} L$ a ring homomorphism such that $\vartheta\defon{M} = \sigma$ shows that $\sigma$ has an extension to an element of $\Aut(L / K)$. % TODO make this rigorous - If $M$ is normal over $K$, it is easily seen to be $\Aut(L / K)$ invariant. - Thus $L^G$ is the union of $M^{\Aut(M / K)}$ over all intermediate fields which are finite and normal over $K$, and it is sufficient to show the proposition for finite normal extensions $L / K$. - - \begin{itemize} - \item Characteristic $0$: The extension is normal, hence Galois, and the assertion follows from Galois theory. - \item Characteristic $p > 0$: Let $l \in L^G$ and $P \in K[T]$ be the minimal polynomial of $l$ over $K$. - We show that $l^{p^n} \in K$ for some $n \in \N$ by induction on $\deg(l / K) \coloneqq \deg(P)$. - - If $\deg(l / K) = 1$, we have $l \in K$. - Otherwise, assume that the assertion has been shown for elements of $L^G$ whose degree over $K$ is smaller than $\deg( l / K)$. - Let $\overline{L}$ be an algebraic closure of $L$ and $\lambda$ a zero of $P$ in $\overline{L}$. - If $M = K(l) \se L$, then there is a ring homomorphism $M - \overline{L}$ sending $l$ to $\lambda$. This can be extended to a ring homomorphism $L \xrightarrow{\sigma} \overline{L}$. We have $\sigma \in G$ because $L / K$ is normal. Hence $\lambda = \sigma(l) = l$, as $l \in L^G$. Thus $l$ is the only zero of $P$ in $\overline{L}$ and because $\deg P >1$ it is a multiple zero. - It is shown in the Galois theory lecture % TODO: link to EinfAlg - that this is possible only when $P(T) = Q(T^p)$ for some $Q \in K[T]$. Then $Q(l^p) = 0$ and the induction assumption can be applied to $x = l^p$ showing $x^{p^m} \in K$ hence $l^{p^{m+1}} \in K$ for some $m \in \N$. - \end{itemize} -\end{proof} -\subsubsection{Integral closure and normal domains} - -\begin{definition}[Integral closure, normal domains] - Let $A$ be a domain with field of quotients $Q(A)$ and let $L$ be a field extension of $Q(A)$. - By \ref{intclosure} the set of elements of $L$ integral over $A$ is a subring of $L$, the \vocab{integral closure} of $A$ in $L$. - $A$ is \vocab{Domain!integrally closed} in $L$ if the integral closure of $A$ in $L$ equals $A$. - $A$ is \vocab{Domain!normal} if it is integrally closed in $Q(A)$. -\end{definition} - -\begin{proposition}\label{ufdnormal} - Any factorial domain (UFD) is normal. -\end{proposition} -\begin{proof} - Let $x \in Q(A)$ be integral over $A$. Then there is a normed polynomial $P \in A[T]$ with $P(x) = 0$. - In \einfalg it was shown that $A[T]$ is a UFD and that the prime elements of $A[T]$ are the elements which are irreducible in $Q(A)[T]$ and for which the $\gcd$ of the coefficients is $\sim 1$. % TODO reference - The prime factors of a normed polynomial are all normed up to multiplicative equivalence. We may thus assume $P$ to be irreducible in $Q(A)[T]$. - But then $\deg P = 1$ as $x$ is a zero of $P$ in $Q(A)$, hence $P(T) = T - x$ and $x \in A$ as $P \in A[T]$. - - - Alternative proof\footnote{\url{http://www.math.lsa.umich.edu/~tfylam/Math221/2.pdf}}: - Let $x = \frac{a}{b} \in Q(A)$ be integral over $A$. \Wlog $\gcd(a,b) = 1$. Then $x^n + c_{n-1} x^{n-1} + \ldots + c_0 = 0$ for some $c_i \in A$. - Multiplication with $b^n$ yields $a^n + c_{n-1} b a^{n-1} + \ldots +c_0 b^n = 0$. Thus $b | a^n$. Since $\gcd(a,b) = 1$ it follows that $b$ is a unit, hence $x \in A$. -\end{proof} - -\begin{remark} - It follows from \ref{cintclosure} and \ref{locandquot} that the integral closure of $A$ in some field extension $L$ of $Q(A)$ is always normal. -\end{remark} -\begin{remark} - A finite field extension of $\Q$ is called an \vocab{algebraic number field} (ANF). If $K$ is an ANF, let $\cO_K$ (the \vocab[Ring of integers in an ANF]{ring of integers in $K$}) be the integral closure of $\Z$ in $K$. - One can show that this is a finitely generated (hence free, by results of \einfalg % EINFALG - ) abelian group. - We have $\cO_{\Q} = \Z$ by the proposiiton. -\end{remark} - -\subsubsection{Action of \texorpdfstring{$\Aut(L / K)$}{Aut(L / K)} on prime ideals of a normal ring extension} - -\begin{theorem}\label{autonprime} - Let $A$ be a normal domain, $L$ a normal field extension of $K \coloneqq Q(A)$, $B$ the integral closure of $A$ in $L$ and $\fp \in \Spec A$. - Then $G \coloneqq \Aut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq \cap A = \fp\}$. -\end{theorem} - - -\begin{proof} - Let $\fq, \fr$ be prime ideals of $B$ above the given $\fp \in \Spec A$. - We must show that there exists $\sigma \in G$ such that $\fq = \sigma(\fr)$. - This is equivalent to $\fq \se \sigma(\fr)$, since the Krull going-up theorem (\ref{cohenseidenberg}) applies to the integral ring extension $B / A$, showing that there are no inclusions between different elements of $\Spec B$ lying above $\fp \in \Spec A$. - - If $L / K$ is finite and there is no such $\sigma$, then by prime avoidance (\ref{primeavoidance}) there is $ x \in \fq \sm \bigcup_{\sigma \in G} \sigma(\fr)$. - As $\fr$ is a prime ideal, $y = \prod_{\sigma \in G} \sigma(x) \in \fq \sm \fr$.\footnote{$\prod_{\sigma \in G} \sigma(x) = \prod_{\sigma \in G} \sigma\inv(x)$} - By the characterization of $L^G$ for normal field extensions (\ref{characfixnormalfe}), there is a positive integer $k$ with $y^k \in K$. - As $A$ is normal, we have $y^k \in K \cap B = A$. - Thus $y^k \in (A \cap \fq) \sm (A \cap \fr) = \fp \sm \fp = \emptyset \lightning$. - - If $L / K$ is not finite, one applies Zorn's lemma to the poset of pairs $(M, \sigma)$ where $M$ is an intermediate field and $\sigma \in \Aut(M / K)$ such that $\sigma(\fr \cap M) = \fq \cap M$. - -\end{proof} -\begin{remark} - The theorem is very important for its own sake. For instance, if $K$ is an ANF which is a Galois extension of $\Q$ it shows that $\Gal(K / \Q)$ transitively acts on the set of prime ideals of $\cO_K$ over a given prime number $p$. More generally, if $L / K$ is a Galois extension of ANF then $\Gal(L / K)$ transitively acts on the set of $\fq \in \Spec \cO_L$ for which $\fq \cap K$ is a given $\fp \in \Spec \cO_K$. -\end{remark} - -\subsubsection{A going-down theorem} -\begin{theorem}[Going-down for integral extensions of normal domains (Krull)]\label{gdkrull} - Let $B$ be a domain which is integral over its subring $A$. If $A$ is a normal domain, then going-down holds for $B / A$. -\end{theorem} - -\begin{proof} - It follows from the assumptions that the field of quotients $Q(B)$ is an algebraic field extension of $Q(A)$. - There is an algebraic extension $L$ of $Q(B)$ such that $L / Q(A)$ is normal (for instance an algebraic closure of $Q(B)$). - Let $C$ be the integral closure of $A$ in $L$. Then $B \se C$ and $C / B$ is integral. -\begin{figure}[H] - \centering - \begin{tikzcd} - Q(A) \arrow[hookrightarrow]{r}{} & Q(B) \arrow[hookrightarrow]{r}{} & L \coloneqq \overline{Q(B)} \\ - A \arrow[hookrightarrow]{u}{}\arrow[hookrightarrow]{r}{} & B \arrow[hookrightarrow]{r}{}\arrow[hookrightarrow]{u}{}& C \arrow[hookrightarrow]{u}{}\\ - \end{tikzcd} -\end{figure} - - \begin{claim} - Going-down holds for $C / A$. - \end{claim} - \begin{subproof} - Let $\fp \se \tilde \fp$ be an inclusion of prime ideals of $A$ and $\tilde \fr \in \Spec C$ with $\tilde \fr \cap A = \tilde \fp$. - By going-up for integral ring extensions (\ref{cohenseidenberg}), $\Spec C \xrightarrow{\cdot \cap A} \Spec A$ is surjectiv. Thus there is $\fr' \in \Spec C$ such that $\fr' \cap A = \fp$. By going up for $C / A$ there is $\tilde \fr' \in \Spec C$ with $\tilde \fr' \cap A = \tilde \fp, \fr' \se \tilde \fr'$. - By the theorem about the action of the automorphism group on prime ideals of a normal ring extension (\ref{autonprime}) there exists a $\sigma \in \Aut(L / Q(A))$ with $\sigma(\tilde \fr') = \tilde \fr$. Then $\fr \coloneqq \sigma(\fr')$ satisfies $\fr \se \tilde \fr$ and $\fr \cap A = \fp$. - \end{subproof} - If $\fp \se \tilde \fp$ is an inclusion of elements of $\Spec A$ and $\tilde \fq \in \Spec B$ with $\tilde \fp \cap A = \tilde \fp$, by the surjectivity of $\Spec C \xrightarrow{\cdot \cap B} \Spec B$ (\ref{cohenseidenberg}) there is $\tilde \fr \in \Spec C$ with $\tilde \fr \cap B = \fq$. - By going-down for $C / A$, there is $\fr \in \Spec C$ with $\fr \se \tilde \fr$ and $\fr \cap A = \fp$. - Then $\fq \coloneqq \fr \cap B \in \Spec B, \fq \se \tilde \fq$ and $\fq \cap A = \fp$. Thus going-down holds for $B / A$. -\end{proof} - -\begin{remark}[Universally Japanese rings] - A Noetherian ring $A$ is called universally Japanese if for every $\fp \in \Spec A$ and every finite field extension $L$ of $\fk(\fp)$, the integral closure of $A / \fp$ in $L$ is a finitely generated $A$-module. This notion was coined by Grothendieck because the condition was extensively studied by the Japanese mathematician Nataga Masayoshji. - By a hard result of Nagata, algebras of finite type over a universally Japanese ring are universally Japanese. - Every field is universally Japanese, as is every PID of characteristic $0$. - There are, however, examples of Noetherian rings which fail to be universally Japanese. -\end{remark} - -\begin{dexample}[Counterexample to going down] - Let $R = \fk[X,Y]$ and $A = \fk[X,Y, \frac{X}{Y}]$. Then going down does not hold for $A / R$: - - For any ideal $Y \in \fq \se A$ we have $X = \frac{X}{Y} \cdot Y \in \fq$. - Consider $(Y)_R \subsetneq (X,Y)_R \se \fq \cap R$. As $(X,Y)_R$ is maximal and the preimage of a prime ideal is prime and thus proper, we have $(X,Y)_R = \fq \cap R$. - The prime ideal $(\frac{X}{Y},Y)_A = (\frac{X}{Y}, X,Y)_A$ is lying over $(X,Y)_R$, so going down is violated. -\end{dexample} - - -\subsubsection{Proof of \texorpdfstring{$\codim(\{y\},Y) = \trdeg(\fK(Y) / \fk)$}{codim(\{y\},Y) = trdeg(K(Y) /k)}} -\label{proofcodimletrdeg} -This is part of the proof of \ref{trdegandkdim}. %TODO: reorder - -\begin{proof} - % DIMT - Let $B = \fk[X_1,\ldots,X_n]$ and $X \se Y = V(\fp) \se \fk^n$ irreducible closed subsets of $\fk^n$. - We want to show that $\codim(X,Y) = \trdeg(\fK(Y) / \fk) - \trdeg(\fK(X) / \fk)$. - $\le $ was shown in \ref{upperboundcodim}. - $\dim Y \ge \trdeg(\fK(Y) / \fk)$ was shown in \ref{lowerbounddimy} by - - Applying Noether normalization to $A \coloneqq B / \fp$, giving us $(f_i)_{i=1}^d \in A^d$ such that the $f_i$ are algebraically independent and $A$ finite over the subalgebra generated by them. - We then used going-up to lift a chain of prime ideals corresponding to $\fk^d \supsetneq \{0\} \times \fk^{n-1} \supsetneq \ldots \supsetneq \{0\}$ under $Y \xrightarrow{F = (f_1,\ldots,f_d)} \fk^d$ to a chain of prime ideals in $A$. - This was done left-to-right as going-up was used to make prime ideals larger. In particular, when $\{0\} \in \fk^d$ has several preimages under $F$ we cannot control to which of them the maximal ideal terminating the lifted chain belongs. Thus, we can show that in the inequality - \[ - \codim(\{y\}, Y) \le d = \trdeg(\fK(Y) \sm \fk) - \] - (see \ref{upperboundcodim}) - equality holds for at least one pint $y \in F\inv(\{0\})$ but cannot rule out that there are other $y \in F\inv(\{0\})$ for which the inequality becomes strict. - However using going-down (\ref{gdkrull}) for $F$, we can use a similar argument, but start lifting of the chain at the right end for the point $y \in Y$ for which we would like to show equality. - From this $\codim(X,Y) \ge \trdeg(\fK(Y) / \fk) - \trdeg(\fK(X) / \fk)$ can be derived similarly to \ref{upperboundcodim}. - Thus - \[ - \codim(X,Y) = \trdeg(\fK(Y) / \fk) - \trdeg(\fK(X) / \fk) - \] - follows (see \ref{htandcodim} and \ref{htandtrdeg}). -\end{proof} -\begin{remark} - The going-down theorem used to prove this is somewhat more general, as it does not depend on $\fk$ being algebraically closed. -\end{remark} - - - - -% Lecture 09 -% i = ic - -\subsection{The height of a prime ideal} -In order to complete the proof of \ref{proofcodimletrdeg} and show $\codim(X,Y) = \trdeg(\fK(Y) / \fk) - \trdeg(\fK(X) / \fk)$, -we need to localize the $\fk$-algebra with