From 14408addd81cb3341fa26aeb399f81c50a8f58b4 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Maximilian=20Ke=C3=9Fler?= Date: Wed, 16 Feb 2022 01:12:27 +0100 Subject: [PATCH] replace fk --- 2021_Algebra_I.tex | 412 ++++++++++++++++++++++----------------------- 1 file changed, 206 insertions(+), 206 deletions(-) diff --git a/2021_Algebra_I.tex b/2021_Algebra_I.tex index 8b2b42d..e37d036 100644 --- a/2021_Algebra_I.tex +++ b/2021_Algebra_I.tex @@ -24,7 +24,7 @@ \noindent The \LaTeX template by \textsc{Maximilian Kessler} is published under the MIT-License and can be obtained from \url{https://github.com/kesslermaximilian/LatexPackages}. % TODO \newline -\noindent $\fk$ is {\color{red} always} an algebraically closed field and $\fk^n$ is equipped with the Zariski-topology. +\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 $\fl$). \pagebreak @@ -313,12 +313,12 @@ is injective. $n$ and the $a_i$ can be chosen such that $A$ is finite over the i \end{proof} \section{The Nullstellensatz and the Zariski topology} \subsection{The Nullstellensatz} %LECTURE 1 -Let $\fk$ be a field, $R \coloneqq \fk[X_1,\ldots,X_n], I \se R$ an ideal. +Let $\mathfrak{k}$ be a field, $R \coloneqq \mathfrak{k}[X_1,\ldots,X_n], I \se R$ an ideal. \begin{definition}[zero] - $x \in \fk^n$ is \vocab[Ideal!zero]{a zero of $I$} if $\A x \in I: P(x) = 0$. Let $\Va(I)$ denote the set of zeros if $I$ in $\fk^n$. + $x \in \mathfrak{k}^n$ is \vocab[Ideal!zero]{a zero of $I$} if $\A x \in I: P(x) = 0$. Let $\Va(I)$ denote the set of zeros if $I$ in $\mathfrak{k}^n$. - The \vocab[Ideal!zero]{zero in a field extension $\mathfrak{i}$ of $\fk$} is defined similarly. + The \vocab[Ideal!zero]{zero in a field extension $\mathfrak{i}$ of $\mathfrak{k}$} is defined similarly. \end{definition} \begin{remark}[Set of zeros and generators] @@ -327,14 +327,14 @@ Let $\fk$ be a field, $R \coloneqq \fk[X_1,\ldots,X_n], I \se R$ an ideal. \end{remark} \begin{theorem}[Hilbert's Nullstellensatz (1)]\label{hns1} - If $\fk$ is algebraically closed and $I \subsetneq R$ a proper ideal, then $I$ has a zero in $\fk^n$. + If $\mathfrak{k}$ is algebraically closed and $I \subsetneq R$ a proper ideal, then $I$ has a zero in $\mathfrak{k}^n$. \end{theorem} \begin{remark} Will be shown later (see proof of \ref{hns1b}). - Trivial if $n = 1$: $R$ is a PID, thus $I = pR$ for some $p \in R$. Since $I \neq R$ $p = 0$ or $P$ is non-constant. $\fk$ algebraically closed $\leadsto$ there exists a zero of $p$.\\ + Trivial if $n = 1$: $R$ is a PID, thus $I = pR$ for some $p \in R$. Since $I \neq R$ $p = 0$ or $P$ is non-constant. $\mathfrak{k}$ algebraically closed $\leadsto$ there exists a zero of $p$.\\ - If $\fk$ is not algebraically closed and $n > 0$, the theorem fails (consider $I = p(X_1) R$). + If $\mathfrak{k}$ is not algebraically closed and $n > 0$, the theorem fails (consider $I = p(X_1) R$). \end{remark} Equivalent\footnote{used in a vague sense here} formulation: @@ -426,14 +426,14 @@ Let $R$ be a ring and $I,J, I_\lambda \se R$ ideals, $\lambda \in \Lambda$. $\bigcap_{\lambda \in \Lambda} I_\lambda$ is an ideal. \end{fact} -Let $R = \fk[X_1,\ldots,X_n]$ where $\fk$ is an algebraically closed field. +Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ where $\mathfrak{k}$ is an algebraically closed field. \begin{fact} \label{fvop} Let $I, J, (I_{\lambda})_{\lambda \in \Lambda}$ be ideals in $R$. $\Lambda$ may be infinite. \begin{enumerate}[A] \item $\Va(I) = \Va(\sqrt{I})$ \item $\sqrt{J} \se \sqrt{I} \implies \Va(I) \se \Va(J)$ - \item $\Va(R) = \emptyset, \Va(\{0\} =\fk^n$ + \item $\Va(R) = \emptyset, \Va(\{0\} =\mathfrak{k}^n$ \item $\Va(I \cap J) = \Va(I\cdot J) = \Va(I) \cup \Va(J)$ \item $\Va(\sum_{\lambda \in \Lambda} I_\lambda) = \bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$ \end{enumerate} @@ -456,16 +456,16 @@ Let $R = \fk[X_1,\ldots,X_n]$ where $\fk$ is an algebraically closed field. For instance if $n = 1, I_k \coloneqq X_1^k R$ then $\bigcap_{k=0}^\infty I_k = \{0\} $ but $\bigcup_{k=0}^{\infty} \Va(I_k) = \{0\}$. \end{remark} \subsubsection{Definition of the Zariski topology} -Let $\fk$ be algebraically closed, $R = \fk[X_1,\ldots,X_n]$. +Let $\mathfrak{k}$ be algebraically closed, $R = \mathfrak{k}[X_1,\ldots,X_n]$. \begin{corollary} (of \ref{fvop}) - There is a topology on $\fk^n$ for which the set of closed sets coincides with the set $\fA$ of subsets of the form $\Va\left(I \right) $ for ideals $I \se R$. + There is a topology on $\mathfrak{k}^n$ for which the set of closed sets coincides with the set $\fA$ of subsets of the form $\Va\left(I \right) $ for ideals $I \se R$. This topology is called the \vocab{Zariski-Topology} \end{corollary} \begin{example}\label{zariskinothd} - Let $n = 1$. Then $R$ is a PID. Hence every ideal is a principal ideal and the Zariski-closed subsets of $\fk$ are the subsets of the form $\Va(P)$ for $P \in R$. -As $\Va(0) = \fk$ and $\Va(P)$ finite for $P \neq 0$ and $\{x_1,\ldots,x_n\} = \Va(\prod_{i=1}^{n}(T-x_i))$ the Zariski-closed subsets of $\fk$ are $\fk$ and the finite subsets. - Because $\fk$ is infinite, this topology is not Hausdorff. + Let $n = 1$. Then $R$ is a PID. Hence every ideal is a principal ideal and the Zariski-closed subsets of $\mathfrak{k}$ are the subsets of the form $\Va(P)$ for $P \in R$. +As $\Va(0) = \mathfrak{k}$ and $\Va(P)$ finite for $P \neq 0$ and $\{x_1,\ldots,x_n\} = \Va(\prod_{i=1}^{n}(T-x_i))$ the Zariski-closed subsets of $\mathfrak{k}$ are $\mathfrak{k}$ and the finite subsets. + Because $\mathfrak{k}$ is infinite, this topology is not Hausdorff. \end{example} \subsubsection{Separation properties of topological spaces} @@ -484,10 +484,10 @@ As $\Va(0) = \fk$ and $\Va(P)$ finite for $P \neq 0$ and $\{x_1,\ldots,x_n\} = \ $T_0 \iff$ every point is closed. \end{fact} \begin{fact} - The Zariski topology on $\fk^n$ is $T_1$ but for $n \ge 1$ not Hausdorff. For $n \ge 1$ the intersection of two non-empty open subsets of $\fk^n$ is always non-empty. + The Zariski topology on $\mathfrak{k}^n$ is $T_1$ but for $n \ge 1$ not Hausdorff. For $n \ge 1$ the intersection of two non-empty open subsets of $\mathfrak{k}^n$ is always non-empty. \end{fact} \begin{proof} - $\{x\} $ is closed, as $\{x\} = V(\Span{X_1 - x_1, \ldots, X_n - x_n}_R)$. If $A = V(I), B = V(J)$ are two proper closed subsets of $\fk^n$ then $I \neq \{0\} , J \neq \{0\} $ and thus $IJ \neq \{0\} $. Therefore $A \cup B = V(IJ)$ is a proper closed subset of $\fk^n$. + $\{x\} $ is closed, as $\{x\} = V(\Span{X_1 - x_1, \ldots, X_n - x_n}_R)$. If $A = V(I), B = V(J)$ are two proper closed subsets of $\mathfrak{k}^n$ then $I \neq \{0\} , J \neq \{0\} $ and thus $IJ \neq \{0\} $. Therefore $A \cup B = V(IJ)$ is a proper closed subset of $\mathfrak{k}^n$. \end{proof} @@ -514,20 +514,20 @@ Let $X$ be a topological space. \end{enumerate} \end{proof} -\subsection{Another form of the Nullstellensatz and Noetherianness of \texorpdfstring{$\fk^n$}{kn}} -Let $\fk$ be algebraically closed, $R = \fk[X_1,\ldots,X_n]$. +\subsection{Another form of the Nullstellensatz and Noetherianness of \texorpdfstring{$\mathfrak{k}^n$}{kn}} +Let $\mathfrak{k}$ be algebraically closed, $R = \mathfrak{k}[X_1,\ldots,X_n]$. For $f \in R$ let $V(f) = V(fR)$. \begin{theorem}[Hilbert's Nullstellensatz (3)] \label{hns3} Let $I \se R$ be an ideal. Then $V(I) \se V(f)$ iff $f \in \sqrt{I}$. \end{theorem} \begin{proof} - Suppose $f$ vanishes on all zeros of $I$. Let $R' \coloneqq \fk[X_1,\ldots,X_n,T]$, + Suppose $f$ vanishes on all zeros of $I$. Let $R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n,T]$, $g(X_1,\ldots,X_n,T) \coloneqq 1 - T \cdot f(X_1,\ldots,X_n)$ and $J \se R'$ the ideal generated by $g$ and the elements of $I$ (viewed as elements of $R'$ which are constant in the $T$-direction). - If $f$ vanishes on all zeros of $I$, then $J$ has no zeros in $\fk^{n+1}$. + If $f$ vanishes on all zeros of $I$, then $J$ has no zeros in $\mathfrak{k}^{n+1}$. - Thus there exist $p_i \in I, i=1,\ldots,n, q_i \in \fk[X_1,\ldots,X_n,T], i = 1,\ldots,n$ and $q \in \fk[X_1,\ldots,X_n,T]$ such that + Thus there exist $p_i \in I, i=1,\ldots,n, q_i \in \mathfrak{k}[X_1,\ldots,X_n,T], i = 1,\ldots,n$ and $q \in \mathfrak{k}[X_1,\ldots,X_n,T]$ such that \[ 1 = g \cdot q + \sum_{i=1}^{n} p_{i}q_i \] @@ -544,17 +544,17 @@ For $f \in R$ let $V(f) = V(fR)$. \begin{corollary}\label{antimonbij} \begin{align} - f: \{I \se R | I \text{ ideal}, I = \sqrt{I} \} &\longrightarrow \{A \se \fk^n | A \text{ Zariski-closed}\} \\ + f: \{I \se R | I \text{ ideal}, I = \sqrt{I} \} &\longrightarrow \{A \se \mathfrak{k}^n | A \text{ Zariski-closed}\} \\ I &\longmapsto V(I)\\ \{f \in R | A \se V(f)\} &\longmapsfrom A \end{align} is a $\se$-antimonotonic bijection. \end{corollary} \begin{corollary} - The topological space $\fk^n$ is Noetherian. + The topological space $\mathfrak{k}^n$ is Noetherian. \end{corollary} \begin{proof} - Because the map from \ref{antimonbij} is antimonotonic, strictly decreasing chains of closed subsets of $\fk^n$ are mapped to strictly increasing chains of ideals in $R$. + Because the map from \ref{antimonbij} is antimonotonic, strictly decreasing chains of closed subsets of $\mathfrak{k}^n$ are mapped to strictly increasing chains of ideals in $R$. By the Basissatz (\ref{basissatz}), $R$ is Noetherian. \end{proof} @@ -586,7 +586,7 @@ Let $X$ be a topological space. Every irreducible topological space is connected. \end{corollary} \begin{example} - $\fk^n$ is irreducible as shown in \ref{zariskinothd}. + $\mathfrak{k}^n$ is irreducible as shown in \ref{zariskinothd}. \end{example} \begin{fact} @@ -651,12 +651,12 @@ Let $X$ be a topological space. \begin{proposition}\label{bijiredprim} By \ref{antimonbij} there exists a bijection \begin{align} - f: \{I \se R | I \text{ ideal}, I = \sqrt{I} \} &\longrightarrow \{A \se \fk^n | A \text{ Zariski-closed}\} \\ + f: \{I \se R | I \text{ ideal}, I = \sqrt{I} \} &\longrightarrow \{A \se \mathfrak{k}^n | A \text{ Zariski-closed}\} \\ I &\longmapsto V(I)\\ \{f \in R | A \se V(f)\} &\longmapsfrom A \end{align} - Under this correspondence $A \se \fk^n$ is irreducible iff $I \coloneqq f\inv(A)$ is a prime ideal. + Under this correspondence $A \se \mathfrak{k}^n$ is irreducible iff $I \coloneqq f\inv(A)$ is a prime ideal. Moreover, $\#A = 1$ iff $I$ is a maximal ideal. \end{proposition} \begin{proof} @@ -665,7 +665,7 @@ Let $X$ be a topological space. On the other hand suppose that $fg \in I, f \notin I, g \not\in I$. By the Nullstellensatz (\ref{hns3}) and $I = \sqrt{I} $ neither $f$ nor $g$ vanishes on all of $A$. Thus $(A \cap V(f)) \cup (A \cap V(g))$ is a decomposition and $A$ fails to be irreducible. - The remaining assertion follows from the fact, that the bijection is $\se$-antimonotonic and thus maximal ideals correspond to minimal irreducible closed subsets, which are the one-point subsets as $\fk^n$ is T${}_1$. + The remaining assertion follows from the fact, that the bijection is $\se$-antimonotonic and thus maximal ideals correspond to minimal irreducible closed subsets, which are the one-point subsets as $\mathfrak{k}^n$ is T${}_1$. \end{proof} \subsection{Krull dimension} \begin{definition} @@ -690,7 +690,7 @@ Let $X$ be a topological space. If $X = \{x\}$, then $\dim X = 0$. \end{fact} \begin{fact} - For every $x \in \fk$, $\codim( \{x\} ,\fk) = 1$. The only other irreducible closed subset of $\fk$ is $\fk$ itself, which has codimension zero. Thus $\dim \fk = 1$. + For every $x \in \mathfrak{k}$, $\codim( \{x\} ,\mathfrak{k}) = 1$. The only other irreducible closed subset of $\mathfrak{k}$ is $\mathfrak{k}$ itself, which has codimension zero. Thus $\dim \mathfrak{k} = 1$. \end{fact} \begin{fact} Let $Y \se X$ be irreducible and $U \se X$ an open subset such that $U \cap Y \neq \emptyset$. Then we have a bijection @@ -729,18 +729,18 @@ In general, these inequalities may be strict. A topological space $T$ is called \vocab[Topological space!catenary]{catenary} if equality holds in \eqref{eq:cdp} whenever $X$ is an irreducible closed subset of $T$. \end{definition} -\subsubsection{Krull dimension of \texorpdfstring{$\fk^n$}{kn}} % from lecture 04 +\subsubsection{Krull dimension of \texorpdfstring{$\mathfrak{k}^n$}{kn}} % from lecture 04 \begin{theorem}\label{kdimkn} - $\dim \fk^n = n$ and $\fk^n$ is catenary. Moreover, if $X$ is an irreducible closed subset of $\fk^n$, then equality occurs in \eqref{eq:dp}. + $\dim \mathfrak{k}^n = n$ and $\mathfrak{k}^n$ is catenary. Moreover, if $X$ is an irreducible closed subset of $\mathfrak{k}^n$, then equality occurs in \eqref{eq:dp}. \end{theorem} \begin{proof} Considering \[ - \{0\} \subsetneq \fk \times \{0\} \subsetneq \fk^2 \times \{0\} \subsetneq \ldots \subsetneq \fk^n + \{0\} \subsetneq \mathfrak{k} \times \{0\} \subsetneq \mathfrak{k}^2 \times \{0\} \subsetneq \ldots \subsetneq \mathfrak{k}^n \] - it is clear that $\codim(\{0\}, \fk^n) \ge n$.Translation by $x \in \fk^n$ gives us $\codim(\{x\} , \fk^n) \ge n$. + it is clear that $\codim(\{0\}, \mathfrak{k}^n) \ge n$.Translation by $x \in \mathfrak{k}^n$ gives us $\codim(\{x\} , \mathfrak{k}^n) \ge n$. - The opposite inequality follows from \ref{upperbounddim} ($Z = \fk^n$ $\dim \fk^n \le \trdeg(\fK(Z) / \fk) = \trdeg(Q(\fk[X_1,\ldots,X_n]) / \fk) = n$). + The opposite inequality follows from \ref{upperbounddim} ($Z = \mathfrak{k}^n$ $\dim \mathfrak{k}^n \le \trdeg(\fK(Z) / \mathfrak{k}) = \trdeg(Q(\mathfrak{k}[X_1,\ldots,X_n]) / \mathfrak{k}) = n$). The theorem is a special case of \ref{htandtrdeg}. % DIMT @@ -756,17 +756,17 @@ In general, these inequalities may be strict. \end{proof} \begin{proposition}[Irreducible subsets of codimension one]\label{irredcodimone} - Let $p \in R = \fk[X_1,\ldots, X_n]$ be a prime element. Then the irreducible subset $X = V(p) \se \fk^n$ has codimension one, and every codimension one subset of $\fk^n$ has this form. + Let $p \in R = \mathfrak{k}[X_1,\ldots, X_n]$ be a prime element. Then the irreducible subset $X = V(p) \se \mathfrak{k}^n$ has codimension one, and every codimension one subset of $\mathfrak{k}^n$ has this form. \end{proposition} \begin{proof} - Since $pR$ is a prime ideal, $X = V(p)$ is irreducible. Since $p \neq 0$, $X$ is a proper subset of $\fk^n$. - If $X \se Y \se \fk^n$ is irreducible and closed, then $Y = V(\fq)$ for some prime ideal $\fp \se pR$. - If $Y \neq \fk^n$, then $\fp \neq \{0\}$. By \ref{ufdprimeideal} there exists a prime element $q \in \fq$. As $\fq \se pR$ we have $p \divides q$. + Since $pR$ is a prime ideal, $X = V(p)$ is irreducible. Since $p \neq 0$, $X$ is a proper subset of $\mathfrak{k}^n$. + If $X \se Y \se \mathfrak{k}^n$ is irreducible and closed, then $Y = V(\fq)$ for some prime ideal $\fp \se pR$. + If $Y \neq \mathfrak{k}^n$, then $\fp \neq \{0\}$. By \ref{ufdprimeideal} there exists a prime element $q \in \fq$. As $\fq \se pR$ we have $p \divides q$. By the irreducibility of $p$ and $q$ it follows that $p \sim q$. Hence $\fq = pR$ and $X = Y$. - Suppose $X = V(\fp) \se \fk^n$ is closed, irreducible and of codimension one. - Then $\fp \neq \{0\}$, hence $X \neq \fk^n$. By \ref{ufdprimeideal} there is a prime element $p \in \fp$. If $\fp \neq pR$, then - $X \subsetneq V(p) \subsetneq \fk^n$ contradicts $\codim(X, \fk^n) = 1$. + Suppose $X = V(\fp) \se \mathfrak{k}^n$ is closed, irreducible and of codimension one. + Then $\fp \neq \{0\}$, hence $X \neq \mathfrak{k}^n$. By \ref{ufdprimeideal} there is a prime element $p \in \fp$. If $\fp \neq pR$, then + $X \subsetneq V(p) \subsetneq \mathfrak{k}^n$ contradicts $\codim(X, \mathfrak{k}^n) = 1$. \end{proof} % Lecture 05 @@ -913,27 +913,27 @@ The Nullstellensatz (\ref{hns2}) can be reduced to the case of \ref{rfuncnft}: \subsection{Transcendence degree and Krull dimension} -Let $R = \fk[X_1,\ldots,X_n]$. +Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. %i = ic \begin{notation} - Let $X \se \fk^n$ be an irreducible closed subset. Then $X = V(\fp)$ for a unique prime ideal $\fp \se R$. + Let $X \se \mathfrak{k}^n$ be an irreducible closed subset. Then $X = V(\fp)$ for a unique prime ideal $\fp \se R$. Let $\fK(X) \coloneqq Q(R / \fp)$ denote the field of quotients of $R / \fp$. \end{notation} \begin{remark} As the elements of $\fp$ vanish on $X$, $R / \fp$ may be