From 13a7e012a9e70a1c5822f9e183c1983e9e7fd0e1 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Sun, 30 Jul 2023 21:24:08 +0200 Subject: [PATCH] fixed some wrong linebreaks (introduced by latexindent?) --- .../nullstellensatz_and_zariski_topology.tex | 2678 ++++++++--------- 1 file changed, 1201 insertions(+), 1477 deletions(-) diff --git a/inputs/nullstellensatz_and_zariski_topology.tex b/inputs/nullstellensatz_and_zariski_topology.tex index 2a67953..59d57ff 100644 --- a/inputs/nullstellensatz_and_zariski_topology.tex +++ b/inputs/nullstellensatz_and_zariski_topology.tex @@ -1,7 +1,7 @@ \subsection{The Nullstellensatz} %LECTURE 1 -Let $\mathfrak{k}$ be a field, $R \coloneqq -\mathfrak{k}[X_1,\ldots,X_n], I -\subseteq R$ an ideal. +Let $\mathfrak{k}$ be a field, +$R \coloneqq \mathfrak{k}[X_1,\ldots,X_n]$, +$I \subseteq R$ an ideal. \begin{definition}[zero] $x \in \mathfrak{k}^n$ is \vocab[Ideal!zero]{a zero of $I$} @@ -17,15 +17,15 @@ Let $\mathfrak{k}$ be a field, $R \coloneqq Then $\{x \in R | \forall s \in S: s(x) = 0\} = \Va(I)$. 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 + 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 $\mathfrak{k}$ is algebraically closed and $I \subsetneq R$ a - proper ideal, + If $\mathfrak{k}$ is algebraically closed + and $I \subsetneq R$ a proper ideal, then $I$ has a zero in $\mathfrak{k}^n$. \end{theorem} @@ -35,7 +35,7 @@ Let $\mathfrak{k}$ be a field, $R \coloneqq $R$ is a PID, thus $I = pR$ for some $p \in R$. Since $I \neq R$ $p = 0$ or $P$ is non-constant. $\mathfrak{k}$ algebraically closed - $\leadsto$ there exists a zero of $p$.\\ + $\leadsto$ there exists a zero of $p$.\\ If $\mathfrak{k}$ is not algebraically closed and $n > 0$, the theorem fails (consider $I = p(X_1) R$). @@ -46,8 +46,8 @@ 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. + 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} @@ -57,7 +57,7 @@ Equivalent\footnote{used in a vague sense here} formulation: \item[$\impliedby$ ] Apply the Noether normalization theorem (\ref{noenort}) to $A = L$. This yields an injective ring homomorphism $\ev_a: - K[X_1,\ldots,X_n] \to A$ + 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 @@ -77,8 +77,8 @@ Equivalent\footnote{used in a vague sense here} formulation: \begin{theorem}[Hilbert's Nullstellensatz (1b)] \label{hns1b} - Let $\mathfrak{l}$ be a field and - $I \subset R = \mathfrak{l}[X_1,\ldots,X_m]$ a proper ideal. + Let $\mathfrak{l}$ be a field + and $I \subset R = \mathfrak{l}[X_1,\ldots,X_m]$ a proper ideal. Then there are a finite field extension $\mathfrak{i}$ of $\mathfrak{l}$ and a zero of $I$ in $\mathfrak{i}^m$. \end{theorem} @@ -99,7 +99,7 @@ Equivalent\footnote{used in a vague sense here} formulation: \[ P(x_1,\ldots,x_m) = \iota(P \mod \mathfrak{m}) \] - Both sides are morphisms $R \to \mathfrak{i}$ of $\mathfrak{l}$-algebras. + Both sides are morphisms $R \to \mathfrak{i}$ of $\mathfrak{l}$-algebras. For for $P = X_i$ the equality is trivial. It follows in general, since the $X_i$ generate $R$ as a $\mathfrak{l}$-algebra. @@ -119,18 +119,19 @@ but will be accepted in the exam. 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. + 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 $I \coloneqq \{\kappa \in K | x_{\kappa} \neq 0\}$ + 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) $. + Let $d \coloneqq \prod_{\kappa \in I} (T - \kappa) $. Then for $\lambda \in I$ we have \[ 0 = (dg)(\lambda) = @@ -163,16 +164,16 @@ but will be accepted in the exam. Let $R$ be a ring and $I,J, I_\lambda \subseteq R$ ideals, $\lambda \in \Lambda$. \begin{definition}[Radical, product and sum of ideals] \[ - \sqrt{I} \coloneqq \bigcap_{n=0}^{\infty} \{f \in R | f^n \in I\} + \sqrt{I} \coloneqq \bigcap_{n=0}^{\infty} \{f \in R | f^n \in I\}, \] \[ - I \cdot J \coloneqq \langle\{ i \cdot j | i \in I , j \in J\}\rangle_R + I \cdot J \coloneqq \langle\{ i \cdot j | i \in I , j \in J\}\rangle_R, \] \[ \sum_{\lambda \in \Lambda} I_\lambda \coloneqq \left\{\sum_{\lambda \in \Lambda'} i_\lambda | - \Lambda' \subseteq \Lambda \text{ finite}\right\} + \Lambda' \subseteq \Lambda \text{ finite}\right\}. \] \end{definition} \begin{fact} @@ -183,14 +184,14 @@ Let $R$ be a ring and $I,J, I_\lambda \subseteq R$ ideals, $\lambda \in \Lambda $\bigcap_{\lambda \in \Lambda} I_\lambda$ is an ideal. \end{fact} -Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ where $\mathfrak{k}$ is an -algebraically closed field. +Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ where $\mathfrak{k}$ +is an algebraically closed field. \begin{fact} \label{fvop} - Let $I, J, (I_{\lambda})_{\lambda \in \Lambda}$ be - ideals in $R$. + Let $I, J, (I_{\lambda})_{\lambda \in \Lambda}$ be ideals in $R$. $\Lambda$ may be infinite. + Then \begin{enumerate}[A] \item $\Va(I) = \Va(\sqrt{I})$, @@ -218,7 +219,7 @@ algebraically closed field. 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 $(f \cdot g)(x) \neq 0 \implies x \not\in \Va(I\cdot J)$. + thus $(f \cdot g)(x) \neq 0 \implies x \not\in \Va(I\cdot J)$. Therefore \[ \Va(I) \cup \Va(J) \subseteq \Va(I \cap J) @@ -232,7 +233,7 @@ algebraically closed field. Thus $\Va(\sum_{\lambda \in \Lambda} I_\lambda) \subseteq \bigcap_{\lambda \in \Lambda} \Va(I_\lambda)$. - On the other hand if $f \in \sum_{\lambda \in \Lambda} I_\lambda$ + 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 $\bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$ @@ -243,37 +244,31 @@ algebraically closed field. \end{proof} \begin{remark} There is no similar way to describe - $\Va(\bigcap_{\lambda \in \Lambda} I_\lambda)$ + $\Va(\bigcap_{\lambda \in \Lambda} I_\lambda)$ 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 $\mathfrak{k}$ be algebraically closed, $R = -\mathfrak{k}[X_1,\ldots,X_n]$. +Let $\mathfrak{k}$ be algebraically closed, $R = \mathfrak{k}[X_1,\ldots,X_n]$. \begin{corollary} - (of - \ref{fvop}) - There is a topology on $\mathfrak{k}^n$ for which the set of closed - sets + (of \ref{fvop}) + There is a topology on $\mathfrak{k}^n$ for which the set of closed sets coincides with the set $\mathfrak{A}$ of subsets of the form - $\Va\left(I - \right) $ for ideals $I \subseteq R$. + $\Va\left(I\right)$ for ideals $I \subseteq R$. This topology is called the \vocab{Zariski-Topology} \end{corollary} \begin{example} - \label{zariskinothd} Let $n = 1$. + \label{zariskinothd} + Let $n = 1$. Then $R$ is a PID. Hence every ideal is a principal ideal and the Zariski-closed subsets of - $\mathfrak{k}$ are the subsets of the form $\Va(P)$ - for $P \in R$. - As $\Va(0) = \mathfrak{k}$ and - $\Va(P)$ finite for $P \neq 0$ and + $\mathfrak{k}$ are the subsets of the form $\Va(P)$ for $P \in R$. + As $\Va(0) = \mathfrak{k}$ 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 $\mathfrak{k}$ are $\mathfrak{k}$ and the finite - subsets. + of $\mathfrak{k}$ are $\mathfrak{k}$ and the finite subsets. Because $\mathfrak{k}$ is infinite, this topology is not Hausdorff. \end{example} @@ -281,7 +276,7 @@ Let $\mathfrak{k}$ be algebraically closed, $R = \begin{definition} Let $X$ be a topological space. $X$ satisfies the separation properties $T_{0-2}$ if for - any $x \neq y \in X$ + any $x \neq y \in X$ \begin{enumerate} \item[$T_0$ ] $\exists U \subseteq X$ open such that $|U \cap \{x,y\}| = 1$ @@ -293,27 +288,24 @@ Let $\mathfrak{k}$ be algebraically closed, $R = \end{enumerate} \end{definition} \begin{remark} - Let $x \sim y :\iff$ the open subsets of $X$ containing $x$ are precisely the + 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 $x \sim y \implies x =y$. + 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 $\mathfrak{k}^n$ is $T_1$ but for $n \ge 1$ - not - Hausdorff. - For $n \ge 1$ the intersection of two non-empty open subsets of + not Hausdorff. + For $n \ge 1$ the intersection of two non-empty open subsets of $\mathfrak{k}^n$ 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 - $\mathfrak{k}^n$ then - $I \neq \{0\} , J \neq \{0\} $ and thus $IJ \neq \{0\} $. - Therefore $A \cup B = V(IJ)$ is a proper closed subset of - $\mathfrak{k}^n$. + If $A = V(I), B = V(J)$ are two proper closed subsets of $\mathfrak{k}^n$ + then $I \neq \{0\}$, $J \neq \{0\}$ and thus $IJ \neq \{0\}$. + Therefore $A \cup B = V(IJ)$ is a proper closed subset of $\mathfrak{k}^n$. \end{proof} @@ -322,18 +314,18 @@ 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. + 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: + $X$ 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. + Every descending sequence $A_0 \supseteq A_1 \supseteq \ldots$ + of closed subsets of $X$ stabilizes. \item Every non-empty set $\mathcal{M}$ of closed subsets of $X$ has a $\subseteq$-minimal element. @@ -345,26 +337,26 @@ Let $X$ be a topological space. \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 $X \setminus A = \bigcup_{j = 0}^{\infty} (X \setminus - A_j)$ has a finite - subcovering. - \item[B $\implies$ C] + If A holds, the covering + $X \setminus A = \bigcup_{j = 0}^{\infty} (X \setminus A_j)$ + has a finite subcovering. + \item[B $\implies$ C] Suppose $\mathcal{M}$ does not have a $\subseteq$-minimal element. - Using DC, one can construct a counterexample $A_1 \subsetneq A_2 \supsetneq - \ldots$ to B. + 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 $U \subseteq X$. - By C, the set $\mathcal{M} \coloneqq \{X \setminus - \bigcup_{i \in F} V_i | F - \subseteq I \text{ finite} \}$ has a $\subseteq$-minimal element. + Let $\bigcup_{i \in I} V_i$ be an open covering + of an open subset $U \subseteq X$. + By C, the set + $\mathcal{M} \coloneqq \{X \setminus \bigcup_{i \in F} V_i | F \subseteq I \text{ finite} \}$ + has a $\subseteq$-minimal element. \end{enumerate} \end{proof} \subsection{Another form of the Nullstellensatz and Noetherianness of \texorpdfstring{$\mathfrak{k}^n$}{kn}} -Let $\mathfrak{k}$ be algebraically closed, $R = -\mathfrak{k}[X_1,\ldots,X_n]$. +Let $\mathfrak{k}$ be algebraically closed, +$R = \mathfrak{k}[X_1,\ldots,X_n]$. For $f \in R$ let $V(f) = V(fR)$. \begin{theorem}[Hilbert's Nullstellensatz (3)] \label{hns3} @@ -373,45 +365,43 @@ For $f \in R$ let $V(f) = V(fR)$. \end{theorem} \begin{proof} Suppose $f$ vanishes on all zeros of $I$. - Let $R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n,T]$, $g(X_1,\ldots,X_n,T) - \coloneqq 1 - T \cdot f(X_1,\ldots,X_n)$ and $J \subseteq R'$ the ideal - generated by $g$ and the elements of $I$ (viewed as elements of $R'$ which are - constant in the $T$-direction). + Let $R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n,T]$, + $g(X_1,\ldots,X_n,T) \coloneqq 1 - T \cdot f(X_1,\ldots,X_n)$ + and $J \subseteq R'$ 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 - $\mathfrak{k}^{n+1}$. + If $f$ vanishes on all zeros of $I$, + then $J$ has no zeros in $\mathfrak{k}^{n+1}$. - Thus there exist $p_i \in I, i=1,\ldots,n, q_i \in - \mathfrak{k}[X_1,\ldots,X_n,T], i = 1,\ldots,n$ and $q \in - \mathfrak{k}[X_1,\ldots,X_n,T]$ such that + Thus there exist $p_i \in I, i=1,\ldots,n$, + $q_i \in \mathfrak{k}[X_1,\ldots,X_n,T], i = 1,\ldots,n$ + and $q \in \mathfrak{k}[X_1,\ldots,X_n,T]$ such that \[ - 1 = g \cdot q + \sum_{i=1}^{n} - p_{i}q_i + 1 = g \cdot q + \sum_{i=1}^{n} p_{i}q_i. \] - Formally substituting $\frac{1}{f(x_1,\ldots,x_n)}$ for $Y$, one - obtains: + Formally substituting $\frac{1}{f(x_1,\ldots,x_n)}$ for $Y$, + one obtains: \[ - 1 = \sum_{i=1}^{n} - p_{i}\left(x_1,\ldots,x_n\right) q_i\left( - x_1,\ldots,x_n, \frac{1}{f(x_1,\ldots,x_n)} \right) + 1 = \sum_{i=1}^{n} p_{i}\left(x_1,\ldots,x_n\right) % + q_i\left(x_1,\ldots,x_n, \frac{1}{f(x_1,\ldots,x_n)} \right) \] - Multiplying by a - sufficient power of $f$, this yields an equation in $R$ : + Multiplying by a sufficient power of $f$, + this yields an equation in $R$ : \[ - f^d = - \sum_{i=1}^{n} p_{i}(x_1,\ldots,_n) \cdot - q_i'(x_1,\ldots,x_n) \in I + f^d = \sum_{i=1}^{n} p_{i}(x_1,\ldots,_n) \cdot q_i'(x_1,\ldots,x_n) % + \in I \] - Thus $f - \in \sqrt{I}$. + Thus $f \in \sqrt{I}$. \end{proof} \begin{corollary} \label{antimonbij} \begin{align*} - f: \{I \subseteq R | I \text{ ideal}, I = \sqrt{I} \} & \longrightarrow \{A \subseteq \mathfrak{k}^n | A \text{ Zariski-closed}\} \\ - I & \longmapsto V(I) \\ - \{f \in R | A \subseteq V(f)\} & \longmapsfrom A + f: \{I \subseteq R | I \text{ideal}, I = \sqrt{I} \} & + \longrightarrow + \{A \subseteq \mathfrak{k}^n | A \text{ Zariski-closed}\} \\ + I & \longmapsto V(I)\\ + \{f \in R | A \subseteq V(f)\}&\longmapsfrom A \end{align*} is a $\subseteq$-antimonotonic bijection. \end{corollary} @@ -419,14 +409,11 @@ For $f \in R$ let $V(f) = V(fR)$. The topological space $\mathfrak{k}^n$ is Noetherian. \end{corollary} \begin{proof} - Because the map from - \ref{antimonbij} is antimonotonic, strictly - decreasing - chains of closed subsets of $\mathfrak{k}^n$ are mapped to strictly - increasing - chains of ideals in $R$. - By the Basissatz ( - \ref{basissatz}), $R$ is Noetherian. + Because the map from \ref{antimonbij} is antimonotonic, + strictly decreasing chains of closed subsets of $\mathfrak{k}^n$ + are mapped to strictly increasing chains of ideals in $R$. + By the Basissatz (\ref{basissatz}), + $R$ is Noetherian. \end{proof} % Lecture 04 @@ -441,7 +428,7 @@ Let $X$ be a topological space. equivalent conditions hold: \begin{enumerate}[A] \item - Every open $\emptyset \neq U \subseteq X$ is dense. + Every open $\emptyset \neq U \subseteq X$ is dense. \item The intersection of non-empty, open subsets $U, V \subseteq X$ is non-empty. \item @@ -510,18 +497,21 @@ Let $X$ be a topological space. $X = \emptyset$ 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 $\overline{A} \cap D \neq D$. + 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 $X$, + 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 $Z \subseteq X$ 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 $Z \subseteq Y \subseteq X$ coincides with $Z$. + \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 $Z \subseteq Y \subseteq X$ coincides with $Z$. \end{definition} \begin{corollary} \begin{enumerate} @@ -540,103 +530,94 @@ Let $X$ be a topological space. \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 $X$. + 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 - $X$. + 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 $\mathfrak{M}$ be the set of closed subsets of $X$ which cannot - be + Let $\mathfrak{M}$ be the set of closed subsets of $X$ which cannot be decomposed as a union of finitely many irreducible subsets. - Suppose $\mathfrak{M} \neq \emptyset$. + Suppose $\mathfrak{M} \neq \emptyset$. Then there exists a $\subseteq$-minimal $Y \in \mathfrak{M}$. $Y$ cannot be empty or irreducible. - Hence $Y = A \cup B$ where $A,B$ are proper closed subsets of $ Y$. + 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 $\lightning$. + 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 $X$, $Y = \bigcup_{i = 1}^n (Y \cap - Z_i)$ + 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 $Y \subseteq Z_i$. Thus the $Z_i$ are irreducible components. - Conversely, if $Y$ is an irreducible component of $X$, $Y \subseteq Z_i$ for - some $i$ and $Y = Z_i$ by the definition of irreducible component. + Conversely, if $Y$ is an irreducible component of $X$, + $Y \subseteq Z_i$ 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 $A \subseteq X$. + 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 $A \subseteq X$. \end{remark} \begin{proposition} \label{bijiredprim} - By - \ref{antimonbij} there exists a bijection + By \ref{antimonbij} there exists a bijection \begin{align*} f: \{I \subseteq R | - I \text{ ideal}, I = \sqrt{I} \} & \longrightarrow \{A \subseteq \mathfrak{k}^n - | A \text{ Zariski-closed}\} \\ I & \longmapsto V(I)\\ \{f \in R | A - \subseteq V(f)\} & \longmapsfrom A + I \text{ ideal}, I = \sqrt{I} \} & + \longrightarrow + \{A \subseteq \mathfrak{k}^n | A \text{ Zariski-closed}\}\\ + I & \longmapsto V(I)\\ + \{f \in R | A \subseteq V(f)\} & \longmapsfrom A \end{align*} - Under this correspondence $A \subseteq \mathfrak{k}^n$ is - irreducible iff $I - \coloneqq f^{-1}(A)$ is a prime ideal. + Under this correspondence $A \subseteq \mathfrak{k}^n$ is irreducible + iff $I \coloneqq f^{-1}(A)$ 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 $f \in J \setminus I, g \in K - \setminus I$. + 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 $f \in J \setminus I, g \in K \setminus I$. $fg$ vanishes on $A = B \cup C$. - By the Nullstellensatz ( - \ref{hns3}) $fg \in - \sqrt{I} = I$ and $I$ fails to be - prime. + 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 $A$. + 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 - $\subseteq$-antimonotonic and thus maximal ideals correspond to minimal - irreducible closed subsets, which are the one-point subsets as - $\mathfrak{k}^n$ - is T${}_1$. + The remaining assertion follows from the fact, + that the bijection is $\subseteq$-antimonotonic + and thus maximal ideals correspond to minimal irreducible closed subsets, + which are the one-point subsets as $\mathfrak{k}^n$ 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 $Z \subseteq Z_0 \subsetneq Z_1 \subsetneq \ldots \subsetneq Z_n$ of - irreducible closed subsets of $X$ containing $Z$ or $\infty$ if such chains can - be found for arbitrary $n$. + Let $Z$ be an irreducible subset of the topological space $X$. + Let $\codim(Z,X)$ be the maximum of the length $n$ of + strictly increasing chains + $Z \subseteq Z_0 \subsetneq Z_1 \subsetneq \ldots \subsetneq Z_n$ + of irreducible closed subsets of $X$ containing $Z$ or $\infty$ + if such chains can be found for arbitrary $n$. Let \[ \dim X \coloneqq \begin{cases} - - \infty & \text{if } X = \emptyset \\ - \sup_{\substack{Z \subseteq X \\ Z \text{ irreducible}}} \codim(Z,X) & - \text{otherwise} + - \infty & \text{if } X = \emptyset,\\ + \sup_{\substack{Z \subseteq X \\ Z \text{ irreducible}}} \codim(Z,X) & \text{otherwise}. \end{cases} \] \end{definition} @@ -645,17 +626,18 @@ Let $X$ be a topological space. \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. + 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 - $\max_{x \in X} \codim(\{x\}, X)$. + 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. + 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} @@ -663,14 +645,14 @@ Let $X$ be a topological space. \end{fact} \begin{fact} For every $x \in \mathfrak{k}$, $\codim( \{x\} ,\mathfrak{k}) = 1$. - The only other irreducible closed subset of $\mathfrak{k}$ is - $\mathfrak{k}$ - itself, which has codimension zero. + The only other irreducible closed subset of $\mathfrak{k}$ + is $\mathfrak{k}$ itself, + which has codimension zero. Thus $\dim \mathfrak{k} = 1$. \end{fact} \begin{fact} - Let $Y \subseteq X$ be irreducible and $U \subseteq X$ an open subset such that - $U \cap Y \neq \emptyset$. + Let $Y \subseteq X$ be irreducible + and $U \subseteq X$ an open subset such that $U \cap Y \neq \emptyset$. Then we have a bijection \begin{IEEEeqnarray*}{rl} f: &\{A \subseteq X | @@ -687,17 +669,15 @@ Let $X$ be a topological space. 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 $A = - \overline{B}$. + 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. + closure operator. \end{proof} \begin{corollary}[Locality of Krull codimension] \label{lockrullcodim} - Let $Y \subseteq X$ be irreducible and $U \subseteq X$ an open subset such that - $U \cap Y \neq \emptyset$. + Let $Y \subseteq X$ be irreducible and $U \subseteq X$ an open subset + such that $U \cap Y \neq \emptyset$. Then $\codim(Y,X) = \codim(Y \cap U, U)$. \end{corollary} \begin{fact} @@ -705,15 +685,14 @@ Let $X$ be a topological space. topological space $X$. Then \[ - \codim(Z,Y) + \codim(Y,X) \le \codim(Z,X) + \codim(Z,Y) + \codim(Y,X) \le \codim(Z,X) \tag{CD+} \label{eq:cdp} \] \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. + 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} @@ -721,41 +700,40 @@ Taking the supremum over all $Z$ we obtain: irreducible closed subset of the topological space $X$, then \[ \dim(Y) + - \codim(Y,X) \le \dim(X) + \codim(Y,X) \le \dim(X) \tag{D+} \label{eq:dp} \] \end{fact} -In general, these -inequalities may be strict. +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 + \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{$\mathfrak{k}^n$}{kn}} % from lecture 04 +\subsubsection{Krull dimension of \texorpdfstring{$\mathfrak{k}^n$}{kn}} % from lecture 04 \begin{theorem} \label{kdimkn} - $\dim \mathfrak{k}^n = n$ and $\mathfrak{k}^n$ is - catenary. - Moreover, if $X$ is an irreducible closed subset of - $\mathfrak{k}^n$, then - equality occurs in \eqref{eq:dp}. + $\dim \mathfrak{k}^n = n$ and $\mathfrak{k}^n$ is catenary. + Moreover, if $X$ is an irreducible closed subset of $\mathfrak{k}^n$, + then equality occurs in \eqref{eq:dp}. \end{theorem} \begin{proof} Considering \[ - \{0\} \subsetneq \mathfrak{k} \times \{0\} \subsetneq + \{0\} \subsetneq \mathfrak{k} \times \{0\} \subsetneq \mathfrak{k}^2 \times \{0\} \subsetneq \ldots \subsetneq \mathfrak{k}^n \] it - is clear that $\codim(\{0\}, \mathfrak{k}^n) \ge n$.Translation by $x \in - \mathfrak{k}^n$ gives us $\codim(\{x\} , \mathfrak{k}^n) \ge n$. + is clear that $\codim(\{0\}, \mathfrak{k}^n) \ge n$. + Translation by $x \in \mathfrak{k}^n$ gives us + $\codim(\{x\}, \mathfrak{k}^n) \ge n$. - The opposite inequality follows from \ref{upperbounddim} ($Z = - \mathfrak{k}^n$ + The opposite inequality follows from \ref{upperbounddim} + ($Z = \mathfrak{k}^n$, $\dim \mathfrak{k}^n \le \trdeg(\mathfrak{K}(Z) / \mathfrak{k}) = \trdeg(Q(\mathfrak{k}[X_1,\ldots,X_n]) / \mathfrak{k}) = n$). @@ -769,43 +747,37 @@ inequalities may be strict. Every non-zero prime ideal $\fp$ of a UFD $R$ contains a prime element. \end{lemma} \begin{proof} - Let $p \in \fp \setminus \{0\} $ with the minimal number of prime factors, + Let $p \in \fp \setminus \{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 $b \in \fp$ + 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 $p \in R = \mathfrak{k}[X_1,\ldots, X_n]$ be a prime element. + Let $p \in R = \mathfrak{k}[X_1,\ldots, X_n]$ be a prime element. Then the irreducible subset $X = V(p) \subseteq \mathfrak{k}^n$ has - codimension - one, and every codimension one subset of $\mathfrak{k}^n$ has this - form. + codimension one, + and every codimension one subset of $\mathfrak{k}^n$ 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 $\mathfrak{k}^n$. - If $X \subseteq Y \subseteq \mathfrak{k}^n$ is irreducible and - closed, then $Y - = V(\fq)$ for some prime ideal $\fp \subseteq pR$. + If $X \subseteq Y \subseteq \mathfrak{k}^n$ is irreducible and closed, + then $Y = V(\fq)$ for some prime ideal $\fp \subseteq pR$. If $Y \neq \mathfrak{k}^n$, then $\fp \neq \{0\}$. - By - \ref{ufdprimeideal} there exists a prime element $q \in \fq$. + By \ref{ufdprimeideal} there exists a prime element $q \in \fq$. As $\fq \subseteq pR$ 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 $X = V(\fp) \subseteq \mathfrak{k}^n$ is closed, irreducible - and of - codimension one. - Then $\fp \neq \{0\}$, hence $X \neq \mathfrak{k}^n$. - By - \ref{ufdprimeideal} there is a prime element $p \in \fp$. - If $\fp \neq pR$, then $X \subsetneq V(p) \subsetneq - \mathfrak{k}^n$ + Suppose $X = V(\fp) \subseteq \mathfrak{k}^n$ is closed, + irreducible and of codimension one. + Then $\fp \neq \{0\}$, hence $X \neq \mathfrak{k}^n$. + By \ref{ufdprimeideal} there is a prime element $p \in \fp$. + If $\fp \neq pR$, then $X \subsetneq V(p) \subsetneq \mathfrak{k}^n$ contradicts $\codim(X, \mathfrak{k}^n) = 1$. \end{proof} @@ -832,43 +804,43 @@ inequalities may be strict. conditions hold: \begin{enumerate} \item[M] - If $m,n \in X$ and $A \subseteq X$ - then $m \in \mathcal{H}( \{n\} \cup A) \setminus \mathcal{H}(A) \iff n - \in - \mathcal{H}(\{m\} \cup A) \setminus \mathcal{H}(A). - $ + If $m,n \in X$ and $A \subseteq X$ then + $m \in \mathcal{H}( \{n\} \cup A) \setminus \mathcal{H}(A) % + \iff n \in \mathcal{H}(\{m\} \cup A) \setminus \mathcal{H}(A)$, \item[F] - $\mathcal{H}(A) = \bigcup_{F \subseteq A \text{ finite}} \mathcal{H}(F)$. + $\mathcal{H}(A) = % + \bigcup_{F \subseteq A \text{ finite}} \mathcal{H}(F)$. \end{enumerate} - In this case, $S \subseteq X$ is called \vocab{Independent subset}, if $s - \not\in \mathcal{H}(S \setminus \{s\})$ for all $s \in S$ and - \vocab[Generating subset]{generating} if $X = \mathcal{H}(S)$. - $S$ is called a \vocab{base}, if it is both generating and - independent. + In this case, $S \subseteq X$ is called \vocab{Independent subset}, + if $s \not\in \mathcal{H}(S \setminus \{s\})$ for all $s \in S$ and + \vocab[Generating subset]{generating} if $X = \mathcal{H}(S)$. + $S$ is called a \vocab{base}, if it is both generating and independent. \end{definition} \begin{theorem} - If $\mathcal{H}$ 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. + If $\mathcal{H}$ 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 $\mathcal{L}(T)$ the - $K$-linear - hull of $T$ for $T \subseteq V$. - Then $\mathcal{L}$ is a matroidal hull operator on $V$. + $K$-linear hull of $T$ for $T \subseteq V$. + Then $\mathcal{L}$ is a matroidal hull operator on $V$. \end{example} \subsubsection{Transcendence degree} \begin{lemma} Let $L / K$ be a field extension and let $\mathcal{H}(T)$ 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 $T$.} + 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 $\mathcal{H}$ is a matroidal hull operator. \end{lemma} \begin{proof} @@ -877,52 +849,44 @@ inequalities may be strict. $\mathcal{H}(M) = M$. Thus $\mathcal{H}(\mathcal{H}(T)) = \mathcal{H}(T)$ (H3). - Let $x,y \in L$, $T \subseteq L$ and $x \in \mathcal{H}(T \cup \{y\}) - \setminus \mathcal{H}(T)$. - We have to show that $y \in \mathcal{H}(T \cup \{x\}) \setminus - \mathcal{H}(T)$. - If $y \in \mathcal{H}(T)$ we have $\mathcal{H}(T \cup \{y\}) \subseteq - \mathcal{H}(\mathcal{H}(T)) = \mathcal{H}(T) \implies x \in - \mathcal{H}(T) - \setminus \mathcal{H}(T) \lightning$. - Hence it is sufficient to show $y \in \mathcal{H}(T \cup \{x\})$. - Without loss of generality loss of generality $T = \emptyset$ (replace $K$ be - the subfield generated by $K \cup T$). + Let $x,y \in L$, $T \subseteq L$ and + $x \in \mathcal{H}(T \cup \{y\}) \setminus \mathcal{H}(T)$. + We have to show that + $y \in \mathcal{H}(T \cup \{x\}) \setminus \mathcal{H}(T)$. + If $y \in \mathcal{H}(T)$ we have + $\mathcal{H}(T \cup \{y\}) \subseteq \mathcal{H}(\mathcal{H}(T)) = \mathcal{H}(T) \implies % + x \in \mathcal{H}(T) \setminus \mathcal{H}(T) \lightning$. + Hence it is sufficient to show $y \in \mathcal{H}(T \cup \{x\})$. + Without loss of generality loss of generality $T = \emptyset$ + (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 $P(x) = - 0$. + 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 $p_i = - \frac{Q_i(y)}{R(y)}$, $R(y) \neq 0$. + that $p_i = \frac{Q_i(y)}{R(y)}$, $R(y) \neq 0$. Let \[ - Q(X,Y) \coloneqq \sum_{i=0}^{\infty} X^i Q_i(Y) = \sum_{i,j=0}^{\infty} - q_{i,j}X^i Y^j = \sum_{j=0}^{\infty} Y^j - \hat{Q_j}(X) \in K[X,Y]. + Q(X,Y) \coloneqq \sum_{i=0}^{\infty} X^i Q_i(Y) % + = \sum_{i,j=0}^{\infty} q_{i,j}X^i Y^j % + = \sum_{j=0}^{\infty} Y^j \hat{Q_j}(X) \in K[X,Y]. \] 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 $x \not\in \mathcal{H}(\emptyset)$. - Thus $\hat{P} \in \hat{M}[X] \setminus \{0\} $, - where $\hat{M}$ is the - subfield of $L$ generated by $K$ and $x$. - Thus $y$ is algebraic over $\hat{M}$ and $y \in - \mathcal{H}(\{x\})$, + 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 $x \not\in \mathcal{H}(\emptyset)$. + Thus $\hat{P} \in \hat{M}[X] \setminus \{0\} $, + where $\hat{M}$ is the subfield of $L$ generated by $K$ and $x$. + Thus $y$ is algebraic over $\hat{M}$ and $y \in \mathcal{H}(\{x\})$, \end{proof} -\begin{definition}[Transcendence Base] Let $L / K$ be a field - extension and $\mathcal{H}(T)$ the algebraic closure in $L$ of the - subfield +\begin{definition}[Transcendence Base] + Let $L / K$ be a field extension and + $\mathcal{H}(T)$ the algebraic closure in $L$ of the subfield generated by $K$ and $T$. A base for $(L, \mathcal{H})$ is called a \vocab{transcendence base} - and the - \vocab{transcendence degree} $\trdeg(L / K)$ is defined as the - cardinality of - any transcendence base of $L / K$. + 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 $\trdeg(L / K) = 0$. @@ -930,8 +894,8 @@ inequalities may be strict. \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. +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). @@ -939,8 +903,9 @@ transcendence degree. \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. + If the ring extension $B / A$ is finite + (i.e.~$B$ finitely generated as an $A$-module) + then $A$ is Noetherian. \end{theorem} \begin{fact}+ \label{noethersubalg} @@ -948,47 +913,41 @@ transcendence degree. Then every $R$-subalgebra $A \subseteq B$ is finite over $R$. \end{fact} \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). + 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. - + If $ B / R$ is of finite type and $B / A$ is finite, + then $A / R$ is also of finite type. \[ \begin{tikzcd} - A \arrow[hookrightarrow]{rr}{\subseteq}& & B \\ &R \arrow{ul}{\alpha} + A \arrow[hookrightarrow]{rr}{\subseteq}& & B \\ &R \arrow{ul}{\alpha} \arrow{ur}{\alpha} \text{~(Noeth.) } \end{tikzcd} \] \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 $(b_i)_{i=1}^{m}$ 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 - $\alpha_{ij}$. + $a_{ijk}$ 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 $B$. + 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 $A \subseteq B, A / \tilde{A}$ is finite - ( - \ref{noethersubalg}). + Since $A \subseteq B, A / \tilde{A}$ 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. + By the transitivity of ``of finite type'', + it follows that $A / R$ is of finite type. \[ \begin{tikzcd} \tilde A \arrow[hookrightarrow]{r}{\subseteq}& A \arrow[hookrightarrow]{r}{\subseteq} & B \\ @@ -1000,79 +959,69 @@ transcendence degree. \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 $R$. + 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 $K$. + $K(X_1,\ldots,X_n)$ 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. + Suppose $(P_i)_{i =1}^m$ is a complete list of prime elements of $R$ + (up to multiplicative equvialence). + Then $m > 0$, 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 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 $P_i - \lightning$. + 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. + Suppose $(f_i)_{i=1}^m$ generate $K(X_1,\ldots,X_n)$ as a $K$-algebra. Let $f_i = \frac{a_i}{b}, a_i \in R, b \in R \setminus \{0\}$. 