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%& -shell-escape -enable-write18 |
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\documentclass{lecture} |
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\begin{document} |
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\section{Regular functions} |
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\begin{lemma} |
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If $U \subseteq k^{n}$ is a Zariski-open set and |
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$f_P \colon k^{n} \to k$ is a polynomial function such that |
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for $x \in U$, $f_P(x) \neq 0$, then the function $\frac{1}{f_P}$ is continuous on $U$. |
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\label{lemma:1overP-is-cont} |
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\end{lemma} |
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\begin{proof} |
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For all $t \in k$, |
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\begin{salign*} |
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\left(\frac{1}{f_P}\right)^{-1}(\{t\}) |
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&= \left\{ x \in U \mid \frac{1}{f_P(x)} = t \right\} \\ |
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&= \left\{ x \in U \mid t f_P(x) -1 = 0 \right\} \\ |
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&= \mathcal{V}(tf_P -1) \cap U |
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\end{salign*} |
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is closed in $U$. |
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\end{proof} |
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\begin{bem} |
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There can be many continous functions with respect to the Zariski topology. For instance, |
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all bijective maps $f\colon k \to k$ are Zariski-continuous. In algebraic geometry, we will |
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consider only functions which are locally defined by a rational function. We will define |
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them on open subsets of algebrai sets $V \subseteq k^{n}$, endowed with the topology |
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induced by the Zariski topology of $k^{n}$. |
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\end{bem} |
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\begin{bem}[] |
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The open subsets of algebraic sets $V \subseteq k^{n}$ are exactly the |
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\emph{locally closed subsets} of $k^{n}$. |
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\end{bem} |
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\begin{definition}[] |
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Let $X \subseteq k^{n}$ be a locally closed subset of $k^{n}$. A function |
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$f \colon X \to k$ is called \emph{regular at $x \in X$}, if |
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there exist an open subset $x \in U \subseteq X$ and two polynomial functions |
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$P_U, Q_U \colon U \to k$ such that for all $y \in U$, $Q_U(y) \neq 0$ and |
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\[ |
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f(y) = \frac{P_U(y)}{Q_U(y)} |
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.\] The function $f\colon X \to k$ is called \emph{regular on $X$} if, for all $x \in X$, |
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$f$ is regular at $x$. |
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\end{definition} |
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\begin{bsp}[] |
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A rational fraction $\frac{P}{Q} \in k(T_1, \ldots, T_n)$ defines a regular |
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function on the standard open set $D(Q)$. |
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\end{bsp} |
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\begin{satz}[] |
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Let $X \subseteq k^{n}$ be a locally closed subset. If $f\colon X \to k$ is regular, |
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then $f$ is continous. |
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\end{satz} |
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\begin{proof} |
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Since continuity is a local property, we may assume $X = \Omega \subseteq k^{n}$ open |
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and $f = \frac{P}{Q}$ for polynomial functions $P, Q\colon \Omega \to k$ such that |
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$Q(y) \neq 0$. By \ref{lemma:1overP-is-cont} it suffices to prove that |
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if $P, R\colon \Omega \to k$ are continuous, then $PR\colon \Omega \to k$, |
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$z \mapsto P(z)R(z)$ is continuous. Let $t \in k$. Then |
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\begin{salign*} |
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(PR)^{-1}(\{t\}) &= \{ z \in \Omega \mid P(z) R(z) - t = 0\} \\ |
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&= \mathcal{V}(PR - t) \cap \Omega |
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\end{salign*} |
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is closed in $\Omega$. |
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\end{proof} |
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\begin{bem} |
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Being a regular function is a local property. |
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\end{bem} |
