Every subgroup scheme of an affine group is a line stabilizer

Choose finitely many ideal generators of $I$ by [F2], and a finite-dimensional right-regular subcomodule $V\subset A$ containing them by [F1]. Set $W=I\cap V$. Choose a basis $(e_j{)}_{j\in J}$ of $W$ and extend it to $(e_i{)}_{i\in J\bigsqcup K}$ of $V$. Write $\mathrm{\Delta }(e_j)={\sum }_i{e}_i\otimes a_{ij}$. The stabilizer of $W$ is cut out by $a_{ij}$ for $j\in J,i\in K$. Indeed these equations say the lower-left matrix block is zero; an invertible block upper triangular matrix over any commutative $R$ has both diagonal blocks invertible, since their determinants have invertible product. Thus inclusion $g{W}_R\subset W_R$ gives equality.

Because $I$ is a Hopf ideal, $\mathrm{\Delta }(I)\subset I\otimes A+A\otimes I$ and $ϵ(I)=0$. In $V\otimes A$, the first condition says that all the displayed $a_{ij}$ with $i\in K,j\in J$ belong to $I$: project the first factor into $A\mathrm{/}I$, where the images of the $e_i$ for $i\in K$ are independent. Conversely the counit identity gives $e_j={\sum }_{i\in K}ϵ(e_i)a_{ij}$ for $j\in J$. Since the $e_j$ span a space containing the ideal generators of $I$, those matrix entries generate $I$. The stabilizer is therefore exactly $H$ as a closed scheme.

Put $d=\dim W$ and $L={\bigwedge }^{d}W\subset {\bigwedge }^{d}V$. This is a line, including $d=0$. For a basis $e_1,\dots ,e_d$ of $W$ let $w=e_1\wedge \cdots \wedge e_d$. Over every $R$, $W_R=\{v\in V_R:w\wedge v=0\}$, by expanding $v$ in a basis extending that of $W$. If an automorphism $g$ stabilizes $L_R$, then $({\bigwedge }^{d}g)w=cw$ for a unit $c\in R$; applying ${\bigwedge }^{d+1}g$ to $w\wedge v$ shows $g{W}_R\subset W_R$, and applying $g^{-1}$ gives equality. The converse follows by taking determinants on $W_R$. Thus the line stabilizer equals the subspace stabilizer from step 2.1, completing the proof on every algebra.