respect to a multiplicative subset and replace the ground field by a larger subfield of that localization which is no longer algebraically closed. -To formulate a result which still applies in this context, we need the following: -\begin{definition}[Height of a prime ideal] - Let $A$ be a ring, $\fp \in \Spec A$. We define the \vocab[Height of a prime ideal]{height of the prime ideal $\fp$}, $\hght(\fp)$, to be the largest $k \in \N$ such that there is a strictly decreasing sequence $\fp = \fp_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_k$ of prime ideals of $A$, or $\infty$ if there is no finite upper bound on the length of such sequences. -\end{definition} -\begin{example} - Let $A = \fk[X_1,\ldots,X_n]$, $X = V(\fp)$ for a prime ideal $\fp$. - By the correspondence between irreducible subsets of $\fk^n$ and prime ideals in $A$ (\ref{bijiredprim}), - the $\fp_i$ correspond to irreducible subsets $X_i \se \fk^n$ containing $X$. Thus $\hght(\fp) = \codim(X, \fk^n)$. -\end{example} - -\begin{example}\label{htandcodim} - Let $B = \fk[X_1,\ldots,X_n], \fq \in \Spec B$ and let $A \coloneqq B / \fp$. - Let $Y \coloneqq V(\fq) \se \fk^n$, $\tilde \fp \coloneqq \pi_{B, \fq}\inv(\fp)$, where $B \xrightarrow{\pi_{B, \fp}} A $ is the projection to the ring of residue classes, and let $X = V(\tilde \fp)$. - By \ref{idealslocbij} we have a bijection between the prime ideals $\fr \se \fp$ of $A$ contained in $\fp$ and the prime ideals and the prime ideals $\tilde \fr \in \Spec B$ with $\fq \se \tilde \fr \se \tilde \fp$: - \begin{align} - f: \{\fr \in \Spec A | \fr \se \fp \} &\longrightarrow \{\tilde \fr \in \Spec B | \fq \se \tilde \fr \se \tilde \fp\} \\ - \fr &\longmapsto \pi_{B, \fq}\inv(\fr)\\ - \tilde \fr / \fq &\longmapsfrom \tilde \fr - \end{align} - By \ref{bijiredprim}, the $\tilde \fr$ are in canonical bijection with the irreducible subsets $Z$ of $Y$ containing $X$. - Thus, the chains $\fp = \fp_0 \supsetneq \ldots \supsetneq \fp_k$ are in canonical bijection with the chains $X = X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_k \se Y$ of irreducible subsets and - $\hght(\fp) = \codim(X,Y)$. -\end{example} - - -\begin{remark} - Let $A$ be an arbitrary ring. One can show that there is a bijection between $\Spec A$ and the set of irreducible subsets $Y \se \Spec A$: - \begin{align} - f: \Spec A &\longrightarrow \{Y \se \Spec A | Y\text{irreducible}\} \\ - \fp &\longmapsto \Vs(\fp)\\ - \bigcup_{\fp \in Y} \fp &\longmapsfrom Y - \end{align} - Thus, the chains $\fp = \fp_0 \supsetneq \ldots \supsetneq \fp_k$ are in canonical bijection with the chains $V(\fp) = X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_k \se \Spec A$ of irreducible subsets, and $\hght(\fp) = \codim(V(\fp), \Spec A)$. -\end{remark} - - -\subsubsection{The relation between \texorpdfstring{$\hght(\fp)$}{ht(p)} and \texorpdfstring{$\trdeg$}{trdeg}} -We will use the following -\begin{lemma}\label{extendtotrbase} - Let $\fl$ be an arbitrary field, $A$ a $\fl$-algebra of finite type which is a domain, $K \coloneqq Q(A)$ the field of quotients and let $(a_i)_{i=1}^n$ be $\fl$-algebraically independent elements of $A$. Then there exist a natural number $m \ge n$ and a transcendence base $(a_i)_{i = 1}^m$ for $K / \fl$ with $a_i \in A$ for $1 \le i \le m$. -\end{lemma} -\begin{proof} - The proof is similar to the proof of \ref{ltrdegresfieldtrbase}. - There are a natural number $m \ge n$ and elements $(a_i)_{i = n+1}^m \in A^{m-n}$ which generate $K$ in the sense of a matroid used in the definition of $\trdeg$. - For instance, one can use generators of the $\fl$-algebra $A$. We assume $m$ to be minimal and claim that $(a_i)_{i=1}^m$ are $\fl$-algebraically independent. - Otherwise there is $j \in \N$, $1 \le j \le m$ such that $a_j$ is algebraic over the subfield of $K$ generated by $\fl$ and the $(a_i)_{i=1}^{j-1}$. We have $j > n$ by the algebraic independence of $(a_i)_{i=1}^n$. - Exchanging $x_j$ and $x_m$, we may assume $j = m$. But then $K$ is algebraic over its subfield generated by $\fl$ and the $(a_i)_{i=1}^{m-1} $, contradicting the minimality of $m$. -\end{proof} - -\begin{theorem}\label{htandtrdeg} - Let $\fl$ be an arbitrary field, $A$ a $\fl$-algebra of finite type which is a domain, and $\fp \in \Spec A$. - Let $K \coloneqq Q(A)$ be the field of quotients of $A$. Then - \[ - \hght(\fp) = \trdeg(K /\fl) - \trdeg(\fk(\fp) / \fl) - \] -\end{theorem} -\begin{remark} - By example \ref{htandcodim}, theorem \ref{trdegandkdim} is a special case of this theorem. %(\ref{htandtrdeg}). -\end{remark} -\begin{proof} - If $\fp = \fp_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_k$ is a chain of prime ideals in $A$, we have $\trdeg(\fk(\fp_i) / \fl) < \trdeg(\fk(\fp_{i+1}) / \fl)$ by \ref{trdegresfield} (``A first result of dimension theory''). -Thus -\[ - k \le \trdeg(\fk(\fp_k) / \fl) - \trdeg(\fk(\fp) / \fl) \le \trdeg(K / \fl) - \trdeg(\fk(\fp) / \fl) -\] -where the last inequality is another application of \ref{trdegresfield} (using $K = Q(A) = Q(A / \{0\}) = \fk(\{0\})$ and the fact that $\{0\} \se \fp_k$ is a prime ideal). -Hence \[ - \hght(\fp) \le \trdeg( K / \fl) - \trdeg(\fk(\fp) / \fl) -\] -and it remains to show the opposite inequality. - -\begin{claim} - For any maximal ideal $\fp \in \mSpec A$ \[ - \hght(\fm) \ge \trdeg(K / \fl) - \] -\end{claim} -\begin{subproof} - By the Noether normalization theorem (\ref{noenort}), there are $(x_i)_{i=1}^d \in A^d$ which are algebraically independent over $\fl$ such that $A$ is finite over the subalgebra $S$ generated by the $x_i$. We have $d = \trdeg(K / \fl)$ as the $x_i$ form a transcendence base of $K / \fl$. -\begin{claim} - We can choose $x_i \in \fm$ -\end{claim} -\begin{subproof} - By the Nullstellensatz (\ref{hns2}), $\fk(\fm) = A / \fm$ is a finite field extension of $\fl$. Hence there exists a normed polynomial $P_i \in \fl[T]$ with $P_i(x_i \mod \fm) = 0$ in $\fk(\fm)$. - Let $\tilde x_i \coloneqq P_i(x_i) \in \fm$ and $\tilde S$ the subalgebra generated by the $\tilde x_i$. As $P_i(x_i) - \tilde x_i = 0$, $x_i$ is integral over $\tilde S$ and so is $S / \tilde S$. It follows that $A / \tilde S$ is integral, hence finite by \ref{ftaiimplf}. Replacing $x_i$ by $\tilde x_i$, we may thus assume that $x_i \in \fm$. - - -\end{subproof} -% TODO: fix names A_1 = A_S, k_1 = R_S - The ring homomorphism $\ev_x : R = \fl[X_1,\ldots,X_d] \xrightarrow{P \mapsto P(x_1,\ldots,x_d)} A$ is injective. Because $R$ is a UFD, $R$ is normal (\ref{ufdnormal}). Thus the going-down theorem (\ref{gdkrull}) applies to the integral $R$-algebra $A$. - For $0 \le i \le d$, let $\fp_i \se R$ be the ideal generated by $(X_j)_{j=i+1}^d$. We have $\fm \sqcap R = \fp_0$ as all $X_i \in \fm$, hence $X_i \in \fm \sqcap R$ and $\fp_0$ is a maximal ideal. - By applying going-down and induction on $i$, there is a chain $\fm = \fq_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_d$ of elements of $\Spec A$ such that $\fq_i \sqcap R = \fp_i$. - It follows that $\hght(\fm) \ge d$. -\end{subproof} -This finishes the proof in the case of $\fp \in \mSpec A$. - -To reduce the general case to that special case, we proceed as in \ref{trdegresfield}: -By lemma \ref{ltrdegresfieldtrbase} there are $a_1,\ldots,a_n \in A$ whose images in $A / \fp$ form a transcendence base for $\fk(\fp) / \fl$. -As these images are $\fl$-algebraically independent, the same holds for the $a_i$ themselves. - -By lemma \ref{extendtotrbase} we can extend $(a_{i})_{i=1}^n$ to a transcendence base $(a_i)_{i=1}^m \in A^m$ of $K / \fl$. -Let $R \se A$ denote the $\fl$-subalgebra generated by $a_1,\ldots,a_n$ and let $S \coloneqq R \sm \{0\}$. -Let $A_1 \coloneqq A_S$ and $\fp_S$ the prime ideal corresponding to $\fp$ under $\Spec(A_1) \cong \{\fr \in \Spec A | \fr \cap S = \emptyset\}$ (\ref{idealslocbij}). -As in \ref{locandquot}, $A_1$ is a domain with $Q(A_1) \cong K = Q(A)$ and by \ref{locandfactor} $A_1 / \fp_S \cong (A / \fp)_{\overline{S}}$, where $\overline{S}$ denotes the image of $S$ in $A / \fp$. -As in \ref{trdegresfield}, $\fk(\fp_S) \cong \fk(\fp)$ is integral over $A_1 / \fp_S$. -From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1 / \fp_S$ is a field. Hence $\fp_S \in \mSpec(A_1)$ and the special case can be applied to $\fp_S$ and $A_1 / \fl_1$, showing that $\hght(\fp_S) \ge e = \trdeg(K / \fl_1)$. We have $\trdeg(K / \fl_1) = m - n$, as $(a_i)_{i = n+1}^m$ is a transcendence base for $K / \fl_1$. By the description of $\Spec A_S$ (\ref{idealslocbij}), a chain $\fp_S = \fq_0 \supsetneq \ldots \supsetneq \fp_e$ of prime ideals in $A_S$ defines a similar chain $\fp_i \coloneqq \fq_i \sqcap A$ in $A$ with $\fp_0 = \fp$. Thus $\hght(\fp) \ge e$. -\end{proof} - -\begin{remark} - As a consequence of his principal ideal theorem, Krull has shown the finiteness of $\hght(\fp)$ for $\fp \in \Spec A$ when $A$ is a Noetherian ring. But $\dim A = \sup_{\fp \in \Spec A} \hght(\fp) = \sup_{\fm \in \mSpec A} \hght(\fm)$, the Krull dimension of the Noetherian topological space $\Spec A$ may nevertheless be infinite. -\end{remark} -\begin{dexample}[Noetherian ring with infinite dimension]\footnote{\url{https://math.stackexchange.com/questions/1109732/noetherian-ring-with-infinite-krull-dimension-nagatas-example}} - Let $A = \fk[X_i | i \in \N]$ and $m_1, m_2, \ldots \in \N$ an increasing sequence such that $m_{i+1}-m_i > m_i - m_{i-1}$. - Let $\fp_i \coloneqq (X_{m_{i}+1},\ldots,X_{m_{i+1}})$ and $S \coloneqq A \sm \bigcup_{i \in \N} \fp_i$. - $S$ is multiplicatively closed. - $A_S$ is Noetherian but $\hght((\fp_i)_S) = m_{i+1}- m_{i}$ hence $\dim(A_S) = \infty$. -\end{dexample} - -% Lecture 10 - - -\subsection{Dimension of products} - - -\begin{proposition}\label{dimprod} - Let $X \se \fk^n$ and $Y \se \fk^n$ be irreducible and closed. Then $X \times Y$ is also an irreducible closed subset of $\fk^{m+n}$. - Moreover, $\dim(X \times Y) = \dim(X) + \dim(Y)$ and $\codim(X \times Y, \fk^{m+n}) = \codim(X, \fk^m) + \codim(Y, \fk^n)$. -\end{proposition} -\begin{proof} - Let $X = V(\fp)$ and $Y = V(\fq)$ where $\fp \in \Spec \fk[X_1,\ldots,X_m]$ and $\fq \in \Spec \fk[X_1,\ldots,X_n]$. - We denote points of $\fk^{m+n}$ as $x = (x',x'')$ with $x' \in \fk^m, x''\in\fk^n$. Then $X \times Y$ is the set of zeroes of the ideal in $\fk[X_1,\ldots,X_{m+n}]$ generated by the polynomials $f(x) = \phi(x')$, with $\phi$ running over $\fp$ and $g(x) = \gamma(x'')$ with $\gamma$ running over $\fq$. - Thus $X \times Y$ is closed in $\fk^{m+n}$. - We must also show irreducibility. $X \times Y \neq \emptyset$ is obvious. - - Assume that $X \times Y = A_1 \cup A_2$, where the $A_i \se \fk^{m+n}$ are closed. - For $x' \in \fk^m, x' \times Y$ is homeomorphic to the irreducible $Y$. Thus $X = X_1 \cup X_2$ where $X_i = \{x \in X | \{x\} \times Y \se A_i\} $. - Because $X_i = \bigcap_{y \in Y} \{x \in X | (x,y) \in A_i\}$, this is closed. As $X$ is irreducible, there is $i \in \{1;2\} $ which $X_i = X$. Then $X \times Y = A_i$ confirming the irreducibility of $X \times Y$. - - Let $a = \dim X$ and $b = \dim Y$ and $X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_a = X$,$Y_0 \subsetneq Y_1 \subsetneq \ldots \subsetneq Y_b = Y$ be chains of irreducible subsets. By the previous result, - $X_0 \times Y_0 \subsetneq X_1 \times Y_0 \subsetneq \ldots \subsetneq X_a \times Y_0 \subsetneq X_a \times Y_1 \subsetneq \ldots \subsetneq X_a \times Y_a = X \times Y$ is a chain of irreducible subsets. - Thus $\dim(X \times Y) \ge a + b = \dim X + \dim Y$. - Similarly one derives $\codim(X \times Y, \fk^{m+n}) \ge \codim(X, \fk^m) + \codim(Y, \fk^n)$. - By \ref{trdegandkdim} we have $\dim(A) + \codim(A, \fk^l) = l$ for irreducible subsets of $\fk^l$. Thus equality must hold in the previous two inequalities. - -\end{proof} -\subsection{The nil radical} -\begin{notation} - Let $\Vspec(I)$ denote the set of $\fp \in \Spec A$ containing $I$. -\end{notation} - -\begin{proposition}[Nil radical] - For a ring $A$, $\bigcap_{\fp \in \Spec A} \fp = \sqrt{\{0\} } = \{a \in A | \E k \in \N ~ a^k = 0\} \text{\reflectbox{$\coloneqq$}} \nil(A)$, the set of nilpotent elements of $A$. - This is called the \vocab{nil radical} of $A$. -\end{proposition} -\begin{proof} - It is clear that elements of $\sqrt{\{0\} } $ must belong to all prime ideals. Conversely, let $a \in A \sm \sqrt{\{0\} }$. Then $S = a^{\N}$ is a multiplicative subset of $A$ not containing $0$. - The localisation $A_S$ of $A$ is thus not the null ring. Hence $\Spec A_S \neq \emptyset$. If $\fq \in \Spec A_S$, then by the description of $\Spec A_S$ (\ref{idealslocbij}), $\fp \coloneqq \fq \sqcap A$ is a prime ideal of $A$ disjoint from $S$, hence $a \not\in \fp$. -\end{proof} - -\begin{corollary}\label{sqandvspec} - For an ideal $I$ of $R$, $\sqrt{I} = \bigcap_{\fp \in \Vspec(I)} \fp$. -\end{corollary} -\begin{proof} - This is obtained by applying the proposition to $A = R / I $ and using the bijection $\Spec( R / I) \cong V(I)$ sending $\fp \in V(I)$ to $\fp \coloneqq \fp / I$ and $\fq \in \Spec(R / I)$ to its inverse image $\fp$ in $R$. -\end{proof} -\subsubsection{Closed subsets of \texorpdfstring{$\Spec R$}{Spec R}} -\begin{proposition}\label{bijspecideal} - There is a bijection - \begin{align} - f: \{A \se \Spec R | A\text{ closed}\} &\longrightarrow \{I \se R | I \text{ ideal and } I = \sqrt{I} \} \\ - A &\longmapsto \bigcap_{\fp \in A} \fp\\ - \Vspec(I) &\longmapsfrom I - \end{align} - Under this bijection, the irreducible subsets correspond to the prime ideals and the closed points $\{\fm\}, \fm \in \Spec A$ to the maximal ideals. -\end{proposition} -\begin{proof} - If $A = \Vspec(I)$, then by \ref{sqandvspec} $\sqrt{I} = \bigcap_{\fp \in A} \fp$. Thus, an ideal with $\sqrt{I} = I$ can be recovered from $\Vspec( I)$. Since $\Vspec(J) = \Vspec(\sqrt{J})$, the map from ideals with $\sqrt{I} = I$ to closed subsets is surjective. - - Sine $R$ corresponds to $\emptyset$, the proper ideals correspond to non-empty subsets of $\Spec R$. Assume that $\Vspec(I) = \Vspec(J_1) \cup \Vspec(J_2)$, where the decomposition is proper and the ideals coincide with their radicals. - Let $g = f_1f_2$ with $f_k \in J_k \sm I$. Since $\Vspec(g) \supseteq \Vspec(f_k) \supseteq \Vspec(I_k), \Vspec(I) \se \Vspec(g)$. Hence $g \in \sqrt{I} = I$. - As $f_k \not\in I$, $I$ fails to be a prime ideal. - Conversely, assume that $f_1f_2 \in I$ while the factors are not in $I$. Since $I = \sqrt{I}, \Vspec(f_k) \not\supseteq \Vspec(I)$. But $\Vspec(f_1) \cup \Vspec(f_2) = \Vspec(f_1f_2) \supseteq \Vspec(I)$. - The proper decomposition $\Vspec(I) = \left( \Vspec(I) \cap \Vspec(f_1) \right) \cup \left( \Vspec(I) \cap \Vspec(f_2) \right) $ now shows that $\Vspec(I)$ fails to be irreducible. - The final assertion is trivial. -\end{proof} - -\begin{corollary} - If $R$ is a Noetherian ring, then $\Spec R$ is a Noetherian topological space. -\end{corollary} -\begin{remark} - It is not particularly hard to come up with examples which show that the converse implication does not hold. -\end{remark} -\begin{dexample} - Let $A = \fk[X_n | n \in \N] / I$ where $I$ denotes the ideal generated by $\{X_i^2 | i \in \N\}$. - $A$ is not Noetherian, since the ideal $J$ generated by $\{X_i | i \in \N\} $ is not finitely generated. - $A / J \cong \fk$, hence $J$ is maximal. As every prime ideal must contain $\nil(A) \supseteq J$, $J$ is the only prime ideal. - Thus $\Spec A$ contains only one element and is hence Noetherian. -\end{dexample} - -\begin{corollary}[About the smallest prime ideals containing $I$ ]\label{smallestprimesvi} - If $R$ is Noetherian and $I \se R$ an ideal, then the set $\Vspec(I) = \{\fp \in \Spec R | I \se \fp\}$ has finitely many $\se$-minimal elements $(\fp_i)_{i=1}^k$ and every element of $V(I)$ contains at least one $\fp_i$. - The $\Vspec(\fp_i)$ are precisely the irreducible components of $V(I)$. Moreover $\bigcap_{i=1}^k \fp_i = \sqrt{I}$ and $k > 0$ if $I$ is a proper ideal. -\end{corollary} -\begin{proof} - If $\Vspec(I) = \bigcup_{i=1}^k \Vspec(\fp_i)$ is the decomposition into irreducible components then every $\fq \in \Vspec(I)$ must belong to at least one $\Vspec(\fp_i)$, hence $\fp_i \se \fq$. Also $\fp_i \in \Vspec(\fp_i) \se \Vspec(I)$. - It follows that the sets of $\se$-minimal elements of $\Vspec(I)$ and of $\{\fp_1,\ldots,\fp_k\} $ coincide. - As there are no non-trivial inclusions between the $\Vspec(\fp_i)$, there are no non-trivial inclusions between the $\fp_i$ and the assertion follows. - The final remark is trivial. -\end{proof} -\begin{corollary} - If $R$ is any ring, $\hght(\fp) = \codim(\Vspec(\fp), \Spec R)$. -\end{corollary} - - -\subsection{The principal ideal theorem} -Krull was able to show: -\begin{theorem}[Principal ideal theorem / Hauptidealsatz]\label{pitheorem} - Let $A$ be a Noetherian ring, $a \in A$ and $\fp \in \Spec A$ a $\se$-minimal element of $\Vspec(a)$. Then $\hght(\fp) \le 1$. -\end{theorem} -\begin{proof} - Probably not relevant for the exam. -\end{proof} -\begin{remark} - Intuitively, the theorem says that by imposing a single equation one ends up in codimension at most $1$. This would not be true in real analysis (or real algebraic geometry) as the equation $\sum_{i=1}^{n} X_i^2 = 0$ shows. By \ref{smallestprimesvi}, if $a$ is a non-unit then a $\fp \in \Spec A$ to which the theorem applies can always be found. - Using induction on $k$, Krull was able to derive: -\end{remark} - -\begin{theorem}[Generalized principal ideal theorem] - Let $A$ be a Noetherian ring, $(a_i)_{i=1}^k \in A$ and $\fp \in \Spec A$ a $\se$-minimal element of $\bigcap_{i=1}^k V(a_i)$, the set of prime ideals containing all $a_i$. - Then $\hght(\fp) \le k$. -\end{theorem} -Modern approaches to the principal ideal theorem usually give a direct proof of this more general theorem. - -\begin{corollary} - If $R$ is a Noetherian ring and $\fp \in \Spec R$, then $\hght(\fp) < \infty$. -\end{corollary} -\begin{proof} - If $\fp$ is generated by $(f_i)_{i=1}^k$, then $\hght(\fp) \le k$. -\end{proof} -\subsubsection{Application to the dimension of intersections} - -\begin{remark}\label{smallestprimeandirredcomp} -Let $R = \fk[X_1,\ldots,X_n]$ and $I \se R$ an ideal. - - If $(\fp_i)_{i=1}^k$ are the smallest prime ideals of $R$ containing $I$, then $(\Va(\fp_i))_{i=1}^k$ are the irreducible components of $\Va(I)$. -\end{remark} -\begin{proof} - The $\Va(\fp_i)$ are irreducible, there are no non-trivial inclusions between them and $ \Va(I) = \Va(\sqrt{I}) = \Va(\bigcap_{i=1}^k \fp_i) = \bigcup_{i=1}^k \Va(\fp_i)$. -\end{proof} - -\begin{corollary}[of the principal ideal theorem] - \label{corpithm} - Let $X \se \fk^n$ be irreducible, $(f_i)_{i=1}^k$ elements of $R = \fk[X_1,\ldots,X_n]$ and $Y$ an irreducible component of $A = X \cap \bigcap_{i=1}^k V(f_i)$. - Then $\codim(Y,X) \le k$. -\end{corollary} -\begin{remark} - This confirms the naive geometric intuition that by imposing $k$ equations one ends up in codimension at most $k$. -\end{remark} -\begin{proof} - If $X = v(\fp), X \cap \bigcap_{i=1}^k V(f_i) = V(I)$ where $I \se R$ is the ideal generated by $\fp$ and the $f_i$. - By \ref{smallestprimeandirredcomp}, $Y = V(\fq)$ where $\fq$ is the smallest prime ideal containing $I$. - Then $\fq / \fp$ is a smallest prime ideal of $R / \fp$ containing all $(f_i \mod \fp)_{i=1}^k$. - By the principal ideal theorem (\ref{pitheorem}), $\hght(\fq / \fp) \le k$ and the assertion follows from example \ref{htandcodim}. -\end{proof} -\begin{remark}\label{affineproblem} - Note that the intersection $X \cap \bigcap_{i=1}^k V(f_i)$ can easily be empty, even when $k$ is much smaller than $\dim X$. -\end{remark} - -\begin{corollary}\label{codimintersection} - Let $A$ and $B$ be irreducible subsets of $\fk^n$. If $C$ is an irreducible component of $A \cap B$, then $\codim(C, \fk^n) \le \codim(A, \fk^n) + \codim(B, \fk^n)$. -\end{corollary} -\begin{dremark} - Equivalently, $\dim(C) \ge \dim(A) + \dim(B)-n$. -\end{dremark} -\begin{proof} - Let $X = A \times B \se \fk^{2n}$, where we use $(X_1,\ldots,X_n,Y_1,\ldots,Y_n)$ as coordinates of $\fk^{2n}$. - Let $\Delta \coloneqq \{(x_1,\ldots,x_n,x_1,\ldots,x_n) | x \in \fk^n\} $ be the diagonal in $\fk^n \times \fk^n$. - The projection $\fk^{2n}\to \fk^n$ to the $X$-coordinates defines a homeomorphism between $(A \times B) \cap \Delta$ and $A \cap B$. - Thus, $C$ is homeomorphic to an irreducible component $C'$ of $(A \times B) \cap \Delta$ and - \begin{align} - \codim(C, \fk^n) = n - \dim(C) = n - \dim(C') = n - \dim(A \times B) + \codim(C', A \times B)\\ - \overset{\text{\ref{corpithm}}}{\le }2n - \dim(A \times B) \overset{\text{\ref{dimprod}}}{=} 2n - \dim(A) - \dim(B) = \codim(A,\fk^n) + \codim(B, \fk^n) - \end{align} - by the general properties of dimension and codimension, \ref{corpithm} applied to $(X_i - Y_i)_{i=1}^n$, - the result about the dimension of products (\ref{dimprod}) and again the general properties of dimension and codimension. - -\end{proof} -\begin{remark} - As in \ref{affineproblem}, $A \cap B$ can easily be empty, even when $A$ and $B$ have codimension $1$ and $n$ is very large. -\end{remark} - -\subsubsection{Application to the property of being a UFD} -\begin{proposition}\limrel - Let $R$ be a Noetherian domain. Then $R$ is a UFD iff every $\fp \in \Spec R$ with $\hght(\fp)= 1$\footnote{In other words, every $\se$-minimal element of the set of non-zero prime ideals of $R$ } is a principal ideal. -\end{proposition} -\begin{proof} - Every element of every Noetherian domain can be written as a product of irreducible elements.\footnote{Consider the set of principal ideals $rR$ where $r$ is not a product of irreducible elements.} - Thus, $R$ is a UFD iff every irreducible element of $R$ is prime. - - - Assume that this is the case. Let $\fp \in \Spec R, \hght(\fp) = 1$. - Let $p \in \fp \sm \{0\}$. Replacing $p$ by a prime factor of $p$, we may assume $p$ to be prime. Thus $\{0\} \subsetneq pR \se \fp$ is a chain of prime ideals and since $\hght(\fp) = 1$ it follows that $\fp = pR$. - - Conversely, assume that every $\fp \in \Spec R$ with $\hght(\fp)=1$ is a principal ideal. Let $f \in R$ be irreducible. - Let $\fp \in \Spec R$ be a $\se$-minimal element of $V(f)$. By the principal ideal theorem (\ref{pitheorem}), $\hght(\fp)=1$. - Thus $\fp = pR$ for some prime element $p$. We have $p | f$ since $f \in \fp$. As $f$ is irreducible, $p$ and $f$ are multiplicatively equivalent. Thus $f$ is a prime element. -\end{proof} - -\subsection{The Jacobson radical}\limrel -\begin{proposition} - For a ring $A, \bigcap_{\fm \in \mSpec A} \fm = \{a \in A | \A x \in A ~ 1 - ax \in A^{\times }\} \text{\reflectbox{$\coloneqq$}} \rad(A)$, the \vocab{Jacobson radical} of $A$. -\end{proposition} -\begin{proof} - Suppose $\fm \in \mSpec A$ and $a \in A \sm \fm$. Then $a \mod \fm \neq 0$ and $A / \fm$ is a field. Hence $a \mod \fm$ has an inverse $x \mod \fm$. - $1 - ax \in \fm$, hence $1 - ax \not\in A^{\times}$ and $a $ is not al element of the RHS. - - Conversely, let $a \in A$ belong to all $\fm \in \mSpec A$. If there exists $x \in A$ such that $1 - ax \not\in A^{\times }$ then $(1-ax) A$ was a proper ideal in $A$, hence contained in a maximal ideal $\fm$. As $a \in \fm, 1 = (1-ax) + ax \in \fm$, a contradiction. - Hence every element of $\bigcap_{\fm \in \mSpec A} \fm$ belongs to the right hand side. -\end{proof} - -\begin{example} - If $A$ is a local ring, then $\rad(A) = \fm_A$. -\end{example} -\begin{example} - If $A$ is a PID with infinitely many multiplicative equivalence classes of prime elements (e.g. $\Z$ of $\fk[X]$), then $\rad(A) = \{0\}$: - Prime ideals of a PID are maximal. Thus if $x \in \rad(A)$, every prime element divides $x$. If $x \neq 0$, it follows that $x$ has infinitely many prime divisors. - However every PID is a UFD. -\end{example} -\begin{example} -If $A$ is a PID for which $p_1,\ldots,p_n$ is a list of representatives of the multiplicative equivalence classes of prime elements, then -$\rad(A) = f A$ where $f = \prod_{i=1}^{n} p_i$. -\end{example} - -% proof