uni/ws2022/rav/lecture/rav2.tex

%& -shell-escape -enable-write18
\documentclass{lecture}

\begin{document}

\section{Regular functions}

\begin{lemma}
    If $U \subseteq k^{n}$ is a Zariski-open set and
    $f_P \colon k^{n} \to k$ is a polynomial function such that
    for $x \in U$, $f_P(x) \neq 0$, then the function $\frac{1}{f_P}$ is continuous on $U$.

    \label{lemma:1overP-is-cont}
\end{lemma}

\begin{proof}
    For all $t \in k$,
    \begin{salign*}
        \left(\frac{1}{f_P}\right)^{-1}(\{t\})
        &= \left\{ x \in U  \mid \frac{1}{f_P(x)} = t \right\} \\
        &= \left\{ x \in U  \mid t f_P(x) -1 = 0 \right\} \\
        &= \mathcal{V}(tf_P -1) \cap U
    \end{salign*}
    is closed in $U$.
\end{proof}

\begin{bem}
    There can be many continous functions with respect to the Zariski topology. For instance,
    all bijective maps $f\colon k \to k$ are Zariski-continuous. In algebraic geometry, we will
    consider only functions which are locally defined by a rational function. We will define
    them on open subsets of algebrai sets $V \subseteq k^{n}$, endowed with the topology
    induced by the Zariski topology of $k^{n}$.
\end{bem}

\begin{bem}[]
    The open subsets of algebraic sets $V \subseteq k^{n}$ are exactly the
    \emph{locally closed subsets} of $k^{n}$.
\end{bem}

\begin{definition}[]
    Let $X \subseteq k^{n}$ be a locally closed subset of $k^{n}$. A function
    $f \colon X \to k$ is called \emph{regular at $x \in X$}, if
    there exist an open subset $x \in U \subseteq X$ and two polynomial functions
    $P_U, Q_U \colon U \to k$ such that for all $y \in U$, $Q_U(y) \neq 0$ and
    \[
    f(y) = \frac{P_U(y)}{Q_U(y)}
    .\] The function $f\colon X \to k$ is called \emph{regular on $X$} if, for all $x \in X$,
    $f$ is regular at $x$.
\end{definition}

\begin{bsp}[]
    A rational fraction $\frac{P}{Q} \in k(T_1, \ldots, T_n)$ defines a regular
    function on the standard open set $D(Q)$.
\end{bsp}

\begin{satz}[]
    Let $X \subseteq k^{n}$ be a locally closed subset. If $f\colon X \to k$ is regular,
    then $f$ is continous.
\end{satz}

\begin{proof}
    Since continuity is a local property, we may assume $X = \Omega \subseteq k^{n}$ open
    and $f = \frac{P}{Q}$ for polynomial functions $P, Q\colon \Omega \to k$ such that
    $Q(y) \neq 0$. By \ref{lemma:1overP-is-cont} it suffices to prove that
    if $P, R\colon \Omega \to k$ are continuous, then $PR\colon \Omega \to k$,
    $z \mapsto P(z)R(z)$ is continuous. Let $t \in k$. Then
    \begin{salign*}
        (PR)^{-1}(\{t\}) &= \{ z \in \Omega  \mid P(z) R(z) - t = 0\} \\
                         &= \mathcal{V}(PR - t) \cap \Omega
    \end{salign*}
    is closed in $\Omega$.
\end{proof}

\begin{bem}
    Being a regular function is a local property.
\end{bem}

\begin{satz}
    Let $X \subseteq k^{n}$ be a locally closed subset of $k^{n}$, endowed
    with the induced topology. The map
    \begin{salign*}
        \mathcal{O}_X\colon \{\text{open sets of }X\} &\longrightarrow k-\text{algebras}\\
        U &\longmapsto \{ \text{regular functions on }U\} 
    \end{salign*}
    defines a sheaf of sheaf of $k$-algebras on $X$, which is a subsheaf of the sheaf of functions.
\end{satz}

\begin{proof}
    Constants, sums and products of regular functions are regular, thus
    $\mathcal{O}_X(U)$ is a sub algebra of the $k$-algebra of functions
    $U \to k$.
    Since restricting a function preserves regularity, $\mathcal{O}_X$ is a presheaf. Since
    being regular is a local property and the presheaf of functions is a sheaf,
    $\mathcal{O}_X$ is also a sheaf.
\end{proof}

