diff --git a/inputs/finiteness_conditions.tex b/inputs/finiteness_conditions.tex index de9d9d4..e8d257f 100644 --- a/inputs/finiteness_conditions.tex +++ b/inputs/finiteness_conditions.tex @@ -5,31 +5,29 @@ Then the following sets coincide \begin{enumerate} \item - $\left\{ \sum_{s \in - S'} r_{s} \cdot s ~ |~ S - \subseteq S' \text{finite}, r_s \in R, \right\}$ + $\left\{ \sum_{s \in S'} r_{s} \cdot s ~ |~ S \subseteq S' \text{finite}, r_s \in R, \right\}$, \item - $\bigcap_{\substack{S \subseteq N \subseteq M\\N \text{submodule}}} N$ + $\bigcap_{\substack{S \subseteq N \subseteq M\\N \text{submodule}}} N$, \item - The $\subseteq$-smallest submodule of $M$ containing $S$ + The $\subseteq$-smallest submodule of $M$ containing $S$. \end{enumerate} - This subset of $N \subseteq M$ is called the \vocab[Module! - Submodule]{submodule of $M $ generated by $S$}. - If $N= M$ we say that \vocab[Module! - generated by subset $S$]{$ M$ is generated by $S$}. - $M$ is finitely generated $:\iff \exists S \subseteq M$ finite such that $M$ is - generated by $S$. + This subset of $N \subseteq M$ is called the + \vocab[Module!Submodule]{submodule of $M $ generated by $S$}. + If $N= M$ we say that + \vocab[Module!generated by subset $S$]{$ M$ is generated by $S$}. + $M$ is finitely generated + $:\iff \exists S \subseteq M$ finite such that $M$ is generated by $S$. \end{definition} \begin{definition}[Noetherian $R$-module] - $M$ is a \vocab{Noetherian} $R$-module if the - following equivalent conditions hold: + $M$ is a \vocab{Noetherian} $R$-module + if the following equivalent conditions hold: \begin{enumerate} \item Every submodule $N \subseteq M$ is finitely generated. \item - Every sequence $N_0 \subset N_1 \subset \ldots$ of submodules terminates + Every sequence $N_0 \subset N_1 \subset \ldots$ of submodules terminates. \item Every set $\mathfrak{M} \neq \emptyset$ of submodules of $M$ has a $\subseteq$-largest element. @@ -37,9 +35,9 @@ \end{definition} \begin{proposition}[Hilbert's Basissatz] \label{basissatz} - If $R$ is a Noetherian ring, then the polynomial rings - $R[X_1,\ldots, X_n]$ in - finitely many variables are Noetherian. + If $R$ is a Noetherian ring, + then the polynomial rings $R[X_1,\ldots, X_n]$ + in finitely many variables are Noetherian. \end{proposition} \subsubsection{Properties of finite generation and Noetherianness} @@ -61,8 +59,8 @@ By definition, $M$ is a submodule of itself. Thus it is finitely generated. \item - Since $M$ is finitely generated, there exists a surjective homomorphism $R^n - \to M$. + Since $M$ is finitely generated, + there exists a surjective homomorphism $R^n \to M$. As $R$ is Noetherian, $R^n$ is Noethrian as well. \item trivial @@ -73,49 +71,47 @@ Let $M, M', M''$ be $R$-modules. \begin{enumerate} \item - Suppose $M \xrightarrow{p} - M''$ is surjective. - If $M$ is finitely generated (resp. - Noetherian), then so is $M''$. + Suppose $M \xrightarrow{p} M''$ is surjective. + If $M$ is finitely generated (resp. Noetherian), + then so is $M''$. \item - Let $M' \xrightarrow{f} - M \xrightarrow{p} M'' \to 0$ be exact. - If $M'$ and $M ''$ are finitely generated (reps. - Noetherian), so is $M$. + Let $M' \xrightarrow{f} M \xrightarrow{p} M'' \to 0$ be exact. + If $M'$ and $M ''$ are finitely generated (reps. Noetherian), + so is $M$. \end{enumerate} \end{fact} \begin{proof} \begin{enumerate} \item Consider a sequence $M_0'' \subset M_1'' \subset \ldots \subset M''$. - Then $p^{-1} M_i''$ yields a strictly ascending - sequence. - If $M$ is generated by $S, |S| < \omega$, then $M''$ is generated by $p(S)$. + Then $p^{-1} M_i''$ yields a strictly ascending sequence. + If $M$ is generated by $S, + |S| < \omega$, then $M''$ is generated by $p(S)$. \item - Because of 1. we can replace $M'$ by $f(M')$ and assume $0 \to M' - \xrightarrow{f} - M \xrightarrow{p} M'' \to 0$ to be exact. - The fact about finite generation follows from EInführung in die Algebra. - - If $M', M''$ are Noetherian, $N \subseteq M$ a submodule, then $N' \coloneqq - f^{-1}(N)$ and $N''\coloneqq p(N)$ are finitely - generated. - Since $0 \to N' \to N \to N'' \to 0$ is exact, $N$ is finitely generated. + Because of 1.~we can replace $M'$ by $f(M')$ + and assume $0 \to M' \xrightarrow{f} M \xrightarrow{p} M'' \to 0$ + to be exact. + The fact about finite generation follows from + Einführung in die Algebra. + If $M', M''$ are Noetherian, $N \subseteq M$ a submodule, + then $N' \coloneqq f^{-1}(N)$ and $N''\coloneqq p(N)$ + are finitely generated. + Since $0 \to N' \to N \to N'' \to 0$ is exact, + $N$ is finitely generated. \end{enumerate} \end{proof} \subsection{Ring extensions of finite type} \begin{definition}[$R$-algebra] Let $R$ be a ring. - An $R$-algebra $(A, \alpha)$ is a ring $A$ with a ring homomorphism $R - \xrightarrow{\alpha} A$. + An $R$-algebra $(A, \alpha)$ is a ring $A$ + with a ring homomorphism $R \xrightarrow{\alpha} A$. $\alpha$ will usually be omitted. In general $\alpha$ is not assumed to be injective. \\ \\ - An $R$-subalgebra is a subring $\alpha(R) \subseteq A' \subseteq - A$.\\ + An $R$-subalgebra is a subring $\alpha(R) \subseteq A' \subseteq A$.\\ A morphism of $R$-algebras $A \xrightarrow{f} \tilde{A}$ is a ring homomorphism with $\tilde{\alpha} = f \alpha$. \end{definition} @@ -123,9 +119,9 @@ \begin{definition}[Generated (sub)algebra, algebra of finite type] Let $(A, \alpha)$ be an $R$-algebra. \begin{align*} - \alpha: R[X_1,\ldots,X_m] & \longrightarrow A[X_1,\ldots,X_m] \\ - P = \sum_{\beta \in \N^m} p_\beta X^{\beta} & \longmapsto \sum_{\beta \in \N^m} \alpha(p_\beta) - X^{\beta} + \alpha: R[X_1,\ldots,X_m] & \longrightarrow A[X_1,\ldots,X_m]\\ + P = \sum_{\beta \in \N^m} p_\beta X^{\beta} + & \longmapsto \sum_{\beta \in \N^m} \alpha(p_\beta) X^{\beta} \end{align*} is a ring homomorphism. We will sometimes write $P(a_1,\ldots,a_m)$ instead of @@ -134,33 +130,31 @@ Fix $a_1,\ldots,a_m \in A^m$. Then we get a ring homomorphism $R[X_1,\ldots,X_m] \to A$. The image of this ring homomorphism is the $R$-subalgebra of $A$ - \vocab[Algebra! - generated subalgebra]{generated by the $a_i$}. - $A$ is \vocab[Algebra!of finite type]{of finite type} if it can be generated by finitely many $a_i - \in I$. + \vocab[Algebra!generated subalgebra]{generated by the $a_i$}. + $A$ is \vocab[Algebra!of finite type]{of finite type} + if it can be generated by finitely many $a_i \in I$. For arbitrary $S \subseteq A$ the subalgebra generated by $S$ is the - intersection of all subalgebras containing $S$ \\ $=$ the union of subalgebras - generated by finite $S' \subseteq S$\\ $= $ the image of - $R[X_s | s \in S]$ - under $P \mapsto (\alpha(P))(S)$. - + intersection of all subalgebras containing $S$ \\ + $=$ the union of subalgebras generated by finite $S' \subseteq S$\\ + $= $ the image of $R[X_s | s \in S]$ under $P \mapsto (\alpha(P))(S)$. \end{definition} \subsection{Finite ring extensions} % LECTURE 2 \begin{definition}[Finite ring extension] - Let $R$ be a ring and $A$ an $R$-algebra. $A$ is a - module over itself and the ringhomomorphism $R \to A$ allows us to derive an - $R$-module structure on $A$. - $A$ \vocab[Algebra!finite over]{is finite over} $R$ / the $R$-algebra $A$ is finite / $A / R$ is - finite if $A$ is finitely generated as an $R$-module. + Let $R$ be a ring and $A$ an $R$-algebra. + $A$ is a module over itself and the ringhomomorphism $R \to A$ + allows us to derive an $R$-module structure on $A$. + $A$ \vocab[Algebra!finite over]{is finite over} $R$ / + the $R$-algebra $A$ is finite / $A / R$ is finite + if $A$ is finitely generated as an $R$-module. \end{definition} \begin{fact}[Basic properties of finiteness] \begin{enumerate}[A] \item Every ring is finite over itself. \item - A field extension is finite as a ring extension iff it is finite as a field - extension. + A field extension is finite as a ring extension + iff it is finite as a field extension. \item $A$ finite $\implies$ $A$ of finite type. \item @@ -177,16 +171,14 @@ Let $A $ be generated by $a_1,\ldots,a_n$ as an $R$-module. Then $A$ is generated by $a_1,\ldots,a_n$ as an $R$-algebra. \item - Let $A$ be generated by $a_1,\ldots,a_m$ as an $R$-module and $B$ by - $b_1,\ldots,b_n$ as an $A$-module. - For every $b$ there exist $\alpha_j \in A$ such that $b = - \sum_{j=1}^{n} - \alpha_j b_j$. - We have $\alpha_j = \sum_{i=1}^{m} \rho_{ij} a_i$ for some - $\rho_{ij} \in R$ - thus $b = \sum_{i=1}^{m} \sum_{j=1}^{n} \rho_{ij} - a_i b_j$ and the $a_ib_j$ - generate $B$ as an $R$-module. + Let $A$ be generated by $a_1,\ldots,a_m$ as an $R$-module + and $B$ by $b_1,\ldots,b_n$ as an $A$-module. + For every $b$ there exist $\alpha_j \in A$ + such that $b = \sum_{j=1}^{n} \alpha_j b_j$. + We have $\alpha_j = \sum_{i=1}^{m} \rho_{ij} a_i$ + for some $\rho_{ij} \in R$ + thus $b = \sum_{i=1}^{m} \sum_{j=1}^{n} \rho_{ij} a_i b_j$ + and the $a_ib_j$ generate $B$ as an $R$-module. \end{enumerate} \end{proof} @@ -194,39 +186,35 @@ \subsection{Determinants and Caley-Hamilton} %LECTURE 2 TODO: move to int. elements? This generalizes some facts about matrices to matrices with elements from commutative rings with $1$. -\footnote{Most of this even works in commutative rings without $ 1$, since $1$ simply can be adjoined.} +\footnote{Most of this even works in commutative rings without $ 1$, + since $1$ simply can be adjoined.} \begin{definition}[Determinant] - Let $A = (a_{ij}) - \Mat(n,n,R)$. + Let $A = (a_{ij}) \in \Mat(n,n,R)$. We define the determinant by the Leibniz formula \[ - \det(A) \coloneqq \sum_{\pi - \in S_n} \sgn(\pi) - \prod_{i=1}^{n} a_{i, \pi(i)} + \det(A) \coloneqq \sum_{\pi \in S_n} \sgn(\pi) \prod_{i=1}^{n} a_{i, \pi(i)}. \] - Define $\text{Adj}(A)$ by $\text{Adj}(A)^{T}_{ij} - \coloneqq (-1)^{i+j} \cdot - M_{ij}$, where $M_{ij}$ is the - determinant of the matrix resulting from $A$ - after deleting the $i^{\text{th}}$ row and the - $j^{\text{th}}$ column. + Define $\text{Adj}(A)$ by $\text{Adj}(A)^{T}_{ij} \coloneqq (-1)^{i+j} \cdot M_{ij}$, + where $M_{ij}$ is the determinant of the matrix resulting from $A$ + after deleting the $i^{\text{th}}$ row and the $j^{\text{th}}$ column. \end{definition} \begin{fact} \begin{enumerate} \item - $\det(AB) = \det(A)\det(B)$ + $\det(AB) = \det(A)\det(B)$. \item Development along a row or column works. \item - Cramer's rule: $A \cdot \text{Adj}(A) = \text{Adj}(A) \cdot - A = \det(A) \cdot \mathbf{1}_n$. $A$ is invertible - iff $\det(A)$ is a unit. + Cramer's rule: + $A \cdot \text{Adj}(A) = \text{Adj}(A) \cdot A = \det(A) \cdot \mathbf{1}_n$. + $A$ is invertible iff $\det(A)$ is a unit. \item - Caley-Hamilton: If $P_A = \det(T \cdot \mathbf{1}_n - A)$ \footnote{$T \cdot \mathbf{1}_n -A \in \Mat(n,n,A[T])$}, then - $P_A(A) = 0$. + Caley-Hamilton: + If $P_A = \det(T \cdot \mathbf{1}_n - A)$% + \footnote{$T \cdot \mathbf{1}_n -A \in \Mat(n,n,A[T])$}, + then $P_A(A) = 0$. \end{enumerate} - \end{fact} \begin{proof} All rules hold for the image of a matrix under a ring homomorphism if they hold @@ -234,20 +222,17 @@ commutative rings with $1$. The converse holds in the case of injective ring homomorphisms. Caley-Hamilton was shown for algebraically closed fields in LA2 using the Jordan normal form. - Fields can be embedded into their algebraic closure, thus Caley-Hamilton holds - for fields. + Fields can be embedded into their algebraic closure, + thus Caley-Hamilton holds for fields. Every domain can be embedded in its field of quotients $\implies$ Caley-Hamilton holds for domains. - In general, $A$ is the image of - $(X_{i,j})_{i,j = 1}^{n} \in - \Mat(n,n,S)$ where - $S \coloneqq \Z[X_{i,j} | 1 \le i, j \le n]$ (this is a domain) under the - morphism $S \to A$ of evaluation defined by $X_{i,j} - \mapsto a_{i,j}$. + In general, $A$ is the image of $(X_{i,j})_{i,j = 1}^{n} \in \Mat(n,n,S)$ + where $S \coloneqq \Z[X_{i,j} | 1 \le i, j \le n]$ (this is a domain) + under the morphism $S \to A$ of evaluation defined by + $X_{i,j} \mapsto a_{i,j}$. Thus Caley-Hamilton holds in general. \end{proof} -%TODO: lernen \subsection{Integral elements and integral ring extensions} %LECTURE 2 \begin{proposition}[on integral elements] @@ -256,88 +241,77 @@ commutative rings with $1$. Then the following are equivalent: \begin{enumerate}[A] \item - $\exists n \in - \N, (r_i)_{i=0}^{n-1}, r_i \in R: a^n = - \sum_{i=0}^{n-1} r_i a^i$ + $\exists n \in \N, (r_i)_{i=0}^{n-1}, r_i \in R: a^n = \sum_{i=0}^{n-1} r_i a^i$. \item - There - exists a subalgebra $B \subseteq A$ finite over $R$ and containing $a$. + There exists a subalgebra $B \subseteq A$ finite over $R$ and containing $a$. \end{enumerate} - If $a_1, \ldots, a_k \in A$ satisfy these conditions, there is a subalgebra of - $A$ finite over $R$ and containing all $a_i$. + If $a_1, \ldots, a_k \in A$ satisfy these conditions, + there is a subalgebra of $A$ finite over $R$ and containing all $a_i$. \end{proposition} \begin{definition} \label{intclosure} - Elements that satisfy the conditions from - \ref{propinte} are called + Elements that satisfy the conditions from \ref{propinte} are called \vocab{integral over} $R$. - $A / R$ is \vocab[Algebra!integral]{integral}, if all $a \in A$ are integral over $R$. + $A / R$ is \vocab[Algebra!integral]{integral}, + if all $a \in A$ are integral over $R$. The set of elements of $A$ integral over $R$ is called the - \vocab{integral - closure} of $R$ in $A$. + \vocab{integral closure} of $R$ in $A$. \end{definition} \begin{proof} \hskip 10pt \begin{enumerate} - {\color{gray} - \item[B $\implies$ A] + {\color{gray} \item[B $\implies$ A] Let $a \in A$ such that there is a subalgebra $B \subseteq A$ containing $a$ and finite over $R$. - Let $(b_i)_{i=1}^{n}$ generate $B$ as an - $R$-module. + Let $(b_i)_{i=1}^{n}$ generate $B$ as an $R$-module. \begin{align*} - q: R^n & \longrightarrow B \\ + q: R^n & \longrightarrow B\\ (r_1,\ldots,r_n) & \longmapsto \sum_{i=1}^{n} r_i b_i \end{align*} is surjective. - Thus there are $\rho_{i} = \left( r_{i,j} \right)_{j=1}^n \in R^n$ - such that - $a b_i = q(\rho_i)$. + Thus there are $\rho_{i} = \left(r_{i,j}\right)_{j=1}^n \in R^n$ + such that $a b_i = q(\rho_i)$. Let $\mathfrak{A}$ be the matrix with the $\rho_i$ as columns. Then for all $v \in R^n: q(\mathfrak{A} \cdot v) = a \cdot q(v)$. - By induction it follows that $q(P(\mathfrak{A}) \cdot v) = P(a)q(v)$ - for all $P - \in R[T]$. - Applying this to $P(T) = \det(T\cdot \mathbf{1}_n - \mathfrak{A})$ and using - Caley-Hamilton, we obtain $P(a) \cdot q(v) = 0$. - $P$ is monic. - Since $q$ is surjective, we find $v \in R^{n} : q(v) = - 1$. - Thus $P(a) = 0$ and $a$ satisfies A. - } + By induction it follows that + $q(P(\mathfrak{A}) \cdot v) = P(a)q(v)$ for all $P \in R[T]$. + Applying this to $P(T) = \det(T\cdot \mathbf{1}_n - \mathfrak{A})$ + and using Caley-Hamilton, + we obtain $P(a) \cdot q(v) = 0$. $P$ is monic. + Since $q$ is surjective, we find $v \in R^{n} : q(v) = 1$. + Thus $P(a) = 0$ and $a$ satisfies A.} \item[B $\implies$ A] if $R$ is Noetherian.\footnote{This suffices in the exam.