of the pitheorem probably won't be relevant in the exam -% last 2 slides are of "limited relevance" (3 option questions), and may improve grade, but 1.0 can be obtained without it - - - -% Lecture 11 - -\section{Projective spaces} -Let $\fl$ be any field. -\begin{definition} - For a $\fl$-vector space $V$, let $\bP(V)$ be the set of one-dimensional subspaces of $V$. - Let $\bP^n(\fl) \coloneqq \bP(\fl^{n+1})$, the \vocab[Projective space]{$n$-dimensional projective space over $\fl$}. - - If $\fl$ is kept fixed, we will often write $\bP^n$ for $\bP^n(\fl)$. - - When dealing with $\bP^n$, the usual convention is to use $0$ as the index of the first coordinate. - - We denote the one-dimensional subspace generated by $(x_0,\ldots,x_n) \in \fk^{n+1} \sm \{0\}$ by $[x_0,\ldots,x_n] \in \bP^n$. - If $x = [x_0,\ldots,x_n] \in \bP^n$, the $(x_{i})_{i=0}^n$ are called \vocab{homogeneous coordinates} of $x$. - At least one of the $x_{i}$ must be $\neq 0$. -\end{definition} -\begin{remark} - There are points $[1,0], [0,1] \in \bP^1$ but there is no point $[0,0] \in \bP^1$. -\end{remark} -\begin{definition}[Infinite hyperplane] - For $0 \le i \le n$ let $U_i \se \bP^n$ denote the set of $[x_0,\ldots,x_{n}]$ with $x_{i}\neq 0$. - This is a correct definition since two different sets $[x_0,\ldots,x_{n}]$ and $[\xi_0,\ldots,\xi_n]$ of homogeneous coordinates for the same point $x \in \bP^n$ differ by scaling with a $\lambda \in \fl^{\times}$, $x_i = \lambda \xi_i$. Since not all $x_i$ may be $0$, $\bP^n = \bigcup_{i=0}^n U_i$. We identify $\bA^n = \bA^n(\fl) = \fl^n$ with $U_0$ by identifying $(x_1,\ldots,x_n) \in \bA^n$ with $[1,x_1,\ldots,x_n] \in \bP^n$. - Then $\bP^1 = \bA^1 \cup \{\infty\} $ where $\infty=[0,1]$. More generally, when $n > 0$ $\bP^n \sm \bA^n$ can be identified with $\bP^{n-1}$ identifying $[0,x_1,\ldots,x_n] \in \bP^n \sm \bA^n$ with $[x_1,\ldots,x_n] \in \bP^{n-1}$. - - Thus $\bP^n$ is $\bA^n \cong \fl^n$ with a copy of $\bP^{n-1}$ added as an \vocab{infinite hyperplane} . -\end{definition} - -\subsubsection{Graded rings and homogeneous ideals} -\begin{notation} - Let $\bI = \N$ or $\bI = \Z$. -\end{notation} -\begin{definition} - By an \vocab[Graded ring]{$\bI$-graded ring} $A_\bullet$ we understand a ring $A$ with a collection $(A_d)_{d \in \bI}$ of subgroups of the additive group $(A, +)$ such that $A_a \cdot A_b \se A_{a + b}$ for $a,b \in \bI$ and such that $A = \bigoplus_{d \in \bI} A_d$ in the sense that every $r \in A$ has a unique decomposition $r = \sum_{d \in \bI} r_d$ with $r_d \in A_d$ and but finitely many $r_d \neq 0$. - - We call the $r_d$ the \vocab{homogeneous components} of $r$. - - An ideal $I \se A$ is called \vocab{homogeneous} if $r \in I \implies \A d \in \bI ~ r_d \in I_d$ where $I_d \coloneqq I \cap A_d$. - - By a \vocab{graded ring} we understand an $\N$-graded ring. Tin this case, $A_{+} \coloneqq \bigoplus_{d=1}^{\infty} A_d = \{r \in A | r_0 = 0\} $ is called the \vocab{augmentation ideal} of $A$. -\end{definition} -\begin{remark}[Decomposition of $1$] - If $1 = \sum_{d \in \bI} \eps_d$ is the decomposition into homogeneous components, then $\eps_a = 1 \cdot \eps_a = \sum_{b \in \bI} \eps_a\eps_b$ with $\eps_a\eps_b \in A_{a+b}$. - By the uniqueness of the decomposition into homogeneous components, $\eps_a \eps_0 = \eps_a$ and $b \neq 0 \implies \eps_a \eps_b = 0$. - Applying the last equation with $a = 0$ gives $b\neq 0 \implies \eps_b = \eps_0 \eps _b = 0$. - Thus $1 = \eps_0 \in A_0$. -\end{remark} -\begin{remark} - The augmentation ideal of a graded ring is a homogeneous ideal. -\end{remark} - -% Graded rings and homogeneous ideals (2) - -\begin{proposition}\footnote{This holds for both $\Z$-graded and $\N$-graded rings.} - \begin{itemize} - \item A principal ideal generated by a homogeneous element is homogeneous. - \item The operations $\sum, \bigcap, \sqrt{}$ preserve homogeneity. - \item An ideal is homogeneous iff it can be generated by a family of homogeneous elements. - \end{itemize} -\end{proposition} -\begin{proof} - Most assertions are trivial. We only show that $J$ homogeneous $\implies \sqrt{J} $ homogeneous. - Let $A$ be $\bI$-graded, $f \in \sqrt{J} $ and $f = \sum_{d \in \bI} f_d$ the decomposition. - To show that all $f_d \in \sqrt{J} $, we use induction on $N_f \coloneqq \# \{d \in \bI | f_d \neq 0\}$. - $N_f = 0$ is trivial. Suppose $N_f > 0$ and $e \in \bI$ is maximal with $f_e \neq 0$. - For $l \in \N$, the $le$-th homogeneous component of $f^l$ is $f_e^l$. Choosing $l$ large enough such that $f^l \in J$ and using the homogeneity of $J$, we find $f_e \in \sqrt{J}$. - As $\sqrt{J} $ is an ideal, $\tilde f \coloneqq f - f_e \in \sqrt{J} $. As $N_{\tilde f} = N_f -1$, the induction assumption may be applied to $\tilde f$ and shows $f_d \in \sqrt{J} $ for $d \neq e$. -\end{proof} -\begin{fact} - A homogeneous ideal is finitely generated iff it can be generated by finitely many of its homogeneous elements. - In particular, this is always the case when $A$ is a Noetherian ring. -\end{fact} - - -\subsubsection{The Zariski topology on $\bP^n$} -\begin{notation} - Recall that for $\alpha \in \N^{n+1}$ $|\alpha| = \sum_{i=0}^{n} \alpha_i$ and $x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}$. -\end{notation} -\begin{definition}[Homogeneous polynomials] - Let $R$ be any ring and $f = \sum_{\alpha \in \N^{n+1}} f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n]$. - We say that $f$ is \vocab{homogeneous of degree $d$} if $|\alpha| \neq d \implies f_\alpha = 0$ . - We denote the subset of homogeneous polynomials of degree $d$ by $R[X_0,\ldots,X_n]_d \se R[X_0,\ldots,X_n]$. -\end{definition} -\begin{remark} - This definition gives $R$ the structure of a graded ring. -\end{remark} -\begin{definition}[Zariski topology on $\bP^n(\fk)$]\label{ztoppn} - Let $A = \fk[X_0,\ldots,X_n]$.\footnote{As always, $\fk$ is algebraically closed} - For $f \in A_d = \fk[X_0,\ldots,X_n]_d$, the validity of the equation $f(x_0,\ldots,x_{n}) = 0$ does not depend on the choice of homogeneous coordinates, as - \[ - f(\lambda x_0,\ldots, \lambda x_n) 0 \lambda^d f(x_0,\ldots,x_n) - \] - Let $\Vp(f) \coloneqq \{x \in \bP^n | f(x) = 0\}$. - - We call a subset $X \se \bP^n$ Zariski-closed if it can be represented as - \[ - X = \bigcap_{i=1}^k \Vp(f_i) - \] - where the $f_i \in A_{d_i}$ are homogeneous polynomials. -\end{definition} -\pagebreak -\begin{fact} - If $X = \bigcap_{i = 1}^k \Vp(f_i) \se \bP^n$ is closed, then $Y = X \cap \bA^n$ can be identified with the closed subset - \[ - \{(x_1,\ldots,x_n) \in \fk^n | f_i(1,x_1,\ldots,x_n) = 0, 1 \le i \le k\} \se \fk^n - \] - Conversely, if $Y \se \fk^n$ is closed it has the form - \[ - \{(x_1,\ldots,x_n) \in \fk^n | g_i(x_1,\ldots,x_n) = 0, 1 \le i \le k\} - \] - and can thus be identified with $X \cap \bA^n$ where $X \coloneqq \bigcap_{i=1}^k \Vp(f_i)$ is given by \[f_i(X_0,\ldots,X_n) \coloneqq X_0^{d_i} g_i(X_1 / X_0,\ldots, X_n / X_0), d_i \ge \deg(g_i)\] - Thus, the Zariski topology on $\fk^n$ can be identified with the topology induced by the Zariski topology on $\bA^n = U_0$, and the same holds for $U_i$ with $0 \le i \le n$. - - In this sense, the Zariski topology on $\bP^n$ can be thought of as gluing the Zariski topologies on the $U_i \cong \fk^n$. -\end{fact} - -% The Zariski topology on P^n (2) - -\begin{definition} - Let $I \se A = \fk[X_0,\ldots,X_n]$ be a homogeneous ideal. - Let $\Vp(I) \coloneqq \{[x_0,\ldots,_n] \in \bP^n | \A f \in I ~ f(x_0,\ldots,x_n) = 0\}$ - As $I$ is homogeneous, it is sufficient to impose this condition for the homogeneous elements $f \in I$. - Because $A$ is Noetherian, $I$ can finitely generated by homogeneous elements $(f_i)_{i=1}^k$ and $\Vp(I)=\bigcap_{i=1}^k \Vp(f_i)$ as in \ref{ztoppn}. - Conversely, if the homogeneous $f_i$ are given, then $I = \langle f_1,\ldots,f_k \rangle_A$ is homogeneous. -\end{definition} -\begin{remark} - Note that $V(A) = V(A_+) = \emptyset$. -\end{remark} -\begin{fact} - For homogeneous ideals in $A$ and $m \in \N$, we have: - \begin{itemize} - \item $\Vp(\sum_{\lambda \in \Lambda} I_\lambda) = \bigcap_{\lambda \in \Lambda} \Vp(I_\lambda)$ - \item $\Vp(\bigcap_{k=1}^m I_k) = \Vp(\prod_{k=1}^{m} I_k) = \bigcup_{k=1}^m \Vp(I_k)$ - \item $\Vp(\sqrt{I}) = \Vp(I)$ - \end{itemize} -\end{fact} -\begin{fact} - If $X = \bigcup_{\lambda \in \Lambda} U_\lambda$ is an open covering of a topological space then $X$ is Noetherian iff there is a finite subcovering and all $U_\lambda$ are Noetherian. -\end{fact} -\begin{proof} - By definition, a topological space is Noetherian $\iff$ all open subsets are quasi-compact. -\end{proof} -\begin{corollary} - The Zariski topology on $\bP^n$ is indeed a topology. - The induced topology on the open set $\bA^n = \bP^n \sm \Vp(X_0) \cong \fk^n$ is the Zariski topology on $\fk^n$. - The same holds for all $U_i = \bP^n \sm \Vp(X_i) \cong \fk^n$. - Moreover, the topological space $\bP^n$ is Noetherian. -\end{corollary} - -\subsection{Noetherianness of graded rings} -\begin{proposition} - For a graded ring $R_{\bullet}$, the following conditions are equivalent: - \begin{enumerate}[A] - \item $R$ is Noetherian. - \item Every homogeneous ideal of $R_{\bullet}$ is finitely generated. - \item Every chain $I_0\se I_1 \se \ldots$ of homogeneous ideals terminates. - \item Every set $\fM \neq \emptyset$ of homogeneous ideals has a $\se$-maximal element. - \item $R_0$ is Noetherian and the ideal $R_+$ is finitely generated. - \item $R_0$ is Noetherian and $R / R_0$ is of finite type. - \end{enumerate} -\end{proposition} -\begin{proof} - \noindent\textbf{A $\implies$ B,C,D} trivial. - - \noindent\textbf{B $\iff$ C $\iff$ D} similar to the proof about Noetherianness. - - \noindent\textbf{B $\land$ C $\implies $E} B implies that $R_+$ is finitely generated. Since $I \oplus R_+$ is homogeneous for any homogeneous ideal $I \se R_0$, C implies the Noetherianness of $R_0$. - -\noindent\textbf{E $\implies$ F} Let $R_+$ be generated by $f_i \in R_{d_i}, d_i > 0$ as an ideal. - \begin{claim} - The $R_0$-subalgebra $\tilde R$ of $R$ generated by the $f_i$ equals $R$. - \end{claim} - \begin{subproof} - It is sufficient to show that every homogeneous $f \in R_d$ belongs to $\tilde R$. We use induction on $d$. The case of $d = 0$ is trivial. - Let $d > 0$ and $R_e \se \tilde R$ for all $e < d$. - as $f \in R_+$, $f = \sum_{i=1}^{k} g_if_i$. Let $f_a = \sum_{i=1}^{k} g_{i, a-d_i} f_i$, where $g_i = \sum_{b=0}^{\infty} g_{i,b}$ is the decomposition into homogeneous components. - Then $f = \sum_{a=0}^{\infty} f_a$ is the decomposition of $f$ into homogeneous components, hence $a \neq d \implies f_a = 0 $. Thus we may assume $g_i \in R_{d-d_i}$. - As $d_i > 0$, the induction assumption may now be applied to $g_i$, hence $g_i \in \tilde R$, hence $f \in \tilde R$. - \end{subproof} - - \noindent\textbf{F $\implies$ A} Hilbert's Basissatz (\ref{basissatz}) - -\end{proof} - - - -% Lecture 12 - -\subsection{The projective form of the Nullstellensatz and the closed subsets of $\bP^n$} -Let $A = \fk[X_0,\ldots,X_n]$. -\begin{proposition}[Projective form of the Nullstellensatz]\label{hnsp} - If $I \se A$ is a homogeneous ideal and $f \in A_d$ with $d>0$, then $\Vp(I) \se \Vp(f) \iff f \in \sqrt{I}$. -\end{proposition} -\begin{proof} - $\impliedby$ is clear. Let $\Vp(I) \se \Vp(f)$. If $x = (x_0,\ldots,x_n) \in \Va(I)$, then either $x = 0$ in which case $f(x) = 0$ since $d > 0$ - or the point $[x_0,\ldots,x_n] \in \bP^n$ is well-defined and belongs to $\Vp(I) \se \Vp(f)$, hence $f(x) = 0$. - Thus $\Va(I) \se \Va(f)$ and $f \in \sqrt{I}$ be the Nullstellensatz (\ref{hns3}). -\end{proof} - -\begin{definition}\footnote{This definition is not too important, the characterization in the following remark suffices.