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