uni/ws2022/rav/lecture/rav6.tex

\documentclass{lecture}

\begin{document}

\section{Abstract affine varieties}

Recall that an isomorphism of spaces with functions is a morphism
$f\colon (X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$ that admits an inverse morphism.

\begin{bem}[]
    As we have seen, a bijective morphism is not necessarily an isomorphism.
\end{bem}

\begin{bem}
    Somewhat more formally, one could also define a morphism of spaces
    with functions (over $k$) to be a pair $(f, \varphi)$ such that
    $f\colon X \to Y$ is a continuous map and $\varphi\colon \mathcal{O}_Y \to f_{*}\mathcal{O}_X$
    is the morphism of sheaves $f^{*}$. The question then arises how to define
    properly the composition $(g, \psi) \circ (f, \varphi)$. The formal answer is
    $(g \circ f, f_{*}(\varphi) \circ \psi)$.
\end{bem}

\begin{definition}[]
    Let $k$ be a field. An (abstract) \emph{affine variety over $k$}
    (also called an affine $k$-variety)
    is a space with functions $(X, \mathcal{O}_X)$
    over $k$ that is isomorphic to the space with functions $(V, \mathcal{O}_V)$, where
    $V$ is an algebraic subset of some affine space $k^{n}$ and $\mathcal{O}_V$ is the
    sheaf of regular functions on $V$.

    A morphism of affine $k$-varieties is a morphism of the underlying spaces with functions.
\end{definition}

\begin{bsp}[]

\begin{enumerate}[(i)]
    \item An algebraic subset $V \subseteq k^{n}$, endowed with its sheaf of regular functions
        $\mathcal{O}_V$, is an affine variety.
    \item It is perhaps not obvious at first, but a standard open set
        $D_V(f)$, where $f\colon V \to k$ is a regular function on an algebraic set
        $V \subseteq k^{n}$, defines an affine variety. Indeed, when
        equipped with its sheaf of regular functions,
        $D_V(f) \simeq \mathcal{V}_{k^{n+1}}(tf(x) - 1)$.
\end{enumerate}

\end{bsp}

\begin{bem}[]
    Let $(X, \mathcal{O}_X)$ be a space with functions. An open subset $U \subseteq X$ defines
    a space with functions $(U, \mathcal{O}_U)$. If
    $(U, \mathcal{O}_U)$ is isomorphic to some standard open set
    $D_V(f)$ of an algebraic set $V \subseteq k^{n}$, we will call
    $U$ an \emph{affine open set}.

    Then the observation is the following: since an algebraic set $V \subseteq k^{n}$
    is a finite union of standard open sets, every point $x$ in an affine variety $X$
    has an affine open neighbourhood.

    Less formally, an affine variety $X$, locally ,,looks like`` a standard open set
    $D_V(f) \subseteq k^{n}$, where $V \subseteq k^{n}$ is an algebraic set. In particular,
    open subsets of an affine variety also locally look like standard open sets. In fact,
    they are finite unions of such sets.
\end{bem}

\begin{bsp}[]
    The algebraic group $\mathrm{GL}(n ; k)$ is an affine variety over $k$.
\end{bsp}

\begin{bem}[]
    An algebraic set $(V, \mathcal{O}_V)$ is a subset $V \subseteq k^{n}$ defined
    by polynomial equations and equipped with its sheaf of regular functions.
    An affine variety $(X, \mathcal{O}_X)$ is
    ,,like an algebraic set`` but without a reference to a particular
    ,,embedding`` in affine space. This is similar to having a finitely generated $k$-Algebra $A$
    without specifying a particular isomorphism
    \[
    A \simeq k[X_1, \ldots, X_n] / I
    .\] The next example will illustrate precisely this fact.
\end{bem}

\begin{bsp}[]
    Let us now give an abstract example of an affine variety.
    We consider a finitely generated $k$-algebra $A$ and define
    $X \coloneqq \operatorname{Hom}_{k-\mathrm{Alg}}(A, k)$. The idea is to think
    of $X$ as points on which we can evaluate elements of $A$, which are thought of
    as functions on $X$. For $x \in \operatorname{Hom}_{k}(A, k)$ and
    $f \in A$ we set $f(x) \coloneqq x(f) \in k$.
    \begin{itemize}
        \item Topology on $X$: for all ideal $I \subseteq A$, set
            \[
            \mathcal{V}_X(I) \coloneqq \{ x \in X  \mid \forall x \in I\colon f(x) = 0\}
            .\] These subsets of $X$ are the closed sets of a topology on $X$, which
            we may call the Zariski topology.
        \item Regular functions on $X$: if $U \subseteq X$ is open,
            a function $h\colon U \to k$ is called regular at $x \in U$ if
            there it exists an open set $x \in U_x$ and elements
            $P, Q \in A$ such that for $y \in U_x$, $Q(y) \neq 0$ and
            $h(y) = \frac{P(y)}{Q(y)}$ in $k$.