\section{Irreducibility}

\begin{definition}
    Let $X$ be a topological space. $X$ is
    \begin{enumerate}[(i)]
        \item \emph{irreducible} if $X \neq \emptyset$ and $X$ is not the union
            of two proper closed subets, i.e. for $X = F_1 \cup F_2$ with $F_1, F_2 \subseteq X$ closed,
            we have $X = F_1$ or $X = F_2$.
        \item \emph{connected} if $X$ is not the union of two disjoint proper closed subets, i.e.
            for $X = F_1 \cup F_2$ with $F_1, F_2 \subseteq X$ closed and $F_1 \cap F_2 = \emptyset$,
            we have $X = F_1$ or $X = F_2$.
    \end{enumerate}
    A space $X$ which is not irreducible, is called \emph{reducible}.
\end{definition}

\begin{lemma}
    If $k$ is infinite, $k$ is irreducible in the Zariski topology.
\end{lemma}

\begin{proof}
    Closed subsets of $k$ are $k$ and finite subsets of $k$.
\end{proof}

\begin{bem}
    If $k$ is finite, $k^{n}$ is the finite union of its points, which are closed, so
    $k^{n}$ is reducible.
\end{bem}

\begin{bem}
    $X$ irreducible $\implies$ $X$ connected, but the converse is false: Let $k$ be infinite and
    consider $X = \mathcal{V}_{k^2}(x^2 - y^2)$ (see figure \ref{fig:reducible-alg-set}).
    Since $x^2 - y^2 = (x-y)(x+y) = 0$ in $k$
    if and only if $x = -y$ or $x = y$, we have
    $X = \mathcal{V}_{k^2}(x - y) \cup \mathcal{V}_{k^2}(x+y)$. Thus $X$ is reducible. But
    $\mathcal{V}_{k^2}(x-y)$ and $\mathcal{V}_{k^2}(x+y)$ are homeomorphic to $k$, in particular
    irreducible and thus connected. Since $\mathcal{V}_{k^2}{x-y} \cap \mathcal{V}_{k^2}(x+y) \neq \emptyset$,
    $X$ is connected.
\end{bem}

\begin{figure}
    \centering
    \begin{tikzpicture}
        \begin{axis}
        \algebraiccurve[red]{x^2 - y^2}
        \end{axis}
    \end{tikzpicture}
    \caption{Reducible connected algebraic set}
    \label{fig:reducible-alg-set}
\end{figure}

\begin{satz}
    Let $X$ be a non-empty topological space. The following conditions are equivalent:
    \begin{enumerate}[(i)]
        \item $X$ is irreducible
        \item If $U_1 \cap U_2 = \emptyset$ with $U_1, U_2$ open subsets of $X$, then
            $U_1 = \emptyset$ or $U_2 = \emptyset$.
        \item If $U \subseteq X$ is open and non-empty, then $U$ is dense in $X$.
    \end{enumerate}
    \label{satz:equiv-irred}
\end{satz}

\begin{proof}
    Left as an exercise to the reader.
\end{proof}

\begin{satz}
    Let $X$ be a topological space and $V \subseteq X$. Then
    $V$ is irreducible if and only if $\overline{V}$ is irreducible.
    \label{satz:closure-irred}
\end{satz}

\begin{proof}
    Since $\emptyset$ is closed in $X$, we have
    $ V = \emptyset \iff \overline{V} = \emptyset$.

    ($\Rightarrow$)
    Let $\overline{V} \subseteq Z_1 \cup Z_2$ with $Z_1, Z_2 \subseteq X$ closed.
    Then $V \subseteq Z_1 \cup Z_2$ and by irreducibility of $V$ we may assume $V \subseteq Z_1$.
    Since $Z_1$ is closed, it follows $\overline{V} \subseteq Z_1$.

    ($\Leftarrow$) Let $V \subseteq Z_1 \cup Z_2$ with $Z_1, Z_2 \subseteq X$ closed.
    Since $Z_1 \cup Z_2$ is closed, we get $\overline{V} \subseteq Z_1 \cup Z_2$. By
    irreducibility of $\overline{V}$ we may assume $\overline{V} \subseteq Z_1$, thus
    $V \subseteq Z_1$.
\end{proof}

\begin{korollar}
    Let $X$ be an irreducible topological space. Then every non-empty open
    subset $U \subseteq X$ is irreducible.
    \label{kor:non-empty-open-of-irred}
\end{korollar}

\begin{proof}
    By \ref{satz:equiv-irred}, $\overline{U} = X$ and thus irreducible. The claim
    follows now from \ref{satz:closure-irred}.
\end{proof}

\begin{lemma}[prime avoidance]
    Let $\mathfrak{p}$ be a prime ideal in a commutative ring $A$. If $I, J \subseteq A$ are
    ideals such that $IJ \subseteq \mathfrak{p}$, then
    $I \subseteq \mathfrak{p}$ or $J \subseteq \mathfrak{p}$.

