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Reminders:
\begin{itemize}
  \item $X \subseteq \AA^n$.
  \item $\cring A(X) = \polyA / \I(X)$.
  \item We defined the notion of \gls{regf}
    function on an \gls{zzopen} subset $U
    \subseteq X$.
  \item $\OX_X(U) \defeq \{f : U \to \KK \st
    \text{$f$ is \gls{regf} on $U$}\}$
\end{itemize}

\begin{lemma*}
  $\OX_X(X) = \cring A(X)$.
\end{lemma*}

\begin{proof}
  Later (we will prove this after proving
  \nameref{hilbert_null}).
\end{proof}

Recall from \courseref{GRM}: Let $A$ be an
integral domain. Then the \emph{field of
fractions} of $A$ (or \emph{fraction field} of
$A$) is
\[ \left\{\frac{f}{g} \st f, g \in A, g \neq 0\right\} / \sim \]
with $\frac{f}{g} \sim \frac{f'}{g'}$ if $fg' = f'g$.

We define addition and
multiplication using:
\[ \frac{f}{g} + \frac{f'}{g'} = \frac{fg' +
gf'}{gg'} \qquad \frac{f}{g} \frac{f'}{g'} =
\frac{ff'}{gg'} \]
and we observe that this is a field since
\[ \left( \frac{f}{g} \right)^{-1} = \frac{g}{f} \]
is an inverse whenever $f \neq 0$.

\begin{remark*}
  If $X \subseteq \AA^n$ is an \gls{affvty}, then
  $\cring A(X) = \polyA / \I(X)$ is an integral
  domain. (This is because for any ring $R$ and
  ideal $P \subseteq R$, $R / P$ is an
  integral domain if and only if $P$ is prime --
  see \courseref{GRM}).
\end{remark*}

\begin{flashcard}[function-field-defn]
\begin{definition*}[Function field]
  \glssymboldefn{KX}{$K(X)$}{$K(X)$}
  \glsnoundefn{funcfield}{function field}{function
  fields}
  \cloze{If $X \subseteq \AA^n$ is a \gls{aavty}, its
  \emph{function field} is $K(X)$, the fraction
  field of $\cring A(X)$. Elements of $K(X)$ are
  called \emph{rational functions} on $X$. Note
  $\frac{g}{h} \in K(X)$ induces a \gls{regf}
  function on $X \setminus \Z(h)$.}
\end{definition*}
\end{flashcard}

\subsection{Morphisms}

\begin{flashcard}[morphism-defn]
\begin{definition*}[Morphism]
  \glsnoundefn{morph}{morphism}{morphisms}
  \cloze{A map $f : X \to Y$ between \glspl{affvty} is
  called a \emph{morphism} if:
  \begin{enumerate}[(1)]
    \item $f$ is continuous in the induced
      \glsref[zopen]{Zariski topology} on $X$ and
      $Y$ ($Z \subseteq X \subseteq \AA^n$ is
      closed in $X$ if and only if it is closed in
      $\AA^n$).
    \item $\forall V \subseteq Y$ be an open
      subset, $\varphi : V \to \KK$ a \gls{regf}
      function, we have that $\varphi \circ f :
      f^{-1}(V) \to \KK$ is a \gls{regf} function
      on $f^{-1}(V)$.
  \end{enumerate}}
\end{definition*}
\end{flashcard}

\glssymboldefn{sh}{$f^\#$}{$f^\#$}
\textbf{Observation:} Let $f : X \to Y$ be a
\gls{morph}. Then for any $\varphi \in \cring
A(Y)$, we get $\varphi \circ f : X \to \KK$ a
\gls{regf} function. Assuming $\KK$ is
algebraically closed, $\OX_X(X) = \cring A(X)$, so
$\varphi \circ f \in \cring A(X)$. This gives a
map $\sh f : \cring A(Y) \to \cring A(X)$. This is
a \Kalgebra{\KK} homomorphism. We first check that
it is indeed a ring homomorphism:
\begin{align*}
  \sh f(\varphi_1 + \varphi_2)
  &= (\varphi_1 + \varphi_2) \circ f \\
  &= \varphi_1 \circ f + \varphi_2 \circ f \\
  &= \sh f(\varphi_1) + \sh f(\varphi_2) \\
  \sh f(\varphi_1 \cdot \varphi_2)
  &= (\varphi \cdot \varphi_2) \circ f \\
  &= (\varphi_1 \circ f) \cdot (\varphi_2 \circ f)
  \\
  &= \sh f(\varphi_1) \cdot \sh f(\varphi_2) \\
  \sh f(1)
  &= 1
\end{align*}
Now we check multiplication by elements of $\KK$.
For $a \in \KK$,
\[ \sh f(a \cdot \varphi) = a \cdot \sh f(\varphi) \]
So this is a \Kalgebra{\KK} homomorphism.

