\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 $G$.
        ´\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}