}. - For a graded ring $R_\bullet$, let $\Proj(R_\bullet)$ be the set of $\fp \in \Spec R$ such that $\fp$ is a homogeneous ideal and $\fp \not\supseteq R_+$. -\end{definition} -\begin{remark}\label{proja} - As the elements of $A_0 \sm \{0\}$ are units in $A$ it follows that for every homogeneous ideal $I$ we have $I \se A_+$ or $I = A$. - In particular, $\Proj(A_\bullet) = \{\fp \in \Spec A \sm A_+ | \fp \text{ is homogeneous}\} $. -\end{remark} -\begin{proposition}\label{bijproj} - There is a bijection - \begin{align} - f: \{I \se A_+ | I \text{ homogeneous ideal}, I = \sqrt{I}\} &\longrightarrow \{X \se \bP^n | X \text{ closed}\} \\ - I &\longmapsto \Vp(I)\\ - \langle \{f \in A_d | d > 0, X \se \Vp(f)\} \rangle & \longmapsfrom X - \end{align} - Under this bijection, the irreducible subsets correspond to the elements of $\Proj(A_\bullet)$. -\end{proposition} -\begin{proof} - From the projective form of the Nullstellensatz it follows that $f$ is injective and that $f\inv(\Vp\left( I \right)) = \sqrt{I} = I$. - If $X \se \bP^n$ is closed, then $X = \Vp(J)$ for some homogeneous ideal $J \se A$. \Wlog $J = \sqrt{J}$. If $J \not\se A_+$, then $J = A$ (\ref{proja}), hence $X = \Vp(J) = \emptyset = \Vp(A_+)$. - Thus we may assume $J \se A_+$, and $f$ is surjective. - - - Suppose $\fp \in \Proj(A_\bullet)$. Then $\fp \neq A_+$ hence $X = \Vp(\fp) \neq \emptyset$ by the proven part of the proposition. - Assume $X = X_1 \cup X_2$ is a decomposition into proper closed subsets, where $X_k = \Vp(I_k)$ for some $I_k \se A_+, I_k = \sqrt{I_k}$. Since $X_k$ is a proper subset of $X$, there is $f_k \in I_k \sm \fp$. - We have $\Vp(f_1f_2) \supseteq \Vp(f_k) \supseteq \Vp(I_k)$ hence $\Vp(f_1f_2) \supseteq \Vp(I_1) \cup \Vp(I_2) = X = \Vp(\fp)$ and it follows that $f_1f_2\in \sqrt{\fp} = \fp \lightning$. - - Assume $X = \Vp(\fp)$ is irreducible, where $\fp = \sqrt{\fp} \in A_+$ is homogeneous. The $\fp \neq A_+$ as $X = \emptyset$ otherwise. Assume that $f_1f_2 \in \fp$ but $f_i \not\in A_{d_i} \sm \fp$. - Then $X \not \se \Vp(f_i)$ by the projective Nullstellensatz when $d_i > 0$ and because $\Vp(1) = \emptyset$ when $d_i = 0$. - Thus $X = (X \cap \Vp\left( f_1 \right)) \cup (X \cap \Vp(f_2))$ is a proper decomposition $\lightning$. - By lemma \ref{homprime}, $\fp$ is a prime ideal. - -\end{proof} -\begin{remark} - It is important that $I \se A_{\color{red} +}$, since $\Vp(A) = \Vp(A_+) = \emptyset$ would be a counterexample. -\end{remark} -\begin{corollary} - $\bP^n$ is irreducible. -\end{corollary} -\begin{proof} - Apply \ref{bijproj} to $\{0\} \in \Proj(A_\bullet)$. -\end{proof} - -\subsection{Some remarks on homogeneous prime ideals} -\begin{lemma}\label{homprime} - Let $R_\bullet$ be an $\bI$ graded ring ($\bI = \N$ or $\bI = \Z$). - A homogeneous ideal $I \se R$ is a prime ideal iff $1 \not\in I$ and for homogeneous elements $f, g \in R , fg \in I \implies f \in I \lor g \in I$. -\end{lemma} -\begin{proof} - $\implies$ is trivial. - It suffices to show that for arbitrary $f,g \in R fg \in I \implies f \in I \lor g \in I$. - Let $f = \sum_{d \in \bI} f_d, g = \sum_{d \in \bI} g_d $ be the decompositions into homogeneous components. - If $f \not\in I$ and $g \not\in I$ there are $d,e \in I$ with $f_d \in I, g_e \in I$, and they may assumed to be maximal with this property. - As $I$ is homogeneous and $fg \in I$, we have $(fg)_{d+e} \in I$ but - \[ - (fg)_{d+e} = f_dg_e + \sum_{\delta = 1}^{\infty} (f_{d + \delta} g_{e - \delta} + f_{d - \delta} g_{e + \delta}) - \] - where $f_dg_e \not\in I$ by our assumption on $I$ and all other summands on the right hand side are $\in I$ (as $f_{d+ \delta} \in I$ and $g_{e + \delta} \in I$ by the maximality of $d$ and $e$), a contradiction. -\end{proof} - -\begin{remark} - If $R_\bullet$ is $\N$-graded and $\fp \in \Spec R_0$, then $\fp \oplus R_+ = \{r \in R | r_0 \in \fp\} $ is a homogeneous prime ideal of $R$. - \[\{\fp \in \Spec R | \fp \text{ is a homogeneous ideal of } R_\bullet\} = \Proj(R_\bullet) \sqcup \{\fp \oplus R_+ | \fp \in \Spec R_0\}\] -\end{remark} - -\subsection{Dimension of $\bP^n$} -\begin{proposition} - \begin{itemize} - \item $\bP^n$ is catenary. - \item $\dim(\bP^n) = n$. Moreover, $\codim(\{x\} ,\bP^n) = n$ for every $x \in \bP^n$. - \item If $X \se \bP^n$ is irreducible and $x \in X$, then $\codim(\{x\}, X) = \dim(X) = n - \codim(X, \bP^n)$. - \item If $X \se Y \se \bP^n$ are irreducible subsets, then $\codim(X,Y) = \dim(Y) - \dim(X)$. - \end{itemize} -\end{proposition} -\begin{proof} - Let $X \se \bP^n$ be irreducible. If $x \in X$, there is an integer $0 \le i \le n$ and $X \in U_i = \bP^n \sm \Vp(X_i)$. - \Wlog $i = 0$. Then $\codim(X, \bP^n) = \codim(X \cap \bA^n, \bA^n)$ by the locality of Krull codimension (\ref{lockrullcodim}). - Applying this with $X = \{x\}$ and our results about the affine case gives the second assertion. - If $Y$ and $Z$ are also irreducible with $X \se Y \se Z$, then $\codim(X,Y) = \codim(X \cap \bA^n, Y \cap \bA^n)$, $\codim(X,Z) = \codim(X \cap \bA^n, Z \cap \bA^n)$ and $\codim(Y,Z) = \codim(Y \cap \bA^n, Z \cap \bA^n)$. - Thus - \begin{align} - \codim(X,Y) + \codim(Y,Z) &= \codim(X \cap \bA^n, Y \cap \bA^n) + \codim(Y \cap \bA^n, Z \cap \bA^n)\\ - &= \codim(X \cap \bA^n, Z \cap \bA^n)\\ - &= \codim(X, Z) - \end{align} - because $\fk^n$ is catenary and the first point follows. - The remaining assertions can easily be derived from the first two. -\end{proof} - -\subsection{The cone $C(X)$} -\begin{definition} - If $X \se \bP^n$ is closed, we define the \vocab{affine cone over $X$} - \[ - C(X) = \{0\} \cup \{(x_0,\ldots,x_n) \in \fk^{n+1} \sm \{0\} | [x_0,\ldots,x_n] \in X\} - \] - If $X = \Vp(I)$ where $I \se A_+ = \fk[X_0,\ldots,X_n]_+$ is homogeneous, then $C(X) = \Va(I)$. -\end{definition} -\begin{proposition}\label{conedim} - \begin{itemize} - \item $C(X)$ is irreducible iff $X$ is irreducible or $X = \emptyset$. - \item If $X$ is irreducible, then - - $\dim(C(X)) = \dim(X) + 1$ and - - $\codim(C(X), \fk^{n+1}) = \codim(X, \bP^n)$ - \end{itemize} -\end{proposition} -\begin{proof} - The first assertion follows from \ref{bijproj} and \ref{bijiredprim} (bijection of irreducible subsets and prime ideals in the projective and affine case). - - Let $d = \dim(X)$ and - \[ - X_0 \subsetneq \ldots \subsetneq X_d = X \subsetneq X_{d+1} \subsetneq \ldots \subsetneq X_n = \bP^n - \] - be a chain of irreducible subsets of $\bP^n$. Then - \[ - \{0\} \subsetneq C(X_0) \subsetneq \ldots \subsetneq C(X_d) = C(X) \subsetneq \ldots \subsetneq C(X_n) = \fk^{n+1} - \] - is a chain of irreducible subsets of $\fk^{n+1}$. Hence $\dim(C(X)) \ge 1 + d$ and $\codim(C(X), \fk^{n+1}) \ge n-d$. Since $\dim(C(X)) + \codim(C(X), \fk^{n+1}) = \dim(\fk^{n+1}) = n+1$, the two inequalities must be equalities. -\end{proof} -\subsubsection{Application to hypersurfaces in $\bP^n$} -\begin{definition}[Hypersurface] - Let $n > 0$. - By a \vocab{hypersurface} in $\bP^n$ or $\bA^n$ we understand an irreducible closed subset of codimension $1$. -\end{definition} - -\begin{corollary} - If $P \in A_d$ is a prime element, then $H = \Vp(P)$ is a hypersurface in $\bP^n$ and every hypersurface $H$ in $\bP^n$ can be obtained in this way. -\end{corollary} -\begin{proof} - If $H = \Vp(P)$ then $C(H) = \Va(P)$ is a hypersurface in $\fk^{n+1}$ by \ref{irredcodimone}. By \ref{conedim}, $H$ is irreducible and of codimension $1$. - - Conversely, let $H$ be a hypersurface in $\bP^n$. By \ref{conedim}, $C(H)$ is a hypersurface in $\fk^{n+1}$, hence $C(H) = \Vp(P)$ for some prime element $P \in A$ (again by \ref{irredcodimone}). - We have $H = \Vp(\fp)$ for some $\fp \in \Proj(A)$ and $C(H) = \Va(\fp)$. By the bijection between closed subsets of $\fk^{n+1}$ and ideals $I = \sqrt{I} \se A$ (\ref{antimonbij}), $\fp = P \cdot A$. -Let $P = \sum_{k=0}^{d}P_k$ with $P_d \neq 0$ be the decomposition into homogeneous components. -If $P_e $ with $e < d$ was $\neq 0$, it could not be a multiple of $P$ contradicting the homogeneity of $\fp = P \cdot A$. Thus, $P$ is homogeneous of degree $d$. -\end{proof} -\begin{definition} - A hypersurface $H \se \bP^n$ has \vocab{degree $d$} if $H = \Vp(P)$ where $P \in A_d$ is an irreducible polynomial. -\end{definition} - -\subsubsection{Application to intersections in $\bP^n$ and Bezout's theorem} -\begin{corollary} - Let $A \se \bP^n$ and $B \se \bP^n$ be irreducible subsets of dimensions $a$ and $b$. If $a+ b \ge n$, then $A \cap B \neq \emptyset$ and every irreducible component of $A \cap B$ as dimension $\ge a + b - n$. -\end{corollary} - -\begin{remark} - This shows that $\bP^n$ indeed fulfilled the goal of allowing for nicer results of algebraic geometry because ``solutions at infinity'' to systems of algebraic equations are present in $\bP^n$ - (see \ref{affineproblem}). -\end{remark} - -\begin{proof} - The lower bound on the dimension of irreducible components of $A \cap B$ is easily derived from the similar affine result (corollary of the principal ideal theorem, \ref{codimintersection}). - From the definition of the affine cone it follows that $C(A \cap B) = C(A) \cap C(B)$. - We have $\dim(C(A)) = a+1$ and $\dim(C(B)) = b + 1$ by \ref{conedim}. - If $A \cap B = \emptyset$, then $C(A) \cap C(B) = \{0\}$ with $\{0\} $ as an irreducible component, contradicting the lower bound $a + b + 1 - n > 0$ for the dimension of irreducible components of $C(A) \cap C(B)$ (again \ref{codimintersection}). -\end{proof} -\begin{remark}[Bezout's theorem] - If $A \neq B$ are hypersurfaces of degree $a$ and $b$ in $\bP^2$, then $A \cap B$ has $ab$ points counted by (suitably defined) multiplicity. -\end{remark} - - -%TODO Proof of "Dimension of P^n" -% SLIDE APPLICATION TO HYPERSURFACES IN $\P^n$ -%ERROR: C(H) = V_A(P) -%If n = 0, P = 0, V_P(P) = \emptyset is a problem! - - - -% Lecture 13 -\section{Varieties} - -\subsection{Sheaves} - -\begin{definition}[Sheaf] - Let $X$ be any topological space. - - A \vocab{presheaf} $\cG$ of sets (or rings, (abelian) groups) on $X$ associates a set (or rings, or (abelian) group) $\cG(U)$ to every open subset $U$ of $X$, and a map (or ring or group homomorphism) $\cG(U) \xrightarrow{r_{U,V}} \cG(V)$ to every inclusion $V \se U$ of open subsets of $X$ such that $r_{U,W} = r_{V,W} r_{U,V}$ for inclusions $U \se V \se W$ of open subsets. - - Elements of $\cG(U)$ are often called \vocab{sections} of $\cG$ on $U$ or \vocab{global sections} when $U = X$. - - Let $U \se X$ be open and $U = \bigcup_{i \in I} U_i$ an open covering. - A family $(f_i)_{i \in I} \in \prod_{i \in I} \cG(U_i)$ is called \vocab[Sections!compatible]{compatible} if $r_{U_i, U_i \cap U_j}(f_i) = r_{U_j, U_i \cap U_j}(f_j)$ for all $i,j \in I$. - - Consider the map - \begin{align} - \phi_{U, (U_i)_{i \in I}}: \cG(U) &\longrightarrow \{(f_i)_{i \in I} \in \prod_{i \in I} \cG(U_i) | r_{U_i, U_i \cap U_j}(f_i) = r_{U_j, U_i \cap U_j}(f_j) \text{ for } i,j \in I \} \\ - f &\longmapsto (r_{U, U_i}( f))_{i \in I} - \end{align} - - A presheaf is called \vocab[Presheaf!separated]{separated} if $\phi_{U, (U_i)_{i \in I}}$ is injective for all such $U$ and $(U_i)_{i \in I}$.\footnote{This also called ``locality''.