            The function $h$ is called regular on $U$
            iff it is regular at $x \in U$. Regular functions then form a sheaf of
            $k$-algebras on $X$.

            Moreover, if $h\colon U \to k$ is regular on $X$, the
            set $D_X(h) \coloneqq \{ x \in X  \mid h(x) \neq 0\} $ is open in $X$
            and the function $\frac{1}{h}$ is regular on $D_X(h)$.
    \end{itemize}
    So, we have defined a space with functions $(X, \mathcal{O}_X)$, at least
    whenever $X \neq \emptyset$. We show that $X$ is an affine variety.

    \begin{proof}
        Fix a system of generators of $A$, i.e.
        \[
        A \simeq k[t_1, \ldots, t_n] / I
        \] where $k[t_1, \ldots, t_n]$ is a polynomial algebra. We denote
        by $\overline{t_1}, \ldots, \overline{t_n}$ the images of $t_1, \ldots, t_n$ in $A$
        and we define
        \begin{salign*}
            \varphi\colon X = \operatorname{Hom}_{k}(A, k)& \to k^{n} \\
            x &\mapsto (x(\overline{t_1}), \ldots, x(\overline{t_n}))
        .\end{salign*}
        Let $P \in I$ and $x \in X$. Then
        \[
        P(\varphi(x)) = P(x(\overline{t_1}), \ldots, x(\overline{t_n}))
        = x(\overline{P}) = 0
        .\] Thus $\varphi(x) \in \mathcal{V}_{k^{n}}(I)$.
        Conversely let $a = (a_1, \ldots, a_n) \in \mathcal{V}_{k^{n}}(I)$, then
        we can define a morphism of $k$-algebras
        \[
        x_a\colon A \to A / (\overline{t_1} -a_1, \ldots, \overline{t_n} - a_n)
        \simeq k
        \] which satisfies $x_a(\overline{t_i}) = a_i$ for all $i$. So
        $(a_1, \ldots, a_n) = \varphi(x_a) \in \text{im } \varphi$.

        In particular, we have defined a map
        \begin{salign*}
            \psi\colon \mathcal{V}_{k^{n}}(I) &\to X = \operatorname{Hom}_k(A, k) \\
            a &\mapsto x_a
        \end{salign*} such that $\varphi \circ \psi = \text{Id}_{\mathcal{V}_{k^{n}}(I)}$. In fact,
        we also have $\psi \circ \varphi = \text{Id}_X$.

        It remains to check that $\varphi$ and $\psi$ are morphisms of spaces with functions, which
        follows from the definition of the topology and the notion of regular function on $X$.
    \end{proof}

    The elements of $X \coloneqq \operatorname{Hom}_k(A, k)$ are also called the
    \emph{characters} of the $k$-algebra $A$, and this is sometimes denoted
    by $\hat{A} \coloneqq \operatorname{Hom}_{k-\text{alg}}(A, k)$. Note that
    $\hat{A}$ is a $k$-subalgebra of the algebra of all functions $f\colon A \to k$.

    The character $x_a$ introduced above and associated to an alemenet $a \in A$ is then
    denoted by $\hat{a}$ and called the \emph{Gelfand transform} of $a$. The
    \emph{Gelfand transformation} is the morphism of $k$-algebras
    \begin{salign*}
        A &\to \hat{A} \\
        a &\mapsto \hat{a}
    .\end{salign*}
\end{bsp}

\begin{aufgabe}
    Let $A$ be a finitely generated $k$-algebra and let
    $X = \operatorname{Hom}_{k\text{-alg}}(A, k)$. Show that the map
    $x \mapsto \text{ker } x$ induces a bijection
    \[
    X \simeq \{ \mathfrak{m} \in \operatorname{Spm} A  \mid A / \mathfrak{m} \simeq k\}
    .\] 
\end{aufgabe}

