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\input{lec01}
\input{lec02}
\input{lec03}
\input{lec04}

\bibliographystyle{alpha}
\bibliography{refs}


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\documentclass{lecture}

\begin{document}

\section{Group schemes over a field}

Let $k$ be a field and $S = \Spec k$.

\begin{lemma}
Let $G$ be a group scheme over $k$. Then $G \to \Spec k$ is separated.
\end{lemma}

\begin{proof}
Let $\pi \colon G \to S$ the structure morphism. Then
$\pi$ is separated if and only if $e\colon S \to G$ is a closed immersion. For
any $x \in \mathrm{im}(e) \in G$, choose an affine open neighbourhood
$x \in U = \Spec A \subseteq G$.
Then $\pi|_{U} \circ e = \mathrm{id}_S$, hence the induced map
$A \xrightarrow{\Gamma(e)} k$ has a section $\Gamma(\pi|_U)$ and is therefore
surjective. Thus $e$ is a closed immersion.
\end{proof}

\begin{satz}
Let $G$ be a group scheme locally of finite type over $k$. Then
$G$ is smooth over $k$ if and only if $G$ is geometrically reduced.
\end{satz}

\begin{proof}
The first direction is immediate, since smoothness is invariant under base change and
smooth over a field implies reduced.
Conversely, for any field extension $\ell / k$ by a prior result
$G$ is smooth over $k$ if and only if $G$ is smooth over $\ell$. Thus
we may assume $k = \bar k$. By \ref{idk} and \ref{idk}, we obtain
$G_{\mathrm{sm}} \neq \emptyset$. By the transitive action
of $G(k)$ on $G$, every closed point is smooth. Since
\[
G_{(0)} = \{ g \in G \mid \mathrm{dim} \overline{\{g\}} = 0 \}
\] is very dense in $G$ and $G_{\mathrm{sm}} \subseteq G$ is open, the result follows.
\end{proof}

\begin{lemma}
Let $k$ be perfect and $G$ a group scheme locally of finite type over $k$. Then
the induced reduced subscheme $G_{\mathrm{red}}$ is a subgroup scheme of $G$.
\end{lemma}

\begin{proof}
Since $(-)_{\mathrm{red}}$ is a functor, we obtain
$i\colon G_{\mathrm{red}} \to G_{\mathrm{red}}$ and
$e\colon S \to G_{\mathrm{red}}$. By \ref{idk},
reduced is equivalent to geometrically reduced since $k$ is perfect. Thus
$G_{\mathrm{red}} \times_k G_{\mathrm{red}}$ is reduced and we obtain
\[
\begin{tikzcd}
G x_k G \arrow{r}{m} & G \\
G_{\mathrm{red}} \times_k G_{\mathrm{red}} \arrow{u}
\arrow[dashed]{r} & G_{\mathrm{red}} \arrow{u}
\end{tikzcd}
.\]
\end{proof}

\begin{korollar}
If $k$ is perfect and $G$ a group scheme locally of finite type over $k$. Then
$G_{\mathrm{red}}$ is smooth over $k$.
\end{korollar}

\begin{lemma}
Let $G$ be locally of finite type over $k$. Then $G$ is geometrically irreducible
if (and only if) $G$ is connected.
\end{lemma}

\begin{proof}
Since $G(k) \neq \emptyset$, we have a morphism
$\Spec k \to G$ and $\Spec k$ is geometrically connected. Thus $G$ is geometrically connected.
We may therefore assume $k = \bar k$. Since the statement is purely topological, we may
further assume that $G$ is reduced and thus smooth over $k$. Hence
$G$ is regular by \ref{idk}, in particular for every $g \in G$ the local ring
$\mathcal{O}_{G,g}$ is regular and hence an integral domain. Since $G$ is locally noetherian
and connected, the claim follows.
\end{proof}

\begin{definition}
An \emph{abelian variety} over $k$ is a connected, geometrically reduced
and proper $k$-group scheme.
\end{definition}

\begin{bem}
Abelian varieties are smooth and geometrically integral.
\end{bem}

\begin{bsp}
Elliptic curves are abelian varieties of dimension $1$.
\end{bsp}

The goal is now to show that abelian varieties are commutative group schemes.

\begin{lemma}
Let $X$ be a proper, geometrically connected and geometrically reduced $k$-scheme and
$Y$ an affine $k$-scheme. Then every morphism $X \xrightarrow{f} Y$ factors over a
$k$-valued point of $Y$.
\label{lemma:constant-of-proper-conn-irred-affine}
\end{lemma}

\begin{proof}
By the Liouville theorem for schemes, the global
sections of $\mathcal{O}_{X_{\bar k}}$ is $\bar k$. Since
$k \to \bar k$ is flat, we obtain
\[
\Gamma(X, \mathcal{O}_X) \otimes_k \bar k
\xlongrightarrow{\simeq} \Gamma(X_{\bar k}, \mathcal{O}_{X_{\bar k}})
.\] Since $k \to \bar k$ is even faithfully flat, we obtain
$\Gamma(X, \mathcal{O}_X) \simeq k$.