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\begin{satz} |
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Let $X \subseteq k^{n}$ be a locally closed subset of $k^{n}$, endowed |
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with the induced topology. The map |
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\begin{salign*} |
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\mathcal{O}_X\colon \{\text{open sets of }X\} &\longrightarrow k-\text{algebras}\\ |
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U &\longmapsto \{ \text{regular functions on }U\} |
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\end{salign*} |
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defines a sheaf of sheaf of $k$-algebras on $X$, which is a subsheaf of the sheaf of functions. |
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\end{satz} |
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\begin{proof} |
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Constants, sums and products of regular functions are regular, thus |
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$\mathcal{O}_X(U)$ is a sub algebra of the $k$-algebra of functions |
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$U \to k$. |
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Since restricting a function preserves regularity, $\mathcal{O}_X$ is a presheaf. Since |
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being regular is a local property and the presheaf of functions is a sheaf, |
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$\mathcal{O}_X$ is also a sheaf. |
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\end{proof} |
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\section{Irreducibility} |
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\begin{definition} |
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Let $X$ be a topological space. $X$ is |
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\begin{enumerate}[(i)] |
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\item \emph{irreducible} if $X \neq \emptyset$ and $X$ is not the union |
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of two proper closed subets, i.e. for $X = F_1 \cup F_2$ with $F_1, F_2 \subseteq X$ closed, |
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we have $X = F_1$ or $X = F_2$. |
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\item \emph{connected} if $X$ is not the union of two disjoint proper closed subets, i.e. |
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for $X = F_1 \cup F_2$ with $F_1, F_2 \subseteq X$ closed and $F_1 \cap F_2 = \emptyset$, |
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we have $X = F_1$ or $X = F_2$. |
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\end{enumerate} |
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A space $X$ which is not irreducible, is called \emph{reducible}. |
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\end{definition} |
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\begin{lemma} |
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If $k$ is infinite, $k$ is irreducible in the Zariski topology. |
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\end{lemma} |
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\begin{proof} |
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Closed subsets of $k$ are $k$ and finite subsets of $k$. |
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\end{proof} |
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\begin{bem} |
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If $k$ is finite, $k^{n}$ is the finite union of its points, which are closed, so |
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$k^{n}$ is reducible. |
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\end{bem} |
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\begin{bem} |
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$X$ irreducible $\implies$ $X$ connected, but the converse is false: Let $k$ be infinite and |
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consider $X = \mathcal{V}_{k^2}(x^2 - y^2)$ (see figure \ref{fig:reducible-alg-set}). |
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Since $x^2 - y^2 = (x-y)(x+y) = 0$ in $k$ |
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if and only if $x = -y$ or $x = y$, we have |
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$X = \mathcal{V}_{k^2}(x - y) \cup \mathcal{V}_{k^2}(x+y)$. Thus $X$ is reducible. But |
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$\mathcal{V}_{k^2}(x-y)$ and $\mathcal{V}_{k^2}(x+y)$ are homeomorphic to $k$, in particular |
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irreducible and thus connected. Since $\mathcal{V}_{k^2}{x-y} \cap \mathcal{V}_{k^2}(x+y) \neq \emptyset$, |
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$X$ is connected. |
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\end{bem} |
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\begin{figure} |
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\centering |
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\begin{tikzpicture} |
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\begin{axis} |
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\algebraiccurve[red]{x^2 - y^2} |
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\end{axis} |
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\end{tikzpicture} |
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\caption{Reducible connected algebraic set} |
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\label{fig:reducible-alg-set} |
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\end{figure} |
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\begin{satz} |
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Let $X$ be a non-empty topological space. The following conditions are equivalent: |
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\begin{enumerate}[(i)] |