} - It satisfies \vocab{gluing} if $\phi_{U, (U_i)_{i \in I}}$ is surjective. - - A presheaf is called a \vocab{sheaf} if it is separated and satisfies gluing. - - The bijectivity of the $\phi_{U, (U_i)_{i \in I}}$ is called the \vocab{sheaf axiom}. -\end{definition} -\begin{dtrivial} - A presheaf is a contravariant functor $\cG : \cO(X) \to C$ where $\cO(X)$ denotes the category of open subsets of $X$ with inclusions as morphisms and $C$ is the category of sets, rings or (abelian) groups. -\end{dtrivial} -\begin{definition} - A subsheaf $\cG'$ is defined by subsets (resp. subrings or subgroups) $\cG'(U) \se \cG(U)$ for all open $U \se X$ such that the sheaf axioms still hold. -\end{definition} -\begin{remark} - If $\cG$ is a sheaf on $X$ and $\Omega \se X$ open, then $\cG\defon{\Omega}(U) \coloneqq \cG(U)$ for open $U \se \Omega$ and $r_{U,V}^{(\cG\defon{\Omega})}(f) \coloneqq r_{U,V}^{(\cG)}(f)$ is a sheaf of the same kind as $\cG$ on $\Omega$. -\end{remark} -\begin{remark} - The notion of restriction of a sheaf to a closed subset, or of preimages under general continuous maps, can be defined but this is a bit harder. -\end{remark} -\begin{notation} - It is often convenient to write $f \defon{V}$ instead of $r_{U,V}(f)$. -\end{notation} -\begin{remark} - Applying the \vocab{sheaf axiom} to the empty covering of $U = \emptyset$, one finds that $\cG(\emptyset) = \{0\} $. -\end{remark} - - - - -\subsubsection{Examples of sheaves} -\begin{example} - Let $G$ be a set and let $\fG(U)$ be the set of arbitrary maps $U \xrightarrow{f} G$. We put $r_{U,V}(f) = f\defon{V}$. - It is easy to see that this defines a sheaf. - If $\cdot $ is a group operation on $G$, then $(f\cdot g)(x) \coloneqq f(x)\cdot g(x)$ defines the structure of a sheaf of group on $\fG$. - Similarly, a ring structure on $G$ can be used to define the structure of a sheaf of rings on $\fG$. -\end{example} -\begin{example} - If in the previous example $G$ carries a topology and $\cG(U) \se \fG(U)$ is the subset (subring, subgroup) of continuous functions $U \xrightarrow{f} G$, then $\cG$ is a subsheaf of $\fG$, called the sheaf of continuous $G$-valued functions on (open subsets of) $X$. -\end{example} - -\begin{example} - If $X = \R^n$, $\bK \in \{\R, \C\}$ and $\cO(U)$ is the sheaf of $\bK$-valued $C^{\infty}$-functions on $U$, then $\cO$ is a subsheaf of the sheaf (of rings) of $\bK$-valued continuous functions on $X$. -\end{example} -\begin{example} - If $X = \C^n$ and $\cO(U)$ the set of holomorphic functions on $X$, then $\cO$ is a subsheaf of the sheaf of $\C$-valued $C^{\infty}$-functions on $X$. -\end{example} -\subsubsection{The structure sheaf on a closed subset of $\fk^n$} - -Let $X \se \fk^n$ be open. Let $R = \fk[X_1,\ldots,X_n]$. -\begin{definition}\label{structuresheafkn} - For open subsets $U \se X$, let $\cO_X(U)$ be the set of functions $U \xrightarrow{\phi} \fk$ such that every $x \in U$ has a neighbourhood $V$ such that there are $f,g \in R$ such that for $y \in V$ we have $g(y) \neq 0$ and $\phi(y) = \frac{f(y)}{g(y)}$. -\end{definition} - -\begin{remark}\label{structuresheafcontinuous} - $\cO_X$ is a subsheaf (of rings) of the sheaf of $\fk$-valued functions on $X$. - The elements of $\cO_X(U)$ are continuous: - Let $M \se \fk$ be closed. We must show the closedness of $N \coloneqq \phi\inv(M)$ in $U$. For $M = \fk$ this is trivial. Otherwise $M$ is finite and we may assume $M = \{t\} $ for some $t \in \fk$. For $x \in U$, there are open $V_x \se U$ and $f_x, g_x \in R$ such that $\phi = \frac{f_x}{g_x}$ on $V_x$. - Then $N \cap V_x = V(f_x - t\cdot g_x) \cap V_x)$ is closed in $V_x$. As the $V_x$ cover $U$ and $U$ is quasi-compact, $N$ is closed in $U$. -\end{remark} - -\begin{proposition}\label{structuresheafri} - Let $X = V(I)$ where $I = \sqrt{I} \se R$ is an ideal. Let $A = R / I$. Then - \begin{align} - \phi: A &\longrightarrow \cO_X(X) \\ - f \mod I &\longmapsto f\defon{X} - \end{align} -is an isomorphism. -\end{proposition} - -\begin{proof} - It is easy to see that the map $A \to \cO_X(X)$ is well-defined and a ring homomorphism. - Its injectivity follows from the Nullstellensatz and $I = \sqrt{I}$ (\ref{hns3}). - - - Let $\phi \in \cO_X(X)$. for $x \in X$, there are an open subset $U_x \se X$ and $f_x, g_x \in R$ such that $\phi = \frac{f_x}{g_x}$ on $U_x$. - \begin{claim} - \Wlog we can assume $U_x = X \sm V(g_x)$. - \end{claim} - \begin{subproof} - The closed subsets $(X \sm U_x) \se \fk^n$ has the form $X\sm U_x = V(J_x)$ for some ideal $J_x \se R$. - As $x \not\in X \sm V_x$ there is $h_x \in J_x$ with $h_x(x) \neq 0$. - Replacing $U_x$ by $X \sm V(h_x)$, $f_x$ by $f_xh_x$ and $g_x$ by $g_xh_x$, we may assume $U_x = X \sm V(g_x)$. - \end{subproof} - \begin{claim} - \Wlog we can assume $V(g_x) \se V(f_x)$. - \end{claim} - \begin{subproof} - Replace $f_x$ by $f_xg_x$ and $g_x$ by $g_x^2$. - \end{subproof} - As $X$ is quasi-compact, there are finitely many points $(x_i)_{i=1}^m$ such that the $U_{x_i}$ cover $X$. - Let $U_i \coloneqq U_{x_i}, f_i \coloneqq f_{x_i}, g_i \coloneqq g_{x_i}$. - - As the $U_i = X \sm V(g_i)$ cover $X$, $V(I) \cap \bigcap_{i=1}^m V(g_i) = X \cap \bigcap_{i=1}^m V(g_i) = \emptyset$. - By the Nullstellensatz (\ref{hns1}) the ideal of $R$ generated by $I$ and the $a_i$ equals $R$. - There are thus $n \ge m \in \N$ and elements $(g_i)_{i = m+1}^n$ of $I$ and $(a_i)_{i=1}^n \in R^n$ such that $1 = \sum_{i=1}^{n} a_ig_i$. - Let for $i > m$ $f_i \coloneqq 0$, $F = \sum_{i=1}^{n} a_if_i = \sum_{i=1}^{m} a_if_i \in R$. - - \begin{claim} - For all $x \in X $ ~ $f_i(x) = \phi(x) g_i(x)$. - \end{claim} - \begin{subproof} - If $x \in V_i$ this follows by our choice of $f_i$ and $g_i$. - If $x \in X \sm V_i$ or $i > m$ both sides are zero. - \end{subproof} - It follows that - \[ - \phi(x) = \phi(x) \cdot 1 = \phi(x) \cdot \sum_{i=1}^{n} a_i(x) g_i(x) = \sum_{i=1}^{n} a_i(x) f_i(x) = F(x) - \] - Hence $\phi = F\defon{X}$. -\end{proof} -\subsubsection{The structure sheaf on closed subsets of $\bP^n$} -Let $X \se \bP^n$ be closed and $R_\bullet = \fk[X_0,\ldots,X_n]$ with its usual grading. - -\begin{definition}\label{structuresheafpn} - For open $U \se X$, let $\cO_X(U)$ be the set of functions $U \xrightarrow{\phi} \fk$ such that for every $x \in U$, there are an open subset $W \se U$, a natural number $d$ and $f,g \in R_d$ such that $W \cap \Vp(g) = \emptyset$ and $\phi(y) = \frac{f(y_0,\ldots,y_n)}{g(y_0,\ldots,y_n)}$ for $y = [y_0,\ldots,y_n] \in W$. -\end{definition} - -\begin{remark} - This is a subsheaf of rings of the sheaf of $\fk$-valued functions on $X$. -Under the identification $\bA^n =\fk^n$ with $\bP^n \sm \Vp(X_0)$, one has $\cO_X \defon{X \sm \Vp(X_0)} = \cO_{X \cap \bA^n}$ as subsheaves of the sheaf of $\fk$-valued functions, where the second sheaf is a sheaf on a closed subset of $\fk^n$: - -Indeed, if $W$ is as in the definition then $\phi([1,y_1,\ldots,y_n]) = \frac{f(1,y_1,\ldots,y_n)}{g(1,y_1,\ldots,y_n)}$ for $[1,y_1,\ldots,y_n] \in W$. -Conversely if $\phi([1,y_1,\ldots,y_n]) = \frac{f(y_1,\ldots,y_n)}{g(y_1,\ldots,y_n)}$ on an open subset $W $ of $X \cap \bA^n$ then -$\phi([y_0,\ldots,y_n]) = \frac{F(y_0,\ldots,y_n)}{G(y_0,\ldots,y_n)}$ on $W$ where $F(X_0,\ldots,X_n) \coloneqq X_0^d f(\frac{X_1}{X_0}, \ldots, \frac{X_n}{X_0})$ and $G(X_0,\ldots,X_n) = X_0^d g(\frac{X_1}{X_0},\ldots, \frac{X_n}{X_0})$ with a sufficiently large $d \in \N$. -\end{remark} -\begin{remark} - It follows from the previous remark and the similar result in the affine case that the elements of $\cO_X(U)$ are continuous on $U \sm V(X_0)$. - Since the situation is symmetric in the homogeneous coordinates, they are continuous on all of $U$. -\end{remark} -The following is somewhat harder than in the affine case: -\begin{proposition} - If $X$ is connected (e.g. irreducible), then the elements of $\cO_X\left( X \right) $ are constant functions on $X$. -\end{proposition} - - - -% Lecture 14 - -\subsection{The notion of a category} -\begin{definition} - A \vocab{category} $\cA$ consists of: - \begin{itemize} - \item A class $\Ob \cA$ of \vocab[Objects]{objects of $\cA$}. - \item For two arbitrary objects $A, B \in \Ob \cA$, a \textbf{set} $\Hom_\cA(A,B)$ of \vocab[Morphism]{morphisms for $A$ to $B$ in $\cA$}. - \item A map $\Hom_\cA(B,C) \times \Hom_\cA(A,B) \xrightarrow{\circ} \Hom_\cA(A,C)$, the composition of morphisms, for arbitrary triples $(A,B,C)$ of objects of $\cA$. - \end{itemize} - The following conditions must be satisfied: - \begin{enumerate}[A] - \item For morphisms $A \xrightarrow{f} B\xrightarrow{g} C \xrightarrow{h} D$, we have $h \circ (g \circ f) = (h \circ g) \circ f$. - \item For every $A \in \Ob(\cA)$, there is an $\Id_A \in \Hom_{\cA}(A,A)$ such that $\Id_A \circ f = f$ (reps. $g \circ \Id_A = g$) for arbitrary morphisms $B \xrightarrow{f} A$ (reps. $A \xrightarrow{g} C).$ - \end{enumerate} - - A morphism $X \xrightarrow{f} Y$ is called an \vocab[Isomorphism]{isomorphism (in $\cA $)} if there is a morphism $Y \xrightarrow{g} X$ (called the \vocab[Inverse morphism]{inverse $f\inv$ of $f$)} such that $g \circ f = \Id_X$ and $f \circ g = \Id_Y$. -\end{definition} -\begin{remark} - \begin{itemize} - \item The distinction between classes and sets is important here. - \item We will usually omit the composition sign $\circ$. - \item It is easy to see that $\Id_A$ is uniquely determined by the above condition $B$, and that the inverse $f\inv$ of an isomorphism $f$ is uniquely determined. - \end{itemize} -\end{remark} -\subsubsection{Examples of categories} -\begin{example} - \begin{itemize} - \item The category of sets. - \item The category of groups. - \item The category of rings. - \item If $R$ is a ring, the category of $R$-modules and the category $\Alg_R$ of $R$-algebras - \item The category of topological spaces - \item The category $\Var_\fk$ of varieties over $\fk$ (see \ref{defvariety}) - \item If $\cA$ is a category, then the \vocab{opposite category} or \vocab{dual category} is defined by $\Ob(\cA\op) = \Ob(\cA)$ and $\Hom_{\cA\op}(X,Y) = \Hom_\cA(Y,X)$. - \end{itemize} - In most of these cases, isomorphisms in the category were just called `isomorphism'. The isomorphisms in the category of topological spaces are the homeomophisms. -\end{example} -\subsubsection{Subcategories} -\begin{definition}[Subcategories] - A \vocab{subcategory} of $\cA$ is a category $\cB$ such that $\Ob(\cB) \se \Ob(\cA)$, such that $\Hom_\cB(X,Y) \se \Hom_\cA(X,Y)$ for objects $X$ and $Y$ of $\cB$, such that for every object $X \in \Ob(\cB)$, the identity $\Id_X$ of $X$ is the same in $\cB$ as in $\cA$, and such that for composable morphisms in $\cB$, their compositions in $\cA$ and $\cB$ coincide. - We call $\cB$ a \vocab{full subcategory} of $\cA$ if in addition $\Hom_\cB(X,Y) = \Hom_\cA(X,Y)$ for arbitrary $X,Y \in \Ob(\cB)$. -\end{definition} -\begin{example} - \begin{itemize} - \item The category of abelian groups is a full subcategory of the category of groups. - It can be identified with the category of $\Z$-modules. - \item The category of finitely generated $R$-modules as a full subcategory of the category of $R$-modules. - \item The category of $R$-algebras of finite type as a full subcategory of $\Alg_R$. - \item The category of affine varieties over $\fk$ as a full subcategory of the category of varieties over $\fk$. - \end{itemize} -\end{example} - -\subsubsection{Functors and equivalences of categories} -\begin{definition} - A \vocab[Functor!covariant]{(covariant) functor} (resp. \vocab[Functor!contravariant]{contravariant functor}) between categories $\cA \xrightarrow{F} \cB$ is a map $\Ob(\cA) \xrightarrow{F} \Ob(\cB)$ with a family of maps $\Hom_\cA(X,Y) \xrightarrow{F} \Hom_\cB(F(X),F(Y))$ (resp. $\Hom_\cA(X,Y) \xrightarrow{F} \Hom_\cB(F(Y),F(X))$ in the case of contravariant functors), where $X$ and $Y$ are arbitrary objects of $\cA$, such that the following conditions hold: - \begin{itemize} - \item $F(\Id_X) = \Id_{F(X)}$ - \item For morphisms $X \xrightarrow{f} Y \xrightarrow{g} Z$ in $\cA$, we have $F(gf) = F(g)F(f)$ ( resp. $F(gf) = F(f)F(g)$) - \end{itemize} - A functor is called \vocab[Functor!essentially surjective]{essentially surjective} if every object of $\cB$ is isomorphic to an element of the image of $\Ob(\cA) \xrightarrow{F} \Ob(\cB)$. - A functor is called \vocab[Functor!full]{full} (resp. \vocab[Functor!faithful]{faithful}) if it induces surjective (resp. injective) maps between sets of morphisms. - It is called an \vocab{equivalence of categories} if it is full, faithful and essentially surjective. -\end{definition} -\begin{example} - \begin{itemize} - \item There are \vocab[Functor!forgetful]{forgetful functors} from rings to abelian groups or from abelian groups to sets which drop the multiplicative structure of a ring or the group structure of a group. - \item If $\fk$ is any vector space there is a contravariant functor from $\fk$-vector spaces to itself sending $V$ to its dual vector space $V\se$ and $V \xrightarrow{f} W$ to the dual linear map $W\st \xrightarrow{f\st} V\st$. - When restricted to the full subcategory of finite-dimensional vector spaces it becomes a contravariant self-equivalence of that category. - \item The embedding of a subcategory is a faithful functor. In the case of a full subcategory it is also full. - \end{itemize} -\end{example} - - - -\subsection{The category of varieties} - -\begin{definition}[Algebraic variety]\label{defvariety} - An \vocab{algebraic variety} or \vocab{prevariety} over $\fk$ is a pair $(X, \cO_X)$, where $X$ is a topological space and $\cO_X$ a subsheaf of the sheaf of $\fk$-valued functions on $X$ such that for every $x \in X$, there are a neighbourhood $U_x$ of $x$ in $X$, an open subset $V_x$ of a closed subset $Y_x$ of $\fk^{n_x}$\footnote{By the result of \ref{affopensubtopbase} it can be assumed that $V_x = Y_x$ without altering the definition.