} Let $a \in A$ satisfy B. Let $B$ be a subalgebra of $A$ containing $b$ and finite over $R$. - Let $M_n \subseteq B$ be the $R$-submodule generated by the $a^i$ with $0 \le - i < n$. - As a finitely generated module over the Noetherian ring $R$, $B$ is a - Noetherian $R$-module. - Thus the ascending sequence $M_n$ stabilizes at some step $d$ and $a^d \in - M_d$. - Thus there are $(r_i)_{i=0}^{d-1} \in R^d$ such - that $a^d = \sum_{i=0}^{d-1} - r_ia^i$. + Let $M_n \subseteq B$ be the $R$-submodule generated by the $a^i$ + with $0 \le i < n$. + As a finitely generated module over the Noetherian ring $R$, + $B$ is a Noetherian $R$-module. + Thus the ascending sequence $M_n$ stabilizes at some step $d$ + and $a^d \in M_d$. + Thus there are $(r_i)_{i=0}^{d-1} \in R^d$ + such that $a^d = \sum_{i=0}^{d-1} r_ia^i$. \item[A $\implies$ B] - Let $a = (a_i)_{i=1}^n$ where all $a_i$ - satisfy A, i.e. $a_i^{d_i} = \sum_{j=0}^{d_i - 1} - r_{i,j}a_i^j$ with $r_{i,j} \in - R$. - Let $B \subseteq A$ be the sub-$R$-module generated by $a^\alpha = - \prod_{i=1}^n a_i^{\alpha_i}$ with $0 \le \alpha_i < d_i$. + Let $a = (a_i)_{i=1}^n$ where all $a_i$ satisfy A, + i.e.~$a_i^{d_i} = \sum_{j=0}^{d_i - 1} r_{i,j}a_i^j$ + with $r_{i,j} \in R$. + Let $B \subseteq A$ be the sub-$R$-module generated by + $a^\alpha = \prod_{i=1}^n a_i^{\alpha_i}$ + with $0 \le \alpha_i < d_i$. $B$ is closed under $a_1 \cdot $ since \[ a_1a^{\alpha} = \begin{cases} - a^{(\alpha_1 + 1, \alpha')} & \text{if } \alpha = (\alpha_1, \alpha'), 0 \le \alpha_1 < d_1 - 1 \\ - \sum_{j=0}^{d_1 - 1} r_{i_1,j} a^{(j, \alpha')} & \text{if } \alpha_1 = d_1 - 1 + a^{(\alpha_1 + 1, \alpha')} & \text{if } \alpha = (\alpha_1, \alpha'), 0 \le \alpha_1 < d_1 - 1, \\ + \sum_{j=0}^{d_1 - 1} r_{i_1,j} a^{(j, \alpha')} & \text{if } \alpha_1 = d_1 - 1. \end{cases} \] By symmetry, this hold for all $a_i$. - By induction on $|\alpha| = \sum_{i=1}^{n} \alpha_i$, $B$ is invariant - under - $a^{\alpha}\cdot $. - Since these generate $B$ as an $R$-module, $B$ is multiplicatively closed. + By induction on $|\alpha| = \sum_{i=1}^{n} \alpha_i$, + $B$ is invariant under $a^{\alpha}\cdot $. + Since these generate $B$ as an $R$-module, + $B$ is multiplicatively closed. Thus A holds. Furthermore we have shown the final assertion of the proposition. \end{enumerate} @@ -348,13 +322,15 @@ commutative rings with $1$. \item[Q] Every finite $R$-algebra $A$ is integral. \item[R] - The integral closure of $R$ in $A$ is an $R$-subalgebra of $A$ + The integral closure of $R$ in $A$ is an $R$-subalgebra of $A$. \item[S] - If $A$ is an $R$-algebra, $B$ an $A$-algebra and $b \in B$ integral over $R$, + If $A$ is an $R$-algebra, + $B$ an $A$-algebra and $b \in B$ integral over $R$, then it is integral over $A$. \item[T] - If $A$ is an integral $R$-algebra and $B$ any $A$-algebra, $b \in B$ integral - over $A$, then $b$ is integral over $R$. + If $A$ is an integral $R$-algebra and $B$ any $A$-algebra, + $b \in B$ integral over $A$, + then $b$ is integral over $R$. \end{enumerate} \end{corollary} \begin{proof} @@ -362,136 +338,126 @@ commutative rings with $1$. \item[Q] Put $ B = A $ in B. \item[R] - For every $r \in R$ $\alpha(r)$ is a solution to $T - r = 0$, hence integral - over $R$. - From B it follows, that the integral closure is closed under ring operations. + For every $r \in R$ $\alpha(r)$ is a solution to $T - r = 0$, + hence integral over $R$. + From B it follows, + that the integral closure is closed under ring operations. \item[S] trivial \item[T] Let $b \in B$ such that $b^n = \sum_{i=0}^{n-1} a_ib^{i}$. - Then there is a subalgebra $\tilde{A} \subseteq A$ finite over - $R$, such that - all $a_i \in \tilde{A}$. - $b$ is integral over $\tilde{A} \implies \exists - \tilde{B} \subseteq B$ finite over $\tilde{A}$ and - $b \in \tilde{B}$. - Since $\tilde{B} / \tilde{A} $ and - $\tilde{A} / R$ are finite, $\tilde{B} / R$ - is finite and $b$ satisfies B. + Then there is a subalgebra $\tilde{A} \subseteq A$ finite over $R$, + such that all $a_i \in \tilde{A}$. + $b$ is integral over $\tilde{A}$ + Hence $\exists \tilde{B} \subseteq B$ finite over $\tilde{A}$ and $b \in \tilde{B}$. + Since $\tilde{B} / \tilde{A} $ and $\tilde{A} / R$ are finite, + $\tilde{B} / R$ is finite and $b$ satisfies B. \end{enumerate} \end{proof} -\subsection{Finiteness, finite generation and integrality} %some more remarks on finiteness, finite generation and integrality +\subsection{Finiteness, finite generation and integrality} +% some more remarks on finiteness, finite generation and integrality \begin{fact}[Finite type and integral $\implies$ finite] \label{ftaiimplf} - If $A$ is an integral $R$-algebra of finite type, then it is a finite - $R$-algebra. + If $A$ is an integral $R$-algebra of finite type, + then it is a finite $R$-algebra. \end{fact} \begin{proof} - Let $A $ be generated by $\left( a_i \right) _{i=1}^{n}$ as an $R$- algebra. - By the proposition on integral elements ( - \ref{propinte}), there is a - finite - $R$-algebra $B \subseteq A$ such that all $a_i \in B$. + Let $A $ be generated by $\left( a_i \right)_{i=1}^{n}$ as an $R$-algebra. + By the proposition on integral elements (\ref{propinte}), + there is a finite $R$-algebra $B \subseteq A$ such that all $a_i \in B$. We have $B = A$, as $A$ is generated by the $a_i$ as an $R$-algebra. \end{proof} \begin{fact}[Finite type in tower] - If $A$ is an $R$-algebra of finite type and $B$ an - $A$-algebra of finite type, then $B$ is an $R$-algebra of finite type. + If $A$ is an $R$-algebra of finite type and $B$ an $A$-algebra + of finite type, then $B$ is an $R$-algebra of finite type. \end{fact} \begin{proof} - If $A / R$ is generated by $(a_i)_{i=1}^m$ and - $B / A$ by $(b_j)_{j=1}^{n}$, + If $A / R$ is generated by $(a_i)_{i=1}^m$ and $B / A$ by $(b_j)_{j=1}^{n}$, then $B /R$ is generated by the $b_j$ and the images of the $a_i$ in $B$. \end{proof} -{\color{red} -\begin{fact}[About integrality and fields] - \label{fintaf} - Let $B$ be a domain integral over its subring $A$. - Then $B$ is a field iff $A$ is a field. -\end{fact} +{ + \color{red} + \begin{fact}[About integrality and fields] + \label{fintaf} + Let $B$ be a domain integral over its subring $A$. + Then $B$ is a field iff $A$ is a field. + \end{fact} } \begin{proof} Let $B$ be a field and $a \in A \setminus \{0\} $. - Then $a^{-1} \in B$ is integral over $A$, hence - $a^{-d} = \sum_{i=0}^{d-1} - \alpha_i a^{-i}$ for some $\alpha_i \in A$. + Then $a^{-1} \in B$ is integral over $A$, + hence $a^{-d} = \sum_{i=0}^{d-1} \alpha_i a^{-i}$ + for some $\alpha_i \in A$. Multiplication by $a^{d-1}$ yields - $a^{-1} = \sum_{i=0}^{d-1} \alpha_i - a^{d-1-i} \in A$. + $a^{-1} = \sum_{i=0}^{d-1} \alpha_i a^{d-1-i} \in A$. On the other hand, let $B$ be integral over the field $A$. Let $b \in B \setminus \{0\}$. - As $B$ is integral over $A$, there is a sub-$A$-algebra $\tilde{B} - \subseteq B, - b \in \tilde{B}$ finitely generated as an $A$-module, i.e. a - finite-dimensional - $A$-vector space. - Since $B$ is a domain, $\tilde{B} \xrightarrow{b\cdot } - \tilde{B}$ is - injective, hence surjective, thus $\exists x \in \tilde{B} : b - \cdot x \cdot - 1$. + As $B$ is integral over $A$, + there is a sub-$A$-algebra $\tilde{B} \subseteq B$, + $b \in \tilde{B}$ finitely generated as an $A$-module, + i.e.~a finite-dimensional $A$-vector space. + Since $B$ is a domain, + $\tilde{B} \xrightarrow{b\cdot } \tilde{B}$ + is injective, + hence surjective, + thus $\exists x \in \tilde{B} : b \cdot x \cdot 1$. \end{proof} \subsection{Noether normalization theorem} \begin{lemma} \label{nntechlemma} Let $S \subseteq \N^n$ be finite. - Then there exists $\vec k \in \N^n$ such that $k_1 =1$ and - $w_{\vec k}(\alpha) - \neq w_{\vec k}(\beta)$ for $\alpha \neq \beta \in S$, - where $w_{\vec - k}(\alpha) = \sum_{i=1}^{n} k_i - \alpha_i$. + Then there exists $\vec k \in \N^n$ such that $k_1 =1$ + and $w_{\vec k}(\alpha) \neq w_{\vec k}(\beta)$ + for $\alpha \neq \beta \in S$, + where $w_{\vec k}(\alpha) = \sum_{i=1}^{n} k_i \alpha_i$. \end{lemma} \begin{proof} - Intuitive: For $\alpha \neq \beta$ the equation - $w_{(1, \vec \kappa)}(\alpha) = - w_{(1, \vec \kappa)}(\beta)$ ($\kappa \in \R^{n-1}$) - defines a codimension $1$ + Intuitive: + For $\alpha \neq \beta$ the equation + $w_{(1, \vec \kappa)}(\alpha) = w_{(1, \vec \kappa)}(\beta)$ + ($\kappa \in \R^{n-1}$) defines a codimension $1$ affine hyperplane in $\R^{n-1}$. - It is possible to choose $\kappa$ such that all $\kappa_i$ are $> - \frac{1}{2}$ - and with Euclidean distance $> \frac{\sqrt{n-1} }{2}$ from the union of these - hyperplanes. - By choosing the closest $\kappa'$ with integral coordinates, each coordinate - will be disturbed by at most $\frac{1}{2}$, thus at Euclidean - distance $\le - \frac{\sqrt{n-1} }{2}$. + It is possible to choose $\kappa$ such that all $\kappa_i$ are + $> \frac{1}{2}$ + and with Euclidean distance $> \frac{\sqrt{n-1} }{2}$ + from the union of these hyperplanes. + By choosing the closest $\kappa'$ with integral coordinates, + each coordinate will be disturbed by at most $\frac{1}{2}$, + thus at Euclidean distance $\le \frac{\sqrt{n-1} }{2}$. - More formally:\footnote{The intuitive version suffices in the exam. - } - Define $M \coloneqq \max \{\alpha_i | \alpha \in S, 1 \le i \le n\} $. + More formally:\footnote{The intuitive version suffices in the exam.} + Define $M \coloneqq \max \{\alpha_i | \alpha \in S, 1 \le i \le n\}$. We can choose $k$ such that $k_i > (i-1) M k_{i-1}$. Suppose $\alpha \neq \beta$. Let $i$ be the maximal index such that $\alpha_i \neq \beta_i$. - Then the contributions of $\alpha_j$ (resp. - $\beta_j$) with $1 \le j < i$ to $w_{\vec k}(\alpha)$ - (resp. $w_{\vec k}(\beta)$) cannot undo the difference - $k_i(\alpha_i - \beta_i)$. + Then the contributions of $\alpha_j$ + (resp.~$\beta_j$) + with $1 \le j < i$ to $w_{\vec k}(\alpha)$ + (resp. $w_{\vec k}(\beta)$) + cannot undo the difference $k_i(\alpha_i - \beta_i)$. \end{proof} \begin{theorem}[Noether normalization] \label{noenort} Let $K$ be a field and $A$ a $K$-algebra of finite type. Then there are $a = (a_i)_{i=1}^{n} \in A$ which - are algebraically independent - over $K$, i.e. the ring homomorphism + are algebraically independent over $K$, + i.e.~the ring homomorphism \begin{align*} - \ev_a: K[X_1,\ldots,X_n] & - \longrightarrow A \\ P & \longmapsto P(a_1,\ldots,a_n) + \ev_a: K[X_1,\ldots,X_n] & \longrightarrow A\\ + P & \longmapsto P(a_1,\ldots,a_n) \end{align*} - is - injective. + is injective. $n$ and the $a_i$ can be chosen such that $A$ is finite over the image of $\ev_a$. \end{theorem} \begin{proof} - - Let $(a_i)_{i=1}^n$ be a minimal number of - elements such that $A$ is integral - over its $K$-subalgebra generated by $a_1, \ldots, a_n$. + Let $(a_i)_{i=1}^n$ be a minimal number of elements + such that $A$ is integral over its $K$-subalgebra + generated by $a_1, \ldots, a_n$. (Such $a_i$ exist, since $A$ is of finite type). Let $\tilde{A}$ be the $K$-subalgebra generated by the $a_i$. If suffices to show that the $a_i$ are algebraically independent. @@ -547,5 +513,4 @@ commutative rings with $1$. This contradicts the minimality of $n$, as $B$ can be generated by $< n$ elements $b_i$. - \end{proof} diff --git a/inputs/preamble.tex b/inputs/preamble.tex index c9c2654..bffbff1 100644 --- a/inputs/preamble.tex +++ b/inputs/preamble.tex @@ -5,13 +5,12 @@ \end{warning} \noindent The \LaTeX template by \textsc{Maximilian Kessler} is published under the -MIT-License and can be obtained from \url{https://github.com/kesslermaximilian/LatexPackages}. -% TODO +MIT-License and can be obtained from +\url{https://gitlab.com/latexci/LatexPackages}. \newline -\noindent $\mathfrak{k}$ is {\color{red} always} an -algebraically closed field and $\mathfrak{k}^n$ is equipped with the -Zariski-topology. +\noindent $\mathfrak{k}$ is {\color{red} always} an algebraically closed field +and $\mathfrak{k}^n$ is equipped with the Zariski-topology. Fields which are not assumed to be algebraically closed have been renamed (usually to $\mathfrak{l}$). diff --git a/inputs/projective_spaces.tex b/inputs/projective_spaces.tex index 901476c..d513500 100644 --- a/inputs/projective_spaces.tex +++ b/inputs/projective_spaces.tex @@ -1,64 +1,50 @@ Let $\mathfrak{l}$ be any field. \begin{definition} - For a $\mathfrak{l}$-vector space $V$, let $\mathbb{P}(V)$ be - the set of + For a $\mathfrak{l}$-vector space $V$, + let $\mathbb{P}(V)$ be the set of one-dimensional subspaces of $V$. - Let $\mathbb{P}^n(\mathfrak{l}) \coloneqq - \mathbb{P}(\mathfrak{l}^{n+1})$, the - \vocab[Projective space]{$n$-dimensional projective space over $\mathfrak{l}$}. + Let $\mathbb{P}^n(\mathfrak{l}) \coloneqq \mathbb{P}(\mathfrak{l}^{n+1})$, + the \vocab[Projective space]% + {$n$-dimensional projective space over $\mathfrak{l}$}. - If $\mathfrak{l}$ is kept fixed, we will often write - $\mathbb{P}^n$ for - $\mathbb{P}^n(\mathfrak{l})$. + If $\mathfrak{l}$ is kept fixed, + we will often write $\mathbb{P}^n$ for $\mathbb{P}^n(\mathfrak{l})$. - When dealing with $\mathbb{P}^n$, the usual convention is to use - $0$ as the - index of the first coordinate. + When dealing with $\mathbb{P}^n$, + the usual convention is to use $0$ as the index of the first coordinate. - We denote the one-dimensional subspace generated by $(x_0,\ldots,x_n) \in - \mathfrak{k}^{n+1} \setminus \{0\}$ by + We denote the one-dimensional subspace generated by + $(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \setminus \{0\}$ by $[x_0,\ldots,x_n] \in \mathbb{P}^n$. - If $x = [x_0,\ldots,x_n] \in \mathbb{P}^n$, the - $(x_{i})_{i=0}^n$ are - called - \vocab{homogeneous coordinates} of $x$. + If $x = [x_0, + ldots,x_n] \in \mathbb{P}^n$, the $(x_{i})_{i=0}^n$ are called + \vocab{homogeneous coordinates} of $x$. At least one of the $x_{i}$ must be $\neq 0$. \end{definition} \begin{remark} - There are points $[1,0], - [0,1] \in \mathbb{P}^1$ but there - is no point $[0,0] - \in \mathbb{P}^1$. + There are points $[1,0], [0,1] \in \mathbb{P}^1$ but there + is no point $[0,0] \in \mathbb{P}^1$. \end{remark} \begin{definition}[Infinite hyperplane] - For $0 \le i \le n$ let $U_i \subseteq - \mathbb{P}^n$ denote the set of $[x_0,\ldots,x_{n}]$ with - $x_{i}\neq 0$. - This is a correct definition since two different sets $[x_0,\ldots,x_{n}]$ and - $[\xi_0,\ldots,\xi_n]$ of homogeneous coordinates for the - same point $x \in - \mathbb{P}^n$ differ by scaling with a $\lambda \in - \mathfrak{l}^{\times}$, + For $0 \le i \le n$ let $U_i \subseteq \mathbb{P}^n$ + denote the set of $[x_0,\ldots,x_{n}]$ with $x_{i}\neq 0$. + This is a correct definition since two different sets $[x_0,\ldots,x_{n}]$ + and $[\xi_0,\ldots,\xi_n]$ of homogeneous coordinates for the + same point $x \in \mathbb{P}^n$ differ by scaling with a + $\lambda \in \mathfrak{l}^{\times}$, $x_i = \lambda \xi_i$. - Since not all $x_i$ may be $0$, $\mathbb{P}^n = - \bigcup_{i=0}^n U_i$. - We identify $\mathbb{A}^n = - \mathbb{A}^n(\mathfrak{l}) = \mathfrak{l}^n$ - with - $U_0$ by identifying $(x_1,\ldots,x_n) \in \mathbb{A}^n$ with + Since not all $x_i$ may be $0$, $\mathbb{P}^n = \bigcup_{i=0}^n U_i$. + We identify $\mathbb{A}^n = \mathbb{A}^n(\mathfrak{l}) = \mathfrak{l}^n$ + with $U_0$ by identifying $(x_1,\ldots,x_n) \in \mathbb{A}^n$ with $[1,x_1,\ldots,x_n] \in \mathbb{P}^n$. Then $\mathbb{P}^1 = \mathbb{A}^1 \cup \{\infty\} $ where $\infty=[0,1]$. - More generally, when $n > 0$ $\mathbb{P}^n \setminus - \mathbb{A}^n$ can be + More generally, when $n > 0$ $\mathbb{P}^n \setminus \mathbb{A}^n$ can be identified with $\mathbb{P}^{n-1}$ identifying - $[0,x_1,\ldots,x_n] \in - \mathbb{P}^n \setminus \mathbb{A}^n$ with - $[x_1,\ldots,x_n] \in - \mathbb{P}^{n-1}$. + $[0,x_1,\ldots,x_n] \in \mathbb{P}^n \setminus \mathbb{A}^n$ with + $[x_1,\ldots,x_n] \in \mathbb{P}^{n-1}$. - Thus $\mathbb{P}^n$ is $\mathbb{A}^n \cong - \mathfrak{l}^n$ with a copy of + Thus $\mathbb{P}^n$ is $\mathbb{A}^n \cong \mathfrak{l}^n$ with a copy of $\mathbb{P}^{n-1}$ added as an \vocab{infinite hyperplane} . \end{definition} @@ -67,40 +53,37 @@ Let $\mathfrak{l}$ be any field. Let $\mathbb{I} = \N$ or $\mathbb{I} = \Z$. \end{notation} \begin{definition} - By an \vocab[Graded ring]{$\mathbb{I}$-graded ring} $A_\bullet$ we understand a - ring $A$ with a collection $(A_d)_{d \in \mathbb{I}}$ of - subgroups of the - additive group $(A, +)$ such that $A_a \cdot A_b \subseteq - A_{a + b}$ for $a,b - \in \mathbb{I}$ and such that $A = \bigoplus_{d \in \mathbb{I}} A_d$ in - the - sense that every $r \in A$ has a unique decomposition $r = - \sum_{d \in - \mathbb{I}} r_d$ with $r_d \in A_d$ and but finitely many $r_d - \neq 0$. + By an \vocab[Graded ring]{$\mathbb{I}$-graded ring} $A_\bullet$ + we understand a ring $A$ with a collection $(A_d)_{d \in \mathbb{I}}$ of + subgroups of the additive group $(A, +)$ + such that $A_a \cdot A_b \subseteq A_{a + b}$ for $a,b \in \mathbb{I}$ + and such that $A = \bigoplus_{d \in \mathbb{I}} A_d$ in + the sense that every $r \in A$ has a unique decomposition + $r = \sum_{d \in \mathbb{I}} r_d$ with $r_d \in A_d$ and but finitely many + $r_d \neq 0$. We call the $r_d$ the \vocab{homogeneous components} of $r$. - An ideal $I \subseteq A$ is called \vocab{homogeneous} if $r \in I - \implies - \forall d \in \mathbb{I} ~ r_d \in I_d$ where $I_d \coloneqq I - \cap A_d$. + An ideal $I \subseteq A$ is called \vocab{homogeneous} if + $r \in I \implies \forall d \in \mathbb{I} ~ r_d \in I_d$ + where $I_d \coloneqq I \cap A_d$. By a \vocab{graded ring} we understand an $\N$-graded ring. - Tin this case, $A_{+} \coloneqq \bigoplus_{d=1}^{\infty} - A_d = \{r \in A | r_0 - = 0\} $ is called the \vocab{augmentation ideal} of $A$. + Tin this case, + $A_{+} \coloneqq \bigoplus_{d=1}^{\infty} A_d = \{r \in A | r_0 = 0\} $ + is called the \vocab{augmentation ideal} of $A$. \end{definition} \begin{remark}[Decomposition of $1$] - If $1 = \sum_{d \in \mathbb{I}} \varepsilon_d$ is the - decomposition into homogeneous components, then $\varepsilon_a = 1 \cdot - \varepsilon_a = \sum_{b \in \mathbb{I}} \varepsilon_a\varepsilon_b$ with - $\varepsilon_a\varepsilon_b \in A_{a+b}$. + If $1 = \sum_{d \in \mathbb{I}} \varepsilon_d$ + is the decomposition into homogeneous components, + then $\varepsilon_a = 1 \cdot \varepsilon_a + = \sum_{b \in \mathbb{I}} \varepsilon_a\varepsilon_b$ + with $\varepsilon_a\varepsilon_b \in A_{a+b}$. By the uniqueness of the decomposition into homogeneous components, - $\varepsilon_a \varepsilon_0 = \varepsilon_a$ and $b \neq 0 \implies - \varepsilon_a \varepsilon_b = 0$. - Applying the last equation with $a = 0$ gives $b\neq 0 \implies \varepsilon_b = - \varepsilon_0 \varepsilon _b = 0$. + $\varepsilon_a \varepsilon_0 = \varepsilon_a$ + and $b \neq 0 \implies \varepsilon_a \varepsilon_b = 0$. + Applying the last equation with $a = 0$ gives + $b\neq 0 \implies \varepsilon_b = \varepsilon_0 \varepsilon _b = 0$. Thus $1 = \varepsilon_0 \in A_0$. \end{remark} \begin{remark} @@ -123,44 +106,41 @@ Let $\mathfrak{l}$ be any field. \end{proposition} \begin{proof} Most assertions are trivial. - We only show that $J$ homogeneous $\implies \sqrt{J} $ - homogeneous. - Let $A$ be $\mathbb{I}$-graded, $f \in \sqrt{J} $ and - $f = \sum_{d \in - \mathbb{I}} f_d$ the decomposition. - To show that all $f_d \in \sqrt{J} $, we use induction on $N_f - \coloneqq \# \{d - \in \mathbb{I} | f_d \neq 0\}$. + We only show that $J$ homogeneous $\implies \sqrt{J}$ homogeneous. + Let $A$ be $\mathbb{I}$-graded, + $f \in \sqrt{J} $ and + $f = \sum_{d \in \mathbb{I}} f_d$ the decomposition. + To show that all $f_d \in \sqrt{J} $, + we use induction on $N_f \coloneqq \# \{d \in \mathbb{I} | f_d \neq 0\}$. $N_f = 0$ is trivial. - Suppose $N_f > 0$ and $e \in \mathbb{I}$ is maximal with $f_e \neq - 0$. + Suppose $N_f > 0$ and $e \in \mathbb{I}$ is maximal with $f_e \neq 0$. For $l \in \N$, the $le$-th homogeneous component of $f^l$ is $f_e^l$. Choosing $l$ large enough such that $f^l \in J$ and using the homogeneity of $J$, we find $f_e \in \sqrt{J}$. - As $\sqrt{J} $ is an ideal, $\tilde f \coloneqq f - f_e \in - \sqrt{J} $. - As $N_{\tilde f} = N_f -1$, the induction assumption may be - applied to $\tilde - f$ and shows $f_d \in \sqrt{J} $ for $d \neq e$. + As $\sqrt{J} $ is an ideal, + $\tilde f \coloneqq f - f_e \in \sqrt{J} $. + As $N_{\tilde f} = N_f -1$, + the induction assumption may be applied to $\tilde f$ + and shows $f_d \in \sqrt{J} $ for $d \neq e$. \end{proof} \begin{fact} - A homogeneous ideal is finitely generated iff it can be generated by finitely - many of its homogeneous elements. + A homogeneous ideal is finitely generated + iff it can be generated by finitely many of its homogeneous elements. In particular, this is always the case when $A$ is a Noetherian ring. \end{fact} \subsubsection{The Zariski topology on $\mathbb{P}^n$} \begin{notation} - Recall that for $\alpha \in \N^{n+1}$ $|\alpha| = - \sum_{i=0}^{n} \alpha_i$ and - $x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}$. + Recall that for + $\alpha \in \N^{n+1}$ $|\alpha| = \sum_{i=0}^{n} \alpha_i$ + and $x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}$. \end{notation} \begin{definition}[Homogeneous polynomials] - Let $R$ be any ring and $f = \sum_{\alpha \in \N^{n+1}} - f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n]$. - We say that $f$ is \vocab{homogeneous of degree $d$} if $|\alpha| \neq d - \implies f_\alpha = 0$ . + Let $R$ be any ring and + $f = \sum_{\alpha \in \N^{n+1}} f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n]$. + We say that $f$ is \vocab{homogeneous of degree $d$} + if $|\alpha| \neq d \implies f_\alpha = 0$ . We denote the subset of homogeneous polynomials of degree $d$ by $R[X_0,\ldots,X_n]_d \subseteq R[X_0,\ldots,X_n]$. \end{definition} @@ -169,57 +149,50 @@ Let $\mathfrak{l}$ be any field. \end{remark} \begin{definition}[Zariski topology on $\mathbb{P}^n(\mathfrak{k})$] \label{ztoppn} - Let $A = \mathfrak{k}[X_0,\ldots,X_n]$. + Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.% \footnote{As always, $\mathfrak{k}$ is algebraically closed} - For $f \in A_d = \mathfrak{k}[X_0,\ldots,X_n]_d$, the validity of the equation - $f(x_0,\ldots,x_{n}) = 0$ does not depend on the choice of homogeneous - coordinates, as + For $f \in A_d = \mathfrak{k}[X_0,\ldots, + _n]_d$, the validity of the equation $f(x_0,\ldots,x_{n}) = 0$ + does not depend on the choice of homogeneous coordinates, as \[ - f(\lambda x_0,\ldots, \lambda x_n) 0 \lambda^d - f(x_0,\ldots,x_n) + f(\lambda x_0,\ldots, \lambda x_n) 0 \lambda^d f(x_0,\ldots,x_n). \] - Let $\Vp(f) \coloneqq \{x \in \mathbb{P}^n | f(x) - = 0\}$. + Let $\Vp(f) \coloneqq \{x \in \mathbb{P}^n | f(x) = 0\}$. We call a subset $X \subseteq \mathbb{P}^n$ Zariski-closed if it - can be - represented as + can be represented as \[ X = \bigcap_{i=1}^k \Vp(f_i) \] - where the $f_i \in A_{d_i}$ - are homogeneous polynomials. + where the $f_i \in A_{d_i}$ are homogeneous polynomials. \end{definition} \pagebreak \begin{fact} - If $X = \bigcap_{i = 1}^k \Vp(f_i) \subseteq - \mathbb{P}^n$ is closed, then $Y - = X \cap \mathbb{A}^n$ can be identified with the closed subset + If $X = \bigcap_{i = 1}^k \Vp(f_i) \subseteq \mathbb{P}^n$ is closed, + then $Y = X \cap \mathbb{A}^n$ can be identified with the closed subset \[ - \{(x_1,\ldots,x_n) \in \mathfrak{k}^n | f_i(1,x_1,\ldots,x_n) = 0, 1 - \le i \le - k\} \subseteq \mathfrak{k}^n + \{(x_1,\ldots,x_n) \in \mathfrak{k}^n | + f_i(1,x_1,\ldots,x_n) = 0, 1 \le i \le k\} + \subseteq \mathfrak{k}^n. \] - Conversely, if $Y \subseteq \mathfrak{k}^n$ is - closed it has the form + Conversely, if $Y \subseteq \mathfrak{k}^n$ is closed it has the form \[ - \{(x_1,\ldots,x_n) \in \mathfrak{k}^n | + \{(x_1,\ldots,x_n) \in \mathfrak{k}^n | g_i(x_1,\ldots,x_n) = 0, 1 \le i \le k\} \] - and can thus be identified with $X - \cap \mathbb{A}^n$ where $X \coloneqq \bigcap_{i=1}^k - \Vp(f_i)$ is given by + and can thus be identified with $X \cap \mathbb{A}^n$ + where $X \coloneqq \bigcap_{i=1}^k \Vp(f_i)$ is given by \[ f_i(X_0,\ldots,X_n) \coloneqq X_0^{d_i} g_i(X_1 / X_0,\ldots, X_n / X_0), d_i - \ge \deg(g_i) + \ge \deg(g_i). \] Thus, the Zariski topology on $\mathfrak{k}^n$ can be identified with the topology induced by the Zariski topology on - $\mathbb{A}^n = - U_0$, and the same holds for $U_i$ with $0 \le i \le n$. + $\mathbb{A}^n = U_0$, + and the same holds for $U_i$ with $0 \le i \le n$. - In this sense, the Zariski topology on $\mathbb{P}^n$ can be - thought of as + In this sense, + the Zariski topology on $\mathbb{P}^n$ can be thought of as gluing the Zariski topologies on the $U_i \cong \mathfrak{k}^n$. \end{fact} @@ -227,18 +200,16 @@ Let $\mathfrak{l}$ be any field. \begin{definition} Let $I \subseteq A = \mathfrak{k}[X_0,\ldots,X_n]$ be a homogeneous ideal. - Let $\Vp(I) \coloneqq - \{[x_0,\ldots,_n] \in \mathbb{P}^n | - \forall f \in I ~ - f(x_0,\ldots,x_n) = 0\}$ As $I$ is homogeneous, it is sufficient to impose this - condition for the homogeneous elements $f \in I$. - Because $A$ is Noetherian, $I$ can finitely generated by homogeneous elements - $(f_i)_{i=1}^k$ and - $\Vp(I)=\bigcap_{i=1}^k \Vp(f_i)$ as - in - \ref{ztoppn}. - Conversely, if the homogeneous $f_i$ are given, then $I = \langle - f_1,\ldots,f_k \rangle_A$ is homogeneous. + Let $\Vp(I) \coloneqq \{[x_0,\ldots,_n] \in \mathbb{P}^n | \forall f \in I ~% + f(x_0,\ldots,x_n) = 0\}$ + As $I$ is homogeneous, + it is sufficient to impose this condition for the homogeneous elements + $f \in I$. + Because $A$ is Noetherian, + $I$ can finitely generated by homogeneous elements $(f_i)_{i=1}^k$ and + $\Vp(I)=\bigcap_{i=1}^k \Vp(f_i)$ as in \ref{ztoppn}. + Conversely, if the homogeneous $f_i$ are given, + then $I = \langle f_1,\ldots,f_k \rangle_A$ is homogeneous. \end{definition} \begin{remark} Note that $V(A) = V(A_+) = \emptyset$. @@ -247,52 +218,47 @@ Let $\mathfrak{l}$ be any field. For homogeneous ideals in $A$ and $m \in \N$, we have: \begin{itemize} \item - $\Vp(\sum_{\lambda \in \Lambda} I_\lambda) = \bigcap_{\lambda \in \Lambda} - \Vp(I_\lambda)$ + $\Vp(\sum_{\lambda \in \Lambda} I_\lambda) = \bigcap_{\lambda \in \Lambda} \Vp(I_\lambda)$. \item - $\Vp(\bigcap_{k=1}^m I_k) = \Vp(\prod_{k=1}^{m} I_k) = - \bigcup_{k=1}^m \Vp(I_k)$ + $\Vp(\bigcap_{k=1}^m I_k) = \Vp(\prod_{k=1}^{m} I_k) = \bigcup_{k=1}^m \Vp(I_k)$. \item - $\Vp(\sqrt{I}) = \Vp(I)$ + $\Vp(\sqrt{I}) = \Vp(I)$. \end{itemize} \end{fact} \begin{fact} - If $X = \bigcup_{\lambda \in \Lambda} U_\lambda$ is an - open covering of a topological space then $X$ is Noetherian iff there is a - finite subcovering and all $U_\lambda$ are Noetherian. + If $X = \bigcup_{\lambda \in \Lambda} U_\lambda$ is an open covering + of a topological space then $X$ is Noetherian + iff there is a finite subcovering and all $U_\lambda$ are Noetherian. \end{fact} \begin{proof} - By definition, a topological space is Noetherian $\iff$ all open subsets are - quasi-compact. + By definition, a topological space is Noetherian + $\iff$ all open subsets are quasi-compact. \end{proof} \begin{corollary} The Zariski topology on $\mathbb{P}^n$ is indeed a topology. - The induced topology on the open set $\mathbb{A}^n = - \mathbb{P}^n \setminus - \Vp(X_0) \cong \mathfrak{k}^n$ is the Zariski - topology on $\mathfrak{k}^n$. - The same holds for all $U_i = \mathbb{P}^n \setminus - \Vp(X_i) \cong - \mathfrak{k}^n$. + The induced topology on the open set + $\mathbb{A}^n = \mathbb{P}^n \setminus \Vp(X_0) \cong \mathfrak{k}^n$ + is the Zariski topology on $\mathfrak{k}^n$. + The same holds for all + $U_i = \mathbb{P}^n \setminus \Vp(X_i) \cong \mathfrak{k}^n$. Moreover, the topological space $\mathbb{P}^n$ is Noetherian. \end{corollary} \subsection{Noetherianness of graded rings} \begin{proposition} - For a graded ring $R_{\bullet}$, the following conditions - are equivalent: + For a graded ring $R_{\bullet}$, + the following conditions are equivalent: \begin{enumerate}[A] \item $R$ is Noetherian. \item - Every homogeneous ideal of $R_{\bullet}$ is finitely - generated. + Every homogeneous ideal of $R_{\bullet}$ is finitely generated. \item - Every chain $I_0\subseteq I_1 \subseteq \ldots$ of homogeneous ideals - terminates. + Every chain $I_0\subseteq I_1 \subseteq \ldots$ of + homogeneous ideals terminates. \item - Every set $\mathfrak{M} \neq \emptyset$ of homogeneous ideals has a - $\subseteq$-maximal element. + Every set $\mathfrak{M} \neq \emptyset$ of homogeneous ideals + has a $\subseteq$-maximal element. \item $R_0$ is Noetherian and the ideal $R_+$ is finitely generated. \item @@ -316,27 +282,27 @@ Let $\mathfrak{l}$ be any field. The $R_0$-subalgebra $\tilde R$ of $R$ generated by the $f_i$ equals $R$. \end{claim} \begin{subproof} - It is sufficient to show that every homogeneous $f \in R_d$ belongs to $\tilde - R$. + It is sufficient to show that every homogeneous $f \in R_d$ + belongs to $\tilde R$. We use induction on $d$. The case of $d = 0$ is trivial. Let $d > 0$ and $R_e \subseteq \tilde R$ for all $e < d$. - as $f \in R_+$, $f = \sum_{i=1}^{k} g_if_i$. - Let $f_a = \sum_{i=1}^{k} g_{i, a-d_i} f_i$, where - $g_i = \sum_{b=0}^{\infty} - g_{i,b}$ is the decomposition into homogeneous - components. + As $f \in R_+$, $f = \sum_{i=1}^{k} g_if_i$. + Let $f_a = \sum_{i=1}^{k} g_{i, + a-d_i} f_i$, where $g_i = \sum_{b=0}^{\infty} g_{i,b}$ + is the decomposition into homogeneous components. Then $f = \sum_{a=0}^{\infty} f_a$ is the decomposition of $f$ into - homogeneous - components, hence $a \neq d \implies f_a = 0 $. + homogeneous components, + hence $a \neq d \implies f_a = 0 $. Thus we may assume $g_i \in R_{d-d_i}$. - As $d_i > 0$, the induction assumption may now be applied to $g_i$, hence $g_i - \in \tilde R$, hence $f \in \tilde R$. + As $d_i > 0$, + the induction assumption may now be applied to $g_i$, + hence $g_i \in \tilde R$, + hence $f \in \tilde R$. \end{subproof} \noindent\textbf{F $\implies$ A} - Hilbert's Basissatz ( - \ref{basissatz}) + Hilbert's Basissatz (\ref{basissatz}) \end{proof} @@ -348,148 +314,130 @@ Let $\mathfrak{l}$ be any field. % Lecture 12 \begin{proposition}[Projective form of the Nullstellensatz] \label{hnsp} - If $I \subseteq A$ is a homogeneous ideal and $f \in A_d$ with $d>0$, then - $\Vp(I) \subseteq \Vp(f) \iff f \in - \sqrt{I}$. + If $I \subseteq A$ is a homogeneous ideal and $f \in A_d$ with $d>0$, + then $\Vp(I) \subseteq \Vp(f) \iff f \in \sqrt{I}$. \end{proposition} \begin{proof} $\impliedby$ is clear. Let $\Vp(I) \subseteq \Vp(f)$. - If $x = (x_0,\ldots,x_n) \in \Va(I)$, then either $x = 0$ in - which case $f(x) = - 0$ since $d > 0$ or the point $[x_0,\ldots,x_n] \in - \mathbb{P}^n$ is - well-defined and belongs to $\Vp(I) \subseteq - \Vp(f)$, hence $f(x) = 0$. - Thus $\Va(I) \subseteq \Va(f)$ and $f \in - \sqrt{I}$ be the Nullstellensatz - ( - \ref{hns3}). + If $x = (x_0, + ldots,x_n) \in \Va(I)$, then either $x = 0$ in + which case $f(x) = 0$ since $d > 0$ + or the point $[x_0,\ldots,x_n] \in \mathbb{P}^n$ is + well-defined and belongs to $\Vp(I) \subseteq \Vp(f)$, + hence $f(x) = 0$. + Thus $\Va(I) \subseteq \Va(f)$ and $f \in \sqrt{I}$ be the Nullstellensatz + (\ref{hns3}). \end{proof} \begin{definition} \footnote{This definition is not too important, the characterization in the following remark suffices.