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\item $X$ is irreducible |
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\item If $U_1 \cap U_2 = \emptyset$ with $U_1, U_2$ open subsets of $X$, then |
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$U_1 = \emptyset$ or $U_2 = \emptyset$. |
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\item If $U \subseteq X$ is open and non-empty, then $U$ is dense in $X$. |
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\end{enumerate} |
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\label{satz:equiv-irred} |
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\end{satz} |
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\begin{proof} |
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Left as an exercise to the reader. |
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\end{proof} |
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\begin{satz} |
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Let $X$ be a topological space and $V \subseteq X$. Then |
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$V$ is irreducible if and only if $\overline{V}$ is irreducible. |
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\label{satz:closure-irred} |
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\end{satz} |
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\begin{proof} |
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Since $\emptyset$ is closed in $X$, we have |
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$ V = \emptyset \iff \overline{V} = \emptyset$. |
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($\Rightarrow$) |
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Let $\overline{V} \subseteq Z_1 \cup Z_2$ with $Z_1, Z_2 \subseteq X$ closed. |
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Then $V \subseteq Z_1 \cup Z_2$ and by irreducibility of $V$ we may assume $V \subseteq Z_1$. |
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Since $Z_1$ is closed, it follows $\overline{V} \subseteq Z_1$. |
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($\Leftarrow$) Let $V \subseteq Z_1 \cup Z_2$ with $Z_1, Z_2 \subseteq X$ closed. |
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Since $Z_1 \cup Z_2$ is closed, we get $\overline{V} \subseteq Z_1 \cup Z_2$. By |
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irreducibility of $\overline{V}$ we may assume $\overline{V} \subseteq Z_1$, thus |
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$V \subseteq Z_1$. |
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\end{proof} |
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\begin{korollar} |
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Let $X$ be an irreducible topological space. Then every non-empty open |
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subset $U \subseteq X$ is irreducible. |
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\label{kor:non-empty-open-of-irred} |
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\end{korollar} |
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\begin{proof} |
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By \ref{satz:equiv-irred}, $\overline{U} = X$ and thus irreducible. The claim |
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follows now from \ref{satz:closure-irred}. |
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\end{proof} |
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\begin{lemma}[prime avoidance] |
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Let $\mathfrak{p}$ be a prime ideal in a commutative ring $A$. If $I, J \subseteq A$ are |
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ideals such that $IJ \subseteq \mathfrak{p}$, then |
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$I \subseteq \mathfrak{p}$ or $J \subseteq \mathfrak{p}$. |
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\label{lemma:prime-avoidance} |
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\end{lemma} |
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\begin{proof} |
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Assume that $I \not\subset \mathfrak{p}$ and $J \not\subset \mathfrak{p}$. Then |
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there exist $a \in I$, such that $a \not\in \mathfrak{p}$ and $b \in J$ such that |
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$b \not\in \mathfrak{p}$. But $ab \in IJ \subseteq \mathfrak{p}$. Since |
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$\mathfrak{p}$ is prime, this implies $a \in \mathfrak{p}$ or |
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$b \in \mathfrak{p}$. Contradiction. |
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\end{proof} |
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\begin{theorem} |
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Let $V \subseteq k^{n}$ be an algebraic set. Then $V$ is irreducible in the Zariski |
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topology if and only if $\mathcal{I}(V)$ is a prime ideal in $k[T_1, \ldots, T_n]$. |
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\end{theorem} |
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\begin{proof} |
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($\Rightarrow$) Since $V \neq \emptyset$, |
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$\mathcal{I}(V) \subsetneq k[T_1, \ldots, T_n]$. |
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Let $P, Q \in k[T_1, \ldots, T_n]$ |
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such that $PQ \in \mathcal{I}(V)$. For $x \in V$, $(PQ)(x) = 0$ in $k$, hence |
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$P(x) = 0$ or $Q(x) = 0$. Thus $x \in \mathcal{V}(P) \cup \mathcal{V}(Q)$. Therefore |
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$V = (\mathcal{V}(P) \cap V) \cup (\mathcal{V}(Q) \cap V)$ is the union of |
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two closed subsets. Since $V$ is irreducible, |