viewed as the ring of polynomials and $\fK(X)$ as the field of rational functions on $X$. \end{remark} \begin{theorem}\label{trdegandkdim} - If $X \se \fk^n$ is irreducible, then $\dim X = \trdeg (\fk(X) / \fk)$ and $\codim(X, \fk^n) = n - \trdeg(\fK(X) / \fk)$. - More generally if $Y \se \fk^n$ is irreducible and $X \se Y$, then $\codim(X,Y) = \trdeg(\fK(Y) / \fk) - \trdeg(\fK(X) / \fk)$. + If $X \se \mathfrak{k}^n$ is irreducible, then $\dim X = \trdeg (\mathfrak{k}(X) / \mathfrak{k})$ and $\codim(X, \mathfrak{k}^n) = n - \trdeg(\fK(X) / \mathfrak{k})$. + More generally if $Y \se \mathfrak{k}^n$ is irreducible and $X \se Y$, then $\codim(X,Y) = \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})$. \end{theorem} \begin{proof} % DIMT One part will be shown in "A first result on dimension theory" (\ref{upperboundcodim}) - and other one in "Aplication to dimension theory: Proof of $\dim Y = \trdeg(\fK(Y) / \fk)$" (\ref{lowerbounddimy}). + and other one in "Aplication to dimension theory: Proof of $\dim Y = \trdeg(\fK(Y) / \mathfrak{k})$" (\ref{lowerbounddimy}). The theorem is a special case of \ref{htandtrdeg}. \end{proof} \begin{remark} - Loosely speaking, the Krull dimension of $X$ is equal to the maximal number of $\fk$-algebraically independent rational functions on $X$. + Loosely speaking, the Krull dimension of $X$ is equal to the maximal number of $\mathfrak{k}$-algebraically independent rational functions on $X$. This is yet another indication that the notion of dimension is the ``correct'' one. \end{remark} \begin{remark} @@ -957,15 +957,15 @@ Let $R = \fk[X_1,\ldots,X_n]$. \end{definition} \begin{remark} - When $R = \fk[X_1,\ldots,X_n]$, the notation $V(I)$ clashes with the previous notation. When several types of $V(I)$ will be in use, they will be distinguished using indices. + When $R = \mathfrak{k}[X_1,\ldots,X_n]$, the notation $V(I)$ clashes with the previous notation. When several types of $V(I)$ will be in use, they will be distinguished using indices. \end{remark} \begin{remark} Let $(I_{\lambda})_{\lambda \in \Lambda}$ and $(l_j)_{j=1}^n$ be ideals in $R$, where $\Lambda$ may be infinite. We have $V(\sum_{\lambda \in \Lambda} I_\lambda ) = \bigcap_{\lambda \in \Lambda} V(I_\lambda)$ and $V(\bigcap_{j=1}^n I_j) = V(\prod_{j=1}^{n} I_j) = \bigcup_{j = 1}^n V(I_j)$. Thus, the Zariski topology on $\Spec R$ is a topology. \end{remark} \begin{remark} - Let $R = \fk[X_1,\ldots,X_n]$. Then there exists a bijection (\ref{antimonbij}, \ref{bijiredprim}) between $\Spec R$ and the set of irreducible closed subsets of $\fk^n$ sending $\fp \in \Spec R$ to $V_{\fk^n}(\fp)$ and identifying the one-point subsets with $\mSpec R$. - This defines a bijection $\fk^n \cong \mSpec R$ which is a homeomorphism if $\mSpec R$ is equipped with the induced topology from the Zariski topology on $\Spec R$. + Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. Then there exists a bijection (\ref{antimonbij}, \ref{bijiredprim}) between $\Spec R$ and the set of irreducible closed subsets of $\mathfrak{k}^n$ sending $\fp \in \Spec R$ to $V_{\mathfrak{k}^n}(\fp)$ and identifying the one-point subsets with $\mSpec R$. + This defines a bijection $\mathfrak{k}^n \cong \mSpec R$ which is a homeomorphism if $\mSpec R$ is equipped with the induced topology from the Zariski topology on $\Spec R$. \end{remark} \subsection{Localization of rings} @@ -1096,7 +1096,7 @@ Because the elements of $S$ become units in $R_S$, $J \sqcap R$ is an $S$-satura \subsection{A first result of dimension theory} \begin{notation} - Let $R$ be a ring, $\fp \in \Spec R$. Let $\fk(\fp)$ denote the field of quotients of the domain $R / \fp$. This is called the \vocab{residue field} of $\fp$. + Let $R$ be a ring, $\fp \in \Spec R$. Let $\mathfrak{k}(\fp)$ denote the field of quotients of the domain $R / \fp$. This is called the \vocab{residue field} of $\fp$. \end{notation} % i = ic @@ -1104,22 +1104,22 @@ Because the elements of $S$ become units in $R_S$, $J \sqcap R$ is an $S$-satura Let $\fl$ be a %% ?? field, $A$ a $\fl$-algebra of finite type and $\fp, \fq \in \Spec A$ with $\fp \subsetneq \fq$. Then \[ - \trdeg(\fk(\fp) / \fl) > \trdeg(\fk(\fq) / \fl) + \trdeg(\mathfrak{k}(\fp) / \fl) > \trdeg(\mathfrak{k}(\fq) / \fl) \] \end{proposition} \begin{proof} - Replacing $A$ by $A / \fp$, we may assume $\fp = \{0\} $ and $A$ to be a domain. Then $\fk(\fp) = Q(A / \fp) = Q(A)$. + Replacing $A$ by $A / \fp$, we may assume $\fp = \{0\} $ and $A$ to be a domain. Then $\mathfrak{k}(\fp) = Q(A / \fp) = Q(A)$. - If $\fq$ is a maximal ideal, $\fk(\fq) = A / \fq$ is of finite type over $\fl$, hence a finite field extension of $\fl$ by the Nullstellensatz (\ref{hns2}). - Thus, $\trdeg(\fk(\fq) / \fl) = 0$. + If $\fq$ is a maximal ideal, $\mathfrak{k}(\fq) = A / \fq$ is of finite type over $\fl$, hence a finite field extension of $\fl$ by the Nullstellensatz (\ref{hns2}). + Thus, $\trdeg(\mathfrak{k}(\fq) / \fl) = 0$. If $\trdeg(Q(A) / \fl) = 0$, $A$ would be integral over $\fl$, hence a field (fact about integrality and fields, \ref{fintaf}). But if $A$ is a field, $\fp = \{0\}$ is a maximal ideal of $A$, hence $\fq = \fp \lightning$. This finishes the proof for $\fq \in \mSpec A$. We will use the following lemma to reduce the general case to this case: \begin{lemma}\label{ltrdegresfieldtrbase} - There are algebraically independent $a_1,\ldots,a_n \in A$ whose images in $A / \fq$ form a transcendence base for $\fk(\fq) / \fl$. + There are algebraically independent $a_1,\ldots,a_n \in A$ whose images in $A / \fq$ form a transcendence base for $\mathfrak{k}(\fq) / \fl$. \end{lemma} \begin{subproof} - There exist $a_1,\ldots,a_n \in A$ such that $\fk(\fq)$ is algebraic over the subfield generated by $\fl$ and their images $\overline{a_i}$ (for instance generators of $A$ as a $\fl$-algebra). + There exist $a_1,\ldots,a_n \in A$ such that $\mathfrak{k}(\fq)$ is algebraic over the subfield generated by $\fl$ and their images $\overline{a_i}$ (for instance generators of $A$ as a $\fl$-algebra). We may assume that $n$ is minimal. If the $a_i$ are $\fl$-algebraically dependent, then w.l.o.g. $\overline{a_n}$ can be assumed to be algebraic over the subfield generated by $\fl$ and the $\overline{a_i}, 1\le i m_i - m_{i-1}$. + Let $A = \mathfrak{k}[X_i | i \in \N]$ and $m_1, m_2, \ldots \in \N$ an increasing sequence such that $m_{i+1}-m_i > m_i - m_{i-1}$. Let $\fp_i \coloneqq (X_{m_{i}+1},\ldots,X_{m_{i+1}})$ and $S \coloneqq A \sm \bigcup_{i \in \N} \fp_i$. $S$ is multiplicatively closed. $A_S$ is Noetherian but $\hght((\fp_i)_S) = m_{i+1}- m_{i}$ hence $\dim(A_S) = \infty$. @@ -1702,24 +1702,24 @@ From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1 \begin{proposition}\label{dimprod} - Let $X \se \fk^n$ and $Y \se \fk^n$ be irreducible and closed. Then $X \times Y$ is also an irreducible closed subset of $\fk^{m+n}$. - Moreover, $\dim(X \times Y) = \dim(X) + \dim(Y)$ and $\codim(X \times Y, \fk^{m+n}) = \codim(X, \fk^m) + \codim(Y, \fk^n)$. + Let $X \se \mathfrak{k}^n$ and $Y \se \mathfrak{k}^n$ be irreducible and closed. Then $X \times Y$ is also an irreducible closed subset of $\mathfrak{k}^{m+n}$. + Moreover, $\dim(X \times Y) = \dim(X) + \dim(Y)$ and $\codim(X \times Y, \mathfrak{k}^{m+n}) = \codim(X, \mathfrak{k}^m) + \codim(Y, \mathfrak{k}^n)$. \end{proposition} \begin{proof} - Let $X = V(\fp)$ and $Y = V(\fq)$ where $\fp \in \Spec \fk[X_1,\ldots,X_m]$ and $\fq \in \Spec \fk[X_1,\ldots,X_n]$. - We denote points of $\fk^{m+n}$ as $x = (x',x'')$ with $x' \in \fk^m, x''\in\fk^n$. Then $X \times Y$ is the set of zeroes of the ideal in $\fk[X_1,\ldots,X_{m+n}]$ generated by the polynomials $f(x) = \phi(x')$, with $\phi$ running over $\fp$ and $g(x) = \gamma(x'')$ with $\gamma$ running over $\fq$. - Thus $X \times Y$ is closed in $\fk^{m+n}$. + Let $X = V(\fp)$ and $Y = V(\fq)$ where $\fp \in \Spec \mathfrak{k}[X_1,\ldots,X_m]$ and $\fq \in \Spec \mathfrak{k}[X_1,\ldots,X_n]$. + We denote points of $\mathfrak{k}^{m+n}$ as $x = (x',x'')$ with $x' \in \mathfrak{k}^m, x''\in\mathfrak{k}^n$. Then $X \times Y$ is the set of zeroes of the ideal in $\mathfrak{k}[X_1,\ldots,X_{m+n}]$ generated by the polynomials $f(x) = \phi(x')$, with $\phi$ running over $\fp$ and $g(x) = \gamma(x'')$ with $\gamma$ running over $\fq$. + Thus $X \times Y$ is closed in $\mathfrak{k}^{m+n}$. We must also show irreducibility. $X \times Y \neq \emptyset$ is obvious. - Assume that $X \times Y = A_1 \cup A_2$, where the $A_i \se \fk^{m+n}$ are closed. - For $x' \in \fk^m, x' \times Y$ is homeomorphic to the irreducible $Y$. Thus $X = X_1 \cup X_2$ where $X_i = \{x \in X | \{x\} \times Y \se A_i\} $. + Assume that $X \times Y = A_1 \cup A_2$, where the $A_i \se \mathfrak{k}^{m+n}$ are closed. + For $x' \in \mathfrak{k}^m, x' \times Y$ is homeomorphic to the irreducible $Y$. Thus $X = X_1 \cup X_2$ where $X_i = \{x \in X | \{x\} \times Y \se A_i\} $. Because $X_i = \bigcap_{y \in Y} \{x \in X | (x,y) \in A_i\}$, this is closed. As $X$ is irreducible, there is $i \in \{1;2\} $ which $X_i = X$. Then $X \times Y = A_i$ confirming the irreducibility of $X \times Y$. Let $a = \dim X$ and $b = \dim Y$ and $X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_a = X$,$Y_0 \subsetneq Y_1 \subsetneq \ldots \subsetneq Y_b = Y$ be chains of irreducible subsets. By the previous result, $X_0 \times Y_0 \subsetneq X_1 \times Y_0 \subsetneq \ldots \subsetneq X_a \times Y_0 \subsetneq X_a \times Y_1 \subsetneq \ldots \subsetneq X_a \times Y_a = X \times Y$ is a chain of irreducible subsets. Thus $\dim(X \times Y) \ge a + b = \dim X + \dim Y$. - Similarly one derives $\codim(X \times Y, \fk^{m+n}) \ge \codim(X, \fk^m) + \codim(Y, \fk^n)$. - By \ref{trdegandkdim} we have $\dim(A) + \codim(A, \fk^l) = l$ for irreducible subsets of $\fk^l$. Thus equality must hold in the previous two inequalities. + Similarly one derives $\codim(X \times Y, \mathfrak{k}^{m+n}) \ge \codim(X, \mathfrak{k}^m) + \codim(Y, \mathfrak{k}^n)$. + By \ref{trdegandkdim} we have $\dim(A) + \codim(A, \mathfrak{k}^l) = l$ for irreducible subsets of $\mathfrak{k}^l$. Thus equality must hold in the previous two inequalities. \end{proof} \subsection{The nil radical} @@ -1770,9 +1770,9 @@ From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1 It is not particularly hard to come up with examples which show that the converse implication does not hold. \end{remark} \begin{dexample} - Let $A = \fk[X_n | n \in \N] / I$ where $I$ denotes the ideal generated by $\{X_i^2 | i \in \N\}$. + Let $A = \mathfrak{k}[X_n | n \in \N] / I$ where $I$ denotes the ideal generated by $\{X_i^2 | i \in \N\}$. $A$ is not Noetherian, since the ideal $J$ generated by $\{X_i | i \in \N\} $ is not finitely generated. - $A / J \cong \fk$, hence $J$ is maximal. As every prime ideal must contain $\nil(A) \supseteq J$, $J$ is the only prime ideal. + $A / J \cong \mathfrak{k}$, hence $J$ is maximal. As every prime ideal must contain $\nil(A) \supseteq J$, $J$ is the only prime ideal. Thus $\Spec A$ contains only one element and is hence Noetherian. \end{dexample} @@ -1819,7 +1819,7 @@ Modern approaches to the principal ideal theorem usually give a direct proof of \subsubsection{Application to the dimension of intersections} \begin{remark}\label{smallestprimeandirredcomp} -Let $R = \fk[X_1,\ldots,X_n]$ and $I \se R$ an ideal. +Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $I \se R$ an ideal. If $(\fp_i)_{i=1}^k$ are the smallest prime ideals of $R$ containing $I$, then $(\Va(\fp_i))_{i=1}^k$ are the irreducible components of $\Va(I)$. \end{remark} @@ -1829,7 +1829,7 @@ Let $R = \fk[X_1,\ldots,X_n]$ and $I \se R$ an ideal. \begin{corollary}[of the principal ideal theorem] \label{corpithm} - Let $X \se \fk^n$ be irreducible, $(f_i)_{i=1}^k$ elements of $R = \fk[X_1,\ldots,X_n]$ and $Y$ an irreducible component of $A = X \cap \bigcap_{i=1}^k V(f_i)$. + Let $X \se \mathfrak{k}^n$ be irreducible, $(f_i)_{i=1}^k$ elements of $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $Y$ an irreducible component of $A = X \cap \bigcap_{i=1}^k V(f_i)$. Then $\codim(Y,X) \le k$. \end{corollary} \begin{remark} @@ -1846,19 +1846,19 @@ Let $R = \fk[X_1,\ldots,X_n]$ and $I \se R$ an ideal. \end{remark} \begin{corollary}\label{codimintersection} - Let $A$ and $B$ be irreducible subsets of $\fk^n$. If $C$ is an irreducible component of $A \cap B$, then $\codim(C, \fk^n) \le \codim(A, \fk^n) + \codim(B, \fk^n)$. + Let $A$ and $B$ be irreducible subsets of $\mathfrak{k}^n$. If $C$ is an irreducible component of $A \cap B$, then $\codim(C, \mathfrak{k}^n) \le \codim(A, \mathfrak{k}^n) + \codim(B, \mathfrak{k}^n)$. \end{corollary} \begin{dremark} Equivalently, $\dim(C) \ge \dim(A) + \dim(B)-n$. \end{dremark} \begin{proof} - Let $X = A \times B \se \fk^{2n}$, where we use $(X_1,\ldots,X_n,Y_1,\ldots,Y_n)$ as coordinates of $\fk^{2n}$. - Let $\Delta \coloneqq \{(x_1,\ldots,x_n,x_1,\ldots,x_n) | x \in \fk^n\} $ be the diagonal in $\fk^n \times \fk^n$. - The projection $\fk^{2n}\to \fk^n$ to the $X$-coordinates defines a homeomorphism between $(A \times B) \cap \Delta$ and $A \cap B$. + Let $X = A \times B \se \mathfrak{k}^{2n}$, where we use $(X_1,\ldots,X_n,Y_1,\ldots,Y_n)$ as coordinates of $\mathfrak{k}^{2n}$. + Let $\Delta \coloneqq \{(x_1,\ldots,x_n,x_1,\ldots,x_n) | x \in \mathfrak{k}^n\} $ be the diagonal in $\mathfrak{k}^n \times \mathfrak{k}^n$. + The projection $\mathfrak{k}^{2n}\to \mathfrak{k}^n$ to the $X$-coordinates defines a homeomorphism between $(A \times B) \cap \Delta$ and $A \cap B$. Thus, $C$ is homeomorphic to an irreducible component $C'$ of $(A \times B) \cap \Delta$ and \begin{align} - \codim(C, \fk^n) = n - \dim(C) = n - \dim(C') = n - \dim(A \times B) + \codim(C', A \times B)\\ - \overset{\text{\ref{corpithm}}}{\le }2n - \dim(A \times B) \overset{\text{\ref{dimprod}}}{=} 2n - \dim(A) - \dim(B) = \codim(A,\fk^n) + \codim(B, \fk^n) + \codim(C, \mathfrak{k}^n) = n - \dim(C) = n - \dim(C') = n - \dim(A \times B) + \codim(C', A \times B)\\ + \overset{\text{\ref{corpithm}}}{\le }2n - \dim(A \times B) \overset{\text{\ref{dimprod}}}{=} 2n - \dim(A) - \dim(B) = \codim(A,\mathfrak{k}^n) + \codim(B, \mathfrak{k}^n) \end{align} by the general properties of dimension and codimension, \ref{corpithm} applied to $(X_i - Y_i)_{i=1}^n$, the result about the dimension of products (\ref{dimprod}) and again the general properties of dimension and codimension. @@ -1901,7 +1901,7 @@ Let $R = \fk[X_1,\ldots,X_n]$ and $I \se R$ an ideal. If $A$ is a local ring, then $\rad(A) = \fm_A$. \end{example} \begin{example} - If $A$ is a PID with infinitely many multiplicative equivalence classes of prime elements (e.g. $\Z$ of $\fk[X]$), then $\rad(A) = \{0\}$: + If $A$ is a PID with infinitely many multiplicative equivalence classes of prime elements (e.g. $\Z$ of $\mathfrak{k}[X]$), then $\rad(A) = \{0\}$: Prime ideals of a PID are maximal. Thus if $x \in \rad(A)$, every prime element divides $x$. If $x \neq 0$, it follows that $x$ has infinitely many prime divisors. However every PID is a UFD. \end{example} @@ -1927,7 +1927,7 @@ Let $\fl$ be any field. When dealing with $\bP^n$, the usual convention is to use $0$ as the index of the first coordinate. - We denote the one-dimensional subspace generated by $(x_0,\ldots,x_n) \in \fk^{n+1} \sm \{0\}$ by $[x_0,\ldots,x_n] \in \bP^n$. + We denote the one-dimensional subspace generated by $(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \sm \{0\}$ by $[x_0,\ldots,x_n] \in \bP^n$. If $x = [x_0,\ldots,x_n] \in \bP^n$, the $(x_{i})_{i=0}^n$ are called \vocab{homogeneous coordinates} of $x$. At least one of the $x_{i}$ must be $\neq 0$. \end{definition} @@ -2000,9 +2000,9 @@ Let $\fl$ be any field. \begin{remark} This definition gives $R$ the structure of a graded ring. \end{remark} -\begin{definition}[Zariski topology on $\bP^n(\fk)$]\label{ztoppn} - Let $A = \fk[X_0,\ldots,X_n]$.