    \label{lemma:prime-avoidance}
\end{lemma}

\begin{proof}
    Assume that $I \not\subset \mathfrak{p}$ and $J \not\subset \mathfrak{p}$. Then
    there exist $a \in I$, such that $a \not\in  \mathfrak{p}$ and $b \in J$ such that
    $b \not\in \mathfrak{p}$. But $ab \in IJ \subseteq \mathfrak{p}$. Since
    $\mathfrak{p}$ is prime, this implies $a \in \mathfrak{p}$ or
    $b \in \mathfrak{p}$. Contradiction.
\end{proof}

\begin{theorem}
    Let $V \subseteq k^{n}$ be an algebraic set. Then $V$ is irreducible in the Zariski
    topology if and only if $\mathcal{I}(V)$ is a prime ideal in $k[T_1, \ldots, T_n]$.
\end{theorem}

\begin{proof}
    ($\Rightarrow$) Since $V \neq \emptyset$,
    $\mathcal{I}(V) \subsetneq k[T_1, \ldots, T_n]$.
    Let $P, Q \in k[T_1, \ldots, T_n]$
    such that $PQ \in \mathcal{I}(V)$. For $x \in V$, $(PQ)(x) = 0$ in $k$, hence
    $P(x) = 0$ or $Q(x) = 0$. Thus $x \in \mathcal{V}(P) \cup \mathcal{V}(Q)$. Therefore
    $V = (\mathcal{V}(P) \cap V) \cup (\mathcal{V}(Q) \cap V)$ is the union of
    two closed subsets. Since $V$ is irreducible,
    we may assume $V = \mathcal{V}(P) \cap V \subseteq \mathcal{V}(P)$, hence
    $P \in \mathcal{I}(V)$ and $\mathcal{I}(V)$ is prime.

    ($\Leftarrow$) $V \neq \emptyset$, since $\mathcal{I}(V)$ is a proper ideal. Let
    $V = V_1 \cup V_2$ with $V_1, V_2$ closed in $V$. Then
    \[
    \mathcal{I}(V) = \mathcal{I}(V_1 \cup V_2) = \mathcal{I}(V_1) \cap \mathcal{I}(V_2)
    \supseteq \mathcal{I}(V_1) \mathcal{I}(V_2)
    .\] By \ref{lemma:prime-avoidance}, we may assume
    $\mathcal{I}(V_1) \subseteq \mathcal{I}(V)$. But then
    \[
    V_1 = \mathcal{V}(\mathcal{I}(V_1)) \supseteq \mathcal{V}(\mathcal{I}(V)) = V
    \] since $V_1$ and $V$ are closed. Therefore $V = V_1$ and $V$ is irreducible.
\end{proof}

\begin{korollar}
    If $k$ is infinite, the affine space $k^{n}$ is irreducible with respect to the Zariski
    topology.
\end{korollar}

\begin{proof}
    Since $k$ is infinite, $\mathcal{I}(k^{n}) = (0)$ by \ref{satz:k-infinite-everywhere-vanish}
    which is a prime ideal in the integral domain $k[T_1, \ldots, T_n]$.
\end{proof}

\begin{theorem}
    Let $V \subseteq k^{n}$ be an algebraic set. Then there exists a decomposition
    \[
    V = V_1 \cup \ldots \cup V_r
    \] such that
    \begin{enumerate}[(i)]
        \item $V_i$ is a closed irreducible subset of $k^{n}$ for all $i$.
        \item $V_{i} \not\subset V_j$ for all $i \neq j$.
    \end{enumerate}
    This decomposition is unique up to permutations.
    \label{thm:decomp-irred}
\end{theorem}

\begin{definition}[]
    For an algebraic set $V \subseteq k^{n}$, the $V_i$'s in the decomposition
    in \ref{thm:decomp-irred} are called the \emph{irreducible components} of $V$.
\end{definition}

\begin{proof}[Proof of \ref{thm:decomp-irred}]
    Existence: Let $A$ be the set of algebraic sets $V \subseteq k^{n}$ that
    admit no finite decomposition into a union of closed irreducible subsets. Assume
    $A \neq \emptyset$. By noetherianity of $k^{n}$,
    there exists a minimal element $V \in A$. In particular
    $V$ is not irreducible, so $V = V_1 \cup V_2$ with $V_1, V_2 \subsetneq V$. By
    minimality of $V$, $V_1, V_2 \not\in A$, thus they admit
    a finite decomposition into a union of closed irreducible subsets. Since
    $V = V_1 \cup V_2$, the same holds for $V$. Contradiction. Removing the
    $V_i's$ for which $V_i \subseteq V_j$ for some $j$, we may assume that
    $V_i \not\subset V_j$ for $i \neq j$.