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we may assume $V = \mathcal{V}(P) \cap V \subseteq \mathcal{V}(P)$, hence |
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$P \in \mathcal{I}(V)$ and $\mathcal{I}(V)$ is prime. |
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($\Leftarrow$) $V \neq \emptyset$, since $\mathcal{I}(V)$ is a proper ideal. Let |
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$V = V_1 \cup V_2$ with $V_1, V_2$ closed in $V$. Then |
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\[ |
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\mathcal{I}(V) = \mathcal{I}(V_1 \cup V_2) = \mathcal{I}(V_1) \cap \mathcal{I}(V_2) |
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\supseteq \mathcal{I}(V_1) \mathcal{I}(V_2) |
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.\] By \ref{lemma:prime-avoidance}, we may assume |
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$\mathcal{I}(V_1) \subseteq \mathcal{I}(V)$. But then |
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\[ |
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V_1 = \mathcal{V}(\mathcal{I}(V_1)) \supseteq \mathcal{V}(\mathcal{I}(V)) = V |
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\] since $V_1$ and $V$ are closed. Therefore $V = V_1$ and $V$ is irreducible. |
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\end{proof} |
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\begin{korollar} |
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If $k$ is infinite, the affine space $k^{n}$ is irreducible with respect to the Zariski |
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topology. |
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\end{korollar} |
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\begin{proof} |
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Since $k$ is infinite, $\mathcal{I}(k^{n}) = (0)$ by \ref{satz:k-infinite-everywhere-vanish} |
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which is a prime ideal in the integral domain $k[T_1, \ldots, T_n]$. |
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\end{proof} |
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\begin{theorem} |
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Let $V \subseteq k^{n}$ be an algebraic set. Then there exists a decomposition |
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\[ |
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V = V_1 \cup \ldots \cup V_r |
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\] such that |
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\begin{enumerate}[(i)] |
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\item $V_i$ is a closed irreducible subset of $k^{n}$ for all $i$. |
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\item $V_{i} \not\subset V_j$ for all $i \neq j$. |
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\end{enumerate} |
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This decomposition is unique up to permutations. |
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\label{thm:decomp-irred} |
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\end{theorem} |
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\begin{definition}[] |
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For an algebraic set $V \subseteq k^{n}$, the $V_i$'s in the decomposition |
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in \ref{thm:decomp-irred} are called the \emph{irreducible components} of $V$. |
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\end{definition} |
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\begin{proof}[Proof of \ref{thm:decomp-irred}] |
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Existence: Let $A$ be the set of algebraic sets $V \subseteq k^{n}$ that |
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admit no finite decomposition into a union of closed irreducible subsets. Assume |
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$A \neq \emptyset$. By noetherianity of $k^{n}$, |
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there exists a minimal element $V \in A$. In particular |
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$V$ is not irreducible, so $V = V_1 \cup V_2$ with $V_1, V_2 \subsetneq V$. By |
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minimality of $V$, $V_1, V_2 \not\in A$, thus they admit |
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a finite decomposition into a union of closed irreducible subsets. Since |
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$V = V_1 \cup V_2$, the same holds for $V$. Contradiction. Removing the |
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$V_i's$ for which $V_i \subseteq V_j$ for some $j$, we may assume that |
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$V_i \not\subset V_j$ for $i \neq j$. |
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Uniqueness: Assume that $V = V_1 \cup \ldots \cup V_r$ |
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and $V = W_1 \cup \ldots \cup W_s$ |
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are decompositions that satisfiy (i) and (ii). Then |
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\[ |
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W_1 = W_1 \cap V = (W_1 \cap V_1) \cup \ldots \cup (W_1 \cap V_r) |
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.\] Since $W_1$ is irreducible and $W_1 \cap V_i$ is closed in $W_1$, |
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there exists $j$ such that $W_1 = W_1 \cap V_j \subseteq V_j$. Likewise, |
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there exists $k$ such that $V_j \subseteq W_k$. Hence $W_1 \subseteq W_k$, |
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which forces $k = 1$ (because for $k \neq 1$, we have $W_1 \subsetneq W_k)$. Thus |
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$W_1 = V_j$ and we can repeat the procedure |
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with $W_2 \cup \ldots \cup W_s = \bigcup_{i \neq j} V_i$. |
