diff --git a/lec.pdf b/lec.pdf index fc17bef..d15e063 100644 Binary files a/lec.pdf and b/lec.pdf differ diff --git a/lec.tex b/lec.tex index 89353b2..d103b82 100644 --- a/lec.tex +++ b/lec.tex @@ -23,6 +23,7 @@ Christian Merten\\ \input{lec02} \input{lec03} \input{lec04} +\input{lec05} \bibliographystyle{alpha} \bibliography{refs} diff --git a/lec05.pdf b/lec05.pdf new file mode 100644 index 0000000..248d360 Binary files /dev/null and b/lec05.pdf differ diff --git a/lec05.tex b/lec05.tex new file mode 100644 index 0000000..3e0c13e --- /dev/null +++ b/lec05.tex @@ -0,0 +1,193 @@ +\documentclass{lecture} + +\begin{document} + +\begin{lemma}[] + Let $X$ be a connected scheme over $k$ and $Y$ a geometrically connected scheme over $k$. If + $\mathrm{Hom}_k(Y, X) \neq \emptyset$, then $X$ is geometrically connected. +\end{lemma} + +\begin{proof} + Use that $X_{\bar k} \to X$ is an open and closed immersion. Let + $\emptyset \neq Z \subseteq X_{\bar k}$ be open and closed. Consider + the commutative diagram + \[ + \begin{tikzcd} + \bar f^{-1}(Z) = Z \times_k Y \arrow{r} \arrow{d} & Y_{\bar k} \arrow{r} \arrow{d}{\bar f} & Y \arrow{d}{f} \\ + Z \arrow[hookrightarrow]{r} & X_{\bar k} \arrow{r}{\pi} & X + \end{tikzcd} + .\] We obtain $\bar f^{-1}(Z) = Y_{\bar k}$. Set $Z' = Y_{\bar k} \setminus Z$. If $Z'$ is + not-empty, then by the same argument $\bar f^{-1}(Z') = Y_{\bar k}$. Contradiction. +\end{proof} + +\begin{satz} + Let $G$ be a group scheme locally of finite type over $k$. + \begin{enumerate} + \item If $U, V \subseteq G$ are open and dense. Then $U V = G$ as topological spaces. + \item If $G$ is irreducible, then $G$ is quasi-compact. + \item Any subgroupscheme $H \subseteq G$ is a closed subscheme. + \end{enumerate} +\end{satz} + +\begin{proof} + We reduce to $k = \bar k$. + \begin{enumerate}[] + \item We know that $G_{\bar k} \to G$ is an open and closed immersion. Taking + pre-images then preserves open and dense (???) and the result follows. + \item By \ref{???} $G$ is geometrically irreducible and $G_{\bar k} \to G$ is surjective, i.e. + the quasi-compactness of $G_{\bar k}$ implies the quasi-compactness of $G$. + ยด\item By \ref{???}, being a closed immersion can be tested by faithfully flat descent. + \end{enumerate} + Now suppose $k = \bar k$. + \begin{enumerate} + \item It suffices to show that $U(k) V(k) = G(k)$, since + $\overline{U(k)V(k)}$ is very dense in $\overline{UV}$. Since + $i\colon G \to G$ is an isomorphism of schemes, $V(k)^{-1} \subseteq G(k)$ is + open and dense. Thus + for all $g \in G$, $g(V(k)^{-1})$ is open and dense. Thus there + exists $u \in g(V(k)^{-1})\cap U(k)$, i.e. there + exists $v \in V(k)$ such that $gv^{-1} = u$, i.e. $g = u v$. + \item Let $U \subseteq G$ be open, dense and quasi-compact. Then $U \times_k U$ is + quasi-compact and $G = \mathrm{im}(U \times_k U \to G)$ is quasi-compact. + \item Put the induced reduced subscheme structure on $\bar H \subseteq G$. By + \ref{???