\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}