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 \[ - b^Ng \in R \tag{+} - \label{bNginR} + b^Ng \in R \tag{+} \label{bNginR} \] - However, if $b = \varepsilon - \prod_{i=1}^{l} P_i$ is a decomposition of $b$ into prime factors $P_i$ - and a - unit $\varepsilon$ 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 $N \in \N$. + However, if $b = \varepsilon \prod_{i=1}^{l} P_i$ + is a decomposition of $b$ into prime factors $P_i$ and a + unit $\varepsilon$ in $R$ and $g = \frac{1}{P}$, + where $P \in R$ is a prime element not multiplicatively equivalent + 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 +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}). + 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 $\tilde R \subseteq L$ 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 $Q(\tilde R)$. + We have $\tilde R \cong R \coloneqq K[X_1,\ldots,X_n]$, + 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 $R \cong \tilde R \implies - K(X_1,\ldots,X_n) \cong Q(\tilde R)$. + 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} @@ -1083,38 +1032,34 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. Let $X \subseteq \mathfrak{k}^n$ be an irreducible closed subset. Then $X = V(\fp)$ for a unique prime ideal $\fp \subseteq R$. Let $\mathfrak{K}(X) \coloneqq Q(R / \fp)$ denote the field of - quotients of $R - / \fp$. + 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 $\mathfrak{K}(X)$ as the field of rational functions - on $X$. + As the elements of $\fp$ vanish on $X$, $R / \fp$ may be viewed as + the ring of polynomials + and $\mathfrak{K}(X)$ as the field of rational functions on $X$. \end{remark} \begin{theorem} \label{trdegandkdim} - If $X \subseteq \mathfrak{k}^n$ is irreducible, then $\dim X = - \trdeg - (\mathfrak{k}(X) / \mathfrak{k})$ and $\codim(X, \mathfrak{k}^n) = n - - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$. - More generally if $Y \subseteq \mathfrak{k}^n$ is irreducible and $X - \subseteq - Y$, then $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$. + If $X \subseteq \mathfrak{k}^n$ is irreducible, + then $\dim X = \trdeg(\mathfrak{k}(X) / \mathfrak{k})$ + and $\codim(X, \mathfrak{k}^n) = n - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$. + More generally if $Y \subseteq \mathfrak{k}^n$ is irreducible + and $X \subseteq Y$, then + $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$. \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 $\dim Y = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$" ( - \ref{lowerbounddimy}). + (\ref{upperboundcodim}) and other one in "Aplication to dimension theory: + Proof of $\dim Y = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$" + (\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 + Loosely speaking, + the Krull dimension of $X$ is equal to the maximal number of $\mathfrak{k}$-algebraically independent rational functions on $X$. This is yet another indication that the notion of dimension is the ``correct'' one. @@ -1134,151 +1079,136 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. Let $R$ be a commutative ring. \begin{itemize} \item - Let $\Spec R$ denote the set of prime ideals and $\MaxSpec R \subseteq \Spec R$ - the set of maximal ideals of $R$. + Let $\Spec R$ denote the set of prime ideals and + $\MaxSpec R \subseteq \Spec R$ the set of maximal ideals of $R$. \item - For an ideal $I \subseteq R$ let $V(I) \coloneqq \{\fp \in \Spec R | I - \subseteq \fp\}$ + For an ideal $I \subseteq R$ let + $V(I) \coloneqq \{\fp \in \Spec R | I \subseteq \fp\}$ \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 $R$. + 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 $R = \mathfrak{k}[X_1,\ldots,X_n]$, 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. + When $R = \mathfrak{k}[X_1,\ldots,X_n]$, + 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 $V(\bigcap_{j=1}^n I_j) = V(\prod_{j=1}^{n} - I_j) = - \bigcup_{j = 1}^n V(I_j)$. + 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 $R = \mathfrak{k}[X_1,\ldots,X_n]$. - Then there exists a bijection ( - \ref{antimonbij}, - \ref{bijiredprim}) between - $\Spec R$ and the set of irreducible closed subsets of - $\mathfrak{k}^n$ sending - $\fp \in \Spec R$ to $V_{\mathfrak{k}^n}(\fp)$ and identifying the - one-point - subsets with $\MaxSpec R$. - This defines a bijection $\mathfrak{k}^n \cong \MaxSpec R$ which is - a - homeomorphism if $\MaxSpec R$ is equipped with the induced topology from the - Zariski topology on $\Spec R$. + Then there exists a bijection (\ref{antimonbij}, \ref{bijiredprim}) + between $\Spec R$ + and the set of irreducible closed subsets of $\mathfrak{k}^n$ + sending $\fp \in \Spec R$ to $V_{\mathfrak{k}^n}(\fp)$ + and identifying the one-point subsets with $\MaxSpec R$. + This defines a bijection $\mathfrak{k}^n \cong \MaxSpec R$ + which is a homeomorphism if $\MaxSpec R$ 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 $S - \subseteq R$ such that $\prod_{i=1}^{n} f_i \in S $ when $n \in \N$ and - all $f_i \in S$. + A \vocab{multiplicative subset} of a ring $R$ is a subset $S \subseteq R$ + 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 $S \subseteq R$ be a multiplicative subset. Then there is a ring homomorphism $R \xrightarrow{i} R_S$ such that - $i(S) - \subseteq R_S^{\times }$ and $i$ has the \vocab{universal property} for such - ring homomorphisms: If $R \xrightarrow{j} T$ is a ring homomorphism - with $j(S) - \subseteq T^{\times }$, then there is a unique ring - homomorphism $R_S - \xrightarrow{\iota} T$ with $j = \iota i$. - + $i(S) \subseteq R_S^{\times }$ + and $i$ has the \vocab{universal property} for such + ring homomorphisms: + If $R \xrightarrow{j} T$ is a ring homomorphism with + $j(S) \subseteq T^{\times }$, + then there is a unique ring homomorphism $R_S \xrightarrow{\iota} T$ + with $j = \iota i$. \[ \begin{tikzcd} R \arrow{r}{i}\arrow{d}{j}& R_S \arrow[dotted]{ld}{\existsone \iota}\\ T \end{tikzcd} \] - \end{proposition} \begin{proof} The construction is similar to the construction of the field of - quotients: + quotients: - Let $R_S \coloneqq (R \times S) / \sim $, where $(r,s) \sim (\rho, \sigma) - : - \iff \exists t \in S ~ t \sigma r = ts\rho$. + Let $R_S \coloneqq (R \times S) / \sim $, where + $(r,s) \sim (\rho, \sigma):\iff \exists t \in S ~ t \sigma r = ts\rho$. \footnote{$t$ does not appear in the construction of the field of quotients, but is important if $S$ contains zero divisors.} - $[r,s] + [\rho, \sigma] - \coloneqq [r\sigma + \rho s, s \sigma]$, $[r,s] - \cdot [\rho, \sigma] \coloneqq [r \cdot \rho, s \cdot \sigma]$. + $[r,s] + [\rho, \sigma] \coloneqq [r\sigma + \rho s, s \sigma]$, + $[r,s] \cdot [\rho, \sigma] \coloneqq [r \cdot \rho, s \cdot \sigma]$. - In order proof the universal property define $\iota([r,s]) \coloneqq - \frac{j(r)}{j(s)}$. + 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 | \exists s - \in S ~ s \cdot r = 0\} $. + $i$ is often not injective and + $\ker(i) = \{r \in R | \exists s \in S ~ s \cdot r = 0\}$. In particular $(r = 1)$, $R_S$ is the null ring iff $0 \in S$. \end{remark} \begin{notation} Let $S \subseteq R$ 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 $X \subseteq R_S$ let $X \sqcap R$ denote - $i^{-1}(X)$. + 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 $X \subseteq R_S$ let $X \sqcap R$ denote $i^{-1}(X)$. \end{notation} \begin{definition}[$S$-saturated ideal] - An ideal $I \subseteq R$ is called - \vocab[Ideal! - S-saturated]{$S$-saturated} if for all $s \in S, r \in R$ $rs \in I \implies r - \in I$. + An ideal $I \subseteq R$ 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 $\fp \subseteq \Spec R$ is $S$-saturated iff $\fp \cap S = - \emptyset$. + A prime ideal $\fp \subseteq \Spec R$ 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 $R_S$. +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 $I \subseteq R$ be an $S$-saturated ideal and let $I_S$ denote the ideal - $\{\frac{r}{s} | r \in R, s \in S\} \subseteq R_S$. - Then for all $r \in R, s \in S$ we have $\frac{r}{s} \in I_S \iff r - \in I$. + $\{\frac{r}{s} | r \in R, s \in S\} \subseteq R_S$. + 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. + 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. + 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 \subseteq R | I \text{ $S$-saturated ideal}\} & \longrightarrow \left\{J \subseteq 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 \\ + f: \{I \subseteq R | I \text{ $S$-saturated ideal}\}& + \longrightarrow \left\{J \subseteq 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. @@ -1286,35 +1216,24 @@ $S$-saturated ideal in $R$ when $J$ is an ideal in $R_S$. \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}, $J \in \Spec R_S \implies - f^{-1}(J) = J \cap R \in - \Spec R_S$. - 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 $I_S \in \Spec R_S$. - + \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}, + $J \in \Spec R_S \implies f^{-1}(J) = J \cap R \in \Spec R_S$. + 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 @@ -1325,8 +1244,7 @@ $S$-saturated ideal in $R$ when $J$ is an ideal in $R_S$. If $S = R \setminus \{0\}$, then $R_S$ is the field of quotients $Q(R)$. If $S \subseteq R \setminus \{0\} $, then \[ - R_S \cong \left\{ \frac{a}{s} \in - K | a \in R, s \in S\right\} + R_S \cong \left\{ \frac{a}{s} \in K | a \in R, s \in S\right\} \] In particular $Q(R) \cong Q(R_S)$. \end{remark} @@ -1334,83 +1252,71 @@ $S$-saturated ideal in $R$ when $J$ is an ideal in $R_S$. \begin{definition}[$S$-saturation] \label{ssaturation} Let $R$ be any ring, $I \subseteq R$ 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 $I$. - - + 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 $R_S / I_S - \cong (R / - I)_{\overline{S}}$. + 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 - $\tau(S) \subseteq - T^{\times }$. - 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 $\tau_1(\overline{S}) = \tau(S) \subseteq - T^{\times }$, 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 $R \to R / I - \to (R / - I)_{\overline{S}} \xrightarrow{\tau_2} T$ + We show that both rings have the universal property + for ring homomorphisms $R \xrightarrow{\tau} T$ + with $\tau(I) = \{0\}$ and $\tau(S) \subseteq T^{\times }$. + For such $\tau$, by the fundamental theorem on homomorphisms% + \footnote{Homomorphiesatz} + there is a unique $R/I \xrightarrow{\tau_1} T$ such that $\tau = \tau_1 \pi_{R,I}$. + We have $\tau_1(\overline{S}) = \tau(S) \subseteq T^{\times }$, + 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 $\tau$. + 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 $\tau$. \[ \begin{tikzcd} - R \arrow{r}{\tau}\arrow[swap]{d}{\pi_{R,I}}& T & - R\arrow[swap]{l}{\tau}\arrow{d}{}\\ R / I \arrow[dotted]{ru}{\existsone - \tau_1}\arrow{d}{} & & R_S \arrow[dotted, swap]{lu}{\existsone - \tau_3}\arrow{d}{\pi_{R_S, I_S}}\\ (R / I)_{\overline{S}} \arrow[dotted,bend - right]{ruu}{\existsone \tau_2} & & R_S / I_S \arrow[dotted, bend left, - swap]{luu}{\existsone \tau_4}\\ + R \arrow{r}{\tau}\arrow[swap]{d}{\pi_{R,I}}& T & R\arrow[swap]{l}{\tau}\arrow{d}{}\\ + R / I \arrow[dotted]{ru}{\existsone \tau_1}\arrow{d}{} & & R_S \arrow[dotted, swap]{lu}{\existsone \tau_3}\arrow{d}{\pi_{R_S, I_S}}\\ + (R / I)_{\overline{S}} \arrow[dotted,bend right]{ruu}{\existsone \tau_2} & & R_S / I_S \arrow[dotted, bend left, swap]{luu}{\existsone \tau_4}\\ \end{tikzcd} \] - \end{proof} -\subsection{A first result of dimension theory} +\subsection{A first result of dimension theory} \begin{notation} Let $R$ be a ring, $\fp \in \Spec R$. - Let $\mathfrak{k}(\fp)$ denote the field of quotients of the domain $R / - \fp$. + Let $\mathfrak{k}(\fp)$ 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 $\mathfrak{l}$ be a field, $A$ a - $\mathfrak{l}$-algebra of finite type and + Let $\mathfrak{l}$ be a field, + $A$ a $\mathfrak{l}$-algebra of finite type and $\fp, \fq \in \Spec A$ with $\fp \subsetneq \fq$. %% ?? Then @@ -1420,132 +1326,116 @@ $S$-saturated ideal in $R$ when $J$ is an ideal in $R_S$. \] \end{proposition} \begin{proof} - Replacing $A$ by $A / \fp$, we - may assume $\fp = \{0\} $ and $A$ to be a domain. + Replacing $A$ by $A / \fp$, + we may assume $\fp = \{0\} $ and $A$ to be a domain. Then $\mathfrak{k}(\fp) = Q(A / \fp) = Q(A)$. - If $\fq$ is a maximal ideal, $\mathfrak{k}(\fq) = A / \fq$ is of finite - type - over $\mathfrak{l}$, hence a finite field extension of - $\mathfrak{l}$ by the - Nullstellensatz ( - \ref{hns2}). + If $\fq$ is a maximal ideal, + $\mathfrak{k}(\fq) = A / \fq$ is of finite type over $\mathfrak{l}$, + hence a finite field extension of $\mathfrak{l}$ by the + Nullstellensatz (\ref{hns2}). Thus, $\trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) = 0$. - If $\trdeg(Q(A) / \mathfrak{l}) = 0$, $A$ would be integral over - $\mathfrak{l}$, 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 $\fq = - \fp \lightning$. + If $\trdeg(Q(A) / \mathfrak{l}) = 0$, + $A$ would be integral over $\mathfrak{l}$, + 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 $\fq \in \MaxSpec A$. 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 + $a_1,\ldots,a_n \in A$ whose images in $A / \fq$ form a transcendence base for $\mathfrak{k}(\fq) / \mathfrak{l}$. \end{lemma} \begin{subproof} There exist $a_1,\ldots,a_n \in A$ such that $\mathfrak{k}(\fq)$ is - algebraic - over the subfield generated by $\mathfrak{l}$ and their images - $\overline{a_i}$ + algebraic over the subfield generated by $\mathfrak{l}$ + and their images $\overline{a_i}$ (for instance generators of $A$ as a $\mathfrak{l}$-algebra). We may assume that $n$ is minimal. - If the $a_i$ are $\mathfrak{l}$-algebraically dependent, then - w.l.o.g. + If the $a_i$ are $\mathfrak{l}$-algebraically dependent, then w.l.o.g. $\overline{a_n}$ can be assumed to be algebraic over the subfield - generated by $\mathfrak{l}$ and the $\overline{a_i}, 1\le - i 0$, then $L^{\Aut(L / K)} = \{l - \in L | \exists n - \in \N ~ l^{p^n} \in K\}$. + \label{characfixnormalfe} Let $L / K$ be a normal field extension. + If the characteristic of the fields is $O$, + then $L^{\Aut( L / K)} = K$. + If the characteristic is $p > 0$, + then $L^{\Aut(L / K)} = \{l \in L | \exists n \in \N ~ l^{p^n} \in K\}$. \end{proposition} \begin{proof} In both cases $L^G \supseteq$ is easy to see. - If $K \subseteq M \subseteq L$ 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 $M \subseteq N \subseteq L$ + If $K \subseteq M \subseteq L$ 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 $M \subseteq N \subseteq L$ and $N \xrightarrow{\vartheta} L$ a ring homomorphism such that $\vartheta\defon{M} = \sigma$ shows that $\sigma$ has an - extension to an - element of $\Aut(L / K)$. + 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 + 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. + 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 $K$. + 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 $\deg(l / K) - \coloneqq \deg(P)$. + 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 $M = K(l) \subseteq L$, then there is a ring homomorphism $M - - \overline{L}$ - sending $l$ to $\lambda$. - This can be extended to a ring homomorphism $L \xrightarrow{\sigma} - \overline{L}$. + 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 $M = K(l) \subseteq L$, 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 that this is possible only when $P(T) - = Q(T^p)$ for some $Q \in K[T]$. + 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 + that this is possible only when + $P(T) = Q(T^p)$ for some $Q \in K[T]$. % TODO: link to EinfAlg - 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 $m \in \N$. + 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} @@ -2083,12 +1942,10 @@ extensions: \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 $A$. + 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} @@ -2100,98 +1957,83 @@ extensions: Let $x \in Q(A)$ be integral over $A$. Then there is a normed polynomial $P \in A[T]$ with $P(x) = 0$. - In EInführung in die Algebra it was shown that $A[T]$ - is a UFD and that the - prime elements of $A[T]$ are the elements which are + In Einführung in die Algebra 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 $P \in A[T]$. + 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 $A$. + 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 $A$. Without loss of generality loss of generality $\gcd(a,b) = 1$. - Then $x^n + c_{n-1} x^{n-1} + - \ldots + c_0 = 0$ for some $c_i \in A$. - Multiplication with $b^n$ yields $a^n + c_{n-1} b - a^{n-1} + \ldots +c_0 b^n = - 0$. + 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 $x \in - A$. + 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. + 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 $\mathcal{O}_K$ (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 - EInführung in die Algebra ) abelian group. - % EINFALG + A finite field extension of $\Q$ is called an + \vocab{algebraic number field} (ANF). + If $K$ is an ANF, let $\mathcal{O}_K$ + (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 Einführung in die Algebra) + abelian group. % EINFALG We have $\mathcal{O}_{\Q} = \Z$ by the proposiiton. \end{remark} -\subsubsection{Action of \texorpdfstring{$\Aut(L / K)$}{Aut(L / K)} on prime ideals of a normal ring extension} +\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 $\{\fq \in - \Spec B | \fq - \cap A = \fp\}$. + 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 $\fq \subseteq \sigma(\fr)$, since the Krull - going-up - theorem ( - \ref{cohenseidenberg}) applies to the integral ring extension $B / - A$, + We must show that there exists $\sigma \in G$ + such that $\fq = \sigma(\fr)$. + This is equivalent to $\fq \subseteq \sigma(\fr)$, + 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 \setminus - \bigcup_{\sigma \in G} - \sigma(\fr)$. - As $\fr$ is a prime ideal, $y = \prod_{\sigma \in G} \sigma(x) - \in \fq - \setminus \fr$. - \footnote{$\prod_{\sigma \in G} \sigma(x) = \prod_{\sigma \in G} - \sigma^{-1}(x)$} + If $L / K$ is finite and there is no such $\sigma$, + then by prime avoidance (\ref{primeavoidance}) + there is $ x \in \fq \setminus \bigcup_{\sigma \in G} \sigma(\fr)$. + As $\fr$ is a prime ideal, + $y = \prod_{\sigma \in G} \sigma(x) \in \fq \setminus \fr$.% + \footnote{$\prod_{\sigma \in G} \sigma(x) = \prod_{\sigma \in G} \sigma^{-1}(x)$} 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 $y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp = - \emptyset \lightning$. + (\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 $y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp = \emptyset \lightning$. - 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 $\sigma(\fr \cap M) = \fq \cap M$. + 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} @@ -2200,17 +2042,16 @@ extensions: that $\Gal(K / \Q)$ transitively acts on the set of prime ideals of $\mathcal{O}_K$ 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 $\fq \in \Spec \mathcal{O}_L$ for - which $\fq - \cap K$ is a given $\fp \in \Spec \mathcal{O}_K$. + $\Gal(L / K)$ transitively acts on the set of + $\fq \in \Spec \mathcal{O}_L$ for which $\fq \cap K$ + is a given $\fp \in \Spec \mathcal{O}_K$. \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 $B / A$. + If $A$ is a normal domain, then going-down holds for $B / A$. \end{theorem} \begin{proof} @@ -2218,7 +2059,7 @@ extensions: 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 $L$. + Let $C$ be the integral closure of $A$ in $L$. Then $B \subseteq C$ and $C / B$ is integral. \[ \begin{tikzcd} @@ -2235,40 +2076,36 @@ extensions: \end{claim} \begin{subproof} Let $\fp \subseteq \tilde \fp$ 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. + $\tilde \fr \in \Spec C$ 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 surjective. 