    Uniqueness: Assume that $V = V_1 \cup \ldots \cup V_r$
    and $V = W_1 \cup \ldots \cup W_s$
    are decompositions that satisfiy (i) and (ii). Then
    \[
    W_1 = W_1 \cap V = (W_1 \cap V_1) \cup \ldots \cup (W_1 \cap V_r)
    .\] Since $W_1$ is irreducible and $W_1 \cap V_i$ is closed in $W_1$,
    there exists $j$ such that $W_1 = W_1 \cap V_j \subseteq V_j$. Likewise,
    there exists $k$ such that $V_j \subseteq W_k$. Hence $W_1 \subseteq W_k$,
    which forces $k = 1$ (because for $k \neq 1$, we have $W_1 \subsetneq W_k)$. Thus
    $W_1 = V_j$ and we can repeat the procedure
    with $W_2 \cup \ldots \cup W_s = \bigcup_{i \neq j} V_i$.
\end{proof}

\begin{korollar}[]
    Let $V \subseteq k^{n}$ be an algebraic set and
    denote by $V_1, \ldots, V_r$ the irreducible components of $V$. Let $W \subseteq V$
    be an irreducible subset. Then $W \subseteq V_i$ for some $i$.
    \label{cor:irred-sub-of-alg-set}
\end{korollar}

\begin{proof}
    We have
    \[
    W = W \cap V = \bigcup_{i=1}^{r} \underbrace{W \cap V_i}_{\text{closed in }W}
    .\] 
    Since $W$ is irreducible, there exists an $i$ such that
    $W = W \cap V_i \subseteq V_i$.
\end{proof}

\begin{bem}
    \begin{enumerate}[(i)]
        \item The $i$ in \ref{cor:irred-sub-of-alg-set} is not unique in general. Consider
        \[
        V = \{ x^2 - y^2 = 0\} = \{ x - y = 0\} \cup \{ x + y = 0\}
        .\] The closed irreducible subset $\{(0, 0)\} $ lies in the intersection of
        the irreducible components of $V$.
    \item In view of the corollary \ref{cor:irred-sub-of-alg-set},
        theorem \ref{thm:decomp-irred} implies that an algebraic
        set $V \subseteq k^{n}$ has a unique minimal decomposition into a union of closed irreducible
        subsets.
    \end{enumerate}
\end{bem}

\begin{korollar}
    Let $V \subseteq k^{n}$ be an algebraic set. The irreducible components of $V$
    are exactly the maximal closed irreducible subsets of $V$. In terms
    of ideals in $k[T_1, \ldots, T_n]$, a closed subset $W \subseteq V$
    is an irreducible component of $V$, if and only if the ideal
    $\mathcal{I}(W)$ is a prime ideal which is minimal among those containing
    $\mathcal{I}(V)$.
\end{korollar}

\begin{proof}
    A closed irreducible subset $W \subseteq V$ is
    contained in an irreducible component $V_j \subseteq V$
    by \ref{cor:irred-sub-of-alg-set}. If $W$ is maximal, then $W = V_j$.

    Conversely, if $V_j$ is an irreducible component of $V$ and
    $V_j \subseteq W$ for some irreducible and closed subset $W \subseteq V$, again
    by \ref{cor:irred-sub-of-alg-set} we have $W \subseteq V_i$ for some $i$, therefore
    $V_j \subseteq V_i$ which implies $i = j$ and $V_j = W$.
\end{proof}

\begin{satz}[Identity theorem for regular functions]
    Let $X \subseteq k^{n}$ be an irreducible algebraic set and let $U \subseteq X$
    be open. Let $f, g \in \mathcal{O}_X(U)$ be regular functions on $U$. If
    there is a non-empty open set $U' \subseteq U$ such that
    $f|_{U'} = g|_{U'}$, then $f = g$ on $U$.
\end{satz}

\begin{proof}
    The set $Y = \mathcal{V}_U(f-g)$ is closed in $U$ and
    contains $U'$. Thus the closure $\overline{U'}^{(U)}$ of $U'$ in $U$
    is also contained in $Y$. By \ref{kor:non-empty-open-of-irred}
    $U$ is irreducible, so $U'$ is dense in $U$. Therefore $Y = U$.
\end{proof}

\begin{bsp}
    If $k$ is infinite and $P \in k[T_1, \ldots, T_n]$ is zero
    outside an algebraic set $V \subseteq k^{n}$, then $P = 0$ on $k^{n}$.
\end{bsp}

\end{document}