}. - For a graded ring $R_\bullet$, let $\Proj(R_\bullet)$ be the set of - $\fp \in - \Spec R$ such that $\fp$ is a homogeneous ideal and $\fp \not\supseteq R_+$. + For a graded ring $R_\bullet$, + let $\Proj(R_\bullet)$ be the set of $\fp \in \Spec R$ + such that $\fp$ is a homogeneous ideal and $\fp \not\supseteq R_+$. \end{definition} \begin{remark} \label{proja} - As the elements of $A_0 \setminus \{0\}$ are units in $A$ it follows that for - every homogeneous ideal $I$ we have $I \subseteq A_+$ or $I = A$. - In particular, $\Proj(A_\bullet) = \{\fp \in \Spec A \setminus A_+ | - \fp - \text{ is homogeneous}\} $. + As the elements of $A_0 \setminus \{0\}$ are units in $A$ + it follows that for every homogeneous ideal $I$ + we have $I \subseteq A_+$ or $I = A$. + In particular, + $\Proj(A_\bullet) = \{\fp \in \Spec A \setminus A_+ | \fp \text{ is homogeneous}\} $. \end{remark} \begin{proposition} \label{bijproj} There is a bijection \begin{align*} - f: \{I \subseteq A_+ | I \text{ homogeneous - ideal}, I = \sqrt{I}\} & \longrightarrow \{X \subseteq \mathbb{P}^n | X \text{ - closed}\} \\ I & \longmapsto \Vp(I)\\ \langle \{f \in A_d | d > 0, X - \subseteq \Vp(f)\} \rangle & \longmapsfrom X + f: \{I \subseteq A_+ | I \text{ homogeneous ideal}, I = \sqrt{I}\} & + \longrightarrow \{X \subseteq \mathbb{P}^n | X \text{ closed}\}\\ + I & + \longmapsto \Vp(I)\\ + \langle \{f \in A_d | d > 0, X \subseteq \Vp(f)\} \rangle & + \longmapsfrom X \end{align*} Under this bijection, the irreducible subsets correspond to the elements of $\Proj(A_\bullet)$. \end{proposition} \begin{proof} - From the projective form of the Nullstellensatz it follows that $f$ is - injective and that $f^{-1}(\Vp\left( I \right)) - = \sqrt{I} = I$. - If $X \subseteq \mathbb{P}^n$ is closed, then $X = - \Vp(J)$ for some homogeneous - ideal $J \subseteq A$. + From the projective form of the Nullstellensatz it follows + that $f$ is injective + and that $f^{-1}(\Vp\left( I \right)) = \sqrt{I} = I$. + If $X \subseteq \mathbb{P}^n$ is closed, + then $X = \Vp(J)$ for some homogeneous ideal $J \subseteq A$. Without loss of generality loss of generality $J = \sqrt{J}$. - If $J \not\subseteq A_+$, then $J = A$ ( - \ref{proja}), hence $X = - \Vp(J) = - \emptyset = \Vp(A_+)$. + If $J \not\subseteq A_+$, then $J = A$ (\ref{proja}), + hence $X = \Vp(J) = \emptyset = \Vp(A_+)$. Thus we may assume $J \subseteq A_+$, and $f$ is surjective. - Suppose $\fp \in \Proj(A_\bullet)$. - Then $\fp \neq A_+$ hence $X = \Vp(\fp) \neq \emptyset$ by the - proven part of - the proposition. - Assume $X = X_1 \cup X_2$ is a decomposition into proper closed subsets, where - $X_k = \Vp(I_k)$ for some $I_k \subseteq A_+, I_k = - \sqrt{I_k}$. + Then $\fp \neq A_+$ hence $X = \Vp(\fp) \neq \emptyset$ by the + proven part of the proposition. + Assume $X = X_1 \cup X_2$ is a decomposition into proper closed subsets, + where $X_k = \Vp(I_k)$ for some $I_k \subseteq A_+, I_k = \sqrt{I_k}$. Since $X_k$ is a proper subset of $X$, there is $f_k \in I_k \setminus \fp$. - We have $\Vp(f_1f_2) \supseteq \Vp(f_k) \supseteq - \Vp(I_k)$ hence $\Vp(f_1f_2) - \supseteq \Vp(I_1) \cup \Vp(I_2) = X = - \Vp(\fp)$ and it follows that $f_1f_2\in - \sqrt{\fp} = \fp \lightning$. + We have $\Vp(f_1f_2) \supseteq \Vp(f_k) \supseteq \Vp(I_k)$ + hence $\Vp(f_1f_2) \supseteq \Vp(I_1) \cup \Vp(I_2) = X = \Vp(\fp)$ + and it follows that $f_1f_2\in \sqrt{\fp} = \fp \lightning$. - Assume $X = \Vp(\fp)$ is irreducible, where $\fp = - \sqrt{\fp} \in A_+$ is - homogeneous. + Assume $X = \Vp(\fp)$ is irreducible, + where $\fp = \sqrt{\fp} \in A_+$ is homogeneous. The $\fp \neq A_+$ as $X = \emptyset$ otherwise. - Assume that $f_1f_2 \in \fp$ but $f_i \not\in A_{d_i} - \setminus \fp$. + Assume that $f_1f_2 \in \fp$ but $f_i \not\in A_{d_i} \setminus \fp$. Then $X \not \subseteq \Vp(f_i)$ by the projective - Nullstellensatz when $d_i > - 0$ and because $\Vp(1) = \emptyset$ when $d_i = 0$. + Nullstellensatz when $d_i > 0$ + and because $\Vp(1) = \emptyset$ when $d_i = 0$. Thus $X = (X \cap \Vp\left( f_1 \right)) \cup (X \cap \Vp(f_2))$ is a - proper - decomposition $\lightning$. - By lemma - \ref{homprime}, $\fp$ is a prime ideal. - + proper decomposition $\lightning$. + By lemma \ref{homprime}, $\fp$ is a prime ideal. \end{proof} \begin{remark} - It is important that $I \subseteq A_{\color{red} +}$, since - $\Vp(A) = \Vp(A_+) - = \emptyset$ would be a counterexample. + It is important that $I \subseteq A_{\color{red} +}$, + since $\Vp(A) = \Vp(A_+) = \emptyset$ would be a counterexample. \end{remark} \begin{corollary} $\mathbb{P}^n$ is irreducible. \end{corollary} \begin{proof} - Apply - \ref{bijproj} to $\{0\} \in \Proj(A_\bullet)$. + Apply \ref{bijproj} to $\{0\} \in \Proj(A_\bullet)$. \end{proof} \subsection{Some remarks on homogeneous prime ideals} \begin{lemma} \label{homprime} Let $R_\bullet$ be an $\mathbb{I}$ graded ring - ($\mathbb{I} = \N$ or - $\mathbb{I} = \Z$). - A homogeneous ideal $I \subseteq R$ is a prime ideal iff $1 \not\in I$ and for - homogeneous elements $f, g \in R , fg \in I \implies f \in I \lor g \in I$. + ($\mathbb{I} = \N$ or $\mathbb{I} = \Z$). + A homogeneous ideal $I \subseteq R$ is a prime ideal + iff $1 \not\in I$ and for all homogeneous elements $f, g \in R$ + \[fg \in I \implies f \in I \lor g \in I.\] \end{lemma} \begin{proof} $\implies$ is trivial. - It suffices to show that for arbitrary $f,g \in R fg \in I \implies f \in I - \lor g \in I$. + It suffices to show that for arbitrary $f,g \in R$ + we have that $fg \in I \implies f \in I \lor g \in I$. Let $f = \sum_{d \in \mathbb{I}} f_d, g = \sum_{d \in \mathbb{I}} g_d $ be the decompositions into homogeneous components. - If $f \not\in I$ and $g \not\in I$ there are $d,e \in I$ with $f_d \in I, g_e - \in I$, and they may assumed to be maximal with this property. - As $I$ is homogeneous and $fg \in I$, we have - $(fg)_{d+e} \in I$ but + If $f \not\in I$ and $g \not\in I$ there are $d,e \in I$ with $f_d \in I$, + $g_e \in I$, and they may assumed to be maximal with this property. + As $I$ is homogeneous and $fg \in I$, we have $(fg)_{d+e} \in I$ but \[ (fg)_{d+e} = f_dg_e + \sum_{\delta = 1}^{\infty} (f_{d + \delta} g_{e - \delta} + f_{d - \delta} g_{e + \delta}) \] where $f_dg_e \not\in I$ by our assumption - on $I$ and all other summands on the right hand side are $\in I$ (as - $f_{d+ - \delta} \in I$ and $g_{e + \delta} \in - I$ by the maximality of $d$ and $e$), a - contradiction. + on $I$ and all other summands on the right hand side are $\in I$ + (as $f_{d+ \delta} \in I$ and $g_{e + \delta} \in I$ by the maximality + of $d$ and $e$), + a contradiction. \end{proof} \begin{remark} - If $R_\bullet$ is $\N$-graded and $\fp \in \Spec R_0$, then $\fp \oplus R_+ = - \{r \in R | r_0 \in \fp\} $ is a homogeneous prime ideal of $R$. + If $R_\bullet$ is $\N$-graded and $\fp \in \Spec R_0$, + then $\fp \oplus R_+ = \{r \in R | r_0 \in \fp\} $ + is a homogeneous prime ideal of $R$. \[ - \{\fp \in \Spec R | \fp \text{ is a homogeneous ideal of } - R_\bullet\} = \Proj(R_\bullet) \sqcup \{\fp \oplus R_+ | \fp \in \Spec - R_0\} + \{\fp \in \Spec R | \fp \text{ is a homogeneous ideal of } R_\bullet\} + = \Proj(R_\bullet) \sqcup \{\fp \oplus R_+ | \fp \in \Spec R_0\}. \] \end{remark} @@ -500,38 +448,35 @@ Let $\mathfrak{l}$ be any field. $\mathbb{P}^n$ is catenary. \item $\dim(\mathbb{P}^n) = n$. - Moreover, $\codim(\{x\} ,\mathbb{P}^n) = n$ for every $x \in - \mathbb{P}^n$. + Moreover, $\codim(\{x\} ,\mathbb{P}^n) = n$ for every $x \in \mathbb{P}^n$. \item - If $X \subseteq \mathbb{P}^n$ is irreducible and $x \in X$, then - $\codim(\{x\}, X) = \dim(X) = n - - \codim(X, \mathbb{P}^n)$. + If $X \subseteq \mathbb{P}^n$ is irreducible and $x \in X$, + then $\codim(\{x\}, X) = \dim(X) = n - \codim(X, \mathbb{P}^n)$. \item If $X \subseteq Y \subseteq \mathbb{P}^n$ are irreducible subsets, - then $\codim(X,Y) = \dim(Y) - - \dim(X)$. + then $\codim(X,Y) = \dim(Y) - \dim(X)$. \end{itemize} \end{proposition} \begin{proof} Let $X \subseteq \mathbb{P}^n$ be irreducible. - If $x \in X$, there is an integer $0 \le i \le n$ and $X \in U_i = - \mathbb{P}^n \setminus \Vp(X_i)$. + If $x \in X$, there is an integer $0 \le i \le n$ and + $X \in U_i = \mathbb{P}^n \setminus \Vp(X_i)$. Without loss of generality loss of generality $i = 0$. Then $\codim(X, \mathbb{P}^n) = \codim(X \cap \mathbb{A}^n, \mathbb{A}^n)$ by - the locality of Krull codimension ( - \ref{lockrullcodim}). - Applying this with $X = \{x\}$ and our results about the affine case gives the - second assertion. - If $Y$ and $Z$ are also irreducible with $X \subseteq Y \subseteq Z$, then - $\codim(X,Y) = \codim(X \cap \mathbb{A}^n, Y \cap \mathbb{A}^n)$, $\codim(X,Z) - = \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$ and $\codim(Y,Z) = - \codim(Y - \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$. + the locality of Krull codimension (\ref{lockrullcodim}). + Applying this with $X = \{x\}$ + and our results about the affine case gives the second assertion. + If $Y$ and $Z$ are also irreducible with $X \subseteq Y \subseteq Z$, + then + $\codim(X,Y) = \codim(X \cap \mathbb{A}^n, Y \cap \mathbb{A}^n)$, + $\codim(X,Z) = \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$ + and $\codim(Y,Z) = \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$. Thus \begin{align*} - \codim(X,Y) + \codim(Y,Z) & = \codim(X \cap \mathbb{A}^n, Y - \cap \mathbb{A}^n) + \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n) \\ & = - \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n) \\ & = \codim(X, Z) + \codim(X,Y) + \codim(Y,Z) + & = \codim(X \cap \mathbb{A}^n, Y \cap \mathbb{A}^n) + \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n) \\ + & = \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)\\ + & = \codim(X, Z) \end{align*} because $\mathfrak{k}^n$ is catenary and the first point follows. The remaining assertions can easily be derived from the first two. @@ -540,15 +485,12 @@ Let $\mathfrak{l}$ be any field. \subsection{The cone $C(X)$} \begin{definition} If $X \subseteq \mathbb{P}^n$ is closed, we define the - \vocab{affine cone over - $X$} + \vocab{affine cone over $X$} \[ - C(X) = \{0\} \cup \{(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \setminus - \{0\} | [x_0,\ldots,x_n] \in X\} + C(X) = \{0\} \cup \{(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \setminus \{0\} | [x_0,\ldots,x_n] \in X\} \] - If $X = \Vp(I)$ where $I \subseteq A_+ = - \mathfrak{k}[X_0,\ldots,X_n]_+$ is homogeneous, then $C(X) = - \Va(I)$. + If $X = \Vp(I)$ where $I \subseteq A_+ = \mathfrak{k}[X_0,\ldots,X_n]_+$ + is homogeneous, then $C(X) = \Va(I)$. \end{definition} \begin{proposition} \label{conedim} @@ -564,17 +506,14 @@ Let $\mathfrak{l}$ be any field. \end{itemize} \end{proposition} \begin{proof} - The first assertion follows from - \ref{bijproj} and - \ref{bijiredprim} + The first assertion follows from \ref{bijproj} and \ref{bijiredprim} (bijection of irreducible subsets and prime ideals in the projective and affine case). Let $d = \dim(X)$ and \[ - X_0 \subsetneq \ldots \subsetneq X_d = X \subsetneq - X_{d+1} \subsetneq \ldots \subsetneq X_n = - \mathbb{P}^n + X_0 \subsetneq \ldots \subsetneq X_d = X + \subsetneq X_{d+1} \subsetneq \ldots \subsetneq X_n = \mathbb{P}^n \] be a chain of irreducible subsets of $\mathbb{P}^n$. Then @@ -586,8 +525,7 @@ Let $\mathfrak{l}$ be any field. Hence $\dim(C(X)) \ge 1 + d$ and $\codim(C(X), \mathfrak{k}^{n+1}) \ge n-d$. Since - $\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) = \dim(\mathfrak{k}^{n+1}) - = n+1$, + $\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) = \dim(\mathfrak{k}^{n+1}) = n+1$, the two inequalities must be equalities. \end{proof} \subsubsection{Application to hypersurfaces in $\mathbb{P}^n$} @@ -605,13 +543,11 @@ Let $\mathfrak{l}$ be any field. \end{corollary} \begin{proof} If $H = \Vp(P)$ then $C(H) = \Va(P)$ - is a hypersurface in $\mathfrak{k}^{n+1}$ - by \ref{irredcodimone}. + is a hypersurface in $\mathfrak{k}^{n+1}$ by \ref{irredcodimone}. By \ref{conedim}, $H$ is irreducible and of codimension $1$. Conversely, let $H$ be a hypersurface in $\mathbb{P}^n$. - By \ref{conedim}, $C(H)$ is a hypersurface in - $\mathfrak{k}^{n+1}$, + By \ref{conedim}, $C(H)$ is a hypersurface in $\mathfrak{k}^{n+1}$, hence $C(H) = \Vp(P)$ for some prime element $P \in A$ (again by \ref{irredcodimone}). We have $H = \Vp(\fp)$ for some $\fp \in \Proj(A)$ and $C(H) = \Va(\fp)$. @@ -636,8 +572,7 @@ Let $\mathfrak{l}$ be any field. \begin{corollary} Let $A \subseteq \mathbb{P}^n$ and $B \subseteq \mathbb{P}^n$ be irreducible subsets of dimensions $a$ and $b$. - If $a+ b \ge n$, - then $A \cap B \neq \emptyset$ + If $a+ b \ge n$, then $A \cap B \neq \emptyset$ and every irreducible component of $A \cap B$ has dimension $\ge a + b - n$. \end{corollary} @@ -656,17 +591,17 @@ Let $\mathfrak{l}$ be any field. From the definition of the affine cone it follows that $C(A \cap B) = C(A) \cap C(B)$. - We have $\dim(C(A)) = a+1$ and $\dim(C(B)) = b + 1$ by - \ref{conedim}. - If $A \cap B = \emptyset$, then $C(A) \cap C(B) = \{0\}$ with $\{0\} $ as an - irreducible component, contradicting the lower bound $a + b + 1 - n > 0$ for + We have $\dim(C(A)) = a+1$ and $\dim(C(B)) = b + 1$ by \ref{conedim}. + If $A \cap B = \emptyset$, + then $C(A) \cap C(B) = \{0\}$ with $\{0\} $ as an irreducible component, + contradicting the lower bound $a + b + 1 - n > 0$ for the dimension of irreducible components of $C(A) \cap C(B)$ (again \ref{codimintersection}). \end{proof} \begin{remark}[Bezout's theorem] If $A \neq B$ are hypersurfaces of degree $a$ and $b$ - in $\mathbb{P}^2$, then $A \cap B$ has $ab$ points counted by - (suitably defined) multiplicity. + in $\mathbb{P}^2$, + then $A \cap B$ has $ab$ points counted by (suitably defined) multiplicity. \end{remark} diff --git a/inputs/varieties.tex b/inputs/varieties.tex index 6dccdb7..fa8471e 100644 --- a/inputs/varieties.tex +++ b/inputs/varieties.tex @@ -3,11 +3,10 @@ \begin{definition}[Sheaf] Let $X$ be any topological space. - A \vocab{presheaf} $\mathcal{G}$ of sets (or rings, - (abelian) groups) on $X$ + A \vocab{presheaf} $\mathcal{G}$ of sets + (or rings, (abelian) groups) on $X$ associates a set (or rings, or (abelian) group) $\mathcal{G}(U)$ to - every open - subset $U$ of $X$, and a map (or ring or group homomorphism) + every open subset $U$ of $X$, and a map (or ring or group homomorphism) $\mathcal{G}(U) \xrightarrow{r_{U,V}} \mathcal{G}(V)$ to every inclusion $V \subseteq U$ of @@ -32,101 +31,86 @@ \begin{align*} \phi_{U, (U_i)_{i \in I}}: \mathcal{G}(U) & \longrightarrow \{(f_i)_{i \in I} \in \prod_{i \in I} \mathcal{G}(U_i) | - r_{U_i, U_i \cap U_j}(f_i) = r_{U_j, U_i \cap U_j}(f_j) \text{ for } i,j \in I - \} \\ f & \longmapsto (r_{U, U_i}( f))_{i \in I} + r_{U_i, U_i \cap U_j}(f_i) = r_{U_j, U_i \cap U_j}(f_j) \text{ for } i,j \in I \}\\ + f & \longmapsto (r_{U, U_i}( f))_{i \in I} \end{align*} - A presheaf is called \vocab[Presheaf! - separated]{separated} if $\phi_{U, (U_i)_{i \in I}}$ is - injective for all such $U$ and + A presheaf is called \vocab[Presheaf!separated]{separated} + if $\phi_{U, (U_i)_{i \in I}}$ is injective for all such $U$ and $(U_i)_{i \in I}$.