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\end{proof} |
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\begin{korollar}[] |
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Let $V \subseteq k^{n}$ be an algebraic set and |
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denote by $V_1, \ldots, V_r$ the irreducible components of $V$. Let $W \subseteq V$ |
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be an irreducible subset. Then $W \subseteq V_i$ for some $i$. |
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\label{cor:irred-sub-of-alg-set} |
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\end{korollar} |
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\begin{proof} |
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We have |
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\[ |
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W = W \cap V = \bigcup_{i=1}^{r} \underbrace{W \cap V_i}_{\text{closed in }W} |
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.\] |
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Since $W$ is irreducible, there exists an $i$ such that |
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$W = W \cap V_i \subseteq V_i$. |
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\end{proof} |
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\begin{bem} |
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\begin{enumerate}[(i)] |
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\item The $i$ in \ref{cor:irred-sub-of-alg-set} is not unique in general. Consider |
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\[ |
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V = \{ x^2 - y^2 = 0\} = \{ x - y = 0\} \cup \{ x + y = 0\} |
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.\] The closed irreducible subset $\{(0, 0)\} $ lies in the intersection of |
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the irreducible components of $V$. |
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\item In view of the corollary \ref{cor:irred-sub-of-alg-set}, |
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theorem \ref{thm:decomp-irred} implies that an algebraic |
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set $V \subseteq k^{n}$ has a unique minimal decomposition into a union of closed irreducible |
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subsets. |
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\end{enumerate} |
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\end{bem} |
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\begin{korollar} |
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Let $V \subseteq k^{n}$ be an algebraic set. The irreducible components of $V$ |
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are exactly the maximal closed irreducible subsets of $V$. In terms |
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of ideals in $k[T_1, \ldots, T_n]$, a closed subset $W \subseteq V$ |
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is an irreducible component of $V$, if and only if the ideal |
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$\mathcal{I}(W)$ is a prime ideal which is minimal among those containing |
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$\mathcal{I}(V)$. |
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\end{korollar} |
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\begin{proof} |
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A closed irreducible subset $W \subseteq V$ is |
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contained in an irreducible component $V_j \subseteq V$ |
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by \ref{cor:irred-sub-of-alg-set}. If $W$ is maximal, then $W = V_j$. |
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Conversely, if $V_j$ is an irreducible component of $V$ and |
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$V_j \subseteq W$ for some irreducible and closed subset $W \subseteq V$, again |
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by \ref{cor:irred-sub-of-alg-set} we have $W \subseteq V_i$ for some $i$, therefore |
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$V_j \subseteq V_i$ which implies $i = j$ and $V_j = W$. |
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\end{proof} |
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\begin{satz}[Identity theorem for regular functions] |
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Let $X \subseteq k^{n}$ be an irreducible algebraic set and let $U \subseteq X$ |
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be open. Let $f, g \in \mathcal{O}_X(U)$ be regular functions on $U$. If |
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there is a non-empty open set $U' \subseteq U$ such that |
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$f|_{U'} = g|_{U'}$, then $f = g$ on $U$. |
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\end{satz} |
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\begin{proof} |
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The set $Y = \mathcal{V}_U(f-g)$ is closed in $U$ and |
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contains $U'$. Thus the closure $\overline{U'}^{(U)}$ of $U'$ in $U$ |
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is also contained in $Y$. By \ref{kor:non-empty-open-of-irred} |
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$U$ is irreducible, so $U'$ is dense in $U$. Therefore $Y = U$. |
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\end{proof} |
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\begin{bsp} |
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If $k$ is infinite and $P \in k[T_1, \ldots, T_n]$ is zero |
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outside an algebraic set $V \subseteq k^{n}$, then $P = 0$ on $k^{n}$. |
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\end{bsp} |
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\end{document} |