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 $\tilde \fr' - \cap A = \tilde \fp, \fr' \subseteq \tilde \fr'$. - 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 $\fr \subseteq \tilde - \fr$ and $\fr - \cap A = \fp$. + By going up for $C / A$ there is $\tilde \fr' \in \Spec C$ + with $\tilde \fr' \cap A = \tilde \fp, \fr' \subseteq \tilde \fr'$. + 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 $\fr \subseteq \tilde \fr$ + and $\fr \cap A = \fp$. \end{subproof} - If $\fp \subseteq \tilde \fp$ 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 $\fr \subseteq - \tilde \fr$ and $\fr \cap A = \fp$. - Then $\fq \coloneqq \fr \cap B \in \Spec B, \fq \subseteq \tilde \fq$ and $\fq - \cap A = \fp$. + If $\fp \subseteq \tilde \fp$ 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 $\fr \subseteq \tilde \fr$ + and $\fr \cap A = \fp$. + Then $\fq \coloneqq \fr \cap B \in \Spec B, \fq \subseteq \tilde \fq$ + 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 - $\mathfrak{k}(\fp)$, the integral closure of $A / \fp$ in $L$ is a + $\mathfrak{k}(\fp)$, 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. @@ -2279,84 +2116,73 @@ extensions: Japanese. \end{remark} -\begin{example} - +[Counterexample to going down] +\begin{align*}+[Counterexample to going down] Let $R = \mathfrak{k}[X,Y]$ and $A = \mathfrak{k}[X,Y, \frac{X}{Y}]$. Then going down does not hold for $A / R$: - For any ideal $Y \in \fq \subseteq A$ we have $X = \frac{X}{Y} - \cdot Y \in - \fq$. + For any ideal $Y \in \fq \subseteq A$ we have + $X = \frac{X}{Y} \cdot Y \in \fq$. Consider $(Y)_R \subsetneq (X,Y)_R \subseteq \fq \cap R$. - 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 + 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{example} +\end{align*} -\subsubsection{Proof of \texorpdfstring{$\codim(\{y\},Y) = \trdeg(\mathfrak{K}(Y) / - \mathfrak{k})$}{codim(\{y\},Y) = trdeg(K(Y) /k)}} +\subsubsection{Proof of \texorpdfstring{$\codim(\{y\},Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$}{codim(\{y\},Y) = trdeg(K(Y) /k)}} \label{proofcodimletrdeg} -This is part of the proof of -\ref{trdegandkdim}. +This is part of the proof of \ref{trdegandkdim}. %TODO: reorder \begin{proof} % DIMT - Let $B = \mathfrak{k}[X_1,\ldots,X_n]$ and $X \subseteq Y = V(\fp) \subseteq - \mathfrak{k}^n$ irreducible closed subsets of - $\mathfrak{k}^n$. - We want to show that $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$. - $\le $ was shown in - \ref{upperboundcodim}. - $\dim Y \ge \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$ 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 $f_i$ + Let $B = \mathfrak{k}[X_1,\ldots,X_n]$ and + $X \subseteq Y = V(\fp) \subseteq \mathfrak{k}^n$ + irreducible closed subsets of $\mathfrak{k}^n$. + We want to show that + $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) % + - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$. + $\le $ was shown in \ref{upperboundcodim}. + $\dim Y \ge \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$ 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 - $\mathfrak{k}^d \supsetneq \{0\} \times \mathfrak{k}^{n-1} - \supsetneq \ldots - \supsetneq \{0\}$ under $Y \xrightarrow{F = (f_1,\ldots,f_d)} \mathfrak{k}^d$ + We then used going-up to lift a chain of prime ideals corresponding to + $\mathfrak{k}^d \supsetneq \{0\} \times \mathfrak{k}^{n-1} % + \supsetneq \ldots \supsetneq \{0\}$ + under $Y \xrightarrow{F = (f_1,\ldots,f_d)} \mathfrak{k}^d$ 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 $\{0\} \in \mathfrak{k}^d$ has several preimages - under $F$ + In particular, when $\{0\} \in \mathfrak{k}^d$ + 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 \[ - \codim(\{y\}, Y) \le d = + \codim(\{y\}, Y) \le d = \trdeg(\mathfrak{K}(Y) \setminus \mathfrak{k}) \] - (see - \ref{upperboundcodim}) - equality holds for at least one pint $y \in - F^{-1}(\{0\})$ but cannot rule out - that there are other $y \in F^{-1}(\{0\})$ 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 $\codim(X,Y) \ge \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$ can be derived similarly to - \ref{upperboundcodim}. + (see \ref{upperboundcodim}) + equality holds for at least one pint $y \in F^{-1}(\{0\})$ + but cannot rule out that there are other $y \in F^{-1}(\{0\})$ + 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 + $\codim(X,Y) \ge \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) % + - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$ + can be derived similarly to \ref{upperboundcodim}. Thus \[ \codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / \mathfrak{k}) \] - follows (see - \ref{htandcodim} and - \ref{htandtrdeg}). + 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 @@ -2365,68 +2191,63 @@ This is part of the proof of - % Lecture 09 % i = ic \subsection{The height of a prime ideal} -In order to complete the proof of -\ref{proofcodimletrdeg} and show -$\codim(X,Y) -= \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / -\mathfrak{k})$, we need to localize -the $\mathfrak{k}$-algebra with respect to +In order to complete the proof of \ref{proofcodimletrdeg} +and show $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$, +we need to localize the $\mathfrak{k}$-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. +\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 $A = \mathfrak{k}[X_1,\ldots,X_n]$, $X = V(\fp)$ for a prime ideal $\fp$. - By the correspondence between irreducible subsets of - $\mathfrak{k}^n$ and prime - ideals in $A$ ( - \ref{bijiredprim}), the $\fp_i$ correspond to irreducible - subsets $X_i \subseteq \mathfrak{k}^n$ containing $X$. + By the correspondence between irreducible subsets of $\mathfrak{k}^n$ + and prime ideals in $A$ (\ref{bijiredprim}), + the $\fp_i$ correspond to irreducible subsets + $X_i \subseteq \mathfrak{k}^n$ containing $X$. Thus $\hght(\fp) = \codim(X, \mathfrak{k}^n)$. \end{example} \begin{example} \label{htandcodim} - Let $B = \mathfrak{k}[X_1,\ldots,X_n], \fq \in \Spec B$ and let $A \coloneqq B - / \fp$. - Let $Y \coloneqq V(\fq) \subseteq \mathfrak{k}^n$, $\tilde \fp - \coloneqq - \pi_{B, \fq}^{-1}(\fp)$, 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 $\fr - \subseteq \fp$ of $A$ contained in $\fp$ and the prime ideals and the prime - ideals $\tilde \fr \in \Spec B$ with $\fq \subseteq \tilde \fr \subseteq \tilde - \fp$: + Let $B = \mathfrak{k}[X_1,\ldots,X_n], \fq \in \Spec B$ + and let $A \coloneqq B / \fp$. + Let $Y \coloneqq V(\fq) \subseteq \mathfrak{k}^n$, + $\tilde \fp \coloneqq \pi_{B, \fq}^{-1}(\fp)$, + 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 + $\fr \subseteq \fp$ of $A$ contained in $\fp$ and the prime ideals + and the prime ideals $\tilde \fr \in \Spec B$ with + $\fq \subseteq \tilde \fr \subseteq \tilde \fp$: \begin{align*} f: \{\fr \in \Spec A | \fr \subseteq \fp \} - & \longrightarrow \{\tilde \fr \in \Spec B | \fq \subseteq \tilde \fr \subseteq - \tilde \fp\} \\ \fr & \longmapsto \pi_{B, \fq}^{-1}(\fr)\\ \tilde \fr / \fq - & \longmapsfrom \tilde \fr + & \longrightarrow \{\tilde \fr \in \Spec B | \fq \subseteq \tilde \fr \subseteq \tilde \fp\}\\ + \fr & \longmapsto \pi_{B, \fq}^{-1}(\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 - $X$. + 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 $X = X_0 \subsetneq X_1 \subsetneq \ldots - \subsetneq X_k \subseteq Y$ of irreducible subsets and - $\hght(\fp) = - \codim(X,Y)$. + canonical bijection with the chains + $X = X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_k \subseteq Y$ + of irreducible subsets and $\hght(\fp) = \codim(X,Y)$. \end{example} @@ -2435,235 +2256,189 @@ following: One can show that there is a bijection between $\Spec A$ and the set of irreducible subsets $Y \subseteq \Spec A$: \begin{align*} - f: \Spec A - & \longrightarrow \{Y \subseteq \Spec A | Y\text{irreducible}\} \\ \fp - & \longmapsto \Vs(\fp) \\ \bigcup_{\fp \in Y} \fp & \longmapsfrom Y + f: \Spec A & + \longrightarrow \{Y \subseteq \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 $V(\fp) = X_0 \subsetneq X_1 \subsetneq - \ldots \subsetneq X_k \subseteq \Spec A$ of irreducible subsets, and - $\hght(\fp) = \codim(V(\fp), \Spec A)$. + canonical bijection with the chains + $V(\fp) = X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_k \subseteq \Spec A$ + 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 - $\mathfrak{l}$ be an arbitrary field, $A$ a - $\mathfrak{l}$-algebra of finite - type which is a domain, $K \coloneqq Q(A)$ the field of quotients and let - $(a_i)_{i=1}^n$ be + \label{extendtotrbase} + Let $\mathfrak{l}$ be an arbitrary field, + $A$ a $\mathfrak{l}$-algebra of finite type which is a domain, + $K \coloneqq Q(A)$ the field of quotients and let $(a_i)_{i=1}^n$ be $\mathfrak{l}$-algebraically independent elements of $A$. Then there exist a natural number $m \ge n$ and a transcendence base - $(a_i)_{i - = 1}^m$ for $K / - \mathfrak{l}$ with $a_i \in A$ for $1 \le i \le m$. + $(a_i)_{i = 1}^m$ for $K / \mathfrak{l}$ + 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}. +\begin{align*} + 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 $\mathfrak{l}$-algebra - $A$. + $(a_i)_{i = n+1}^m \in A^{m-n}$ which generate $K$ + in the sense of a matroid used in the definition of $\trdeg$. + For instance, one can use generators of the $\mathfrak{l}$-algebra $A$. We assume $m$ to be minimal and claim that - $(a_i)_{i=1}^m$ are - $\mathfrak{l}$-algebraically independent. + $(a_i)_{i=1}^m$ are $\mathfrak{l}$-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 $\mathfrak{l}$ and the - $(a_i)_{i=1}^{j-1}$. - We have $j > n$ by the algebraic independence of - $(a_i)_{i=1}^n$. + over the subfield of $K$ generated by $\mathfrak{l}$ + 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 - $\mathfrak{l}$ and the - $(a_i)_{i=1}^{m-1} $, contradicting the minimality - of $m$. -\end{proof} + But then $K$ is algebraic over its subfield generated by $\mathfrak{l}$ + and the $(a_i)_{i=1}^{m-1}$, + contradicting the minimality of $m$. +\end{align*} \begin{theorem} \label{htandtrdeg} - Let $\mathfrak{l}$ be an arbitrary field, $A$ a - $\mathfrak{l}$-algebra of - finite type which is a domain, and $\fp \in \Spec A$. + Let $\mathfrak{l}$ be an arbitrary field, + $A$ a $\mathfrak{l}$-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 \[ - \hght(\fp) = \trdeg(K /\mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / - \mathfrak{l}) + \hght(\fp) = \trdeg(K /\mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}). \] \end{theorem} \begin{remark} - By example - \ref{htandcodim}, - theorem - \ref{trdegandkdim} is a special case of this theorem. + 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 $\trdeg(\mathfrak{k}(\fp_i) / \mathfrak{l}) < - \trdeg(\mathfrak{k}(\fp_{i+1}) / \mathfrak{l})$ by - \ref{trdegresfield} (``A - first result of dimension theory''). + of prime ideals in $A$, + we have $\trdeg(\mathfrak{k}(\fp_i) / \mathfrak{l}) % + < \trdeg(\mathfrak{k}(\fp_{i+1}) / \mathfrak{l})$ + by \ref{trdegresfield} (``A first result of dimension theory''). Thus \[ - k \le \trdeg(\mathfrak{k}(\fp_k) / \mathfrak{l}) - - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) \le \trdeg(K / \mathfrak{l}) - - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) + k \le \trdeg(\mathfrak{k}(\fp_k) / \mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) + \le \trdeg(K / \mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}), \] - where the last inequality is - another application of \ref{trdegresfield} (using $K = Q(A) = Q(A / \{0\}) = - \mathfrak{k}(\{0\})$ and the fact that $\{0\} \subseteq \fp_k$ is a prime - ideal). + where the last inequality is another application of \ref{trdegresfield} + (using $K = Q(A) = Q(A / \{0\}) = \mathfrak{k}(\{0\})$ + and the fact that $\{0\} \subseteq \fp_k$ is a prime ideal). Hence \[ - \hght(\fp) \le \trdeg( K / \mathfrak{l}) - - \trdeg(\mathfrak{k}(\fp) / - \mathfrak{l}) + \hght(\fp) \le \trdeg( K / \mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) \] and it remains to show the opposite inequality. \begin{claim} For any maximal ideal $\fp \in \MaxSpec A$ \[ - \hght(\mathfrak{m}) \ge \trdeg(K - / \mathfrak{l}) + \hght(\mathfrak{m}) \ge \trdeg(K / \mathfrak{l}). \] \end{claim} \begin{subproof} - By the Noether normalization - theorem ( - \ref{noenort}), there are - $(x_i)_{i=1}^d \in A^d$ which are + By the Noether normalization theorem (\ref{noenort}), + there are $(x_i)_{i=1}^d \in A^d$ which are algebraically independent over $\mathfrak{l}$ such that $A$ is - finite over the - subalgebra $S$ generated by the $x_i$. + finite over the subalgebra $S$ generated by the $x_i$. We have $d = \trdeg(K / \mathfrak{l})$ as the $x_i$ form a transcendence base of $K / \mathfrak{l}$. \begin{claim} - We can choose $x_i \in \mathfrak{m}$ + We can choose $x_i \in \mathfrak{m}$. \end{claim} \begin{subproof} - By the - Nullstellensatz ( - \ref{hns2}), $\mathfrak{k}(\mathfrak{m}) = A / - \mathfrak{m}$ + By the Nullstellensatz (\ref{hns2}), + $\mathfrak{k}(\mathfrak{m}) = A / \mathfrak{m}$ is a finite field extension of $\mathfrak{l}$. - Hence there exists a normed polynomial $P_i \in \mathfrak{l}[T]$ with + Hence there exists a normed polynomial $P_i \in \mathfrak{l}[T]$ with $P_i(x_i \mod \mathfrak{m}) = 0$ in $\mathfrak{k}(\mathfrak{m})$. - Let $\tilde x_i \coloneqq P_i(x_i) \in \mathfrak{m}$ 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 $x_i \in - \mathfrak{m}$. - - + Let $\tilde x_i \coloneqq P_i(x_i) \in \mathfrak{m}$ + 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 $x_i \in \mathfrak{m}$. \end{subproof} % TODO: fix names A_1 = A_S, k_1 = R_S - The ring homomorphism $\ev_x : R = \mathfrak{l}[X_1,\ldots,X_d] - \xrightarrow{P - \mapsto P(x_1,\ldots,x_d)} A$ 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 $\fp_i \subseteq R$ be the ideal generated by - $(X_j)_{j=i+1}^d$. - We have $\mathfrak{m} \sqcap R = \fp_0$ as all $X_i \in - \mathfrak{m}$, hence - $X_i \in \mathfrak{m} \sqcap R$ and $\fp_0$ is a maximal ideal. + The ring homomorphism + $\ev_x : R = \mathfrak{l}[X_1,\ldots,X_d] \xrightarrow{P \mapsto P(x_1,\ldots,x_d)} A$ + 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 $\fp_i \subseteq R$ be the ideal generated by $(X_j)_{j=i+1}^d$. + We have $\mathfrak{m} \sqcap R = \fp_0$ as all $X_i \in \mathfrak{m}$, + hence $X_i \in \mathfrak{m} \sqcap R$ and $\fp_0$ is a maximal ideal. By applying going-down and induction on $i$, there is a chain - $\mathfrak{m} = - \fq_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_d$ of elements of - $\Spec A$ such that $\fq_i \sqcap R = \fp_i$. - It follows that $\hght(\mathfrak{m}) \ge d$. + $\mathfrak{m} = \fq_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_d$ + of elements of $\Spec A$ such that $\fq_i \sqcap R = \fp_i$. + It follows that $\hght(\mathfrak{m}) \ge d$. \end{subproof} This finishes the proof in the case of $\fp \in \MaxSpec A$. To reduce the general case to that special case, we proceed as in - \ref{trdegresfield}: By lemma - \ref{ltrdegresfieldtrbase} there are + \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 $\mathfrak{k}(\fp) / \mathfrak{l}$. - As these images are $\mathfrak{l}$-algebraically independent, the - same holds - for the $a_i$ themselves. + As these images are $\mathfrak{l}$-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 $K / \mathfrak{l}$. - Let $R \subseteq A$ denote the $\mathfrak{l}$-subalgebra generated - by - $a_1,\ldots,a_n$ and let $S \coloneqq R \setminus \{0\}$. + 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 $K / \mathfrak{l}$. + Let $R \subseteq A$ denote the $\mathfrak{l}$-subalgebra generated + by $a_1,\ldots,a_n$ and let $S \coloneqq R \setminus \{0\}$. 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}, $\mathfrak{k}(\fp_S) \cong - \mathfrak{k}(\fp)$ 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 $\fp_S \in \MaxSpec(A_1)$ and the special case can be applied - to $\fp_S$ - and $A_1 / \mathfrak{l}_1$, showing that $\hght(\fp_S) - \ge e = \trdeg(K / - \mathfrak{l}_1)$. - We have $\trdeg(K / \mathfrak{l}_1) = m - n$, as - $(a_i)_{i = n+1}^m$ is a + 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}, + $\mathfrak{k}(\fp_S) \cong \mathfrak{k}(\fp)$ 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 $\fp_S \in \MaxSpec(A_1)$ + and the special case can be applied to $\fp_S$ and $A_1 / \mathfrak{l}_1$, + showing that $\hght(\fp_S) \ge e = \trdeg(K / \mathfrak{l}_1)$. + We have $\trdeg(K / \mathfrak{l}_1) = m - n$, + as $(a_i)_{i = n+1}^m$ is a transcendence base for $K / \mathfrak{l}_1$. - 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 + 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 $\dim A = \sup_{\fp \in \Spec A} \hght(\fp) = - \sup_{\mathfrak{m} \in - \MaxSpec A} \hght(\mathfrak{m})$, the Krull dimension of the - Noetherian - topological space $\Spec A$ may nevertheless be infinite. + 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 $\dim A = \sup_{\fp \in \Spec A} \hght(\fp) % + = \sup_{\mathfrak{m} \in \MaxSpec A} \hght(\mathfrak{m})$, + the Krull dimension of the Noetherian topological space $\Spec A$ + may nevertheless be infinite. \end{remark} -\begin{example} - +[Noetherian ring with infinite dimension]\footnote{\url{https://math.stackexchange.com/questions/1109732/noetherian-ring-with-infinite-krull-dimension-nagatas-example}} - Let $A = \mathfrak{k}[X_i | i \in \N]$ 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 $S \coloneqq A - \setminus \bigcup_{i \in \N} \fp_i$. +\begin{example}+% + [Noetherian ring with infinite dimension]% + \footnote{\url{https://math.stackexchange.com/questions/1109732/noetherian-ring-with-infinite-krull-dimension-nagatas-example}} + Let $A = \mathfrak{k}[X_i | i \in \N]$ + 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 $S \coloneqq A \setminus \bigcup_{i \in \N} \fp_i$. $S$ is multiplicatively closed. - $A_S$ is Noetherian but $\hght((\fp_i)_S) = m_{i+1}- - m_{i}$ hence $\dim(A_S) = \infty$. + $A_S$ is Noetherian but $\hght((\fp_i)_S) = m_{i+1}- m_{i}$ + hence $\dim(A_S) = \infty$. \end{example} % Lecture 10 @@ -2671,96 +2446,90 @@ We will use the following \subsection{Dimension of products} - \begin{proposition} \label{dimprod} - Let $X \subseteq \mathfrak{k}^n$ and $Y \subseteq - \mathfrak{k}^n$ be - irreducible and closed. - Then $X \times Y$ is also an irreducible closed subset of - $\mathfrak{k}^{m+n}$. - Moreover, $\dim(X \times Y) = \dim(X) + - \dim(Y)$ and $\codim(X \times Y, - \mathfrak{k}^{m+n}) = \codim(X, \mathfrak{k}^m) - + \codim(Y, \mathfrak{k}^n)$. + Let $X \subseteq \mathfrak{k}^n$ and $Y \subseteq \mathfrak{k}^n$ + be irreducible and closed. + Then $X \times Y$ is also an irreducible closed subset of $\mathfrak{k}^{m+n}$. + Moreover, $\dim(X \times Y) = \dim(X) + \dim(Y)$ + and $\codim(X \times Y, \mathfrak{k}^{m+n}) % + = \codim(X, \mathfrak{k}^m) + \codim(Y, \mathfrak{k}^n)$. \end{proposition} \begin{proof} - Let $X = V(\fp)$ and $Y = V(\fq)$ where $\fp \in \Spec - \mathfrak{k}[X_1,\ldots,X_m]$ and $\fq \in \Spec \mathfrak{k}[X_1,\ldots,X_n]$. - We denote points of $\mathfrak{k}^{m+n}$ as $x = (x',x'')$ with $x' \in - \mathfrak{k}^m, x''\in\mathfrak{k}^n$. - Then $X \times Y$ is the set of zeroes of the ideal in - $\mathfrak{k}[X_1,\ldots,X_{m+n}]$ generated by the polynomials $f(x) = - \phi(x')$, with $\phi$ running over $\fp$ and $g(x) = - \gamma(x'')$ with - $\gamma$ running over $\fq$. + Let $X = V(\fp)$ and $Y = V(\fq)$ + where $\fp \in \Spec \mathfrak{k}[X_1,\ldots,X_m]$ + and $\fq \in \Spec \mathfrak{k}[X_1,\ldots,X_n]$. + We denote points of $\mathfrak{k}^{m+n}$ as $x = (x',x'')$ + with $x' \in \mathfrak{k}^m, x''\in\mathfrak{k}^n$. + Then $X \times Y$ is the set of zeroes of the ideal in + $\mathfrak{k}[X_1,\ldots,X_{m+n}]$ 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 $\mathfrak{k}^{m+n}$. We must also show irreducibility. $X \times Y \neq \emptyset$ is obvious. - Assume that $X \times Y = A_1 \cup A_2$, where the $A_i \subseteq - \mathfrak{k}^{m+n}$ are closed. - For $x' \in \mathfrak{k}^m, x' \times Y$ is homeomorphic to the - irreducible - $Y$. - Thus $X = X_1 \cup X_2$ where $X_i = \{x \in X | \{x\} \times Y \subseteq - 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 $X \times Y$. + Assume that $X \times Y = A_1 \cup A_2$, + where the $A_i \subseteq \mathfrak{k}^{m+n}$ are closed. + For $x' \in \mathfrak{k}^m$, + $x' \times Y$ is homeomorphic to the irreducible $Y$. + Thus $X = X_1 \cup X_2$ + where $X_i = \{x \in X | \{x\} \times Y \subseteq 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 $X_0 \subsetneq X_1 \subsetneq \ldots - \subsetneq X_a = X$,$Y_0 \subsetneq Y_1 \subsetneq \ldots \subsetneq Y_b = Y$ + 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 $\codim(X \times Y, \mathfrak{k}^{m+n}) \ge \codim(X, - \mathfrak{k}^m) + - \codim(Y, \mathfrak{k}^n)$. - By - \ref{trdegandkdim} we have $\dim(A) + - \codim(A, \mathfrak{k}^l) = l$ for + 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 + \[ + \codim(X \times Y, \mathfrak{k}^{m+n}) % + \ge \codim(X, \mathfrak{k}^m) + \codim(Y, \mathfrak{k}^n). + \] + By \ref{trdegandkdim} we have $\dim(A) + \codim(A, \mathfrak{k}^l) = l$ for irreducible subsets of $\mathfrak{k}^l$. 