\footnote{This also called ``locality''.} - It satisfies \vocab{gluing} if $\phi_{U, (U_i)_{i \in I}}$ is - surjective. + It satisfies \vocab{gluing} if $\phi_{U, (U_i)_{i \in I}}$ is surjective. - A presheaf is called a \vocab{sheaf} if it is separated and + A presheaf is called a \vocab{sheaf} if it is separated and satisfies gluing. The bijectivity of the $\phi_{U, (U_i)_{i \in I}}$ is called the - \vocab{sheaf - axiom}. + \vocab{sheaf axiom}. \end{definition} -\begin{trivial} - + - A presheaf is a contravariant functor $\mathcal{G} : - \mathcal{O}(X) \to C$ where $\mathcal{O}(X)$ denotes the category - of open subsets of $X$ with inclusions as morphisms and $C$ is the category of - sets, rings or (abelian) groups. +\begin{trivial}+ + A presheaf is a contravariant functor $\mathcal{G} :\mathcal{O}(X) \to C$ + where $\mathcal{O}(X)$ denotes the category of open subsets of $X$ + with inclusions as morphisms + and $C$ is the category of sets, rings or (abelian) groups. \end{trivial} \begin{definition} - A subsheaf $\mathcal{G}'$ is defined by subsets (resp. - subrings or subgroups) $\mathcal{G}'(U) \subseteq - \mathcal{G}(U)$ for all open $U \subseteq X$ such that the sheaf axioms - still hold. + A subsheaf $\mathcal{G}'$ is defined by subsets + (resp.~subrings or subgroups) $\mathcal{G}'(U) \subseteq \mathcal{G}(U)$ + for all open $U \subseteq X$ such that the sheaf axioms still hold. \end{definition} \begin{remark} If $\mathcal{G}$ is a sheaf on $X$ and $\Omega \subseteq X$ open, then $\mathcal{G}\defon{\Omega}(U) \coloneqq \mathcal{G}(U)$ - for open $U \subseteq - \Omega$ and $r_{U,V}^{(\mathcal{G}\defon{\Omega})}(f) \coloneqq - r_{U,V}^{(\mathcal{G})}(f)$ is a sheaf of the same kind as - $\mathcal{G}$ on - $\Omega$. + for open $U \subseteq \Omega$ + and $r_{U,V}^{(\mathcal{G}\defon{\Omega})}(f) \coloneqq r_{U,V}^{(\mathcal{G})}(f)$ + is a sheaf of the same kind as $\mathcal{G}$ on $\Omega$. \end{remark} \begin{remark} The notion of restriction of a sheaf to a closed subset, or of preimages under general continuous maps, can be defined but this is a bit harder. \end{remark} \begin{notation} - It is often convenient to write $f \defon{V}$ instead of - $r_{U,V}(f)$. + It is often convenient to write $f \defon{V}$ instead of $r_{U,V}(f)$. \end{notation} \begin{remark} Applying the \vocab{sheaf axiom} to the empty covering of $U = \emptyset$, - one - finds that $\mathcal{G}(\emptyset) = \{0\} $. + one finds that $\mathcal{G}(\emptyset) = \{0\} $. \end{remark} - \subsubsection{Examples of sheaves} \begin{example} Let $G$ be a set and let $\mathfrak{G}(U)$ be the set of arbitrary maps - $U - \xrightarrow{f} G$. + $U \xrightarrow{f} G$. We put $r_{U,V}(f) = f\defon{V}$. It is easy to see that this defines a sheaf. - If $\cdot $ is a group operation on $G$, then $(f\cdot g)(x) \coloneqq - f(x)\cdot g(x)$ defines the structure of a sheaf of group on - $\mathfrak{G}$. - Similarly, a ring structure on $G$ can be used to define the structure of a - sheaf of rings on $\mathfrak{G}$. + If $\cdot $ is a group operation on $G$, + then $(f\cdot g)(x) \coloneqq f(x)\cdot g(x)$ defines the structure + of a sheaf of group on $\mathfrak{G}$. + Similarly, a ring structure on $G$ can be used to define the structure + of a sheaf of rings on $\mathfrak{G}$. \end{example} \begin{example} - If in the previous example $G$ carries a topology and $\mathcal{G}(U) - \subseteq - \mathfrak{G}(U)$ is the subset (subring, subgroup) of continuous - functions $U - \xrightarrow{f} G$, then $\mathcal{G}$ is a subsheaf of - $\mathfrak{G}$, called - the sheaf of continuous $G$-valued functions on (open subsets of) $X$. + If in the previous example $G$ carries a topology + and $\mathcal{G}(U) \subseteq \mathfrak{G}(U)$ is the subset + (subring, subgroup) of continuous functions $U \xrightarrow{f} G$, + then $\mathcal{G}$ is a subsheaf of $\mathfrak{G}$, + called the sheaf of continuous $G$-valued functions on + (open subsets of) $X$. \end{example} \begin{example} - If $X = \R^n$, $\mathbb{K} \in \{\R, \C\}$ and - $\mathcal{O}(U)$ is the sheaf - of $\mathbb{K}$-valued $C^{\infty}$-functions on - $U$, then $\mathcal{O}$ is a - subsheaf of the sheaf (of rings) of $\mathbb{K}$-valued continuous - functions on - $X$. + If $X = \R^n$, $\mathbb{K} \in \{\R, \C\}$ + and $\mathcal{O}(U)$ is the sheaf of + $\mathbb{K}$-valued $C^{\infty}$-functions on $U$, + then $\mathcal{O}$ is a subsheaf of the sheaf (of rings) + of $\mathbb{K}$-valued continuous functions on $X$. \end{example} \begin{example} - If $X = \C^n$ and $\mathcal{O}(U)$ the set of holomorphic functions - on $X$, - then $\mathcal{O}$ is a subsheaf of the sheaf of $\C$-valued - $C^{\infty}$-functions on $X$. + If $X = \C^n$ and $\mathcal{O}(U)$ the set of holomorphic functions on $X$, + then $\mathcal{O}$ is a subsheaf of the sheaf of + $\C$-valued $C^{\infty}$-functions on $X$. \end{example} \subsubsection{The structure sheaf on a closed subset of $\mathfrak{k}^n$} @@ -134,26 +118,26 @@ Let $X \subseteq \mathfrak{k}^n$ be open. Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. \begin{definition} \label{structuresheafkn} - For open subsets $U \subseteq X$, let $\mathcal{O}_X(U)$ be the set - of - functions $U \xrightarrow{\phi} \mathfrak{k}$ such that every $x - \in U$ has a - neighbourhood $V$ such that there are $f,g \in R$ such that for $y \in V$ we - have $g(y) \neq 0$ and $\phi(y) = \frac{f(y)}{g(y)}$. + For open subsets $U \subseteq X$, let $\mathcal{O}_X(U)$ be the set of + functions $U \xrightarrow{\phi} \mathfrak{k}$ + such that every $x \in U$ has a neighbourhood $V$ + such that there are $f,g \in R$ + such that for $y \in V$ we have $g(y) \neq 0$ + and $\phi(y) = \frac{f(y)}{g(y)}$. \end{definition} \begin{remark} \label{structuresheafcontinuous} $\mathcal{O}_X$ is a subsheaf (of rings) of the sheaf of $\mathfrak{k}$-valued functions on $X$. - The elements of $\mathcal{O}_X(U)$ are continuous: Let $M \subseteq - \mathfrak{k}$ be closed. + The elements of $\mathcal{O}_X(U)$ are continuous: + Let $M \subseteq \mathfrak{k}$ be closed. We must show the closedness of $N \coloneqq \phi^{-1}(M)$ in $U$. For $M = \mathfrak{k}$ this is trivial. - Otherwise $M$ is finite and we may assume $M = \{t\} $ for some $t \in - \mathfrak{k}$. - For $x \in U$, there are open $V_x \subseteq U$ and $f_x, g_x \in R$ such that - $\phi = \frac{f_x}{g_x}$ on $V_x$. + Otherwise $M$ is finite and we may assume $M = \{t\} $ + for some $t \in \mathfrak{k}$. + For $x \in U$, there are open $V_x \subseteq U$ and $f_x, g_x \in R$ + such that $\phi = \frac{f_x}{g_x}$ on $V_x$. Then $N \cap V_x = V(f_x - t\cdot g_x) \cap V_x)$ is closed in $V_x$. As the $V_x$ cover $U$ and $U$ is quasi-compact, $N$ is closed in $U$. \end{remark} @@ -164,62 +148,55 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. Let $A = R / I$. Then \begin{align*} - \phi: A & \longrightarrow \mathcal{O}_X(X) \\ f \mod I - & \longmapsto f\defon{X} + \phi: A & \longrightarrow \mathcal{O}_X(X) \\ + f \mod I & \longmapsto f\defon{X} \end{align*} is an isomorphism. \end{proposition} \begin{proof} It is easy to see that the map $A \to \mathcal{O}_X(X)$ is - well-defined and a - ring homomorphism. - Its injectivity follows from the Nullstellensatz and $I = - \sqrt{I}$ + well-defined and a ring homomorphism. + Its injectivity follows from the Nullstellensatz and $I = \sqrt{I}$ (\ref{hns3}). - Let $\phi \in \mathcal{O}_X(X)$. - for $x \in X$, there are an open subset $U_x \subseteq X$ and $f_x, g_x \in R$ - such that $\phi = \frac{f_x}{g_x}$ on $U_x$. + For $x \in X$, there are an open subset $U_x \subseteq X$ + and $f_x, g_x \in R$ such that $\phi = \frac{f_x}{g_x}$ on $U_x$. \begin{claim} - Without loss of generality loss of generality we can assume $U_x = X \setminus - V(g_x)$. + Without loss of generality loss of generality we can assume + $U_x = X \setminus V(g_x)$. \end{claim} \begin{subproof} - The closed subsets $(X \setminus U_x) \subseteq \mathfrak{k}^n$ has - the form - $X\setminus U_x = V(J_x)$ for some ideal $J_x \subseteq R$. - As $x \not\in X \setminus V_x$ there is $h_x \in J_x$ with $h_x(x) \neq 0$. - Replacing $U_x$ by $X \setminus V(h_x)$, $f_x$ by $f_xh_x$ and $g_x$ by - $g_xh_x$, we may assume $U_x = X \setminus V(g_x)$. + The closed subsets $(X \setminus U_x) \subseteq \mathfrak{k}^n$ + has the form $X\setminus U_x = V(J_x)$ + for some ideal $J_x \subseteq R$. + As $x \not\in X \setminus V_x$ + there is $h_x \in J_x$ with $h_x(x) \neq 0$. + Replacing $U_x$ by $X \setminus V(h_x)$, $f_x$ by $f_xh_x$ + and $g_x$ by $g_xh_x$, we may assume $U_x = X \setminus V(g_x)$. \end{subproof} \begin{claim} - Without loss of generality loss of generality we can assume $V(g_x) \subseteq - V(f_x)$. + Without loss of generality loss of generality we can assume + $V(g_x) \subseteq V(f_x)$. \end{claim} \begin{subproof} Replace $f_x$ by $f_xg_x$ and $g_x$ by $g_x^2$. \end{subproof} As $X$ is quasi-compact, there are finitely many points - $(x_i)_{i=1}^m$ such - that the $U_{x_i}$ cover $X$. - Let $U_i \coloneqq U_{x_i}, f_i \coloneqq - f_{x_i}, g_i \coloneqq g_{x_i}$. + $(x_i)_{i=1}^m$ such that the $U_{x_i}$ cover $X$. + Let $U_i \coloneqq U_{x_i}, f_i \coloneqq f_{x_i}, g_i \coloneqq g_{x_i}$. - As the $U_i = X \setminus V(g_i)$ cover $X$, $V(I) \cap - \bigcap_{i=1}^m V(g_i) - = X \cap \bigcap_{i=1}^m V(g_i) = \emptyset$. + As the $U_i = X \setminus V(g_i)$ cover $X$, + $V(I) \cap \bigcap_{i=1}^m V(g_i) = X \cap \bigcap_{i=1}^m V(g_i) = \emptyset$. By the Nullstellensatz (\ref{hns1}) the ideal of $R$ generated by - $I$ and the - $a_i$ equals $R$. - There are thus $n \ge m \in \N$ and elements - $(g_i)_{i = m+1}^n$ of $I$ and - $(a_i)_{i=1}^n \in R^n$ such that $1 = - \sum_{i=1}^{n} a_ig_i$. - Let for $i > m$ $f_i \coloneqq 0$, $F = \sum_{i=1}^{n} a_if_i = - \sum_{i=1}^{m} - a_if_i \in R$. + $I$ and the $a_i$ equals $R$. + There are thus $n \ge m \in \N$ + and elements $(g_i)_{i = m+1}^n$ of $I$ + and $(a_i)_{i=1}^n \in R^n$ + such that $1 = \sum_{i=1}^{n} a_ig_i$. + Let for $i > m$ $f_i \coloneqq 0$, + $F = \sum_{i=1}^{n} a_if_i = \sum_{i=1}^{m} a_if_i \in R$. \begin{claim} For all $x \in X $ ~ $f_i(x) = \phi(x) g_i(x)$. @@ -234,62 +211,53 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. \phi(x) \cdot \sum_{i=1}^{n} a_i(x) g_i(x) = \sum_{i=1}^{n} a_i(x) f_i(x) = F(x) \] - Hence $\phi = - F\defon{X}$. + Hence $\phi = F\defon{X}$. \end{proof} \subsubsection{The structure sheaf on closed subsets of $\mathbb{P}^n$} -Let $X \subseteq \mathbb{P}^n$ be closed and $R_\bullet = -\mathfrak{k}[X_0,\ldots,X_n]$ with its usual grading. +Let $X \subseteq \mathbb{P}^n$ be closed +and $R_\bullet = \mathfrak{k}[X_0,\ldots,X_n]$ with its usual grading. \begin{definition} \label{structuresheafpn} - For open $U \subseteq X$, let $\mathcal{O}_X(U)$ be the set of - functions $U - \xrightarrow{\phi} \mathfrak{k}$ such that for every $x \in U$, - there are an - open subset $W \subseteq U$, a natural number $d$ and $f,g \in R_d$ such that - $W \cap \Vp(g) = \emptyset$ and $\phi(y) = - \frac{f(y_0,\ldots,y_n)}{g(y_0,\ldots,y_n)}$ for $y = [y_0,\ldots,y_n] \in W$. + For open $U \subseteq X$, let $\mathcal{O}_X(U)$ be the set of functions + $U \xrightarrow{\phi} \mathfrak{k}$ + such that for every $x \in U$, + there are an open subset $W \subseteq U$, + a natural number $d$ + and $f,g \in R_d$ such that + $W \cap \Vp(g) = \emptyset$ and $\phi(y) = \frac{f(y_0,\ldots,y_n)}{g(y_0,\ldots,y_n)}$ + for $y = [y_0,\ldots,y_n] \in W$. \end{definition} \begin{remark} - This is a subsheaf of rings of the sheaf of $\mathfrak{k}$-valued - functions on - $X$. + This is a subsheaf of rings of the sheaf of $\mathfrak{k}$-valued functions + on $X$. Under the identification $\mathbb{A}^n =\mathfrak{k}^n$ - with $\mathbb{P}^n - \setminus \Vp(X_0)$, one has $\mathcal{O}_X - \defon{X \setminus \Vp(X_0)} = - \mathcal{O}_{X \cap \mathbb{A}^n}$ as subsheaves of the - sheaf of - $\mathfrak{k}$-valued functions, where the second sheaf is a sheaf on - a closed - subset of $\mathfrak{k}^n$: + with $\mathbb{P}^n \setminus \Vp(X_0)$, + one has $\mathcal{O}_X \defon{X \setminus \Vp(X_0)} = \mathcal{O}_{X \cap \mathbb{A}^n}$ + as subsheaves of the sheaf of $\mathfrak{k}$-valued functions, + where the second sheaf is a sheaf on a closed subset of $\mathfrak{k}^n$: - Indeed, if $W$ is as in the definition then $\phi([1,y_1,\ldots,y_n]) = - \frac{f(1,y_1,\ldots,y_n)}{g(1,y_1,\ldots,y_n)}$ for $[1,y_1,\ldots,y_n] \in - W$. - Conversely if $\phi([1,y_1,\ldots,y_n]) = - \frac{f(y_1,\ldots,y_n)}{g(y_1,\ldots,y_n)}$ on an open subset $W $ of $X \cap - \mathbb{A}^n$ then $\phi([y_0,\ldots,y_n]) = - \frac{F(y_0,\ldots,y_n)}{G(y_0,\ldots,y_n)}$ on $W$ where $F(X_0,\ldots,X_n) - \coloneqq X_0^d f(\frac{X_1}{X_0}, \ldots, \frac{X_n}{X_0})$ and - $G(X_0,\ldots,X_n) = X_0^d g(\frac{X_1}{X_0},\ldots, - \frac{X_n}{X_0})$ with a - sufficiently large $d \in \N$. + Indeed, if $W$ is as in the definition then + $\phi([1,y_1,\ldots,y_n]) = \frac{f(1,y_1,\ldots,y_n)}{g(1,y_1,\ldots,y_n)}$ + for $[1,y_1,\ldots,y_n] \in W$. + Conversely if $\phi([1,y_1,\ldots,y_n]) = \frac{f(y_1,\ldots,y_n)}{g(y_1,\ldots,y_n)}$ + on an open subset $W $ of $X \cap \mathbb{A}^n$ + then $\phi([y_0,\ldots,y_n]) = \frac{F(y_0,\ldots,y_n)}{G(y_0,\ldots,y_n)}$ + on $W$ where $F(X_0,\ldots,X_n) \coloneqq X_0^d f(\frac{X_1}{X_0}, \ldots, \frac{X_n}{X_0})$ + and $G(X_0,\ldots,X_n) = X_0^d g(\frac{X_1}{X_0},\ldots, \frac{X_n}{X_0})$ + with a sufficiently large $d \in \N$. \end{remark} \begin{remark} It follows from the previous remark and the similar result in the affine case - that the elements of $\mathcal{O}_X(U)$ are continuous on $U - \setminus V(X_0)$. - Since the situation is symmetric in the homogeneous coordinates, they are - continuous on all of $U$. + that the elements of $\mathcal{O}_X(U)$ are continuous on $U \setminus V(X_0)$. + Since the situation is symmetric in the homogeneous coordinates, + they are continuous on all of $U$. \end{remark} The following is somewhat harder than in the affine case: \begin{proposition} If $X$ is connected (e.g. irreducible), then the elements of - $\mathcal{O}_X\left( X \right) $ are constant functions on - $X$. + $\mathcal{O}_X\left( X \right) $ are constant functions on $X$. \end{proposition} @@ -298,43 +266,40 @@ The following is somewhat harder than in the affine case: \subsection{The notion of a category} \begin{definition} - A \vocab{category} $\mathcal{A}$ consists of: + A \vocab{category} $\mathcal{A}$ consists of: \begin{itemize} \item - A class - $\Ob \mathcal{A}$ of \vocab[Objects]{objects of $\mathcal{A}$}. + A class $\Ob \mathcal{A}$ of + \vocab[Objects]{objects of $\mathcal{A}$}. \item - For two arbitrary objects $A, B \in \Ob \mathcal{A}$, a - \textbf{set} $\Hom_\mathcal{A}(A,B)$ of + For two arbitrary objects $A, B \in \Ob \mathcal{A}$, + a \textbf{set} $\Hom_\mathcal{A}(A,B)$ of \vocab[Morphism]{morphisms for $A$ to $B$ in $\mathcal{A}$}. \item - A map $\Hom_\mathcal{A}(B,C) \times \Hom_\mathcal{A}(A,B) - \xrightarrow{\circ} \Hom_\mathcal{A}(A,C)$, the composition of - morphisms, for arbitrary triples $(A,B,C)$ of objects of - $\mathcal{A}$. + A map $\Hom_\mathcal{A}(B,C) \times \Hom_\mathcal{A}(A,B) \xrightarrow{\circ} \Hom_\mathcal{A}(A,C)$, + the composition of morphisms, + for arbitrary triples $(A,B,C)$ of objects of $\mathcal{A}$. \end{itemize} The following conditions must be satisfied: \begin{enumerate}[A] \item - For - morphisms $A \xrightarrow{f} B\xrightarrow{g} C - \xrightarrow{h} D$, we have $h - \circ (g \circ f) = (h \circ g) \circ f$. + For morphisms + $A \xrightarrow{f} B\xrightarrow{g} C \xrightarrow{h} D$, + we have $h \circ (g \circ f) = (h \circ g) \circ f$. \item - For every $A \in \Ob(\mathcal{A})$, there is an $\Id_A \in - \Hom_{\mathcal{A}}(A,A)$ such that $\Id_A \circ f = f$ (reps. $g \circ \Id_A = - g$) for arbitrary morphisms $B \xrightarrow{f} - A$ (reps. - $A \xrightarrow{g} - C). - $ + For every $A \in \Ob(\mathcal{A})$, + there is an $\Id_A \in \Hom_{\mathcal{A}}(A,A)$ + such that $\Id_A \circ f = f$ + (reps. $g \circ \Id_A = g$) + for arbitrary morphisms $B \xrightarrow{f} A$ + (reps. $A \xrightarrow{g} C$). \end{enumerate} - A morphism $X \xrightarrow{f} Y$ is called an \vocab[Isomorphism]{isomorphism - (in $\mathcal{A} $)} - if there is a morphism $Y \xrightarrow{g} X$ (called the - \vocab[Inverse morphism]{inverse $f^{-1}$ of $f$)} such that $g \circ f = - \Id_X$ and $f \circ g = \Id_Y$. + A morphism $X \xrightarrow{f} Y$ is called an + \vocab[Isomorphism]{isomorphism (in $\mathcal{A} $)} + if there is a morphism $Y \xrightarrow{g} X$ + (called the \vocab[Inverse morphism]{inverse $f^{-1}$ of $f$}) + such that $g \circ f = \Id_X$ and $f \circ g = \Id_Y$. \end{definition} \begin{remark} \begin{itemize} @@ -343,8 +308,9 @@ The following is somewhat harder than in the affine case: \item We will usually omit the composition sign $\circ$. \item - It is easy to see that $\Id_A$ is uniquely determined by the above condition - $B$, and that the inverse $f^{-1}$ of an isomorphism + It is easy to see that $\Id_A$ is uniquely determined by + the above condition $B$, + and that the inverse $f^{-1}$ of an isomorphism $f$ is uniquely determined. \end{itemize} \end{remark} @@ -358,18 +324,19 @@ The following is somewhat harder than in the affine case: \item The category of rings. \item - If $R$ is a ring, the category of $R$-modules and the category $\Alg_R$ of + If $R$ is a ring, + the category of $R$-modules and the category $\Alg_R$ of $R$-algebras \item - The category of topological spaces + The category of topological spaces. \item The category $\Var_\mathfrak{k}$ of varieties over - $\mathfrak{k}$ (see \ref{defvariety}) + $\mathfrak{k}$ (see \ref{defvariety}). \item - If $\mathcal{A}$ is a category, then the \vocab{opposite category} - or \vocab{dual category} is defined by $\Ob(\mathcal{A}\op) = - \Ob(\mathcal{A})$ and $\Hom_{\mathcal{A}\op}(X,Y) = - \Hom_\mathcal{A}(Y,X)$. + If $\mathcal{A}$ is a category, + then the \vocab{opposite category} or \vocab{dual category} + is defined by $\Ob(\mathcal{A}\op) = \Ob(\mathcal{A})$ + and $\Hom_{\mathcal{A}\op}(X,Y) = \Hom_\mathcal{A}(Y,X)$. \end{itemize} In most of these cases, isomorphisms in the category were just called `isomorphism'. @@ -377,31 +344,31 @@ The following is somewhat harder than in the affine case: \end{example} \subsubsection{Subcategories} \begin{definition}[Subcategories] - A \vocab{subcategory} of - $\mathcal{A}$ is a category $\mathcal{B}$ such that - $\Ob(\mathcal{B}) \subseteq \Ob(\mathcal{A})$, such that - $\Hom_\mathcal{B}(X,Y) \subseteq \Hom_\mathcal{A}(X,Y)$ for - objects $X$ and $Y$ of $\mathcal{B}$, such that for every object $X - \in \Ob(\mathcal{B})$, the identity $\Id_X$ of $X$ is the same in - $\mathcal{B}$ as in $\mathcal{A}$, and such that for - composable morphisms in $\mathcal{B}$, their compositions in - $\mathcal{A}$ and $\mathcal{B}$ coincide. - We call $\mathcal{B}$ a \vocab{full subcategory} of - $\mathcal{A}$ if in + A \vocab{subcategory} of $\mathcal{A}$ is a category $\mathcal{B}$ + such that $\Ob(\mathcal{B}) \subseteq \Ob(\mathcal{A})$, + such that $\Hom_\mathcal{B}(X,Y) \subseteq \Hom_\mathcal{A}(X,Y)$ + for objects $X$ and $Y$ of $\mathcal{B}$, + such that for every object $X \in \Ob(\mathcal{B})$, + the identity $\Id_X$ of $X$ is the same in + $\mathcal{B}$ as in $\mathcal{A}$, + and such that for composable morphisms in $\mathcal{B}$, + their compositions in $\mathcal{A}$ and $\mathcal{B}$ coincide. + We call $\mathcal{B}$ a \vocab{full subcategory} of $\mathcal{A}$ if in addition $\Hom_\mathcal{B}(X,Y) = \Hom_\mathcal{A}(X,Y)$ for - arbitrary $X,Y \in - \Ob(\mathcal{B})$. + arbitrary $X,Y \in \Ob(\mathcal{B})$. \end{definition} \begin{example} \begin{itemize} \item - The category of abelian groups is a full subcategory of the category of groups. + The category of abelian groups is a full subcategory of the + category of groups. It can be identified with the category of $\Z$-modules. \item - The category of finitely generated $R$-modules as a full subcategory of the - category of $R$-modules. + The category of finitely generated $R$-modules as a full + subcategory of the category of $R$-modules. \item - The category of $R$-algebras of finite type as a full subcategory of $\Alg_R$. + The category of $R$-algebras of finite type as a full + subcategory of $\Alg_R$. \item The category of affine varieties over $\mathfrak{k}$ as a full subcategory of the category of varieties over $\mathfrak{k}$. @@ -410,49 +377,52 @@ The following is somewhat harder than in the affine case: \subsubsection{Functors and equivalences of categories} \begin{definition} - A \vocab[Functor! - covariant]{(covariant) functor} (resp. \vocab[Functor!contravariant]{contravariant functor}) between categories $\mathcal{A} - \xrightarrow{F} \mathcal{B}$ is a map - $\Ob(\mathcal{A}) \xrightarrow{F} \Ob(\mathcal{B})$ with - a family of maps $\Hom_\mathcal{A}(X,Y) \xrightarrow{F} - \Hom_\mathcal{B}(F(X),F(Y))$ (resp. $\Hom_\mathcal{A}(X,Y) - \xrightarrow{F} \Hom_\mathcal{B}(F(Y),F(X))$ in the case of - contravariant functors), where $X$ and $Y$ are arbitrary objects of - $\mathcal{A}$, such that the following conditions hold: + A \vocab[Functor!covariant]{(covariant) functor} + (resp.~\vocab[Functor!contravariant]{contravariant functor}) + between categories $\mathcal{A} \xrightarrow{F} \mathcal{B}$ is a map + $\Ob(\mathcal{A}) \xrightarrow{F} \Ob(\mathcal{B})$ + with a family of maps + $\Hom_\mathcal{A}(X,Y) \xrightarrow{F} \Hom_\mathcal{B}(F(X),F(Y))$ + (resp. $\Hom_\mathcal{A}(X,Y) \xrightarrow{F} \Hom_\mathcal{B}(F(Y),F(X))$ + in the case of contravariant functors), + where $X$ and $Y$ are arbitrary objects of $\mathcal{A}$, + such that the following conditions hold: \begin{itemize} \item - $F(\Id_X) = \Id_{F(X)}$ + $F(\Id_X) = \Id_{F(X)}$. \item - For morphisms $X \xrightarrow{f} - Y \xrightarrow{g} Z$ in $\mathcal{A}$, we have $F(gf) = - F(g)F(f)$ ( resp. - $F(gf) = F(f) - F(g)$) + For morphisms $X \xrightarrow{f} Y \xrightarrow{g} Z$ + in $\mathcal{A}$, + we have $F(gf) = F(g)F(f)$ + (resp.~$F(gf) = F(f) F(g)$). \end{itemize} - A functor is called \vocab[Functor! - essentially surjective]{essentially surjective} if every object of - $\mathcal{B}$ is isomorphic to an element of the image of + A functor is called + \vocab[Functor!essentially surjective]{essentially surjective} + if every object of $\mathcal{B}$ is isomorphic to + an element of the image of $\Ob(\mathcal{A}) \xrightarrow{F} \Ob(\mathcal{B})$. - A functor is called \vocab[Functor! - full]{full} (resp. \vocab[Functor!faithful]{faithful}) if it induces surjective (resp. - injective) maps between sets of morphisms. - It is called an \vocab{equivalence of categories} if it is full, faithful and - essentially surjective. + A functor is called \vocab[Functor!full]{full} + (resp.~\vocab[Functor!faithful]{faithful}) + if it induces surjective (resp.~injective) maps between sets of morphisms. + It is called an \vocab{equivalence of categories} if it is full, + faithful and essentially surjective. \end{definition} \begin{example} \begin{itemize} \item - There are \vocab[Functor!forgetful]{forgetful functors} from rings to abelian groups or from abelian - groups to sets which drop the multiplicative structure of a ring or the group + There are \vocab[Functor!forgetful]{forgetful functors} + from rings to abelian groups or from abelian groups to sets + which drop the multiplicative structure of a ring or the group structure of a group. \item If $\mathfrak{k}$ is any vector space there is a contravariant functor from $\mathfrak{k}$-vector spaces to itself sending $V$ to - its dual vector space $V\subseteq$ and $V \xrightarrow{f} - W$ to the dual linear map $W^{\ast} - \xrightarrow{f^{\ast}} V^{\ast}$. - When restricted to the full subcategory of finite-dimensional vector spaces it - becomes a contravariant self-equivalence of that category. + its dual vector space $V\subseteq$ and $V \xrightarrow{f} W$ + to the dual linear map + $W^{\ast} \xrightarrow{f^{\ast}} V^{\ast}$. + When restricted to the full subcategory of finite-dimensional + vector spaces it becomes a contravariant self-equivalence of + that category. \item The embedding of a subcategory is a faithful functor. In the case of a full subcategory it is also full. @@ -465,57 +435,52 @@ The following is somewhat harder than in the affine case: \begin{definition}[Algebraic variety] \label{defvariety} - An \vocab{algebraic variety} or \vocab{prevariety} over - $\mathfrak{k}$ is a pair $(X, \mathcal{O}_X)$, - where $X$ is a topological space and $\mathcal{O}_X$ - a subsheaf of the sheaf of $\mathfrak{k}$-valued functions on $X$ + An \vocab{algebraic variety} or \vocab{prevariety} over $\mathfrak{k}$ + is a pair $(X, \mathcal{O}_X)$, + where $X$ is a topological space + and $\mathcal{O}_X$ a subsheaf of the sheaf of $\mathfrak{k}$-valued + functions on $X$ such that for every $x \in X$, there are a neighbourhood $U_x$ of $x$ in $X$, an open subset $V_x$ of a closed subset $Y_x$ of $\mathfrak{k}^{n_x}$% \footnote{By the result of \ref{affopensubtopbase}, it can be assumed that $V_x = Y_x$ without altering the definition.} - and a homeomorphism $V_x - \xrightarrow{\iota_x} - U_x$ such that for every open subset $V \subseteq U_x$ and every function - $V\xrightarrow{f} \mathfrak{k}$, we have $f \in - \mathcal{O}_X(V) \iff - \iota^{\ast}_x(f) \in - \mathcal{O}_{Y_x}(\iota_x^{-1}(V))$. + and a homeomorphism $V_x \xrightarrow{\iota_x} U_x$ + such that for every open subset $V \subseteq U_x$ + and every function $V\xrightarrow{f} \mathfrak{k}$, we have + $f \in \mathcal{O}_X(V) \iff \iota^{\ast}_x(f) \in \mathcal{O}_{Y_x}(\iota_x^{-1}(V))$. In this, the \vocab{pull-back} $\iota_x^{\ast}(f)$ of $f$ is - defined by - $(\iota_x^{\ast}(f))(\xi) \coloneqq f(\iota_x(\xi))$. + defined by $(\iota_x^{\ast}(f))(\xi) \coloneqq f(\iota_x(\xi))$. - A morphism $(X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$ of - varieties is a - continuous map $X \xrightarrow{\phi} Y$ such that for all open $U - \subseteq Y$ - and $f \in \mathcal{O}_Y(U)$, $\phi^{\ast}(f) \in - \mathcal{O}_X(\phi^{-1}(U))$. + A morphism $(X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$ of varieties is a + continuous map $X \xrightarrow{\phi} Y$ such that for all open + $U \subseteq Y$ + and $f \in \mathcal{O}_Y(U)$, + $\phi^{\ast}(f) \in \mathcal{O}_X(\phi^{-1}(U))$. An isomorphism is a morphism such that $\phi$ is bijective and - $\phi^{-1}$ also - is a morphism of varieties. + $\phi^{-1}$ also is a morphism of varieties. \end{definition} \begin{example} \begin{itemize} \item - If $(X, \mathcal{O}_X)$ is a variety and $U \subseteq X$ open, then - $(U, \mathcal{O}_X\defon{U})$ is a variety (called an - \vocab{open subvariety} of $X$), and the embedding $U \to X$ is a morphism of - varieties. + If $(X, \mathcal{O}_X)$ is a variety and $U \subseteq X$ open, + then $(U, \mathcal{O}_X\defon{U})$ is a variety + (called an \vocab{open subvariety} of $X$), + and the embedding $U \to X$ is a morphism of varieties. \item - If $X$ is a closed subset of $\mathfrak{k}^n$ or - $\mathbb{P}^n$, then $(X, \mathcal{O}_X)$ is a - variety, where $\mathcal{O}_X$ is the structure sheaf on $X$ + If $X$ is a closed subset of $\mathfrak{k}^n$ or $\mathbb{P}^n$, + then $(X, \mathcal{O}_X)$ is a variety, + where $\mathcal{O}_X$ is the structure sheaf on $X$ (\ref{structuresheafkn}, reps. \ref{structuresheafpn}). A variety is called \vocab[Variety!affine]{affine} - (resp. \vocab[Variety!projective]{projective}) + (resp.~\vocab[Variety!projective]{projective}) if it is isomorphic to a variety of this form, with $X $ closed in $\mathfrak{k}^n$ (resp. $\mathbb{P}^n$). A variety which is isomorphic to and open subvariety of $X$ is called \vocab[Variety!quasi-affine]{quasi-affine} - (resp. \vocab[Variety!quasi-projective]{quasi-projective}). + (resp.