\footnote{As always, $\fk$ is algebraically closed} - For $f \in A_d = \fk[X_0,\ldots,X_n]_d$, the validity of the equation $f(x_0,\ldots,x_{n}) = 0$ does not depend on the choice of homogeneous coordinates, as +\begin{definition}[Zariski topology on $\bP^n(\mathfrak{k})$]\label{ztoppn} + 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 \[ f(\lambda x_0,\ldots, \lambda x_n) 0 \lambda^d f(x_0,\ldots,x_n) \] @@ -2018,22 +2018,22 @@ Let $\fl$ be any field. \begin{fact} If $X = \bigcap_{i = 1}^k \Vp(f_i) \se \bP^n$ is closed, then $Y = X \cap \bA^n$ can be identified with the closed subset \[ - \{(x_1,\ldots,x_n) \in \fk^n | f_i(1,x_1,\ldots,x_n) = 0, 1 \le i \le k\} \se \fk^n + \{(x_1,\ldots,x_n) \in \mathfrak{k}^n | f_i(1,x_1,\ldots,x_n) = 0, 1 \le i \le k\} \se \mathfrak{k}^n \] - Conversely, if $Y \se \fk^n$ is closed it has the form + Conversely, if $Y \se \mathfrak{k}^n$ is closed it has the form \[ - \{(x_1,\ldots,x_n) \in \fk^n | g_i(x_1,\ldots,x_n) = 0, 1 \le i \le k\} + \{(x_1,\ldots,x_n) \in \mathfrak{k}^n | g_i(x_1,\ldots,x_n) = 0, 1 \le i \le k\} \] and can thus be identified with $X \cap \bA^n$ where $X \coloneqq \bigcap_{i=1}^k \Vp(f_i)$ is given by \[f_i(X_0,\ldots,X_n) \coloneqq X_0^{d_i} g_i(X_1 / X_0,\ldots, X_n / X_0), d_i \ge \deg(g_i)\] - Thus, the Zariski topology on $\fk^n$ can be identified with the topology induced by the Zariski topology on $\bA^n = U_0$, and the same holds for $U_i$ with $0 \le i \le n$. + Thus, the Zariski topology on $\mathfrak{k}^n$ can be identified with the topology induced by the Zariski topology on $\bA^n = U_0$, and the same holds for $U_i$ with $0 \le i \le n$. - In this sense, the Zariski topology on $\bP^n$ can be thought of as gluing the Zariski topologies on the $U_i \cong \fk^n$. + In this sense, the Zariski topology on $\bP^n$ can be thought of as gluing the Zariski topologies on the $U_i \cong \mathfrak{k}^n$. \end{fact} % The Zariski topology on P^n (2) \begin{definition} - Let $I \se A = \fk[X_0,\ldots,X_n]$ be a homogeneous ideal. + Let $I \se A = \mathfrak{k}[X_0,\ldots,X_n]$ be a homogeneous ideal. Let $\Vp(I) \coloneqq \{[x_0,\ldots,_n] \in \bP^n | \A f \in I ~ f(x_0,\ldots,x_n) = 0\}$ As $I$ is homogeneous, it is sufficient to impose this condition for the homogeneous elements $f \in I$. Because $A$ is Noetherian, $I$ can finitely generated by homogeneous elements $(f_i)_{i=1}^k$ and $\Vp(I)=\bigcap_{i=1}^k \Vp(f_i)$ as in \ref{ztoppn}. @@ -2058,8 +2058,8 @@ Let $\fl$ be any field. \end{proof} \begin{corollary} The Zariski topology on $\bP^n$ is indeed a topology. - The induced topology on the open set $\bA^n = \bP^n \sm \Vp(X_0) \cong \fk^n$ is the Zariski topology on $\fk^n$. - The same holds for all $U_i = \bP^n \sm \Vp(X_i) \cong \fk^n$. + The induced topology on the open set $\bA^n = \bP^n \sm \Vp(X_0) \cong \mathfrak{k}^n$ is the Zariski topology on $\mathfrak{k}^n$. + The same holds for all $U_i = \bP^n \sm \Vp(X_i) \cong \mathfrak{k}^n$. Moreover, the topological space $\bP^n$ is Noetherian. \end{corollary} @@ -2103,7 +2103,7 @@ Let $\fl$ be any field. % Lecture 12 \subsection{The projective form of the Nullstellensatz and the closed subsets of $\bP^n$} -Let $A = \fk[X_0,\ldots,X_n]$. +Let $A = \mathfrak{k}[X_0,\ldots,X_n]$. \begin{proposition}[Projective form of the Nullstellensatz]\label{hnsp} If $I \se A$ is a homogeneous ideal and $f \in A_d$ with $d>0$, then $\Vp(I) \se \Vp(f) \iff f \in \sqrt{I}$. \end{proposition} @@ -2197,7 +2197,7 @@ Let $A = \fk[X_0,\ldots,X_n]$. &= \codim(X \cap \bA^n, Z \cap \bA^n)\\ &= \codim(X, Z) \end{align} - because $\fk^n$ is catenary and the first point follows. + because $\mathfrak{k}^n$ is catenary and the first point follows. The remaining assertions can easily be derived from the first two. \end{proof} @@ -2205,9 +2205,9 @@ Let $A = \fk[X_0,\ldots,X_n]$. \begin{definition} If $X \se \bP^n$ is closed, we define the \vocab{affine cone over $X$} \[ - C(X) = \{0\} \cup \{(x_0,\ldots,x_n) \in \fk^{n+1} \sm \{0\} | [x_0,\ldots,x_n] \in X\} + C(X) = \{0\} \cup \{(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \sm \{0\} | [x_0,\ldots,x_n] \in X\} \] - If $X = \Vp(I)$ where $I \se A_+ = \fk[X_0,\ldots,X_n]_+$ is homogeneous, then $C(X) = \Va(I)$. + If $X = \Vp(I)$ where $I \se A_+ = \mathfrak{k}[X_0,\ldots,X_n]_+$ is homogeneous, then $C(X) = \Va(I)$. \end{definition} \begin{proposition}\label{conedim} \begin{itemize} @@ -2216,7 +2216,7 @@ Let $A = \fk[X_0,\ldots,X_n]$. $\dim(C(X)) = \dim(X) + 1$ and - $\codim(C(X), \fk^{n+1}) = \codim(X, \bP^n)$ + $\codim(C(X), \mathfrak{k}^{n+1}) = \codim(X, \bP^n)$ \end{itemize} \end{proposition} \begin{proof} @@ -2228,9 +2228,9 @@ Let $A = \fk[X_0,\ldots,X_n]$. \] be a chain of irreducible subsets of $\bP^n$. Then \[ - \{0\} \subsetneq C(X_0) \subsetneq \ldots \subsetneq C(X_d) = C(X) \subsetneq \ldots \subsetneq C(X_n) = \fk^{n+1} + \{0\} \subsetneq C(X_0) \subsetneq \ldots \subsetneq C(X_d) = C(X) \subsetneq \ldots \subsetneq C(X_n) = \mathfrak{k}^{n+1} \] - is a chain of irreducible subsets of $\fk^{n+1}$. Hence $\dim(C(X)) \ge 1 + d$ and $\codim(C(X), \fk^{n+1}) \ge n-d$. Since $\dim(C(X)) + \codim(C(X), \fk^{n+1}) = \dim(\fk^{n+1}) = n+1$, the two inequalities must be equalities. + is a chain of irreducible subsets of $\mathfrak{k}^{n+1}$. Hence $\dim(C(X)) \ge 1 + d$ and $\codim(C(X), \mathfrak{k}^{n+1}) \ge n-d$. Since $\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) = \dim(\mathfrak{k}^{n+1}) = n+1$, the two inequalities must be equalities. \end{proof} \subsubsection{Application to hypersurfaces in $\bP^n$} \begin{definition}[Hypersurface] @@ -2242,10 +2242,10 @@ Let $A = \fk[X_0,\ldots,X_n]$. If $P \in A_d$ is a prime element, then $H = \Vp(P)$ is a hypersurface in $\bP^n$ and every hypersurface $H$ in $\bP^n$ can be obtained in this way. \end{corollary} \begin{proof} - If $H = \Vp(P)$ then $C(H) = \Va(P)$ is a hypersurface in $\fk^{n+1}$ by \ref{irredcodimone}. By \ref{conedim}, $H$ is irreducible and of codimension $1$. + If $H = \Vp(P)$ then $C(H) = \Va(P)$ 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 $\bP^n$. By \ref{conedim}, $C(H)$ is a hypersurface in $\fk^{n+1}$, hence $C(H) = \Vp(P)$ for some prime element $P \in A$ (again by \ref{irredcodimone}). - We have $H = \Vp(\fp)$ for some $\fp \in \Proj(A)$ and $C(H) = \Va(\fp)$. By the bijection between closed subsets of $\fk^{n+1}$ and ideals $I = \sqrt{I} \se A$ (\ref{antimonbij}), $\fp = P \cdot A$. + Conversely, let $H$ be a hypersurface in $\bP^n$. 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 $C(H) = \Va(\fp)$. By the bijection between closed subsets of $\mathfrak{k}^{n+1}$ and ideals $I = \sqrt{I} \se A$ (\ref{antimonbij}), $\fp = P \cdot A$. Let $P = \sum_{k=0}^{d}P_k$ with $P_d \neq 0$ be the decomposition into homogeneous components. If $P_e $ with $e < d$ was $\neq 0$, it could not be a multiple of $P$ contradicting the homogeneity of $\fp = P \cdot A$. Thus, $P$ is homogeneous of degree $d$. \end{proof} @@ -2348,17 +2348,17 @@ If $P_e $ with $e < d$ was $\neq 0$, it could not be a multiple of $P$ contradic \begin{example} If $X = \C^n$ and $\cO(U)$ the set of holomorphic functions on $X$, then $\cO$ is a subsheaf of the sheaf of $\C$-valued $C^{\infty}$-functions on $X$. \end{example} -\subsubsection{The structure sheaf on a closed subset of $\fk^n$} +\subsubsection{The structure sheaf on a closed subset of $\mathfrak{k}^n$} -Let $X \se \fk^n$ be open. Let $R = \fk[X_1,\ldots,X_n]$. +Let $X \se \mathfrak{k}^n$ be open. Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. \begin{definition}\label{structuresheafkn} - For open subsets $U \se X$, let $\cO_X(U)$ be the set of functions $U \xrightarrow{\phi} \fk$ such that every $x \in U$ has a neighbourhood $V$ such that there are $f,g \in R$ such that for $y \in V$ we have $g(y) \neq 0$ and $\phi(y) = \frac{f(y)}{g(y)}$. + For open subsets $U \se X$, let $\cO_X(U)$ 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} - $\cO_X$ is a subsheaf (of rings) of the sheaf of $\fk$-valued functions on $X$. + $\cO_X$ is a subsheaf (of rings) of the sheaf of $\mathfrak{k}$-valued functions on $X$. The elements of $\cO_X(U)$ are continuous: - Let $M \se \fk$ be closed. We must show the closedness of $N \coloneqq \phi\inv(M)$ in $U$. For $M = \fk$ this is trivial. Otherwise $M$ is finite and we may assume $M = \{t\} $ for some $t \in \fk$. For $x \in U$, there are open $V_x \se U$ and $f_x, g_x \in R$ such that $\phi = \frac{f_x}{g_x}$ on $V_x$. + Let $M \se \mathfrak{k}$ be closed. We must show the closedness of $N \coloneqq \phi\inv(M)$ 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 $V_x \se U$ and $f_x, g_x \in R$ such that $\phi = \frac{f_x}{g_x}$ on $V_x$. Then $N \cap V_x = V(f_x - t\cdot g_x) \cap V_x)$ is closed in $V_x$. As the $V_x$ cover $U$ and $U$ is quasi-compact, $N$ is closed in $U$. \end{remark} @@ -2381,7 +2381,7 @@ is an isomorphism. \Wlog we can assume $U_x = X \sm V(g_x)$. \end{claim} \begin{subproof} - The closed subsets $(X \sm U_x) \se \fk^n$ has the form $X\sm U_x = V(J_x)$ for some ideal $J_x \se R$. + The closed subsets $(X \sm U_x) \se \mathfrak{k}^n$ has the form $X\sm U_x = V(J_x)$ for some ideal $J_x \se R$. As $x \not\in X \sm V_x$ there is $h_x \in J_x$ with $h_x(x) \neq 0$. Replacing $U_x$ by $X \sm V(h_x)$, $f_x$ by $f_xh_x$ and $g_x$ by $g_xh_x$, we may assume $U_x = X \sm V(g_x)$. \end{subproof} @@ -2413,15 +2413,15 @@ is an isomorphism. Hence $\phi = F\defon{X}$. \end{proof} \subsubsection{The structure sheaf on closed subsets of $\bP^n$} -Let $X \se \bP^n$ be closed and $R_\bullet = \fk[X_0,\ldots,X_n]$ with its usual grading. +Let $X \se \bP^n$ be closed and $R_\bullet = \mathfrak{k}[X_0,\ldots,X_n]$ with its usual grading. \begin{definition}\label{structuresheafpn} - For open $U \se X$, let $\cO_X(U)$ be the set of functions $U \xrightarrow{\phi} \fk$ such that for every $x \in U$, there are an open subset $W \se U$, a natural number $d$ and $f,g \in R_d$ such that $W \cap \Vp(g) = \emptyset$ and $\phi(y) = \frac{f(y_0,\ldots,y_n)}{g(y_0,\ldots,y_n)}$ for $y = [y_0,\ldots,y_n] \in W$. + For open $U \se X$, let $\cO_X(U)$ be the set of functions $U \xrightarrow{\phi} \mathfrak{k}$ such that for every $x \in U$, there are an open subset $W \se U$, a natural number $d$ and $f,g \in R_d$ such that $W \cap \Vp(g) = \emptyset$ and $\phi(y) = \frac{f(y_0,\ldots,y_n)}{g(y_0,\ldots,y_n)}$ for $y = [y_0,\ldots,y_n] \in W$. \end{definition} \begin{remark} - This is a subsheaf of rings of the sheaf of $\fk$-valued functions on $X$. -Under the identification $\bA^n =\fk^n$ with $\bP^n \sm \Vp(X_0)$, one has $\cO_X \defon{X \sm \Vp(X_0)} = \cO_{X \cap \bA^n}$ as subsheaves of the sheaf of $\fk$-valued functions, where the second sheaf is a sheaf on a closed subset of $\fk^n$: + This is a subsheaf of rings of the sheaf of $\mathfrak{k}$-valued functions on $X$. +Under the identification $\bA^n =\mathfrak{k}^n$ with $\bP^n \sm \Vp(X_0)$, one has $\cO_X \defon{X \sm \Vp(X_0)} = \cO_{X \cap \bA^n}$ as subsheaves of the sheaf of $\mathfrak{k}$-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 $X \cap \bA^n$ then @@ -2471,7 +2471,7 @@ 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 $R$-algebras \item The category of topological spaces - \item The category $\Var_\fk$ of varieties over $\fk$ (see \ref{defvariety}) + \item The category $\Var_\mathfrak{k}$ of varieties over $\mathfrak{k}$ (see \ref{defvariety}) \item If $\cA$ is a category, then the \vocab{opposite category} or \vocab{dual category} is defined by $\Ob(\cA\op) = \Ob(\cA)$ and $\Hom_{\cA\op}(X,Y) = \Hom_\cA(Y,X)$. \end{itemize} In most of these cases, isomorphisms in the category were just called `isomorphism'. The isomorphisms in the category of topological spaces are the homeomophisms. @@ -2487,7 +2487,7 @@ The following is somewhat harder than in the affine case: It can be identified with the category of $\Z$-modules. \item The category of finitely generated $R$-modules as a full subcategory of the category of $R$-modules. \item The category of $R$-algebras of finite type as a full subcategory of $\Alg_R$. - \item The category of affine varieties over $\fk$ as a full subcategory of the category of varieties over $\fk$. + \item The category of affine varieties over $\mathfrak{k}$ as a full subcategory of the category of varieties over $\mathfrak{k}$. \end{itemize} \end{example} @@ -2505,7 +2505,7 @@ The following is somewhat harder than in the affine case: \begin{example} \begin{itemize} \item There are \vocab[Functor!forgetful]{forgetful functors} from rings to abelian groups or from abelian groups to sets which drop the multiplicative structure of a ring or the group structure of a group. - \item If $\fk$ is any vector space there is a contravariant functor from $\fk$-vector spaces to itself sending $V$ to its dual vector space $V\se$ and $V \xrightarrow{f} W$ to the dual linear map $W\st \xrightarrow{f\st} V\st$. + \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 $V\se$ and $V \xrightarrow{f} W$ to the dual linear map $W\st \xrightarrow{f\st} V\st$. When restricted to the full subcategory of finite-dimensional vector spaces it becomes a contravariant self-equivalence of that category. \item The embedding of a subcategory is a faithful functor. In the case of a full subcategory it is also full. \end{itemize} @@ -2516,7 +2516,7 @@ The following is somewhat harder than in the affine case: \subsection{The category of varieties} \begin{definition}[Algebraic variety]\label{defvariety} - An \vocab{algebraic variety} or \vocab{prevariety} over $\fk$ is a pair $(X, \cO_X)$, where $X$ is a topological space and $\cO_X$ a subsheaf of the sheaf of $\fk$-valued functions on $X$ such that for every $x \in X$, there are a neighbourhood $U_x$ of $x$ in $X$, an open subset $V_x$ of a closed subset $Y_x$ of $\fk^{n_x}$\footnote{By the result of \ref{affopensubtopbase} it can be assumed that $V_x = Y_x$ without altering the definition.} and a homeomorphism $V_x \xrightarrow{\iota_x} U_x$ such that for every open subset $V \se U_x$ and every function $V\xrightarrow{f} \fk$, we have $f \in \cO_X(V) \iff \iota\st_x(f) \in \cO_{Y_x}(\iota_x\inv(V))$, + An \vocab{algebraic variety} or \vocab{prevariety} over $\mathfrak{k}$ is a pair $(X, \cO_X)$, where $X$ is a topological space and $\cO_X$ 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 $V \se U_x$ and every function $V\xrightarrow{f} \mathfrak{k}$, we have $f \in \cO_X(V) \iff \iota\st_x(f) \in \cO_{Y_x}(\iota_x\inv(V))$, In this, the \vocab{pull-back} $\iota_x\st(f)$ of $f$ is defined by $(\iota_x\st(f))(\xi) \coloneqq f(\iota_x(\xi))$. @@ -2527,10 +2527,10 @@ The following is somewhat harder than in the affine case: \begin{example} \begin{itemize} \item If $(X, \cO_X)$ is a variety and $U \se X$ open, then $(U, \cO_X\defon{U})$ is a variety (called an \vocab{open subvariety} of $X$), and the embedding $U \to X$ is a morphism of varieties. - \item If $X$ is a closed subset of $\fk^n$ or $\bP^n$, then $(X, \cO_X)$ is a variety, where $\cO_X$ is the structure sheaf on $X$ (\ref{structuresheafkn}, reps. \ref{structuresheafpn}). - A variety is called \vocab[Variety!affine]{affine} (resp. \vocab[Variety!projective]{projective}) if it is isomorphic to a variety of this form, with $X $ closed in $\fk^n$ (resp. $\bP^n$). + \item If $X$ is a closed subset of $\mathfrak{k}^n$ or $\bP^n$, then $(X, \cO_X)$ is a variety, where $\cO_X$ is the structure sheaf on $X$ (\ref{structuresheafkn}, reps. \ref{structuresheafpn}). + A variety is called \vocab[Variety!affine]{affine} (resp. \vocab[Variety!projective]{projective}) if it is isomorphic to a variety of this form, with $X $ closed in $\mathfrak{k}^n$ (resp. $\bP^n$). A variety which is isomorphic to and open subvariety of $X$ is called \vocab[Variety!quasi-affine]{quasi-affine} (resp. \vocab[Variety!quasi-projective]{quasi-projective}). - \item If $X = V(X^2 - Y^3) \se \fk^2$ then $\fk \xrightarrow{t \mapsto (t^3,t^2)} X$ is a morphism which is a homeomorphism of topological spaces but not an isomorphism of varieties. + \item If $X = V(X^2 - Y^3) \se \mathfrak{k}^2$ then $\mathfrak{k} \xrightarrow{t \mapsto (t^3,t^2)} X$ is a morphism which is a homeomorphism of topological spaces but not an isomorphism of varieties. % TODO \item The composition of two morphisms $X \to Y \to Z$ of varieties is a morphism of varieties. @@ -2540,10 +2540,10 @@ The following is somewhat harder than in the affine case: \subsubsection{The category of affine varieties} \begin{lemma}\label{localinverse} - Let $X$ be any $\fk$-variety and $U \se X$ open. + Let $X$ be any $\mathfrak{k}$-variety and $U \se X$ open. \begin{enumerate}[i)] \item All elements of $\cO_X(U)$ are continuous. - \item If $U \se X$ is open, $U \xrightarrow{\lambda} \fk$ any function and every $x \in U$ has a neighbourhood $V_x \se U$ such that $\lambda \defon{V_x} \in \cO_X(V_x)$, then $\lambda \in \cO_X(U)$. + \item If $U \se X$ is open, $U \xrightarrow{\lambda} \mathfrak{k}$ any function and every $x \in U$ has a neighbourhood $V_x \se U$ such that $\lambda \defon{V_x} \in \cO_X(V_x)$, then $\lambda \in \cO_X(U)$. \item If $\vartheta \in \cO_X(U)$ and $\vartheta(x) \neq 0$ for all $x \in U$, then $\vartheta \in \cO_X(U)^{\times }$. \end{enumerate} \end{lemma} @@ -2553,8 +2553,8 @@ The following is somewhat harder than in the affine case: \item For the second part, let $\lambda_x \coloneqq \lambda \defon{V_x} $. We have $\lambda_x\defon{V_x \cap V_y} = \lambda \defon{V_x \cap V_y} = \lambda_y \defon{V_x \cap V_y} $. The $V_x$ cover $U$. By the sheaf axiom for $\cO_X$ there is $\ell \in \cO_X(U)$ with $\ell\defon{V_x} =\lambda_x$. It follows that $\ell=\lambda$. - \item By the definition of variety, every $x \in U$ has a quasi-affine neighbourhood $V \se U$. We can assume $U$ to be quasi-affine and $X = V(I) \se \fk^n$, as the general assertion follows by an application of ii). - If $x \in U$ there are a neighbourhood $x \in W \se U$ and $a,b \in R = \fk[X_1,\ldots,X_n]$ such that $\vartheta(y) = \frac{a(y)}{b(y)}$ for $y \in W$, with $b(y) \neq 0$. + \item By the definition of variety, every $x \in U$ has a quasi-affine neighbourhood $V \se U$. We can assume $U$ to be quasi-affine and $X = V(I) \se \mathfrak{k}^n$, as the general assertion follows by an application of ii). + If $x \in U$ there are a neighbourhood $x \in W \se U$ and $a,b \in R = \mathfrak{k}[X_1,\ldots,X_n]$ 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 $W \sm V(a)$, we may assume that $a$ has no zeroes on $W$. Then $\lambda(y) = \frac{b(y)}{a(y)}$ for $y \in W$ has a non-vanishing denominator and $\lambda \in \cO_X(U)$. We have $\lambda \cdot \vartheta = 1$, thus $\vartheta \in \cO_X(U)^{\times}$. @@ -2565,29 +2565,29 @@ The following is somewhat harder than in the affine case: \begin{proposition}[About affine varieties] \label{propaffvar} \begin{itemize} - \item Let $X,Y$ be varieties over $\fk$. Then the map + \item Let $X,Y$ be varieties over $\mathfrak{k}$. Then the map \begin{align} - \phi: \Hom_{\Var_\fk}(X,Y) &\longrightarrow \Hom_{\Alg_\fk}(\cO_Y(Y), \cO_X(X)) \\ + \phi: \Hom_{\Var_\mathfrak{k}}(X,Y) &\longrightarrow \Hom_{\Alg_\mathfrak{k}}(\cO_Y(Y), \cO_X(X)) \\ (X \xrightarrow{f} Y) &\longmapsto (\cO_Y(Y) \xrightarrow{f\st} \cO_X(X)) \end{align} is injective when $Y$ is quasi-affine and bijective when $Y$ is affine. \item The contravariant functor \begin{align} - F: \Var_\fk &\longrightarrow \Alg_\fk \\ + F: \Var_\mathfrak{k} &\longrightarrow \Alg_\mathfrak{k} \\ X &\longmapsto \cO_X(X)\\ (X\xrightarrow{f} Y) &\longmapsto (\cO_X(X) \xrightarrow{f\st} \cO_Y(Y)) \end{align} - restricts to an equivalence of categories between the category of affine varieties over $\fk$ and the full subcategory $\cA$ of $\Alg_\fk$, - having the $\fk$-algebras $A$ of finite type with $\nil A = \{0\} $ as objects. + restricts to an equivalence of categories between the category of affine varieties over $\mathfrak{k}$ and the full subcategory $\cA$ 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 $\nil(\cO_X(X)) = \{0\}$ for arbitrary varieties. For general varieties it is however not true that $\cO_X(X)$ is a $\fk$-algebra of finite type. + It is clear that $\nil(\cO_X(X)) = \{0\}$ for arbitrary varieties. For general varieties it is however not true that $\cO_X(X)$ is a $\mathfrak{k}$-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 $I = \sqrt{I} \se R$ is an ideal with $R = \fk[X_1,\ldots,X_n]$. - Then $\cO_X(X) \cong R / I$ (see \ref{structuresheafri}) is a $\fk$-algebra of finite type. + If, however, $X$ is affine, we may assume w.l.o.g. that $X = V(I)$ where $I = \sqrt{I} \se R$ is an ideal with $R = \mathfrak{k}[X_1,\ldots,X_n]$. + Then $\cO_X(X) \cong R / I$ (see \ref{structuresheafri}) is a $\mathfrak{k}$-algebra of finite type. \end{remark} \begin{proof} @@ -2595,19 +2595,19 @@ The following is somewhat harder than in the affine case: - It suffices to investigate $\phi$ when $Y$ is an open subset of $V(I) \se \fk^n$, where $I = \sqrt{I} \se R$ is an ideal and $Y = V(I)$ when $Y$ is affine. - Let $(f_1,\ldots,f_n)$ be the components of $X \xrightarrow{f} Y \se \fk^n$. Let $Y \xrightarrow{\xi_i} \fk$ be the $i$-th coordinate. + It suffices to investigate $\phi$ when $Y$ is an open subset of $V(I) \se \mathfrak{k}^n$, where $I = \sqrt{I} \se R$ is an ideal and $Y = V(I)$ when $Y$ is affine. + Let $(f_1,\ldots,f_n)$ be the components of $X \xrightarrow{f} Y \se \mathfrak{k}^n$. Let $Y \xrightarrow{\xi_i} \mathfrak{k}$ be the $i$-th coordinate. By definition $f_i = f\st(\xi_i) $. Thus $f$ is uniquely determined by $\cO_Y(Y) \xrightarrow{f\st} \cO_X(X)$. - Conversely, let $Y = V(I)$ and $\cO_Y(Y) \xrightarrow{\phi} \cO_X(X)$ be a morphism of $\fk$-algebras. Define $f_i \coloneqq \phi(\xi_i)$ and consider $X \xrightarrow{f = (f_1,\ldots,f_n)} Y\se \fk^n$. + Conversely, let $Y = V(I)$ and $\cO_Y(Y) \xrightarrow{\phi} \cO_X(X)$ be a morphism of $\mathfrak{k}$-algebras. Define $f_i \coloneqq \phi(\xi_i)$ and consider $X \xrightarrow{f = (f_1,\ldots,f_n)} Y\se \mathfrak{k}^n$. \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 $\fk$-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 $\Var_\fk$ + $f$ is a morphism in $\Var_\mathfrak{k}$ \end{claim} \begin{subproof} For open $\Omega \se Y, U = f\inv(\Omega) = \{x \in X | \A \lambda \in J ~ (\phi(\lambda))(x) \neq 0\}$ is open in $X$, where $Y \sm \Omega = V(J)$. @@ -2626,7 +2626,7 @@ The following is somewhat harder than in the affine case: \begin{subproof} It follows from the remark that the functor maps affine varieties to objects of $\cA$. - If $A \in \Ob(\cA)$ then $ A /\fk$ is of finite type, thus $A \cong R / I$ for some $n$. + If $A \in \Ob(\cA)$ 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 $x \in \sqrt{I}$, $x \mod I \in \nil(R / I) \cong \nil(A) = \{0\}$. Thus $A \cong\cO_X(X)$ where $X = V(I)$. \end{subproof} @@ -2634,7 +2634,7 @@ The following is somewhat harder than in the affine case: \end{proof} \begin{remark} - Note that giving a contravariant functor $\cC \to \cD$ is equivalent to giving a functor $\cC \to \cD\op$. We have thus shown that the category of affine varieties is equivalent to $\cA\op$, where $\cA \subsetneq \Alg_\fk$ is the full subcategory of $\fk$-algebras $A$ of finite type with $\nil(A) = \{0\}$. + Note that giving a contravariant functor $\cC \to \cD$ is equivalent to giving a functor $\cC \to \cD\op$. We have thus shown that the category of affine varieties is equivalent to $\cA\op$, where $\cA \subsetneq \Alg_\mathfrak{k}$ is the full subcategory of $\mathfrak{k}$-algebras $A$ of finite type with $\nil(A) = \{0\}$. \end{remark} \subsubsection{Affine open subsets are a topology base} @@ -2650,11 +2650,11 @@ If $X$ is a set, then $\cB \se \cP(X)$ is a base for some topology on $X$ iff $X \end{definition} \begin{proposition}\label{oxulocaf} - Let $X$ be an affine variety over $\fk$, $\lambda \in \cO_X(X)$ and $U = X \sm V(\lambda)$. + Let $X$ be an affine variety over $\mathfrak{k}$, $\lambda \in \cO_X(X)$ and $U = X \sm V(\lambda)$. Then $U$ is an affine variety and the morphism $\phi: \cO_X(X)_\lambda \to \cO_X(U)$ defined by the restriction $\cO_X(X) \xrightarrow{\cdot |_U } \cO_X(U)$ and the universal property of the localization is an isomorphism. \end{proposition} \begin{proof} - Let $X$ be an affine variety over $\fk, \lambda \in \cO_X(X)$ and $U = X \sm V(\lambda)$. The fact that $\lambda\defon{U} \in \cO_x(U)^{\times}$ follows from \ref{localinverse}. + Let $X$ be an affine variety over $\mathfrak{k}, \lambda \in \cO_X(X)$ and $U = X \sm V(\lambda)$. The fact that $\lambda\defon{U} \in \cO_x(U)^{\times}$ follows from \ref{localinverse}. Thus the universal property of the localization $\cO_X(X)_\lambda$ can be applied to $\cO_X(X) \xrightarrow{\cdot |_U} \cO_X(U)$. \begin{figure}[H] \centering @@ -2668,11 +2668,11 @@ If $X$ is a set, then $\cB \se \cP(X)$ is a base for some topology on $X$ iff $X X \arrow[hookrightarrow]{r}{}& U \arrow[swap]{u}{\sigma} & \cO_X(U) \end{tikzcd} \end{figure} - For the rest of the proof, we may assume $X = V(I) \se \fk^n$ where $I = \sqrt{I} \se R \coloneqq\fk[X_1,\ldots,X_n]$ is an ideal. + For the rest of the proof, we may assume $X = V(I) \se \mathfrak{k}^n$ where $I = \sqrt{I} \se R \coloneqq\mathfrak{k}[X_1,\ldots,X_n]$ is an ideal. Then $A \coloneqq \cO_X(X) \cong R / I$ and there is $\ell \in R$ such that $\ell\defon{X} = \lambda$. - Let $Y = V(J) \se \fk^{n+1}$ where $J \se \fk[Z,X_1,\ldots,X_n]$ is generated by the elements of $I$ and $1 - Z\ell(X_1,\ldots,X_n)$. + Let $Y = V(J) \se \mathfrak{k}^{n+1}$ where $J \se \mathfrak{k}[Z,X_1,\ldots,X_n]$ is generated by the elements of $I$ and $1 - Z\ell(X_1,\ldots,X_n)$. - Then $\cO_Y(Y) \cong \fk[Z,X_1,\ldots,X_n] / J \cong A[Z] / (1 -\lambda Z) \cong A_\lambda$. + Then $\cO_Y(Y) \cong \mathfrak{k}[Z,X_1,\ldots,X_n] / J \cong A[Z] / (1 -\lambda Z) \cong A_\lambda$. By the proposition about affine varieties (\ref{propaffvar}), the morphism $\fs: \cO_Y(Y) \cong A_\lambda \to \cO_X(U)$ corresponds to a morphism $U \xrightarrow{\sigma} Y$. We have $\fs(Z \mod J) = \lambda\inv$ and $\fs(X_i \mod J) = X_i \mod I$. Thus $\sigma(x) = (\lambda(x)\inv, x)$ for $x \in U$. @@ -2687,7 +2687,7 @@ If $X$ is a set, then $\cB \se \cP(X)$ is a base for some topology on $X$ iff $X The affine open subsets of a variety $X$ are a topology base on $X$. \end{corollary} \begin{proof} - Let $X = V(I) \se \fk^n$ with $I = \sqrt{I}$. If $U \se X$ is open then $X \sm U = V(J)$ with $J \supseteq I$ and $U = \bigcup_{f \in J} (X \sm V(f))$. + Let $X = V(I) \se \mathfrak{k}^n$ with $I = \sqrt{I}$. If $U \se X$ is open then $X \sm U = V(J)$ with $J \supseteq I$ and $U = \bigcup_{f \in J} (X \sm V(f))$. Thus $U$ is a union of affine open subsets. The same then holds for arbitrary quasi-affine varieties. Let $X$ be any variety, $U \se X$ open and $x \in U$. @@ -2759,7 +2759,7 @@ If $X$ is a set, then $\cB \se \cP(X)$ is a base for some topology on $X$ iff $X \end{proof} \begin{proposition}\label{proplocalring} - Let $X = \Va(I) \se \fk^n$ be equipped with its usual structure sheaf, where $I = \sqrt{I} \se R = \fk[X_1,\ldots,X_n]$ . Let $x \in X$ and $A = \cO_X(X) \cong R / I$. + Let $X = \Va(I) \se \mathfrak{k}^n$ be equipped with its usual structure sheaf, where $I = \sqrt{I} \se R = \mathfrak{k}[X_1,\ldots,X_n]$ . Let $x \in X$ and $A = \cO_X(X) \cong R / I$. $\{P \in R | P(x) = 0\} \text{\reflectbox{$\coloneqq$}} \fn_x \se R$ is maximal, $I \se \fn_x$ and $\fm_x \coloneqq \fn_x / I$ is the maximal ideal of elements of $A$ vanishing at $x$. If $\lambda \in A \sm \fm_x$, we have $\lambda_x \in \cO_{X,x}^{\times}$, where $\lambda_x$ denotes the image under $A \cong \cO_X(X) \to \cO_{X,x}$. By the universal property of the localization, there exists a unique ring homomorphism $A_{\fm_x} \xrightarrow{\iota} \cO_{X,x}$ @@ -2791,13 +2791,13 @@ If $X$ is a set, then $\cB \se \cP(X)$ is a base for some topology on $X$ iff $X \end{proof} \subsubsection{Intersection multiplicities and Bezout's theorem} \begin{definition} - Let $R = \fk[X_0,X_1,X_2]$ equipped with its usual grading and let $x \in \bP^{2}$. + Let $R = \mathfrak{k}[X_0,X_1,X_2]$ equipped with its usual grading and let $x \in \bP^{2}$. Let $G \in R_g, H \in R_h$ be homogeneous polynomials with $x \in V(G) \cap V(h)$. Let $\ell\in R_1$ such that $\ell(x) \neq 0$. Then $x \in U = \bP^2 \sm V(\ell)$ and the rational functions $\gamma = \ell^{-g}G, \eta = \ell^{-h}H$ are elements of $\cO_{\bP^2}(U)$. Let $I_x(G,H) \se \cO_{\bP^2,x}$ denote the ideal generated by $\gamma_x$ and $\eta_x$. -\noindent The dimension $\dim_{\fk}(\cO_{X,x} / I_x(G,H)) \text{\reflectbox{$\coloneqq$}} i_x(G,H)$ is called the \vocab{intersection multiplicity} of $G$ and $H$ at $x$. +\noindent The dimension $\dim_{\mathfrak{k}}(\cO_{X,x} / I_x(G,H)) \text{\reflectbox{$\coloneqq$}} i_x(G,H)$ is called the \vocab{intersection multiplicity} of $G$ and $H$ at $x$. \end{definition} \begin{remark} If $\tilde \ell \in R_1$ also satisfies $\tilde \ell(x) \neq 0$, then the image of $\tilde \ell / \ell$ under $\cO_{\bP^2}(U) \to \cO_{\bP^2,x}$ is a unit, showing that the image of $\tilde \gamma = \tilde \ell^{-g} G$ in $\cO_{\bP^2,x}$ is multiplicatively equivalent to $\gamma_x$, and similarly for $\eta_x$. @@ -2839,7 +2839,7 @@ Let $I_x(G,H) \se \cO_{\bP^2,x}$ denote the ideal generated by $\gamma_x$ and $\ $\ev_a(K[X_1,\ldots,X_n])$ is a field by fact about integrality and fields \ref{fintaf}. Hence $n = 0$ and $L / K$ is finite. \item[UNCHNS2] $K$ uncountable, $L / K$ fin. type. Then $\dim_K L$ is countable. Suppose $l \in L$ is not integral. Then $K(l) \cong K(T)$ and $\dim_K L \ge \dim_K K(T) \ge \aleph_1$. Thus $L / K$ algebraic $\implies$ integral $\implies$ finite. - \item[HNS3] ($V(I) \se V(f) \iff f \in \sqrt{I} $). Suppose $V(I) \se V(f)$. $R' \coloneqq \fk[X_1,\ldots,X_n, T], J \se R'$ the ideal generated by $I$ and $g(X_1,\ldots,X_n,T) \coloneqq 1 - Tf(X_1,\ldots,X_n)$. + \item[HNS3] ($V(I) \se V(f) \iff f \in \sqrt{I} $). Suppose $V(I) \se V(f)$. $R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n, T], J \se R'$ the ideal generated by $I$ and $g(X_1,\ldots,X_n,T) \coloneqq 1 - Tf(X_1,\ldots,X_n)$. \end{itemize} @@ -2873,21 +2873,21 @@ Thus $Q(T) = p_\alpha T^{w_k(\alpha)} + \ldots$ where $\alpha \in S$ such that $ % A first result of dimension theory: -$A \fl$-algebra of finite type, $\fp, \fq \in \Spec A, \fp \subsetneq \fq$. Then $\trdeg(\fk(\fp) /\fl) > \trdeg(\fk(\fq) / \fl)$: +$A \fl$-algebra of finite type, $\fp, \fq \in \Spec A, \fp \subsetneq \fq$. Then $\trdeg(\mathfrak{k}(\fp) /\fl) > \trdeg(\mathfrak{k}(\fq) / \fl)$: \Wlog $\fp = \{0\}$ and $A$ a domain ($A' \coloneqq A / \fp$). -For $\fq \in \mSpec A$, $\fk(\fq) = A / \fq$ finite type, hence finite (HNS) $\implies \trdeg(\fk(\fq) / \fl) = 0$. -$\trdeg(Q(A) / \fl) = 0 \implies A$ integral over $\fk$ $\implies$ $A$ a field $\implies \fp = \fq \lightning$. +For $\fq \in \mSpec A$, $\mathfrak{k}(\fq) = A / \fq$ finite type, hence finite (HNS) $\implies \trdeg(\mathfrak{k}(\fq) / \fl) = 0$. +$\trdeg(Q(A) / \fl) = 0 \implies A$ integral over $\mathfrak{k}$ $\implies$ $A$ a field $\implies \fp = \fq \lightning$. -If $\fq \not\in \mSpec A$, let $a_1,\ldots,a_n \in A$ alg. independent such that the $\overline{a_i}$ are a transcendence base for $\fk(\fq) / \fk$ +If $\fq \not\in \mSpec A$, let $a_1,\ldots,a_n \in A$ alg. independent such that the $\overline{a_i}$ are a transcendence base for $\mathfrak{k}(\fq) / \mathfrak{k}$ Let $R$ be the ring generated by $\fl$ and the $a_i$. Localize with respect to $S \coloneqq R \sm \{0\}$. %TODO % TODO: LERNEN % Dim k^n -$\dim(\fk^n)$ +$\dim(\mathfrak{k}^n)$ $ \ge n$ build chian -$\le n$ a first result in dim T ($\fp \subsetneq \fq \implies \trdeg(\fk(\fq) / \fl) < \trdeg(\fk(\fp) / \fl)$. Thus $\codim(X,Y) \le \trdeg(\fK(Y) / \fl) - \trdeg(\fK(X) / \fl)$. +$\le n$ a first result in dim T ($\fp \subsetneq \fq \implies \trdeg(\mathfrak{k}(\fq) / \fl) < \trdeg(\mathfrak{k}(\fp) / \fl)$. Thus $\codim(X,Y) \le \trdeg(\fK(Y) / \fl) - \trdeg(\fK(X) / \fl)$. TODO % List of proofs of HNS @@ -2897,7 +2897,7 @@ TODO % TODO proof of dim Y = trdeg(K(Y) / k) -$\dim Y \ge \trdeg(\fk(Y) / \fk)$: Noether normalization. Subalgebra $\cong \fk[X_1,\ldots,X_d]$. Lift chain of prime ideals using going up. +$\dim Y \ge \trdeg(\mathfrak{k}(Y) / \mathfrak{k})$: Noether normalization. Subalgebra $\cong \mathfrak{k}[X_1,\ldots,X_d]$. Lift chain of prime ideals using going up. % TODO prime avoidance @@ -2923,8 +2923,8 @@ The ht p and trdeg % TODO % TODO % TODO % % Definitions -Zariski-Topology, Spec, $\fk^n$ -Residue field $\fk(\fp) \coloneqq Q(A / \fp), \fK(V(\fp)) \coloneqq \fk(\fp)$. TODO? +Zariski-Topology, Spec, $\mathfrak{k}^n$ +Residue field $\mathfrak{k}(\fp) \coloneqq Q(A / \fp), \fK(V(\fp)) \coloneqq \mathfrak{k}(\fp)$. TODO? % Counterexamples no going-up % list of definitions of codim, dim, trdeg, ht