\begin{bem}[]
    Note that we have not assumed $A$ to be reduced and that, if we
    set $A_{\text{red}} \coloneqq A / \sqrt{(0)}$, then
    $A_{\text{red}}$ is reduced and
    $\hat{A_{\text{red}}} = \hat{A}$, because a maximal ideal of $A$ necessarily
    contains $\sqrt{(0)}$ and the quotient field is ,,the same``.
\end{bem}

\begin{bem}
    Let $(X, \mathcal{O}_X)$ be an affine variety. One can associate the $k$-algebra
    $\mathcal{O}_X(X)$ of globally defined regular functions on $X$:
    \[
    \mathcal{O}_X(X) = \{ f \colon X \to k  \mid f \text{ regular on } X\} 
    .\]
    Moreover, if $\varphi\colon (X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$ is
    a morphism between two affine varieties, we have a $k$-algebra homomorphism
    \begin{salign*}
        \varphi^{*}\colon \mathcal{O}_Y(Y) &\to \mathcal{O}_X(X) \\
        f &\mapsto f \circ \varphi
    .\end{salign*}
    Also, $(\text{id}_X)^{*} = \text{id}_{\mathcal{O}_X(X)}$ and
    $(\psi \circ \varphi)^{*} = \varphi^{*} \circ \psi^{*}$ whenever
    $\psi\colon (Y, \mathcal{O}_Y) \to (Z, \mathcal{O}_Z)$ is a morphism of
    affine varieties. In other words, we have defined a (contravariant) functor
    $k$-Aff $\to k$-Alg.
\end{bem}

\begin{satz}
    Let $k$ be a field. The functor
    \begin{salign*}
        k\text{-Aff} &\to k\text{-Alg} \\
        (X, \mathcal{O}_X) &\mapsto \mathcal{O}_X(X)
    \end{salign*}
    is fully faithful.
\end{satz}

\begin{proof}
    Since $X$ and $Y$ are affine, we may assume $X = V \subseteq k^{n}$
    and $Y = W \subseteq k^{m}$. Then $\varphi\colon V \to W$
    is given by $m$ regular functions $(\varphi_1, \ldots, \varphi_m)$
    on $V$. On $k^{m}$, let us denote by $y_i$ the projection to the $i$-th factor.
    Its restriction to $W$ is a regular function
    \[
    y_i|_W \colon W \to k
    \] that satisfies $\varphi^{*}(y_i|_W) = \varphi_i$.

    Since for all regular functions $f\colon W \to k$ one has
    \[
    \varphi^{*}f = f \circ \varphi = f(\varphi_1, \ldots, \varphi_m)
    ,\] we see that the morphism
    \[
    \varphi^{*}\colon \mathcal{O}_W(W) \to \mathcal{O}_V(V)
    \] is entirely determined by the $m$ regular functions $\varphi^{*}(y_i|_W) = \varphi_i$
    on $V$. In particular, if $\varphi^{*} = \psi^{*}$, then
    $\varphi_i = \varphi^{*}(y_i|_W) = \psi^{*}(y_i|_W) = \psi_i$, so $\varphi = \psi$,
    which proves that $\varphi \mapsto \varphi^{*}$ is injective.

    Surjectivity: Let $h\colon \mathcal{O}_W(W) \to \mathcal{O}_V(V)$ be a morphism
    of $k$-algebras. Let
    \[
    \varphi \coloneqq (h(y_1|_W), \ldots, h(y_m|_W))
    \] which is a morphism from $V$ to $k^{m}$, because its components are regular functions
    on $V$. It satisfies $\varphi^{*}(y_i|_W) = \varphi_i = h(y_i|_W)$, so $\varphi^{*} = h$.

    It remains to show, that $\varphi(V) \subseteq W$. Let $W = \mathcal{V}(P_1, \ldots, P_r)$
    with $P_j \in k[Y_1, \ldots, Y_m]$. Then for all $j \in \{1, \ldots, r\} $
    and $x \in V$
    \[
    P_j(\varphi(x)) = P_j(h(y_1|_W), \ldots, h(y_m|_W))(x)
    .\] Since $h$ is a morphism of $k$-algebras and $P_j$ is a polynomial, we have
    \[
    P_j(h(y_1|_W), \ldots, h(y_m|_W)) = h(P_j(y_1|_W), \ldots, P_j(y_m|_W))
    .\] But $P_j \in \mathcal{I}(W)$, so
    \[
    P_j(y_1|_W, \ldots, y_m|_W) = P_j(y_1, \ldots, y_m)|_W = 0
    ,\] which proves that for $x \in V$, $\varphi(x) \in W$.
\end{proof}

\end{document}