} and a homeomorphism $V_x \xrightarrow{\iota_x} U_x$ such that for every open subset $V \se U_x$ and every function $V\xrightarrow{f} \fk$, we have $f \in \cO_X(V) \iff \iota\st_x(f) \in \cO_{Y_x}(\iota_x\inv(V))$, - - In this, the \vocab{pull-back} $\iota_x\st(f)$ of $f$ is defined by $(\iota_x\st(f))(\xi) \coloneqq f(\iota_x(\xi))$. - - - A morphism $(X, \cO_X) \to (Y, \cO_Y)$ of varieties is a continuous map $X \xrightarrow{\phi} Y$ such that for all open $U \se Y$ and $f \in \cO_Y(U)$, $\phi\st(f) \in \cO_X(\phi\inv(U))$. - An isomorphism is a morphism such that $\phi$ is bijective and $\phi\inv$ also is a morphism of varieties. -\end{definition} -\begin{example} - \begin{itemize} - \item If $(X, \cO_X)$ is a variety and $U \se X$ open, then $(U, \cO_X\defon{U})$ is a variety (called an \vocab{open subvariety} of $X$), and the embedding $U \to X$ is a morphism of varieties. - \item If $X$ is a closed subset of $\fk^n$ or $\bP^n$, then $(X, \cO_X)$ is a variety, where $\cO_X$ is the structure sheaf on $X$ (\ref{structuresheafkn}, reps. \ref{structuresheafpn}). - A variety is called \vocab[Variety!affine]{affine} (resp. \vocab[Variety!projective]{projective}) if it is isomorphic to a variety of this form, with $X $ closed in $\fk^n$ (resp. $\bP^n$). - A variety which is isomorphic to and open subvariety of $X$ is called \vocab[Variety!quasi-affine]{quasi-affine} (resp. \vocab[Variety!quasi-projective]{quasi-projective}). - \item If $X = V(X^2 - Y^3) \se \fk^2$ then $\fk \xrightarrow{t \mapsto (t^3,t^2)} X$ is a morphism which is a homeomorphism of topological spaces but not an isomorphism of varieties. - % TODO - - \item The composition of two morphisms $X \to Y \to Z$ of varieties is a morphism of varieties. - \item $X\xrightarrow{\Id_X} X$ is a morphism of varieties. - \end{itemize} -\end{example} - -\subsubsection{The category of affine varieties} -\begin{lemma}\label{localinverse} - Let $X$ be any $\fk$-variety and $U \se X$ open. - \begin{enumerate}[i)] - \item All elements of $\cO_X(U)$ are continuous. - \item If $U \se X$ is open, $U \xrightarrow{\lambda} \fk$ any function and every $x \in U$ has a neighbourhood $V_x \se U$ such that $\lambda \defon{V_x} \in \cO_X(V_x)$, then $\lambda \in \cO_X(U)$. - \item If $\vartheta \in \cO_X(U)$ and $\vartheta(x) \neq 0$ for all $x \in U$, then $\vartheta \in \cO_X(U)^{\times }$. - \end{enumerate} -\end{lemma} -\begin{proof} - \begin{enumerate}[i)] - \item The property is local on $U$, hence it is sufficient to show it in the quasi-affine case. This was done in \ref{structuresheafcontinuous}. - \item For the second part, let $\lambda_x \coloneqq \lambda \defon{V_x} $. - We have $\lambda_x\defon{V_x \cap V_y} = \lambda \defon{V_x \cap V_y} = \lambda_y \defon{V_x \cap V_y} $. - The $V_x$ cover $U$. By the sheaf axiom for $\cO_X$ there is $\ell \in \cO_X(U)$ with $\ell\defon{V_x} =\lambda_x$. It follows that $\ell=\lambda$. - \item By the definition of variety, every $x \in U$ has a quasi-affine neighbourhood $V \se U$. We can assume $U$ to be quasi-affine and $X = V(I) \se \fk^n$, as the general assertion follows by an application of ii). - If $x \in U$ there are a neighbourhood $x \in W \se U$ and $a,b \in R = \fk[X_1,\ldots,X_n]$ such that $\vartheta(y) = \frac{a(y)}{b(y)}$ for $y \in W$, with $b(y) \neq 0$. - Then $a(x) \neq 0$ as $\vartheta(x) \neq 0$. Replacing $W$ by $W \sm V(a)$, we may assume that $a$ has no zeroes on $W$. - Then $\lambda(y) = \frac{b(y)}{a(y)}$ for $y \in W$ has a non-vanishing denominator and $\lambda \in \cO_X(U)$. - We have $\lambda \cdot \vartheta = 1$, thus $\vartheta \in \cO_X(U)^{\times}$. - \end{enumerate} - - -\end{proof} -\begin{proposition}[About affine varieties] - \label{propaffvar} - \begin{itemize} - \item Let $X,Y$ be varieties over $\fk$. Then the map - \begin{align} - \phi: \Hom_{\Var_\fk}(X,Y) &\longrightarrow \Hom_{\Alg_\fk}(\cO_Y(Y), \cO_X(X)) \\ - (X \xrightarrow{f} Y) &\longmapsto (\cO_Y(Y) \xrightarrow{f\st} \cO_X(X)) - \end{align} - is injective when $Y$ is quasi-affine and bijective when $Y$ is affine. - \item The contravariant functor - \begin{align} - F: \Var_\fk &\longrightarrow \Alg_\fk \\ - X &\longmapsto \cO_X(X)\\ - (X\xrightarrow{f} Y) &\longmapsto (\cO_X(X) \xrightarrow{f\st} \cO_Y(Y)) - \end{align} - restricts to an equivalence of categories between the category of affine varieties over $\fk$ and the full subcategory $\cA$ of $\Alg_\fk$, - having the $\fk$-algebras $A$ of finite type with $\nil A = \{0\} $ as objects. - \end{itemize} -\end{proposition} - -\begin{remark} - It is clear that $\nil(\cO_X(X)) = \{0\}$ for arbitrary varieties. For general varieties it is however not true that $\cO_X(X)$ is a $\fk$-algebra of finite type. - There are counterexamples even for quasi-affine $X$. %TODO - - If, however, $X$ is affine, we may assume w.l.o.g. that $X = V(I)$ where $I = \sqrt{I} \se R$ is an ideal with $R = \fk[X_1,\ldots,X_n]$. - Then $\cO_X(X) \cong R / I$ (see \ref{structuresheafri}) is a $\fk$-algebra of finite type. -\end{remark} - -\begin{proof} - - - - - It suffices to investigate $\phi$ when $Y$ is an open subset of $V(I) \se \fk^n$, where $I = \sqrt{I} \se R$ is an ideal and $Y = V(I)$ when $Y$ is affine. - Let $(f_1,\ldots,f_n)$ be the components of $X \xrightarrow{f} Y \se \fk^n$. Let $Y \xrightarrow{\xi_i} \fk$ be the $i$-th coordinate. - By definition $f_i = f\st(\xi_i) $. Thus $f$ is uniquely determined by $\cO_Y(Y) \xrightarrow{f\st} \cO_X(X)$. - Conversely, let $Y = V(I)$ and $\cO_Y(Y) \xrightarrow{\phi} \cO_X(X)$ be a morphism of $\fk$-algebras. Define $f_i \coloneqq \phi(\xi_i)$ and consider $X \xrightarrow{f = (f_1,\ldots,f_n)} Y\se \fk^n$. - \begin{claim} - $f$ has image contained in $Y$. - \end{claim} - \begin{subproof} - For $x \in X, \lambda \in I$ we have $\lambda(f(x)) = (\phi(\lambda \mod I))(x) = 0$ as $\phi$ is a morphism of $\fk$-algebras. - Thus $f(x) \in V(I) = Y$. - \end{subproof} - \begin{claim} - $f$ is a morphism in $\Var_\fk$ - \end{claim} - \begin{subproof} - For open $\Omega \se Y, U = f\inv(\Omega) = \{x \in X | \A \lambda \in J ~ (\phi(\lambda))(x) \neq 0\}$ is open in $X$, where $Y \sm \Omega = V(J)$. - If $\lambda \in \cO_Y(\Omega)$ and $x \in U$, then $f(x)$ has a neighbourhood $V$ such that there are $a,b \in R$ with $\lambda(v) = \frac{a(v)}{b(v)}$ and $b(v) \neq 0$ for all $v \in V$. - Let $W \coloneqq f\inv(V)$. Then $\alpha \coloneqq \phi(a)\defon{W} \in \cO_X(W)$, $\beta \coloneqq \phi(b)\defon{W} \in \cO_X(W)$. - By the second part of \ref{localinverse} $\beta \in \cO_X(W)^{\times}$ and $f\st(\lambda)\defon{W} = \frac{\alpha}{\beta} \in \cO_X(W)$. - The first part of \ref{localinverse} shows that $f\st(\lambda) \in \cO_X(U)$. - \end{subproof} - By definition of $f$, we have $f\st = \phi$. This finished the proof of the first point. - - - - \begin{claim} - The functor in the second part maps affine varieties to objects of $\cA$ and is essentially surjective. - \end{claim} - \begin{subproof} - It follows from the remark that the functor maps affine varieties to objects of $\cA$. - - If $A \in \Ob(\cA)$ then $ A /\fk$ is of finite type, thus $A \cong R / I$ for some $n$. - Since $\nil(A) = \{0\}$ we have $I = \sqrt{I}$, as for $x \in \sqrt{I}$, $x \mod I \in \nil(R / I) \cong \nil(A) = \{0\}$. - Thus $A \cong\cO_X(X)$ where $X = V(I)$. - \end{subproof} - Fullness and faithfulness of the functor follow from the first point. -\end{proof} - -\begin{remark} - Note that giving a contravariant functor $\cC \to \cD$ is equivalent to giving a functor $\cC \to \cD\op$. We have thus shown that the category of affine varieties is equivalent to $\cA\op$, where $\cA \subsetneq \Alg_\fk$ is the full subcategory of $\fk$-algebras $A$ of finite type with $\nil(A) = \{0\}$. -\end{remark} -\subsubsection{Affine open subsets are a topology base} - -\begin{definition} - A set $\cB$ of open subsets of a topological space $X$ is called a \vocab{topology base} for $X$ if every open subset of $X$ can be written as a (possibly empty) union of elements of $\cB$. -\end{definition} -\begin{fact} -If $X$ is a set, then $\cB \se \cP(X)$ is a base for some topology on $X$ iff $X = \bigcup_{U \in \cB} U$ and for arbitrary $U, V \in \cB, U \cap V$ is a union of elements of $\cB$. -\end{fact} -\begin{definition} - Let $X$ be a variety. - An \vocab{affine open subset} of $X$ is a subset which is an affine variety. - -\end{definition} -\begin{proposition}\label{oxulocaf} - Let $X$ be an affine variety over $\fk$, $\lambda \in \cO_X(X)$ and $U = X \sm V(\lambda)$. - Then $U$ is an affine variety and the morphism $\phi: \cO_X(X)_\lambda \to \cO_X(U)$ defined by the restriction $\cO_X(X) \xrightarrow{\cdot |_U } \cO_X(U)$ and the universal property of the localization is an isomorphism. -\end{proposition} -\begin{proof} - Let $X$ be an affine variety over $\fk, \lambda \in \cO_X(X)$ and $U = X \sm V(\lambda)$. The fact that $\lambda\defon{U} \in \cO_x(U)^{\times}$ follows from \ref{localinverse}. - Thus the universal property of the localization $\cO_X(X)_\lambda$ can be applied to $\cO_X(X) \xrightarrow{\cdot |_U} \cO_X(U)$. - \begin{figure}[H] - \centering - \begin{tikzcd} - \cO_X(X) \arrow{d}{\cdot |_U}\arrow{r}{x \mapsto \frac{x}{1}} & \cO_X(X)_\lambda \arrow[dotted, bend left]{dl}{\Eone \phi} \\ - \cO_X(U) & - \end{tikzcd} - \hspace{50pt} - \begin{tikzcd} - &Y \arrow[bend right, swap]{ld}{\pi_0} \arrow[bend right, swap]{d}{\pi}&\cO_Y(Y) \cong A_\lambda \arrow{d}{\fs}& \\ - X \arrow[hookrightarrow]{r}{}& U \arrow[swap]{u}{\sigma} & \cO_X(U) - \end{tikzcd} - \end{figure} - For the rest of the proof, we may assume $X = V(I) \se \fk^n$ where $I = \sqrt{I} \se R \coloneqq\fk[X_1,\ldots,X_n]$ is an ideal. - Then $A \coloneqq \cO_X(X) \cong R / I$ and there is $\ell \in R$ such that $\ell\defon{X} = \lambda$. - Let $Y = V(J) \se \fk^{n+1}$ where $J \se \fk[Z,X_1,\ldots,X_n]$ is generated by the elements of $I$ and $1 - Z\ell(X_1,\ldots,X_n)$. - - Then $\cO_Y(Y) \cong \fk[Z,X_1,\ldots,X_n] / J \cong A[Z] / (1 -\lambda Z) \cong A_\lambda$. - By the proposition about affine varieties (\ref{propaffvar}), the morphism $\fs: \cO_Y(Y) \cong A_\lambda \to \cO_X(U)$ corresponds to a morphism $U \xrightarrow{\sigma} Y$. - We have $\fs(Z \mod J) = \lambda\inv$ and $\fs(X_i \mod J) = X_i \mod I$. - Thus $\sigma(x) = (\lambda(x)\inv, x)$ for $x \in U$. - Moreover, the projection $Y \xrightarrow{\pi_0} X$ dropping the $Z$-coordinate has image contained in $U$, as for $(z,x) \in Y$ the equation - \[ - 1 = z\lambda(x) - \] - implies $\lambda(x) \neq 0$. It thus defines a morphism $Y \xrightarrow{\pi} U$ and by the description of $\sigma$ it follows that $\sigma \pi = \Id_U$. - Similarly it follows that $\sigma \pi = \Id_Y$. Thus, $\sigma$ and $\pi$ are inverse to each other. -\end{proof} -\begin{corollary}\label{affopensubtopbase} - The affine open subsets of a variety $X$ are a topology base on $X$. -\end{corollary} -\begin{proof} - Let $X = V(I) \se \fk^n$ with $I = \sqrt{I}$. If $U \se X$ is open then $X \sm U = V(J)$ with $J \supseteq I$ and $U = \bigcup_{f \in J} (X \sm V(f))$. - Thus $U$ is a union of affine open subsets. The same then holds for arbitrary quasi-affine varieties. - - Let $X$ be any