}, the maps $H \to \Spec k$ and $\bar H \to \Spec k$ are universally open. + Since $H \subseteq \bar H$ is dense, we obtain + \[ + H \times_k H \subseteq H \times_k \bar H \subseteq \bar H \times_k \bar H + \] is dense. Since + $H \times_k H \subseteq m^{-1}(H) \subseteq m^{-1}(\bar H) \hookrightarrow G \times G$, + we obtain topologically $\bar H \times \bar H \subseteq m^{-1}(\bar H)$. Since + the objects in the lower row are reduced, we therefore obtain a factorisation + \[ + \begin{tikzcd} + G \times G \arrow{r} & G \\ + \bar H \times_{k} \bar H \arrow[hookrightarrow]{u} + \arrow[dashed]{r} & \bar H \arrow[hookrightarrow]{u} + \end{tikzcd} + .\] Thus $\bar H \subseteq G$ is a subgroupscheme. Thus + $H = H \times H = \bar H$ where the last equality follows from 1. + \end{enumerate} +\end{proof} + +\begin{definition} + Let $G$ be a group scheme locally of finite type over $k$ and $e\colon \Spec k \to G$ is the unit. + Then denote by $G^{0}$ the connected component of $G$ that contains $\mathrm{im}(e)$. We call + $G^{0}$ the \emph{unit component} of $G$. +\end{definition} + +\begin{bem} + Since $G$ is locally noetherian, $G^{0}$ is open and closed. +\end{bem} + +\begin{satz} + Let $G$ be a group scheme locally of finite type over $k$. + \begin{enumerate}[] + \item $G^{0}$ is a quasi-compact, geometrically-irreducible and normal subgroupscheme of $G$. + \item Any group morphism $G \to H$ with $H$ locally of finite type over $k$ induces + a group homomorphism $G^{0} \to H^{0}$. + \item For any field extension $\ell / k$, we have + \[ + (G \times_k \ell)^{0} = G^{0} \times_k \ell + .\] + \end{enumerate} +\end{satz} + +\begin{proof} + \begin{enumerate} + \item Since $G^{0}$ is connected and contains a $k$-rational point, by \ref{???} $G^{0}$ is + geometrically connected. Then $G_0 \times_k G_0$ is connected + and + \[ + \begin{tikzcd} + G \times_k G \arrow{r} & G \\ + G^{0} \times_k G^{0} \arrow{u} \arrow[dashed]{r} & G^{0} \arrow{u} + \end{tikzcd} + .\] Since $G^{0} \hookrightarrow G \xrightarrow{i} G$ factors + over $G^{0} \hookrightarrow G$, $G^{0}$ is a subgroupscheme. + By \ref{???}, $G^{0}$ is geometrically irreducible and therefore + by \ref{???} it is quasi-compact. + For normality consider a connected component $G'$ of $G$. Then we have a commutative diagram + \[ + \begin{tikzcd} + G \times_k G^{0} \arrow{r}{(g, h) \mapsto g h g^{-1}} & G \\ + G' \times_k G^{0} \arrow[hookrightarrow]{u} + \arrow[dashed]{r} & G^{0} \arrow[hookrightarrow]{u} + \end{tikzcd} + .\] Since $G' \times G^{0}$ is connected, the image of the upper horizontal arrow is + in $G^{0}$. + \item Any group homomorphism sends the identity to the identity, i.e. the composition + $G^{0} \hookrightarrow G \to H$ factors via $H^{0} \hookrightarrow H$. + \item Since $G^{0}$ is geometrically connected, the scheme + $G^{0} \times_k \ell$ is connected. Moreover + $G^{0} \times_k \ell \subseteq G \times_k \ell$ is open and closed. Finally, + the identity of $G \times_k \ell$ is contained in $G^{0} \times_k \ell$ by the universal + property of the fibre product. + \end{enumerate} +\end{proof} + +The