Choose an embedding $Y \hookrightarrow \mathbb{A}_k^{(I)}$. Then a
morphism $f\colon X \to Y$ is equivalent to a morphism
$X \xrightarrow{f} Y \hookrightarrow \mathbb{A}_k^{(I)}$, which is equivalent
to the datum of a family of $e_i \in \Gamma(X, \mathcal{O}_X)$ which
corresponds to a morphism
$\Spec k \xrightarrow{e} \mathbb{A}_k^{(I)}$. Thus by construction we obtain
a factorisation
\[
\begin{tikzcd}
X \arrow{r}{f} \arrow[dashed]{d} & Y \arrow{r} & \mathbb{A}^{(I)} \\
\Spec k \arrow{rru}
\end{tikzcd}
\] where the dashed arrow is induced from the isomorphism $\Gamma(X, \mathcal{O}_X) \simeq k$.
\end{proof}

\begin{lemma}[Rigidity]
Let $X$ be a geometrically reduced, geometrically connected and proper $k$-scheme
with $X(k) \neq \emptyset$. Let further $Y$ be an integral scheme over $k$, $Z$
be a separated $k$-scheme and $f\colon X \times_k Y \to Z$ a morphism such that
there exists $y \in Y(k)$ such that
$f|_{X_{y}}$ factors via a $k$-point $z \in Z(k)$. Then
$f$ factors via $\mathrm{pr}_2$.
\label{lemma:rigidity}
\end{lemma}

\begin{proof}
Consider the composition
\[
g\colon X \times_k Y \xrightarrow{pr_2} Y \simeq \Spec k \times_k Y
\xrightarrow{(x_0, \mathrm{id})} X \times_k Y \xrightarrow{f} Z
\] where $x_0$ is an arbitrarily chosen $k$-rational point of $X$.
It remains to show that $f = g$. Choose an open affine
neighbourhood $z \in U \subseteq Z$. Then
$X_y = \mathrm{pr}_2^{-1}(y) \subseteq f^{-1}(U)$. Since
$X$ is proper, $\mathrm{pr}_2$ is a closed map. Thus there
exists a $y \in V \subseteq Y$ open
with $\mathrm{pr}_2^{-1}(V) \subseteq f^{-1}(U)$. For
any $y' \in V$, we obtain
\[
\begin{tikzcd}
X \times_k Y \arrow{r}{f} & Z \\
X_{y'} \arrow[dashed, swap]{d}{\alpha(y')}
\arrow[hookrightarrow]{u} \arrow[dashed]{r} & U \arrow[hookrightarrow]{u} \\
U \times_k \kappa(y') \arrow{ur}
\end{tikzcd}
.\] By \ref{lemma:constant-of-proper-conn-irred-affine}, the morphism
$\alpha(y')$ factors over a $\kappa(y')$-valued point. Thus
$f$ and $g$ agree on the dense open subset $X \times_k V$. By reduced-to-separated,
the result follows.
\end{proof}

\begin{korollar}
Let $A$ and $B$ be abelian varieties over $k$
and $f$ a morphism of $k$-schemes $A \to B$. If under the induced
map $f(k)\colon A(k) \to B(k)$ the identity $e_A$ is mapped to $e_B$.
\label{cor:av-group-homs}
\end{korollar}

\begin{proof}
Consider the composition
\[
g\colon A \times_k A \xrightarrow{(f \circ m_A) \times (i_B \circ m_A \circ (f \times f))}
B \times_k B
\xrightarrow{m_B}
B
.\] It remains to show that the image of $g$ is precisely $\{e_B\} $. By
assumption $f(e_A) = e_B$ and thus
\[
g(\{e_A\} \times_k A) = \{ e_B\} = g(A \times_k \{e_A\})
.\] By repeated application of \ref{lemma:rigidity}, $g$ factors
via $\mathrm{pr}_1$ and $\mathrm{pr}_2$. Thus $g$ is constant and $e_B$ is in the image.
\end{proof}

\begin{korollar}
Every abelian variety is commutative.
\end{korollar}

\begin{proof}
Apply \ref{cor:av-group-homs} on $i\colon A \to A$.
\end{proof}

\end{document}

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