variety, $U \se X$ open and $x \in U$. - By the definition of variety, $x$ has a neighbourhood $V_x$ which is quasi-affine, and replacing $V_x$ by $U \cap V_x$ which is also quasi-affine we may assume $V_x \se U$. - $V_x$ is a union of its affine open subsets. Because $U$ is the union of the $V_x$, $U$ as well is a union of affine open subsets. -\end{proof} - - -% Lecture 14A TODO? - -% Lecture 15 - -% CRTPROG - -\subsection{Stalks of sheaves} - -\begin{definition}[Stalk] - Let $\cG$ be a presheaf of sets on the topological space $X$, and let $x \in X$. - The \vocab{stalk} (\vocab[Stalk]{Halm}) of $\cG$ at $x$ is the set of equivalence classes of pairs $(U, \gamma)$, where $U$ is an open neighbourhood of $x$ and $\gamma \in \cG(U)$ - and the equivalence relation $\sim $ is defined as follows: - $( U , \gamma) \sim (V, \delta)$ iff there exists an open neighbourhood $W \se U \cap V$ of $x$ such that $\gamma \defon{W} = \delta \defon{W}$. - - - If $\cG$ is a presheaf of groups, one can define a groups structure on $\cG_x$ by - \[ - ((U, \gamma) / \sim ) \cdot \left( (V,\delta) / \sim \right) = (U \cap V, \gamma \defon{U \cap V} \cdot \delta\defon{U \cap V}) / \sim - \] - - If $\cG$ is a presheaf of rings, one can similarly define a ring structure on $\cG_x$. - - - If $U$ is an open neighbourhood of $x \in X$, then we have a map (resp. homomorphism) - \begin{align} - \cdot_x : \cG(U) &\longrightarrow \cG_x \\ - \gamma &\longmapsto \gamma_x \coloneqq (U, \gamma) / \sim - \end{align} - -\end{definition} -\begin{fact} - Let $\gamma,\delta \in \cG(U)$. If $\cG$ is a sheaf\footnote{or, more generally, a separated presheaf} and if for all $x \in U$, we have $\gamma_x = \delta_x$, then $\gamma = \delta$. - - In the case of a sheaf, the image of the injective map $\cG(U) \xrightarrow{\gamma \mapsto (\gamma_x)_{x \in U}} \prod_{x \in U} \cG_x$ - is the set of all $(g_x)_{x \in U} \in \prod_{x \in U} \cG_x $ satisfying the following \vocab{coherence condition}: - For every $x \in U$, there are an open neighbourhood $W_x \se U$ of $x$ and $g^{(x)} \in \cG(W_x)$ with $g_y^{(x)} = g_y$ for all $y \in W_x$. -\end{fact} -\begin{proof} - Because of $\gamma_x = \delta_x$, there is $x \in W_x \se U$ open such that $\gamma\defon{W_x} = \delta\defon{W_x}$. As the $W_x$ cover $U$, $\gamma = \delta$ by the sheaf axiom. -\end{proof} -\begin{definition} - Let $\cG$ be a sheaf of functions. - Then $\gamma_x$ is called the \vocab{germ} of the function $\gamma$ at $x$. - The \vocab[Germ!value at $x$]{value at $x$ } of $g = (U, \gamma) / \sim \in \cG_x$ defined as $g(x) \coloneqq \gamma(x)$, which is independent of the choice of the representative $\gamma$. -\end{definition} -\begin{remark} - If $\cG$ is a sheaf of $C^{\infty}$-functions (resp. holomorphic functions), then $\cG_x$ is called the ring of germs of $C^\infty$-functions (resp. of holomorphic functions) at $x$. - -\end{remark} -\subsubsection{The local ring of an affine variety} -\begin{definition} - If $X$ is a variety, the stalk $\cO_{X,x}$ of the structure sheaf at $x$ is called the \vocab{local ring} of $X$ at $x$. - This is indeed a local ring, with maximal ideal $\fm_x = \{f \in \cO_{X,x} | f(x) = 0\}$. -\end{definition} -\begin{proof} - By \ref{localring} it suffices to show that $\fm_x$ is a proper ideal, which is trivial, and that the elements of $\cO_{X,x} \sm \fm_x$ are units in $\cO_{X,x}$. - Let $g = (U, \gamma)/\sim \in \cO_{X,x}$ and $g(x) \neq 0$. - $\gamma$ is Zariski continuous (first point of \ref{localinverse}). Thus $V(\gamma)$ is closed. By replacing $U$ by $U \sm V(\gamma)$ we may assume that $\gamma$ vanishes nowhere on $U$. - By the third point of \ref{localinverse} we have $\gamma \in \cO_X(U)^{\times}$. - $(\gamma\inv)_x$ is an inverse to $g$. -\end{proof} - -\begin{proposition}\label{proplocalring} - Let $X = \Va(I) \se \fk^n$ be equipped with its usual structure sheaf, where $I = \sqrt{I} \se R = \fk[X_1,\ldots,X_n]$ . Let $x \in X$ and $A = \cO_X(X) \cong R / I$. - $\{P \in R | P(x) = 0\} \text{\reflectbox{$\coloneqq$}} \fn_x \se R$ is maximal, $I \se \fn_x$ and $\fm_x \coloneqq \fn_x / I$ is the maximal ideal of elements of $A$ vanishing at $x$. - If $\lambda \in A \sm \fm_x$, we have $\lambda_x \in \cO_{X,x}^{\times}$, where $\lambda_x$ denotes the image under $A \cong \cO_X(X) \to \cO_{X,x}$. - By the universal property of the localization, there exists a unique ring homomorphism $A_{\fm_x} \xrightarrow{\iota} \cO_{X,x}$ - such that - \begin{figure}[H] - \centering - \begin{tikzcd} - A \arrow{r}{} \arrow{d}{\lambda \mapsto \lambda_x} & A_{\fm_x} \arrow[dotted, bend left]{ld}{\Eone \iota} \\ - \cO_{X,x} - \end{tikzcd} - \end{figure} - commutes. - - The morphism $A_{\fm_x}\xrightarrow{\iota} \cO_{X,x}$ is an isomorphism. - -\end{proposition} -\begin{proof} - To show surjectivity, let $\ell = (U, \lambda) / \sim \in \cO_{X,x}$, where $U$ is an open neighbourhood of $x$ in $X$. - We have $X \sm U = V(J)$ where $J \se A$ is an ideal. As $x \in U$ there is $f \in J$ with $f(x) \neq 0$. Replacing $U $ by $X \sm V(f)$ we may assume $U = X \sm V(f)$. - By \ref{oxulocaf}, $\cO_X(U) \cong A_f$, and $\lambda = f^{-n}\vartheta$ for some $n \in \N$ and $\vartheta \in A$. - Then $\ell = \iota(f^{-n} \vartheta)$ where the last fraction is taken in $A_{\fm_x}$. - - - Let $\lambda = \frac{\vartheta}{g} \in A_{\fm_x}$ with $\iota(\lambda) = 0$. - It is easy to see that $\iota(\lambda) = (X \sm V(g), \frac{\vartheta}{g}) / \sim $. - Thus there is an open neighbourhood $U$ of $x$ in $X \sm V(g)$ such that $\vartheta$ vanishes on $U$. - Similar as before there is $h \in A$ with $h(x) \neq 0$ and $W = X \sm V(h) \se U$. - By the isomorphism $\cO_X(W) \cong A_h$, there is $n \in \N$ with $h^{n}\vartheta = 0$ in $A$. Since $h \not\in \fm_x$, $h$ is a unit and the image of $\vartheta$ in $A_{\fm_x}$ vanishes, implying $\lambda = 0$. -\end{proof} -\subsubsection{Intersection multiplicities and Bezout's theorem} -\begin{definition} - Let $R = \fk[X_0,X_1,X_2]$ equipped with its usual grading and let $x \in \bP^{2}$. -Let $G \in R_g, H \in R_h$ be homogeneous polynomials with $x \in V(G) \cap V(h)$. -Let $\ell\in R_1$ such that $\ell(x) \neq 0$. Then $x \in U = \bP^2 \sm V(\ell)$ and the rational functions $\gamma = \ell^{-g}G, \eta = \ell^{-h}H$ are elements of $\cO_{\bP^2}(U)$. -Let $I_x(G,H) \se \cO_{\bP^2,x}$ denote the ideal generated by $\gamma_x$ and $\eta_x$. - - -\noindent The dimension $\dim_{\fk}(\cO_{X,x} / I_x(G,H)) \text{\reflectbox{$\coloneqq$}} i_x(G,H)$ is called the \vocab{intersection multiplicity} of $G$ and $H$ at $x$. -\end{definition} -\begin{remark} - If $\tilde \ell \in R_1$ also satisfies $\tilde \ell(x) \neq 0$, then the image of $\tilde \ell / \ell$ under $\cO_{\bP^2}(U) \to \cO_{\bP^2,x}$ is a unit, showing that the image of $\tilde \gamma = \tilde \ell^{-g} G$ in $\cO_{\bP^2,x}$ is multiplicatively equivalent to $\gamma_x$, and similarly for $\eta_x$. - Thus $I_x(G,H)$ does not depend on the choice of $\ell \in R_1$ with $\ell(x) \neq 0$. -\end{remark} -\begin{theorem}[Bezout's theorem] - In the above situation, assume that $V(H)$ and $V(G)$ intersect properly in the sense that $V(G) \cap V(H) \se \bP^2$ has no irreducible component of dimension $\ge 1$. - Then - \[ - \sum_{x \in V(G) \cap V(H)} i_x(G,H) = gh - \] - Thus, $V(G) \cap V(H)$ has $gh$ elements counted by multiplicity. -\end{theorem} -\printvocabindex -\end{document} -\if\false - -% TODO REMARK ABOUT ZORNS LEMMA (LECTURE 1) - -% TODO REMARK ABOUT FIN PRESENTED MODULES (LECTURE 2) - - -% TODO: LECTURE 9 LEMMA - - - - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -% ÜBERSICHT % -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% - - -% List of forms of HNS - - -\begin{itemize} - \item[HNS2 $\implies$ HNS1b] Let $I \se \fl[X_1,\ldots,X_n]$. $I \se \fm$ maximal. $R / \fm$ is isomorphic to a field extension of $\fl$. Finite by HNS2. - \item[NNT $\implies$ HNS2] Apply NNT to $L / K$ $\leadsto$ alg. independent $a_i$ such that $L$ is finite over the image of $K[X_1,\ldots,X_n] \xrightarrow{\ev_a} L$ ($\ev_a : P \to P(a_1,\ldots,a_n)$). - $\ev_a(K[X_1,\ldots,X_n])$ is a field by fact about integrality and fields \ref{fintaf}. Hence $n = 0$ and $L / K$ is finite. - \item[UNCHNS2] $K$ uncountable, $L / K$ fin. type. Then $\dim_K L$ is countable. Suppose $l \in L$ is not integral. Then $K(l) \cong K(T)$ and $\dim_K L \ge \dim_K K(T) \ge \aleph_1$. - Thus $L / K$ algebraic $\implies$ integral $\implies$ finite. - \item[HNS3] ($V(I) \se V(f) \iff f \in \sqrt{I} $). Suppose $V(I) \se V(f)$. $R' \coloneqq \fk[X_1,\ldots,X_n, T], J \se R'$ the ideal generated by $I$ and $g(X_1,\ldots,X_n,T) \coloneqq 1 - Tf(X_1,\ldots,X_n)$. -\end{itemize} - - -% Proofs -Def of integrality (<=>) - - -Fact about integrality and field: - % TODO - - -Technical lemma for Noether normalization: For $S \se \N^n$ finite, there exists $k \in \N^n$ such that $k_1 = 1$ and $s_1 \neq s_2 \in S \implies \langle k, s_1 \rangle \neq \langle k, s_2 \rangle$: -For $s_1 \neq s_2$, % TODO - -Noether normalization: -$a_i \in A$ minimal such that $A$ is integral over the subalgebra genereted by the $a_i$. -Suppose $\E P \in K[X_1,\ldots,X_n] \sm \{ 0\} ~ P(a_1,\ldots,a_n) = 0$. $P = \sum_{\alpha \in \N^n} p_\alpha X^\alpha, S \coloneqq \{ \alpha \in \N^n | p_\alpha \neq 0\}$. -Choose $k$ as in the lemma. -$b_i \coloneqq a_{i+1} - a_1^{k_{i+1}}, 1 \le i \trdeg(\fk(\fq) / \fl)$: -\Wlog $\fp = \{0\}$ and $A$ a domain ($A' \coloneqq A / \fp$). -For $\fq \in \mSpec A$, $\fk(\fq) = A / \fq$ finite type, hence finite (HNS) $\implies \trdeg(\fk(\fq) / \fl) = 0$. -$\trdeg(Q(A) / \fl) = 0 \implies A$ integral over $\fk$ $\implies$ $A$ a field $\implies \fp = \fq \lightning$. - -If $\fq \not\in \mSpec A$, let $a_1,\ldots,a_n \in A$ alg. independent such that the $\overline{a_i}$ are a transcendence base for $\fk(\fq) / \fk$ -Let $R$ be the ring generated by $\fl$ and the $a_i$. Localize with respect to $S \coloneqq R \sm \{0\}$. -%TODO -% TODO: LERNEN - - -% Dim k^n -$\dim(\fk^n)$ -$ \ge n$ build chian -$\le n$ a first result in dim T ($\fp \subsetneq \fq \implies \trdeg(\fk(\fq) / \fl) < \trdeg(\fk(\fp) / \fl)$. Thus $\codim(X,Y) \le \trdeg(\fK(Y) / \fl) - \trdeg(\fK(X) / \fl)$. - -TODO -% List of proofs of HNS - - -% Going up - - -% TODO proof of dim Y = trdeg(K(Y) / k) -$\dim Y \ge \trdeg(\fk(Y) / \fk)$: Noether normalization. Subalgebra $\cong \fk[X_1,\ldots,X_d]$. Lift chain of prime ideals using going up. - -% TODO prime avoidance - - -Action of $\Aut(L/K)$ on prime ideals of a normal ring extension. $A$ normal domain, $L / Q(A)$ normal field extension, $B$ int closure of $A$ in $L$, $\fp \in \Spec A$. -Then $\Aut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq \cap A = \fp\}$ : - -\begin{itemize} - \item $\fq, \fr \in \Spec B$ lying over $\fp$. - \item only need to show $\fq \se \sigma(\fr)$ for some $\sigma \in G$ (Krull going-up, no inclusions) - \item Suppose not. Then $x \in \fq \sm \bigcup_{\sigma \in G} \sigma(\fr)$ (prime aviodance) - \item $y = \prod_{\sigma \in G} \sigma(x) \in \fq \sm \fr$ ($\fr$ prime ideal) - \item $\E k \in \N$ s.t. $y^k \in K$ ($y \in L^G$) - \item $y^k \in K \cap B = A $ ($A$ normal). Thus $y^k \in (A \cap \fq) \sm (A \cap \fr) = \fp \sm \fp$. - \item $L / K$ infinite: Apply Zorn to pairs $(M, \sigma)$ where $K \se M \se L$ and $\sigma \in \Aut(M /K)$ s.t. $\sigma(\fr \cap M) = \fq \cap M$. -\end{itemize} - - -Going down Krull %TODO - -The ht p and trdeg -================== -% TODO % TODO % TODO % - -% Definitions -Zariski-Topology, Spec, $\fk^n$ -Residue field $\fk(\fp) \coloneqq Q(A / \fp), \fK(V(\fp)) \coloneqq \fk(\fp)$. TODO? -% Counterexamples - no going-up -% list of definitions of codim, dim, trdeg, ht -Original (Noether normalization) -Artin-Tate -Uncountable fields -\begin{landscape} -\section{Übersicht} -{\rowcolors{2}{gray!10}{white} - \begin{longtable}{lll} - \end{longtable} -} - -\end{landscape} -\end{document}

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