proof of the following lemma is left as an exercise to the reader. + +\begin{lemma} + Let $G$ be a group scheme locally of finite type over $k$. Then every connected component + of $G$ is quasi-compact and geometrically irreducible and $G$ is equidimensional. +\end{lemma} + +\begin{satz} + Let $f\colon G \to H$ be a group homomorphism of group schemes locally of finite type over $k$. Then + \begin{enumerate}[] + \item $\mathrm{im}(f) \subseteq H$ is closed. + \item $\mathrm{dim}(G) = \mathrm{dim}(\mathrm{im}(f)) + \mathrm{dim}(\mathrm{ker}(f))$. + \item Is $H$ smooth over $k$ and $f$ surjective, then $f$ is faithfully flat. + \end{enumerate} +\end{satz} + +\begin{bem} + For any integral morphism $f\colon X \to Y$ and $Z \subseteq X$ closed the image + $f(Z)$ is closed in $Y$ and $\mathrm{dim}(Z) = \mathrm{dim}(f(Z))$. +\end{bem} + +\begin{proof} + Since $H_{\bar k} \xrightarrow{\pi} H$ is integral and surjective + and $\mathrm{dim}(Z) = \mathrm{dim}(\pi(Z))$ for any closed subset $Z \subseteq H_{\bar k}$, + we may assume $k = \bar k$. + \begin{enumerate} + \setcounter{enumi}{2} + \item Since smooth implies reduced, $H^{0}$ is reduced and by \ref{???} $H^{0}$ is irreducible. Thus + $H^{0}$ is integral. By generic flatness, we have a + $V \subseteq H^{0}$ that is open and dense such that + $f^{-1}(V) \to V$ is flat. Thus for all $h \in H(k)$, the map + $f^{-1}(hV) \xrightarrow{f} hV$ is flat. By covering $H$ with + translates of $V$, we obtain $f$ is flat. + \setcounter{enumi}{0} + \item We may assume that $G$ is reduced and thus $G$ is smooth over $k$ by \ref{???}. Let + $C$ be $C_{\mathrm{red}} = \overline{f(G)}^{H}$. We claim + that $C$ is a subgroupscheme of $H$. Then $G \to C$ is quasi-compact and dominant. Thus + we have a factorisation + \[ + \begin{tikzcd} + G \times_k G \arrow{r} \arrow{d}{m_G} + & C \times_k C \arrow{r} \arrow{d}[dashed]{m_C} & H \times_k H \arrow{d}{m_H} \\ + G \arrow{r}{f} & C \arrow[hookrightarrow]{r} & H + \end{tikzcd} + .\] Analogously one obtains + \[ + \begin{tikzcd} + C \arrow[hookrightarrow]{d} \arrow[dashed]{r} & C \arrow[hookrightarrow]{d} \\ + H \arrow{r} & H + \end{tikzcd} + .\] Thus we may assume that $f$ is dominant. + By the theorem of Chevalley, $f(G)$ is constructible and is therefore dense. Hence + there exists an open $U \subseteq H$ such that $U \subseteq f(G)$. Thus + $H = U \cdot U \subseteq f(G)$ and $f(G) = H$ is closed. + \item We may assume that also $H$ is reduced and that $f(G) = H$. Then + $H$ is smooth over $k$ and $f$ is flat. By \ref{???} we have $f(G^{0}) \subseteq H$ is open + and by 1) also closed. Thus $G^{0} \xrightarrow{f} H^{0}$ is surjective. + We have $\mathrm{dim}(G^{0}) = \mathrm{dim}(G)$, + $\mathrm{dim}(H^{0}) = \mathrm{dim}(H)$ and + $\mathrm{dim}(\mathrm{ker}(f^{0})) = \mathrm{dim}(\mathrm{ker}(f)^{0})$. Now the result follows since + all fibres are isomorphic and dimension is additive under flat morphism in non-empty fibres + (\cite{gw} 14.119). + \end{enumerate} +\end{proof} + +\end{document}