\begin{flashcard}[sharp-bijection-thm]
\begin{theorem*}
  For $\KK$ algebraically closed, there is a $1-1$
  correspondence between
  \glspl{morph} $f : X \to Y$ and \Kalgebra{\KK}
  homomorphisms $\sh f : \cring A(Y) \to \cring A(X)$.
\end{theorem*}

\begin{proof}
  \cloze{We have already constructed $\sh f$ from $f$.
  Suppose $X \subseteq \AA^n$, $Y \subseteq
  \AA^m$. Then
  \[ \cring A(X) = \frac{\KK[X_1, \ldots,
  X_n]}{\I(X)} \qquad
  \cring A(Y) = \frac{\KK[Y_1, \ldots,
  Y_m]}{\I(Y)} \]
  \[ \AA^n \supseteq X \stackrel[f]{(f_1, \ldots,
  f_m)}{\longrightarrow} Y \subseteq \AA^m
  \stackrel[y_i]{}{\to} \KK \]
  $f_i = y_i \circ f$. Suppose given $\sh f :
  \cring A(Y) \to \cring A(X)$. Set $f_i = \sh
  f(\ol{y}_i)$ ($\ol{y}_i$ is the image of $y_i$
  in $\cring A(Y)$). We now define $f : X \to
  \AA^m$ by $f(p) = (f_1(p), \ldots, f_m(p))$.

  \textbf{Claim:} $f(X) \subseteq Y$.

  Proof: Let $g \in \I(Y)$, and $p \in X$. We need
  to show that $g(f(p)) = 0$. This will show $f(p)
  \in Y$. Consider the map
  \[ \KK[Y_1, \ldots, Y_m] \to \cring A(Y)
  \stackrel{\sh f}{\to} \cring A(X) \]
  \[ Y_i \mapsto \ol{Y_i} \mapsto f_i \]
  Thus
  \[ g(Y_1, \ldots, Y_m) \mapsto g(\ol{Y}_1,
  \ldots, \ol{Y}_m) \mapsto g(f_1, \ldots, f_m) \]
  The right arrow uses $\sh f$ being a
  \Kalgebra{\KK}. The middle expression is the image of
  $g$ under quotient map, hence $0$ since $g \in
  \I(Y)$. Thus $g(f(p)) = g(f_1, \ldots, f_m)(p) = 0$.

  Thus $f(X) \subseteq Y$. This completes the
  proof of the claim.

  Note: If $\varphi \in \cring A(Y)$, can write $\varphi
  = g(\ol{Y}_1, \ldots, \ol{Y}_m)$ and $\sh
  f(\varphi) = g(f_1, \ldots, f_m) = \varphi \circ
  f$.

  \textbf{Claim:} $f$ is a \gls{morph}:
  \begin{enumerate}[(1)]
    \item \textit{$f$ is continuous:} We will show
      $f^{-1}(Z)$ is \gls{zzclosed} for $Z \subseteq Y$
      \gls{zzclosed}. Note $\I(Z) \supseteq
      \I(Y)$, so $\ol{\I(Z)} = \frac{\I(Z)}{\I(Y)}
      \subseteq \cring A(Y)$ is an ideal in
      $\cring A(Y)$. Then define
      \[ Z(\sh f(\ol{\I(Z)})) = \{p \in X \st
      \varphi(p) = 0 ~\forall \varphi \in \sh
      f(\ol{\I(Z)})\} \]
      This is a \gls{zzclosed} subset of $X$ since
      it coincides with
      \[ \Z(\pi_X^{-1}(\sh f(\ol{\I(Z)}))) \]
      where $\pi_X : \KK[X_1, \ldots, X_n] \to
      \cring A(X)$. But
      \begin{align*}
        \Z(\sh f(\ol{I(Z)}))
        &= \{p \in X \st \psi
        \circ f = 0 ~\forall \psi \in \ol{\I(Z)}\}
        \\
        &= \{p \in X \st f(p) \in Z\} \\
        &= f^{-1}(Z)
      \end{align*}
      Thus $f^{-1}(Z)$ is \gls{zzclosed}.
    \item Let $U \subseteq Y$ be an \gls{zzopen}
      subset, $\varphi \in \OX_Y(U)$. We need to
      show $\varphi \circ f : f^{-1}(U) \to \KK$
      is \gls{regf}. Let $p \in f^{-1}(U)$, and
      let $V \subseteq U$ be an \gls{zzopen}
      neighbourhood of $f(p)$ for which we can
      write $f = \frac{g}{h}$, $g, h \in \cring
      A(Y)$, $h$ nowhere vanishing on $V$. Then
      \[ \varphi \circ f|_{f^{-1}(V)} = \frac{g
      \circ f}{h \circ f} = \frac{\sh f(g)}{\sh f
      (h)} .\]
      Now $\sh f(g), \sh f(h)$ lie in $\cring
      A(X)$, and $\sh f(h) = h \circ f$ doesn't
      vanish on $f^{-1}(V)$. Thus $\varphi \circ
      f$ is \gls{regf}. \qedhere
  \end{enumerate}
  }
\end{proof}
\end{flashcard}