~\vocab[Variety!quasi-projective]{quasi-projective}). \item If $X = V(X^2 - Y^3) \subseteq \mathfrak{k}^2$ then $\mathfrak{k} \xrightarrow{t \mapsto (t^3,t^2)} @@ -540,49 +505,49 @@ The following is somewhat harder than in the affine case: All elements of $\mathcal{O}_X(U)$ are continuous. \item If $U \subseteq X$ is open, $U \xrightarrow{\lambda} \mathfrak{k}$ - any function and every $x \in U$ has a neighbourhood $V_x \subseteq U$ such - that $\lambda \defon{V_x} \in \mathcal{O}_X(V_x)$, + any function and every $x \in U$ has a neighbourhood $V_x \subseteq U$ + such that $\lambda \defon{V_x} \in \mathcal{O}_X(V_x)$, then $\lambda \in \mathcal{O}_X(U)$. \item - If $\vartheta \in \mathcal{O}_X(U)$ and $\vartheta(x) - \neq 0$ for all $x \in U$, then $\vartheta \in - \mathcal{O}_X(U)^{\times }$. + If $\vartheta \in \mathcal{O}_X(U)$ and $\vartheta(x) \neq 0$ + for all $x \in U$, + then $\vartheta \in \mathcal{O}_X(U)^{\times }$. \end{enumerate} \end{lemma} \begin{proof} \begin{enumerate}[i)] \item - The property is local on $U$, hence it is sufficient to show it in the - quasi-affine case. + The property is local on $U$, + hence it is sufficient to show it in the quasi-affine case. This was done in \ref{structuresheafcontinuous}. \item - For the second part, let $\lambda_x \coloneqq \lambda \defon{V_x} - $. - We have $\lambda_x\defon{V_x \cap V_y} = \lambda \defon{V_x \cap V_y} = - \lambda_y \defon{V_x \cap V_y} $. + For the second part, + let $\lambda_x \coloneqq \lambda \defon{V_x} $. + We have + $\lambda_x\defon{V_x \cap V_y} + = \lambda \defon{V_x \cap V_y} + = \lambda_y \defon{V_x \cap V_y} $. The $V_x$ cover $U$. - By the sheaf axiom for $\mathcal{O}_X$ there is $\ell \in - \mathcal{O}_X(U)$ - with $\ell\defon{V_x} =\lambda_x$. + By the sheaf axiom for $\mathcal{O}_X$ there is + $\ell \in \mathcal{O}_X(U)$ with $\ell\defon{V_x} =\lambda_x$. It follows that $\ell=\lambda$. \item - By the definition of variety, every $x \in U$ has a quasi-affine neighbourhood - $V \subseteq U$. - We can assume $U$ to be quasi-affine and $X = V(I) \subseteq - \mathfrak{k}^n$, + By the definition of variety, + every $x \in U$ has a quasi-affine neighbourhood $V \subseteq U$. + We can assume $U$ to be quasi-affine + and $X = V(I) \subseteq \mathfrak{k}^n$, as the general assertion follows by an application of ii). - If $x \in U$ there are a neighbourhood $x \in W \subseteq U$ and $a,b \in R = - \mathfrak{k}[X_1,\ldots,X_n]$ such that $\vartheta(y) = - \frac{a(y)}{b(y)}$ for - $y \in W$, with $b(y) \neq 0$. + If $x \in U$ there are a neighbourhood + $x \in W \subseteq U$ and $a,b \in R = \mathfrak{k}[X_1,\ldots,X_n]$ + such that $\vartheta(y) = \frac{a(y)}{b(y)}$ + for $y \in W$, with $b(y) \neq 0$. Then $a(x) \neq 0$ as $\vartheta(x) \neq 0$. - Replacing $W$ by $W \setminus V(a)$, we may assume that $a$ has no zeroes on - $W$. + Replacing $W$ by $W \setminus V(a)$, + we may assume that $a$ has no zeroes on $W$. Then $\lambda(y) = \frac{b(y)}{a(y)}$ for $y \in W$ has a - non-vanishing - denominator and $\lambda \in \mathcal{O}_X(U)$. - We have $\lambda \cdot \vartheta = 1$, thus $\vartheta \in - \mathcal{O}_X(U)^{\times}$. + non-vanishing denominator and $\lambda \in \mathcal{O}_X(U)$. + We have $\lambda \cdot \vartheta = 1$, + thus $\vartheta \in \mathcal{O}_X(U)^{\times}$. \end{enumerate} @@ -594,191 +559,157 @@ The following is somewhat harder than in the affine case: Let $X,Y$ be varieties over $\mathfrak{k}$. Then the map \begin{align*} - \phi: \Hom_{\Var_\mathfrak{k}}(X,Y) & \longrightarrow - \Hom_{\Alg_\mathfrak{k}}(\mathcal{O}_Y(Y), \mathcal{O}_X(X)) \\ (X - \xrightarrow{f} Y) & \longmapsto (\mathcal{O}_Y(Y) \xrightarrow{f^{\ast}} - \mathcal{O}_X(X)) + \phi: \Hom_{\Var_\mathfrak{k}}(X,Y) + & \longrightarrow \Hom_{\Alg_\mathfrak{k}}(\mathcal{O}_Y(Y), \mathcal{O}_X(X))\\ + (X \xrightarrow{f} Y) + & \longmapsto (\mathcal{O}_Y(Y) \xrightarrow{f^{\ast}} \mathcal{O}_X(X)) \end{align*} - is injective when $Y$ is quasi-affine and - bijective when $Y$ is affine. + is injective when $Y$ is quasi-affine + and bijective when $Y$ is affine. \item The contravariant functor \begin{align*} - F: \Var_\mathfrak{k} & \longrightarrow \Alg_\mathfrak{k} \\ X & \longmapsto - \mathcal{O}_X(X) \\ (X\xrightarrow{f} Y) & \longmapsto (\mathcal{O}_X(X) - \xrightarrow{f^{\ast}} \mathcal{O}_Y(Y)) + F: \Var_\mathfrak{k} & \longrightarrow \Alg_\mathfrak{k} \\ + X & \longmapsto \mathcal{O}_X(X)\\ + (X\xrightarrow{f} Y) + & \longmapsto + (\mathcal{O}_X(X) \xrightarrow{f^{\ast}} \mathcal{O}_Y(Y)) \end{align*} - restricts to an - equivalence of categories between the category of affine varieties over - $\mathfrak{k}$ and the full subcategory $\mathcal{A}$ of - $\Alg_\mathfrak{k}$, - having the $\mathfrak{k}$-algebras $A$ of finite type with $\nil A = - \{0\} $ as - objects. + restricts to an equivalence of categories between the category + of affine varieties over $\mathfrak{k}$ + and the full subcategory $\mathcal{A}$ of $\Alg_\mathfrak{k}$, + having the $\mathfrak{k}$-algebras $A$ of finite type + with $\nil A = \{0\} $ as objects. \end{itemize} \end{proposition} \begin{remark} - It is clear that $\nil(\mathcal{O}_X(X)) = \{0\}$ for arbitrary - varieties. - For general varieties it is however not true that - $\mathcal{O}_X(X)$ is a + It is clear that $\nil(\mathcal{O}_X(X)) = \{0\}$ for arbitrary varieties. + For general varieties it is however not true that $\mathcal{O}_X(X)$ is a $\mathfrak{k}$-algebra of finite type. There are counterexamples even for quasi-affine $X$. %TODO - If, however, $X$ is affine, we may assume w.l.o.g. - that $X = V(I)$ where $I = \sqrt{I} \subseteq R$ is an ideal + If, however, $X$ is affine, we may assume w.l.o.g.~that $X = V(I)$ + where $I = \sqrt{I} \subseteq R$ is an ideal with $R = \mathfrak{k}[X_1,\ldots,X_n]$. - Then $\mathcal{O}_X(X) \cong R / I$ (see - \ref{structuresheafri}) - is a - $\mathfrak{k}$-algebra of finite type. + Then $\mathcal{O}_X(X) \cong R / I$ (see \ref{structuresheafri}) + is a $\mathfrak{k}$-algebra of finite type. \end{remark} \begin{proof} - - - - - It suffices to investigate $\phi$ when $Y$ is an open subset of $V(I) \subseteq - \mathfrak{k}^n$, where $I = \sqrt{I} \subseteq R$ is - an ideal and $Y = V(I)$ + It suffices to investigate $\phi$ when $Y$ is an open subset of + $V(I) \subseteq \mathfrak{k}^n$, + where $I = \sqrt{I} \subseteq R$ is an ideal and $Y = V(I)$ when $Y$ is affine. - Let $(f_1,\ldots,f_n)$ be the components of $X \xrightarrow{f} Y - \subseteq - \mathfrak{k}^n$. - Let $Y \xrightarrow{\xi_i} \mathfrak{k}$ be the $i$-th - coordinate. - By definition $f_i = f^{\ast}(\xi_i) $. - Thus $f$ is uniquely determined by $\mathcal{O}_Y(Y) - \xrightarrow{f^{\ast}} - \mathcal{O}_X(X)$. - Conversely, let $Y = V(I)$ and $\mathcal{O}_Y(Y) - \xrightarrow{\phi} - \mathcal{O}_X(X)$ be a morphism of - $\mathfrak{k}$-algebras. - Define $f_i \coloneqq \phi(\xi_i)$ and consider $X - \xrightarrow{f = - (f_1,\ldots,f_n)} Y\subseteq \mathfrak{k}^n$. + Let $(f_1,\ldots,f_n)$ be the components of + $X \xrightarrow{f} Y \subseteq \mathfrak{k}^n$. + Let $Y \xrightarrow{\xi_i} \mathfrak{k}$ be the $i$-th coordinate. + By definition $f_i = f^{\ast}(\xi_i)$. + Thus $f$ is uniquely determined by + $\mathcal{O}_Y(Y) \xrightarrow{f^{\ast}} \mathcal{O}_X(X)$. + Conversely, let $Y = V(I)$ + and $\mathcal{O}_Y(Y) \xrightarrow{\phi} \mathcal{O}_X(X)$ + be a morphism of $\mathfrak{k}$-algebras. + Define $f_i \coloneqq \phi(\xi_i)$ and consider + $X \xrightarrow{f = (f_1,\ldots,f_n)} Y\subseteq \mathfrak{k}^n$. \begin{claim} $f$ has image contained in $Y$. \end{claim} \begin{subproof} - For $x \in X, \lambda \in I$ we have $\lambda(f(x)) = - (\phi(\lambda \mod I))(x) - = 0$ as $\phi$ is a morphism of $\mathfrak{k}$-algebras. + For $x \in X, \lambda \in I$ we have + $\lambda(f(x)) = (\phi(\lambda \mod I))(x) = 0$ + as $\phi$ is a morphism of $\mathfrak{k}$-algebras. Thus $f(x) \in V(I) = Y$. \end{subproof} \begin{claim} $f$ is a morphism in $\Var_\mathfrak{k}$ \end{claim} \begin{subproof} - For open $\Omega \subseteq Y, U = f^{-1}(\Omega) = \{x - \in X | \forall \lambda - \in J ~ (\phi(\lambda))(x) \neq 0\}$ is open in $X$, where $Y - \setminus \Omega - = V(J)$. - If $\lambda \in \mathcal{O}_Y(\Omega)$ and $x \in U$, then $f(x)$ - has a - neighbourhood $V$ such that there are $a,b \in R$ with $\lambda(v) - = - \frac{a(v)}{b(v)}$ and $b(v) \neq 0$ for all $v \in V$. + For open $\Omega \subseteq Y$, + $U = f^{-1}(\Omega) = \{x \in X | \forall \lambda \in J ~ (\phi(\lambda))(x) \neq 0\}$ + is open in $X$, where $Y \setminus \Omega = V(J)$. + If $\lambda \in \mathcal{O}_Y(\Omega)$ and $x \in U$, + then $f(x)$ has a neighbourhood $V$ + such that there are $a,b \in R$ with $\lambda(v) = \frac{a(v)}{b(v)}$ + and $b(v) \neq 0$ for all $v \in V$. Let $W \coloneqq f^{-1}(V)$. - Then $\alpha \coloneqq \phi(a)\defon{W} \in - \mathcal{O}_X(W)$, $\beta - \coloneqq \phi(b)\defon{W} \in - \mathcal{O}_X(W)$. + Then $\alpha \coloneqq \phi(a)\defon{W} \in \mathcal{O}_X(W)$, + $\beta \coloneqq \phi(b)\defon{W} \in \mathcal{O}_X(W)$. By the second part of - \ref{localinverse} $\beta \in - \mathcal{O}_X(W)^{\times}$ - and $f^{\ast}(\lambda)\defon{W} = - \frac{\alpha}{\beta} \in \mathcal{O}_X(W)$. - The first part of - \ref{localinverse} shows that - $f^{\ast}(\lambda) \in - \mathcal{O}_X(U)$. + \ref{localinverse} $\beta \in \mathcal{O}_X(W)^{\times}$ and + $f^{\ast}(\lambda)\defon{W} = \frac{\alpha}{\beta} \in \mathcal{O}_X(W)$. + The first part of \ref{localinverse} shows that + $f^{\ast}(\lambda) \in \mathcal{O}_X(U)$. \end{subproof} By definition of $f$, we have $f^{\ast} = \phi$. This finished the proof of the first point. - \begin{claim} The functor in the second part maps affine varieties to objects of $\mathcal{A}$ and is essentially surjective. \end{claim} \begin{subproof} - It follows from the remark that the functor maps affine varieties to objects of - $\mathcal{A}$. + It follows from the remark that the functor maps affine varieties to + objects of $\mathcal{A}$. - If $A \in \Ob(\mathcal{A})$ then $ A /\mathfrak{k}$ is of - finite type, thus $A - \cong R / I$ for some $n$. + If $A \in \Ob(\mathcal{A})$ then $ A /\mathfrak{k}$ is of finite type, + thus $A \cong R / I$ for some $n$. Since $\nil(A) = \{0\}$ we have $I = \sqrt{I}$, - as for $x \in \sqrt{I}$, $x - \mod I \in \nil(R / I) \cong \nil(A) = \{0\}$. + as for $x \in \sqrt{I}$, + $x \mod I \in \nil(R / I) \cong \nil(A) = \{0\}$. Thus $A \cong\mathcal{O}_X(X)$ where $X = V(I)$. \end{subproof} Fullness and faithfulness of the functor follow from the first point. \end{proof} \begin{remark} - Note that giving a contravariant functor $\mathcal{C} \to - \mathcal{D}$ is - equivalent to giving a functor $\mathcal{C} \to - \mathcal{D}\op$. + Note that giving a contravariant functor $\mathcal{C} \to \mathcal{D}$ is + equivalent to giving a functor $\mathcal{C} \to \mathcal{D}\op$. We have thus shown that the category of affine varieties is equivalent to - $\mathcal{A}\op$, where $\mathcal{A} \subsetneq - \Alg_\mathfrak{k}$ is the full + $\mathcal{A}\op$, + where $\mathcal{A} \subsetneq \Alg_\mathfrak{k}$ is the full subcategory of $\mathfrak{k}$-algebras $A$ of finite type with - $\nil(A) = - \{0\}$. + $\nil(A) = \{0\}$. \end{remark} \subsubsection{Affine open subsets are a topology base} \begin{definition} A set $\mathcal{B}$ of open subsets of a topological space $X$ is - called a - \vocab{topology base} for $X$ if every open subset of $X$ can be written as a - (possibly empty) union of elements of $\mathcal{B}$. + called a \vocab{topology base} for $X$ + if every open subset of $X$ can be written as a (possibly empty) + union of elements of $\mathcal{B}$. \end{definition} \begin{fact} - If $X$ is a set, then $\mathcal{B} \subseteq - \mathcal{P}(X)$ is a base for some - topology on $X$ iff $X = \bigcup_{U \in \mathcal{B}} U$ and for arbitrary $U, V - \in \mathcal{B}, U \cap V$ is a union of elements of - $\mathcal{B}$. + If $X$ is a set, + then $\mathcal{B} \subseteq \mathcal{P}(X)$ is a base for some topology + on $X$ iff $X = \bigcup_{U \in \mathcal{B}} U$ + and for arbitrary $U, V \in \mathcal{B}$, + $U \cap V$ is a union of elements of $\mathcal{B}$. \end{fact} \begin{definition} Let $X$ be a variety. An \vocab{affine open subset} of $X$ is a subset which is an affine variety. - \end{definition} \begin{proposition} \label{oxulocaf} - Let $X$ be an affine variety over $\mathfrak{k}$, $\lambda \in - \mathcal{O}_X(X)$ and $U = X \setminus V(\lambda)$. - Then $U$ is an affine variety and the morphism $\phi: - \mathcal{O}_X(X)_\lambda - \to \mathcal{O}_X(U)$ defined by the restriction - $\mathcal{O}_X(X) - \xrightarrow{\cdot |_U } \mathcal{O}_X(U)$ and the universal - property of the - localization is an isomorphism. + Let $X$ be an affine variety over $\mathfrak{k}$, + $\lambda \in \mathcal{O}_X(X)$ and $U = X \setminus V(\lambda)$. + Then $U$ is an affine variety + and the morphism $\phi: \mathcal{O}_X(X)_\lambda \to \mathcal{O}_X(U)$ + defined by the restriction + $\mathcal{O}_X(X) \xrightarrow{\cdot |_U } \mathcal{O}_X(U)$ + and the universal property of the localization is an isomorphism. \end{proposition} \begin{proof} - Let $X$ be an affine variety over $\mathfrak{k}, \lambda \in - \mathcal{O}_X(X)$ - and $U = X \setminus V(\lambda)$. - The fact that $\lambda\defon{U} \in - \mathcal{O}_x(U)^{\times}$ follows - from - \ref{localinverse}. + Let $X$ be an affine variety over $\mathfrak{k}$, + $\lambda \in \mathcal{O}_X(X)$ and $U = X \setminus V(\lambda)$. + The fact that $\lambda\defon{U} \in \mathcal{O}_x(U)^{\times}$ follows + from \ref{localinverse}. Thus the universal property of the localization - $\mathcal{O}_X(X)_\lambda$ can - be applied to $\mathcal{O}_X(X) \xrightarrow{\cdot |_U} - \mathcal{O}_X(U)$. + $\mathcal{O}_X(X)_\lambda$ can be applied to + $\mathcal{O}_X(X) \xrightarrow{\cdot |_U} \mathcal{O}_X(U)$. \[ \begin{tikzcd} \mathcal{O}_X(X) \arrow{d}{\cdot |_U}\arrow{r}{x \mapsto \frac{x}{1}} & \mathcal{O}_X(X)_\lambda \arrow[dotted, bend left]{dl}{\existsone \phi} \\ @@ -791,40 +722,35 @@ The following is somewhat harder than in the affine case: X \arrow[hookrightarrow]{r}{}& U \arrow[swap]{u}{\sigma} & \mathcal{O}_X(U) \end{tikzcd} \] - For the rest of the proof, we may assume $X = V(I) \subseteq - \mathfrak{k}^n$ where $I = \sqrt{I} \subseteq R - \coloneqq\mathfrak{k}[X_1,\ldots,X_n]$ is an ideal. - Then $A \coloneqq \mathcal{O}_X(X) \cong R / I$ and there is $\ell - \in R$ such - that $\ell\defon{X} = \lambda$. - Let $Y = V(J) \subseteq \mathfrak{k}^{n+1}$ where $J \subseteq - \mathfrak{k}[Z,X_1,\ldots,X_n]$ is generated by the elements of $I$ and $1 - - Z\ell(X_1,\ldots,X_n)$. + For the rest of the proof, + we may assume $X = V(I) \subseteq \mathfrak{k}^n$ + where $I = \sqrt{I} \subseteq R \coloneqq\mathfrak{k}[X_1,\ldots,X_n]$ + is an ideal. + Then $A \coloneqq \mathcal{O}_X(X) \cong R / I$ + and there is $\ell \in R$ + such that $\ell\defon{X} = \lambda$. + Let $Y = V(J) \subseteq \mathfrak{k}^{n+1}$ + where $J \subseteq \mathfrak{k}[Z,X_1,\ldots,X_n]$ is generated by the + elements of $I$ and $1 - Z\ell(X_1,\ldots,X_n)$. - Then $\mathcal{O}_Y(Y) \cong \mathfrak{k}[Z,X_1,\ldots,X_n] / J \cong - A[Z] / (1 - -\lambda Z) \cong A_\lambda$. - By the proposition about affine varieties ( - \ref{propaffvar}), the - morphism - $\mathfrak{s}: \mathcal{O}_Y(Y) \cong A_\lambda \to - \mathcal{O}_X(U)$ + Then $\mathcal{O}_Y(Y) \cong \mathfrak{k}[Z,X_1,\ldots,X_n] / J \cong A[Z] / (1 -\lambda Z) \cong A_\lambda$. + By the proposition about affine varieties (\ref{propaffvar}), + the morphism + $\mathfrak{s}: \mathcal{O}_Y(Y) \cong A_\lambda \to \mathcal{O}_X(U)$ corresponds to a morphism $U \xrightarrow{\sigma} Y$. We have $\mathfrak{s}(Z \mod J) = \lambda^{-1}$ and - $\mathfrak{s}(X_i \mod J) = - X_i \mod I$. + $\mathfrak{s}(X_i \mod J) = X_i \mod I$. Thus $\sigma(x) = (\lambda(x)^{-1}, x)$ for $x \in U$. Moreover, the projection $Y \xrightarrow{\pi_0} X$ dropping the - $Z$-coordinate - has image contained in $U$, as for $(z,x) \in Y$ the equation + $Z$-coordinate has image contained in $U$, + as for $(z,x) \in Y$ the equation \[ - 1 = - z\lambda(x) + 1 = z\lambda(x) \] implies $\lambda(x) \neq 0$. - It thus defines a morphism $Y \xrightarrow{\pi} U$ and by the description - of - $\sigma$ it follows that $\sigma \pi = \Id_U$. + It thus defines a morphism $Y \xrightarrow{\pi} U$ + and by the description of $\sigma$ + it follows that $\sigma \pi = \Id_U$. Similarly it follows that $\sigma \pi = \Id_Y$. Thus, $\sigma$ and $\pi$ are inverse to each other. \end{proof} @@ -833,20 +759,21 @@ The following is somewhat harder than in the affine case: The affine open subsets of a variety $X$ are a topology base on $X$. \end{corollary} \begin{proof} - Let $X = V(I) \subseteq \mathfrak{k}^n$ with $I = - \sqrt{I}$. - If $U \subseteq X$ is open then $X \setminus U = V(J)$ with $J \supseteq I$ and - $U = \bigcup_{f \in J} (X \setminus V(f))$. + Let $X = V(I) \subseteq \mathfrak{k}^n$ + with $I = \sqrt{I}$. + If $U \subseteq X$ is open then $X \setminus U = V(J)$ + with $J \supseteq I$ and $U = \bigcup_{f \in J} (X \setminus V(f))$. Thus $U$ is a union of affine open subsets. The same then holds for arbitrary quasi-affine varieties. Let $X$ be any variety, $U \subseteq X$ open and $x \in U$. By the definition of variety, $x$ has a neighbourhood $V_x$ which is - quasi-affine, and replacing $V_x$ by $U \cap V_x$ which is also quasi-affine we - may assume $V_x \subseteq U$. + quasi-affine, + and replacing $V_x$ by $U \cap V_x$ which is also quasi-affine + we may assume $V_x \subseteq U$. $V_x$ is a union of its affine open subsets. - Because $U$ is the union of the $V_x$, $U$ as well is a union of affine open - subsets. + Because $U$ is the union of the $V_x$, + $U$ as well is a union of affine open subsets. \end{proof} @@ -859,36 +786,30 @@ The following is somewhat harder than in the affine case: \subsection{Stalks of sheaves} \begin{definition}[Stalk] - Let $\mathcal{G}$ be a presheaf of sets on - the topological space $X$, and let $x \in X$. - The \vocab{stalk} (\vocab[Stalk]{Halm}) of $\mathcal{G}$ at $x$ is the set of - equivalence classes of pairs $(U, \gamma)$, where $U$ is an open neighbourhood + Let $\mathcal{G}$ be a presheaf of sets on the topological space $X$, + and let $x \in X$. + The \vocab{stalk} (\vocab[Stalk]{Halm}) of $\mathcal{G}$ at $x$ + is the set of equivalence classes of pairs $(U, + \gamma)$, where $U$ is an open neighbourhood of $x$ and $\gamma \in \mathcal{G}(U)$ and the equivalence relation - $\sim $ is - defined as follows: $( U , \gamma) \sim (V, \delta)$ iff there exists - an open - neighbourhood $W \subseteq U \cap V$ of $x$ such that $\gamma - \defon{W} = - \delta \defon{W}$. + $\sim $ is defined as follows: + $( U , \gamma) \sim (V, \delta)$ iff there exists an open + neighbourhood $W \subseteq U \cap V$ of $x$ + such that $\gamma \defon{W} = \delta \defon{W}$. - - If $\mathcal{G}$ is a presheaf of groups, one can define a groups - structure on - $\mathcal{G}_x$ by + If $\mathcal{G}$ is a presheaf of groups, + one can define a groups structure on $\mathcal{G}_x$ by \[ - ((U, \gamma) / \sim ) \cdot \left( (V,\delta) / \sim - \right) = (U \cap V, \gamma \defon{U \cap V} \cdot - \delta\defon{U \cap V}) / - \sim + ((U, \gamma) / \sim ) \cdot \left( (V,\delta) / \sim \right) + = (U \cap V, \gamma \defon{U \cap V} \cdot \delta\defon{U \cap V}) / \sim. \] - If $\mathcal{G}$ is a presheaf of rings, one can similarly define a - ring - structure on $\mathcal{G}_x$. + If $\mathcal{G}$ is a presheaf of rings, + one can similarly define a ring structure on $\mathcal{G}_x$. - If $U$ is an open neighbourhood of $x \in X$, then we have a map (resp. - homomorphism) + If $U$ is an open neighbourhood of $x \in X$, + then we have a map (resp. homomorphism) \begin{align*} \cdot_x : \mathcal{G}(U) & \longrightarrow \mathcal{G}_x\\ \gamma & \longmapsto \gamma_x \coloneqq (U, \gamma) / \sim @@ -896,177 +817,161 @@ The following is somewhat harder than in the affine case: \end{definition} \begin{fact} - Let $\gamma,\delta \in \mathcal{G}(U)$. - If $\mathcal{G}$ is a sheaf\footnote{or, more generally, a separated presheaf} - and if for all $x \in U$, we have $\gamma_x = \delta_x$, then $\gamma = - \delta$. + Let $\gamma,\delta \in \mathcal{G}(U)$. + If $\mathcal{G}$ is a sheaf% + \footnote{or, more generally, a separated presheaf} + and if for all $x \in U$, + we have $\gamma_x = \delta_x$, + then $\gamma = \delta$. - In the case of a sheaf, the image of the injective map $\mathcal{G}(U) - \xrightarrow{\gamma \mapsto (\gamma_x)_{x \in U}} \prod_{x \in U} - \mathcal{G}_x$ is the set of all - $(g_x)_{x \in U} \in \prod_{x \in U} - \mathcal{G}_x $ satisfying the following \vocab{coherence condition}: - For - every $x \in U$, there are an open neighbourhood $W_x \subseteq U$ of $x$ and - $g^{(x)} \in \mathcal{G}(W_x)$ with $g_y^{(x)} = - g_y$ for all $y \in W_x$. + In the case of a sheaf, + the image of the injective map + $\mathcal{G}(U) \xrightarrow{\gamma \mapsto (\gamma_x)_{x \in U}} + \prod_{x \in U} \mathcal{G}_x$ + is the set of all $(g_x)_{x \in U} \in \prod_{x \in U} \mathcal{G}_x$ + satisfying the following \vocab{coherence condition}: + For every $x \in U$, there are an open neighbourhood $W_x \subseteq U$ of $x$ and + $g^{(x)} \in \mathcal{G}(W_x)$ with $g_y^{(x)} = g_y$ for all $y \in W_x$. \end{fact} \begin{proof} - Because of $\gamma_x = \delta_x$, there is $x \in W_x \subseteq U$ open such - that $\gamma\defon{W_x} = \delta\defon{W_x}$. + Because of $\gamma_x = \delta_x$, + there is $x \in W_x \subseteq U$ open + such that $\gamma\defon{W_x} = \delta\defon{W_x}$. As the $W_x$ cover $U$, $\gamma = \delta$ by the sheaf axiom. \end{proof} \begin{definition} Let $\mathcal{G}$ be a sheaf of functions. - Then $\gamma_x$ is called the \vocab{germ} of the function $\gamma$ - at $x$. - The \vocab[Germ! - value at $x$]{value at $x$ } of $g = (U, \gamma) / \sim \in - \mathcal{G}_x$ defined as $g(x) \coloneqq \gamma(x)$, + Then $\gamma_x$ is called the \vocab{germ} of the function $\gamma$ at $x$. + The \vocab[Germ!value at $x$]{value at $x$} + of $g = (U, \gamma) / \sim \in \mathcal{G}_x$ + defined as $g(x) \coloneqq \gamma(x)$, which is independent of the choice of the representative $\gamma$. \end{definition} \begin{remark} - If $\mathcal{G}$ is a sheaf of - $C^{\infty}$-functions (resp. - holomorphic functions), then $\mathcal{G}_x$ is called the ring of - germs of $C^\infty$-functions (resp. of holomorphic functions) at $x$. - + If $\mathcal{G}$ is a sheaf of $C^{\infty}$-functions + (resp.~holomorphic functions), + then $\mathcal{G}_x$ is called the ring of germs of $C^\infty$-functions + (resp.~of holomorphic functions) at $x$. \end{remark} \subsubsection{The local ring of an affine variety} \begin{definition} - If $X$ is a variety, the stalk $\mathcal{O}_{X,x}$ of the structure sheaf - at - $x$ is called the \vocab{local ring} of $X$ at $x$. - This is indeed a local ring, with maximal ideal $\mathfrak{m}_x = - \{f \in - \mathcal{O}_{X,x} | f(x) = 0\}$. + If $X$ is a variety, + the stalk $\mathcal{O}_{X,x}$ of the structure sheaf at $x$ + is called the \vocab{local ring} of $X$ at $x$. + This is indeed a local ring, + with maximal ideal + $\mathfrak{m}_x = \{f \in \mathcal{O}_{X,x} | f(x) = 0\}$. \end{definition} \begin{proof} - By - \ref{localring} it suffices to show that $\mathfrak{m}_x$ - is a proper ideal, - which is trivial, and that the elements of $\mathcal{O}_{X,x} \setminus - \mathfrak{m}_x$ are units in $\mathcal{O}_{X,x}$. + By \ref{localring} it suffices to show that $\mathfrak{m}_x$ + is a proper ideal, which is trivial, + and that the elements of $\mathcal{O}_{X,x} \setminus \mathfrak{m}_x$ + are units in $\mathcal{O}_{X,x}$. Let $g = (U, \gamma)/\sim \in \mathcal{O}_{X,x}$ and $g(x) \neq 0$. - $\gamma$ is Zariski continuous (first point of - \ref{localinverse}). + $\gamma$ is Zariski continuous (first point of \ref{localinverse}). Thus $V(\gamma)$ is closed. - By replacing $U$ by $U \setminus V(\gamma)$ we may assume that $\gamma$ - vanishes nowhere on $U$. - By the third point of - \ref{localinverse} we have $\gamma \in - \mathcal{O}_X(U)^{\times}$. + By replacing $U$ by $U \setminus V(\gamma)$ + we may assume that $\gamma$ vanishes nowhere on $U$. + By the third point of \ref{localinverse} we have + $\gamma \in \mathcal{O}_X(U)^{\times}$. $(\gamma^{-1})_x$ is an inverse to $g$. \end{proof} \begin{proposition} \label{proplocalring} - Let $X = \Va(I) \subseteq \mathfrak{k}^n$ be - equipped with its usual structure - sheaf, where $I = \sqrt{I} \subseteq R = - \mathfrak{k}[X_1,\ldots,X_n]$ . + Let $X = \Va(I) \subseteq \mathfrak{k}^n$ be equipped with its + usual structure sheaf, + where $I = \sqrt{I} \subseteq R = \mathfrak{k}[X_1,\ldots,X_n]$. Let $x \in X$ and $A = \mathcal{O}_X(X) \cong R / I$. - $\{P \in R | P(x) = 0\} \text{\reflectbox{$\coloneqq$}} \fn_x \subseteq R$ is maximal, - $I \subseteq \fn_x$ and $\mathfrak{m}_x \coloneqq \fn_x / I$ is the - maximal ideal of elements of $A$ vanishing at $x$. - If $\lambda \in A \setminus \mathfrak{m}_x$, we have $\lambda_x \in - \mathcal{O}_{X,x}^{\times}$, where $\lambda_x$ denotes the image under $A \cong - \mathcal{O}_X(X) \to \mathcal{O}_{X,x}$. - By the universal property of the localization, there exists a unique ring - homomorphism $A_{\mathfrak{m}_x} \xrightarrow{\iota} - \mathcal{O}_{X,x}$ such - that + $\{P \in R | P(x) = 0\} \text{\reflectbox{$\coloneqq$}} \fn_x \subseteq R$ + is maximal, $I \subseteq \fn_x$ + and $\mathfrak{m}_x \coloneqq \fn_x / I$ is the maximal ideal of elements + of $A$ vanishing at $x$. + If $\lambda \in A \setminus \mathfrak{m}_x$, + we have $\lambda_x \in \mathcal{O}_{X,x}^{\times}$, + where $\lambda_x$ denotes the image under + $A \cong \mathcal{O}_X(X) \to \mathcal{O}_{X,x}$. + By the universal property of the localization, + there exists a unique ring homomorphism + $A_{\mathfrak{m}_x} \xrightarrow{\iota} \mathcal{O}_{X,x}$ + such that \[ \begin{tikzcd} A \arrow{r}{} \arrow{d}{\lambda \mapsto \lambda_x} & - A_{\mathfrak{m}_x} \arrow[dotted, bend left]{ld}{\existsone \iota} \\ + A_{\mathfrak{m}_x} \arrow[dotted, bend left]{ld}{\existsone \iota} \\ \mathcal{O}_{X,x} \end{tikzcd} \] commutes. - The morphism $A_{\mathfrak{m}_x}\xrightarrow{\iota} - \mathcal{O}_{X,x}$ is an - isomorphism. - + The morphism $A_{\mathfrak{m}_x}\xrightarrow{\iota} \mathcal{O}_{X,x}$ + is an isomorphism. \end{proposition} \begin{proof} - To show surjectivity, let $\ell = (U, \lambda) / \sim \in - \mathcal{O}_{X,x}$, + To show surjectivity, + let $\ell = (U, \lambda) / \sim \in \mathcal{O}_{X,x}$, where $U$ is an open neighbourhood of $x$ in $X$. We have $X \setminus U = V(J)$ where $J \subseteq A$ is an ideal. As $x \in U$ there is $f \in J$ with $f(x) \neq 0$. Replacing $U $ by $X \setminus V(f)$ we may assume $U = X \setminus V(f)$. - By - \ref{oxulocaf}, $\mathcal{O}_X(U) \cong A_f$, and - $\lambda = - f^{-n}\vartheta$ for some $n \in \N$ and $\vartheta \in - A$. - Then $\ell = \iota(f^{-n} \vartheta)$ where the last fraction is taken in - $A_{\mathfrak{m}_x}$. + By \ref{oxulocaf}, + $\mathcal{O}_X(U) \cong A_f$, + and $\lambda = f^{-n}\vartheta$ for some $n \in \N$ and $\vartheta \in A$. + Then $\ell = \iota(f^{-n} \vartheta)$ + where the last fraction is taken in $A_{\mathfrak{m}_x}$. - - Let $\lambda = \frac{\vartheta}{g} \in A_{\mathfrak{m}_x}$ - with - $\iota(\lambda) = 0$. - It is easy to see that $\iota(\lambda) = (X \setminus V(g), - \frac{\vartheta}{g}) / \sim $. - Thus there is an open neighbourhood $U$ of $x$ in $X \setminus V(g)$ such that - $\vartheta$ vanishes on $U$. - Similar as before there is $h \in A$ with $h(x) \neq 0$ and $W = X \setminus - V(h) \subseteq U$. - By the isomorphism $\mathcal{O}_X(W) \cong A_h$, there is $n \in - \N$ with - $h^{n}\vartheta = 0$ in $A$. - Since $h \not\in \mathfrak{m}_x$, $h$ is a unit and the image of - $\vartheta$ in - $A_{\mathfrak{m}_x}$ vanishes, implying $\lambda = 0$. + Let $\lambda = \frac{\vartheta}{g} \in A_{\mathfrak{m}_x}$ + with $\iota(\lambda) = 0$. + It is easy to see that + $\iota(\lambda) = (X \setminus V(g), \frac{\vartheta}{g}) / \sim $. + Thus there is an open neighbourhood $U$ of $x$ in $X \setminus V(g)$ + such that $\vartheta$ vanishes on $U$. + Similar as before there is $h \in A$ + with $h(x) \neq 0$ and $W = X \setminus V(h) \subseteq U$. + By the isomorphism $\mathcal{O}_X(W) \cong A_h$, + there is $n \in \N$ with $h^{n}\vartheta = 0$ in $A$. + Since $h \not\in \mathfrak{m}_x$, + $h$ is a unit + and the image of $\vartheta$ in $A_{\mathfrak{m}_x}$ vanishes, + implying $\lambda = 0$. \end{proof} \subsubsection{Intersection multiplicities and Bezout's theorem} \begin{definition} - Let $R = \mathfrak{k}[X_0,X_1,X_2]$ equipped with its usual grading and let $x - \in \mathbb{P}^{2}$. - Let $G \in R_g, H \in R_h$ be homogeneous polynomials with $x \in V(G) \cap - V(h)$. + Let $R = \mathfrak{k}[X_0,X_1,X_2]$ equipped with its usual grading + and let $x \in \mathbb{P}^{2}$. + Let $G \in R_g, H \in R_h$ be homogeneous polynomials + with $x \in V(G) \cap V(h)$. Let $\ell\in R_1$ such that $\ell(x) \neq 0$. - Then $x \in U = \mathbb{P}^2 \setminus V(\ell)$ and the rational - functions - $\gamma = \ell^{-g}G, \eta = \ell^{-h}H$ are - elements of + Then $x \in U = \mathbb{P}^2 \setminus V(\ell)$ and the rational functions + $\gamma = \ell^{-g}G, \eta = \ell^{-h}H$ are elements of $\mathcal{O}_{\mathbb{P}^2}(U)$. - Let $I_x(G,H) \subseteq \mathcal{O}_{\mathbb{P}^2,x}$ - denote the ideal + Let $I_x(G,H) \subseteq \mathcal{O}_{\mathbb{P}^2,x}$ denote the ideal generated by $\gamma_x$ and $\eta_x$. - - \noindent The dimension $\dim_{\mathfrak{k}}(\mathcal{O}_{X,x} - / I_x(G,H)) \text{\reflectbox{$\coloneqq$}} i_x(G,H)$ is called the - \vocab{intersection multiplicity} of $G$ and $H$ at $x$. + \noindent The dimension + $\dim_{\mathfrak{k}}(\mathcal{O}_{X,x} / I_x(G,H)) \text{\reflectbox{$\coloneqq$}} i_x(G,H)$ + is called the \vocab{intersection multiplicity} of $G$ and $H$ at $x$. \end{definition} \begin{remark} - If $\tilde \ell \in R_1$ also satisfies $\tilde \ell(x) \neq - 0$, then the image - of $\tilde \ell / \ell$ under - $\mathcal{O}_{\mathbb{P}^2}(U) \to - \mathcal{O}_{\mathbb{P}^2,x}$ is a unit, showing that - the image of $\tilde - \gamma = \tilde \ell^{-g} G$ in - $\mathcal{O}_{\mathbb{P}^2,x}$ is - multiplicatively equivalent to $\gamma_x$, and similarly for $\eta_x$. - Thus $I_x(G,H)$ does not depend on the choice of $\ell \in R_1$ with - $\ell(x) - \neq 0$. + If $\tilde \ell \in R_1$ also satisfies $\tilde \ell(x) \neq 0$, + then the image of $\tilde \ell / \ell$ under + $\mathcal{O}_{\mathbb{P}^2}(U) \to \mathcal{O}_{\mathbb{P}^2,x}$ is a unit, + showing that the image of $\tilde \gamma = \tilde \ell^{-g} G$ in + $\mathcal{O}_{\mathbb{P}^2,x}$ is multiplicatively equivalent to $\gamma_x$, + and similarly for $\eta_x$. + Thus $I_x(G,H)$ does not depend on the choice of $\ell \in R_1$ + with $\ell(x) \neq 0$. \end{remark} \begin{theorem}[Bezout's theorem] - In the above situation, assume that $V(H)$ and $V(G)$ - intersect properly in the sense that $V(G) \cap V(H) \subseteq - \mathbb{P}^2$ has no irreducible component of dimension $\ge 1$. + In the above situation, + assume that $V(H)$ and $V(G)$ intersect properly + in the sense that $V(G) \cap V(H) \subseteq \mathbb{P}^2$ + has no irreducible component of dimension $\ge 1$. Then \[ - \sum_{x \in V(G) \cap V(H)} i_x(G,H) = gh + \sum_{x \in V(G) \cap V(H)} i_x(G,H) = gh. \] - Thus, $V(G) \cap V(H)$ has - $gh$ elements counted by multiplicity. + Thus, $V(G) \cap V(H)$ has $gh$ elements counted by multiplicity. \end{theorem} \printvocabindex \end{document}