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 $I$. + Let $\Vspec(I)$ denote the set of $\fp \in \Spec A$ containing $I$. \end{notation} \begin{proposition}[Nil radical] - For a ring $A$, $\bigcap_{\fp \in \Spec A} \fp = - \sqrt{\{0\} } = \{a \in A | \exists k \in \N ~ a^k = 0\} - \text{\reflectbox{$\coloneqq$}} \nil(A)$, the set of nilpotent elements - of $A$. + For a ring $A$, + $\bigcap_{\fp \in \Spec A} \fp = \sqrt{\{0\} } = \{a \in A | \exists k \in \N ~ a^k = 0\} \text{\reflectbox{$\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 $a \in A \setminus \sqrt{\{0\} }$. - Then $S = a^{\N}$ is a multiplicative subset of $A$ - not containing $0$. + It is clear that elements of $\sqrt{\{0\}}$ must belong to all prime ideals. + Conversely, let $a \in A \setminus \sqrt{\{0\}}$. + 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 $a \not\in \fp$. + If $\fq \in \Spec A_S$, then by the description of $\Spec A_S$ + (\autoref{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 $R$, $\sqrt{I} = \bigcap_{\fp \in \Vspec(I)} - \fp$. + 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 + 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}} @@ -2769,94 +2538,83 @@ We will use the following There is a bijection \begin{align*} f: \{A \subseteq \Spec R | A\text{ closed}\} - & \longrightarrow \{I \subseteq R | I \text{ ideal and } I = \sqrt{I} \} \\ A - & \longmapsto \bigcap_{\fp \in A} \fp \\ \Vspec(I) & \longmapsfrom I + & \longrightarrow \{I \subseteq 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 $\{\mathfrak{m}\}, \mathfrak{m} - \in \Spec A$ to the - maximal ideals. + Under this bijection, + the irreducible subsets correspond to the prime ideals and the closed points + $\{\mathfrak{m}\}, \mathfrak{m} \in \Spec A$ 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 $\sqrt{I} = I$ + 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 $f_k \in J_k \setminus I$. - Since $\Vspec(g) \supseteq \Vspec(f_k) \supseteq - \Vspec(I_k), \Vspec(I) - \subseteq \Vspec(g)$. + 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 $f_k \in J_k \setminus I$. + Since $\Vspec(g) \supseteq \Vspec(f_k) \supseteq \Vspec(I_k)$, + $\Vspec(I) \subseteq \Vspec(g)$. Hence $g \in \sqrt{I} = I$. - As $f_k \not\in I$, $I$ fails to be a prime ideal. + 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. + 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. + 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{example} - + - Let $A = \mathfrak{k}[X_n | n \in \N] / I$ 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\} $ +\begin{example}+ + Let $A = \mathfrak{k}[X_n | n \in \N] / I$m + 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. $A / J \cong \mathfrak{k}$, hence $J$ is maximal. - As every prime ideal must contain $\nil(A) \supseteq J$, $J$ is - the only prime - ideal. + 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{example} \begin{corollary}[About the smallest prime ideals containing $I$ ] \label{smallestprimesvi} - If $R$ is Noetherian and $I \subseteq R$ an ideal, then the set - $\Vspec(I) = - \{\fp \in \Spec R | I \subseteq \fp\}$ has finitely many $\subseteq$-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. + If $R$ is Noetherian and $I \subseteq R$ an ideal, + then the set $\Vspec(I) = \{\fp \in \Spec R | I \subseteq \fp\}$ + has finitely many $\subseteq$-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 + 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 $\fp_i \subseteq \fq$. - Also $\fp_i \in \Vspec(\fp_i) \subseteq \Vspec(I)$. + to at least one $\Vspec(\fp_i)$, + hence $\fp_i \subseteq \fq$. + Also $\fp_i \in \Vspec(\fp_i) \subseteq \Vspec(I)$. It follows that the sets of $\subseteq$-minimal elements of - $\Vspec(I)$ and of - $\{\fp_1,\ldots,\fp_k\} $ coincide. + $\Vspec(I)$ 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. + there are no non-trivial inclusions between the $\fp_i$ + and the assertion follows. The final remark is trivial. \end{proof} \begin{corollary} @@ -2866,39 +2624,38 @@ We will use the following \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 $\subseteq$-minimal element of $\Vspec(a)$. +\begin{theorem}[Principal ideal theorem / Hauptidealsatz]\label{pitheorem} + Let $A$ be a Noetherian ring, $a \in A$ + and $\fp \in \Spec A$ a $\subseteq$-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 $1$. + 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 + 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 $\subseteq$-minimal element - of $\bigcap_{i=1}^k V(a_i)$, the set of prime ideals containing all - $a_i$. +\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 $\subseteq$-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 $\hght(\fp) < \infty$. + If $R$ is a Noetherian ring and $\fp \in \Spec R$, + then $\hght(\fp) < \infty$. \end{corollary} \begin{proof} If $\fp$ is generated by $(f_i)_{i=1}^k$, @@ -2910,132 +2667,108 @@ this more general theorem. \label{smallestprimeandirredcomp} Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $I \subseteq R$ 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 $\Va(I)$. + 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) = + 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 $X \subseteq \mathfrak{k}^n$ be irreducible, - $(f_i)_{i=1}^k$ elements of $R - = \mathfrak{k}[X_1,\ldots,X_n]$ and $Y$ an irreducible component of $A = X \cap - \bigcap_{i=1}^k V(f_i)$. - Then $\codim(Y,X) \le k$. + $(f_i)_{i=1}^k$ elements of $R = \mathfrak{k}[X_1,\ldots,X_n]$ + 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 $k$. + 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 $I - \subseteq R$ 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 + If $X = v(\fp), X \cap \bigcap_{i=1}^k V(f_i) = V(I)$ + where $I \subseteq R$ 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 $\dim X$. + 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 $\mathfrak{k}^n$. - If $C$ is an irreducible component of $A \cap B$, then $\codim(C, - \mathfrak{k}^n) - \le \codim(A, \mathfrak{k}^n) + \codim(B, \mathfrak{k}^n)$. + If $C$ is an irreducible component of $A \cap B$, + then $\codim(C, \mathfrak{k}^n) \le \codim(A, \mathfrak{k}^n) + \codim(B, \mathfrak{k}^n)$. \end{corollary} -\begin{remark} - + +\begin{remark}+ Equivalently, $\dim(C) \ge \dim(A) + \dim(B)-n$. \end{remark} \begin{proof} - Let $X = A \times B \subseteq \mathfrak{k}^{2n}$, where we use - $(X_1,\ldots,X_n,Y_1,\ldots,Y_n)$ as coordinates of $\mathfrak{k}^{2n}$. - Let $\Delta \coloneqq \{(x_1,\ldots,x_n,x_1,\ldots,x_n) | x \in - \mathfrak{k}^n\} $ be the diagonal in $\mathfrak{k}^n - \times \mathfrak{k}^n$. - The projection $\mathfrak{k}^{2n}\to \mathfrak{k}^n$ to the - $X$-coordinates + Let $X = A \times B \subseteq \mathfrak{k}^{2n}$, + where we use $(X_1,\ldots,X_n,Y_1,\ldots,Y_n)$ + as coordinates of $\mathfrak{k}^{2n}$. + Let $\Delta \coloneqq \{(x_1,\ldots,x_n,x_1,\ldots,x_n) | x \in \mathfrak{k}^n\}$ + be the diagonal in $\mathfrak{k}^n \times \mathfrak{k}^n$. + The projection $\mathfrak{k}^{2n}\to \mathfrak{k}^n$ 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 + Thus, $C$ is homeomorphic to an irreducible component $C'$ of + $(A \times B) \cap \Delta$ and \begin{align*} - \codim(C, \mathfrak{k}^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,\mathfrak{k}^n) + \codim(B, \mathfrak{k}^n) + \codim(C, \mathfrak{k}^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,\mathfrak{k}^n) + \codim(B, \mathfrak{k}^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 + \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 + 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} 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 $\subseteq$-minimal element of the set of non-zero prime - ideals of $R$ } is a principal ideal. + Then $R$ is a UFD iff every $\fp \in \Spec R$ with $\hght(\fp)= 1$% + \footnote{in other words, every $\subseteq$-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.} + \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 $p \in \fp \setminus \{0\}$. + Let $p \in \fp \setminus \{0\}$. Replacing $p$ by a prime factor of $p$, we may assume $p$ to be prime. - Thus $\{0\} \subsetneq pR \subseteq \fp$ is a chain of prime ideals and since - $\hght(\fp) = 1$ it follows that $\fp = pR$. + Thus $\{0\} \subsetneq pR \subseteq \fp$ 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. + $\hght(\fp)=1$ is a principal ideal. Let $f \in R$ be irreducible. - Let $\fp \in \Spec R$ be a $\subseteq$-minimal element of $V(f)$. - By the principal ideal theorem ( - \ref{pitheorem}), - $\hght(\fp)=1$. + Let $\fp \in \Spec R$ be a $\subseteq$-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. @@ -3044,51 +2777,42 @@ this more general theorem. \subsection{The Jacobson radical} \begin{proposition} - For a ring $A, \bigcap_{\mathfrak{m} \in \MaxSpec A} \mathfrak{m} = \{a \in A | - \forall x \in A ~ 1 - ax \in A^{\times }\} - \text{\reflectbox{$\coloneqq$}} - \rad(A)$, the \vocab{Jacobson radical} of $A$. + For a ring $A$, + $\bigcap_{\mathfrak{m} \in \MaxSpec A} \mathfrak{m} = \{a \in A | \forall x \in A ~ 1 - ax \in A^{\times }\} \text{\reflectbox{$\coloneqq$}} \rad(A)$, + the \vocab{Jacobson radical} of $A$. \end{proposition} \begin{proof} - Suppose $\mathfrak{m} \in \MaxSpec A$ and $a \in A \setminus - \mathfrak{m}$. - Then $a \mod \mathfrak{m} \neq 0$ and $A / \mathfrak{m}$ - is a field. - Hence $a \mod \mathfrak{m}$ has an inverse $x \mod - \mathfrak{m}$. - $1 - ax \in \mathfrak{m}$, hence $1 - ax \not\in - A^{\times}$ and $a $ is not al element of the RHS. + Suppose $\mathfrak{m} \in \MaxSpec A$ and $a \in A \setminus \mathfrak{m}$. + Then $a \mod \mathfrak{m} \neq 0$ and $A / \mathfrak{m}$ is a field. + Hence $a \mod \mathfrak{m}$ has an inverse $x \mod \mathfrak{m}$. + $1 - ax \in \mathfrak{m}$, hence $1 - ax \not\in A^{\times}$ + and $a $ is not al element of the RHS. - Conversely, let $a \in A$ belong to all $\mathfrak{m} \in \MaxSpec - A$. - 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 - $\mathfrak{m}$. - As $a \in \mathfrak{m}, 1 = (1-ax) + ax \in - \mathfrak{m}$, a contradiction. + Conversely, let $a \in A$ belong to all $\mathfrak{m} \in \MaxSpec A$. + 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 $\mathfrak{m}$. + As $a \in \mathfrak{m}, 1 = (1-ax) + ax \in \mathfrak{m}$, a contradiction. Hence every element of $\bigcap_{\mathfrak{m} \in \MaxSpec A} \mathfrak{m}$ belongs to the right hand side. \end{proof} \begin{example} - If $A$ is a local ring, then $\rad(A) = - \mathfrak{m}_A$. + If $A$ is a local ring, then $\rad(A) = \mathfrak{m}_A$. \end{example} \begin{example} If $A$ is a PID with infinitely many multiplicative equivalence classes of - prime elements (e.g. $\Z$ of $\mathfrak{k}[X]$), then - $\rad(A) = \{0\}$: Prime - ideals of a PID are maximal. + prime elements (e.g. $\Z$ of $\mathfrak{k}[X]$), + 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 $f = \prod_{i=1}^{n} p_i$. + 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