Chevalley: every closed subgroup is a line stabilizer

Since $\mathfrak{a}$ is finitely generated as an ideal by [F2], choose a finite generating set $S\subseteq \mathfrak{a}$. By [F3] there is a finite-dimensional subcomodule $V\subseteq A$ under $\mathrm{\Delta }$ with $S\subseteq V$; put $W=\mathfrak{a}\cap V$, choose a basis $(e_j{)}_{j\in J}$ of $W$ and extend it to a basis $(e_i{)}_{i\in J\cup I}$ of $V$, so that $I$ indexes a complement of $W$ in $V$. Write $\mathrm{\Delta }(e_j)={\sum }_{i\in J\cup I}{e}_i\otimes a_{ij}$ for $j\in J$ and let ${\mathfrak{a}}^{\mathrm{\prime }}$ be the ideal of $A$ generated by the elements $a_{ij}$ with $j\in J$, $i\in I$.

For a $k$-algebra $R$ and $g\in G(R)$ one has $g\cdot e_j={\sum }_{i\in J\cup I}{e}_i{a}_{ij}(g)$ under the action associated with the coaction $\mathrm{\Delta }$, and the $e_i$ form an $R$-basis of $V_R$; hence $g\cdot W_R\subseteq W_R$ if and only if $a_{ij}(g)=0$ for all $j\in J$, $i\in I$, that is, if and only if $g$ vanishes on ${\mathfrak{a}}^{\mathrm{\prime }}$. Since $g$ acts invertibly and $W_R$ is a direct summand of $V_R$, the inclusion $g\cdot W_R\subseteq W_R$ is equivalent to $g\cdot W_R=W_R$. Therefore the stabilizer functor of $W$ is represented by the closed subscheme $\mathrm{Spec}(A\mathrm{/}{\mathfrak{a}}^{\mathrm{\prime }})$ of $G=\mathrm{Spec}A$.

${\mathfrak{a}}^{\mathrm{\prime }}\subseteq \mathfrak{a}$: as $\mathfrak{a}$ is a Hopf ideal, $\mathrm{\Delta }(e_j)\in A\otimes \mathfrak{a}+\mathfrak{a}\otimes A$ for $j\in J$ by [F1], so applying $\mathrm{id}\otimes \pi$ for the quotient $\pi :A\to A\mathrm{/}\mathfrak{a}$ and then $q\otimes \mathrm{id}$ for the quotient $q:A\to A\mathrm{/}\mathfrak{a}$ gives ${\sum }_{i\in I}q(e_i)\otimes \pi (a_{ij})=0$ in $(A\mathrm{/}\mathfrak{a})\otimes (A\mathrm{/}\mathfrak{a})$, because $\pi (e_j)=0$ for $j\in J$ and $\mathfrak{a}\cap V=W$; the elements $q(e_i)$, $i\in I$, are linearly independent, so $\pi (a_{ij})=0$, that is, $a_{ij}\in \mathfrak{a}$ for all $j\in J$, $i\in I$.

$\mathfrak{a}\subseteq {\mathfrak{a}}^{\mathrm{\prime }}$: for $j\in J$ one has $\epsilon (e_j)=0$ by [F1], and the counit axiom gives $e_j=(\epsilon \otimes \mathrm{id})\mathrm{\Delta }(e_j)={\sum }_{i\in J\cup I}\epsilon (e_i)a_{ij}={\sum }_{i\in I}\epsilon (e_i)a_{ij}\in {\mathfrak{a}}^{\mathrm{\prime }}$, because the terms with $i\in J$ have $\epsilon (e_i)=0$. The elements $e_j$, $j\in J$, span $W\supseteq S$, and $S$ generates $\mathfrak{a}$ as an ideal, so $\mathfrak{a}$ is contained in the ideal ${\mathfrak{a}}^{\mathrm{\prime }}$.

By steps 2.2 and 2.3, ${\mathfrak{a}}^{\mathrm{\prime }}=\mathfrak{a}$, so $\mathrm{Spec}(A\mathrm{/}{\mathfrak{a}}^{\mathrm{\prime }})=\mathrm{Spec}(A\mathrm{/}\mathfrak{a})=H$. Step 2.1 identifies $\mathrm{Spec}(A\mathrm{/}{\mathfrak{a}}^{\mathrm{\prime }})$ with the stabilizer of $W$, so $H={\mathrm{Stab}}_G(W)$ scheme-theoretically, where $V$ is a finite-dimensional rational representation of $G$ and $W\subseteq V$.

Let $d=\dim W$ and $L={\mathrm{\Lambda }}^{d}W\subseteq {\mathrm{\Lambda }}^{d}V$; this is a line, ${\mathrm{\Lambda }}^{d}V$ is a finite-dimensional rational representation of $G$ by [F4], and [F4] gives ${\mathrm{Stab}}_G(W)={\mathrm{Stab}}_G(L)$ scheme-theoretically. Combined with step 3.1 this produces the required pair $(V,L)$ with $H={\mathrm{Stab}}_G(L)$.

If $L=kv$, then $\{x\in \mathrm{Lie}(G):xL\subseteq L\}=\{x:xv\in kv\}$ and [F5] applied to the line $L$ gives $\mathrm{Lie}(H)=\mathrm{Lie}({\mathrm{Stab}}_G(L))=\{x\in \mathrm{Lie}(G):xv\in kv\}$.

Steps 4.1 and 5.1 prove both assertions of the theorem for the closed subgroup scheme $H$ of $G$.