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Let $k$ be an algebraically closed field and $G$ an algebraic group over $k$ which is also a $k$-variety (so $G$ is integral, etc). Let $I$ be the ideal defining the identity $e \in G$ and let $\{ t_1, \ldots, t_n \} \subseteq I$ be a set of local parameters at $e$. Since $e$ is a smooth point, the $t_i$ are algebraically independent and we naturally obtain a $k$-algebra embedding $k[t_1, \ldots, t_n] \hookrightarrow k[G]$. Let $A$ denote the subalgebra of $k[G]$ thus defined.

I do not expect that $A$ will be a sub-Hopf algebra of $k[G]$ (in particular, I see no reason why the comultiplication should preserve $A$), but I don't know any concrete examples. So I have two questions: (1) Is there a specific example of an algebraic group $G$ as above such that $A$ is not a sub-Hopf algebra of $k[G]$? (2) Are there nice conditions on $G$ under which $A$ will be a sub-Hopf algebra of $k[G]$?

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This never happens for a reductive group $G$ of dimension $n>0$. The conditions you give induce a surjective finite group homomorphism $G\to \mathbb{A}^n$, for some mysterious group structure on $\mathbb{A}^n$. In particular, $G$ would act transitively on $\mathbb{A}^n$, which a reductive group cannot do.

On the other hand, there are many unipotent groups whose underlying schemes are affine spaces, e.g. unipotent upper triangular matrices. Any extension of such a group by a finite group will satisfy the conditions of your question.

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    $\begingroup$ All algebraic group structures on an affine space are unipotent, so even if $G$ is simply not unipotent there cannot be such a Hopf subalgebra. $\endgroup$
    – user29720
    Commented Dec 21, 2012 at 3:25
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    $\begingroup$ @Daniel Litt. You say `This never happens.' That is what Armand Borel wrote. But see Friedrich Knop, Homogeneous varieties for semisimple groups of rank one. Compositio Mathematica, 98 (1995), 77-89. The conclusion is that it is extremely rare, but it does happen. $\endgroup$
    – Wilberd van der Kallen
    Commented Dec 21, 2012 at 10:58
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    $\begingroup$ @kreck: See the remark on p. 85-86 of the paper of Knopp cited in Wilberd's comment. Knopp observe that $\operatorname{SL}_2$ acts transitively on $\mathbf{A}^2$ in char. 2. Knopp points out that Theorem 1.1(i) of Borel's paper is correct, but that Theorem 1.1(ii) is not (the issue seems to be that Borel's argument does not work when the isotropy subgroup is disconnected). $\endgroup$
    – George McNinch
    Commented Dec 21, 2012 at 15:32
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    $\begingroup$ @All: Char $p>0$ can indeed be tricky for group actions. Borel's 1985 Arch. Math. note is reprinted in his collected papers IV (1983-1999), with a note at the end clarifying the error in his Thm. 1.1(ii) pointed out early by Knop [then at Rutgers]. Though he doesn't cite Knop's 1995 paper, Borel describes Knop's counterexample $G=\mathrm{SL}_2(k))$ acting transitively on $\mathbb{A}^2$ with normalizers of maximal tori as isotropy groups. Borel points out that this (and other Knop examples) can work only for $p=2$; otherwise 1.1(ii) and its proof are correct. $\endgroup$
    – Jim Humphreys
    Commented Dec 21, 2012 at 16:12
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    $\begingroup$ @George & Jim: When reading Borel's paper I had noticed that disconnectedness wasn't handled clearly; e.g., I was confused by the proof of 2.5 unless $H$ is connected. I'm glad to know I was confused for a good reason! My main interest was 1.1(i) (with connected $H = R_u(G)$), so I missed the problem with 1.1(ii). (In the proof of 1.2(ii), which is 1.1.(ii), he applies 1.1(2), whose proof in 2.8 goes via 2.5(ii) -- that appeal to 2.5 is where the error creeps in when the isotropy group $H$ is disconnected.) Fortunately, this doesn't affect 2.6, which I invoked above instead of 1.1(ii). :) $\endgroup$
    – user29720
    Commented Dec 21, 2012 at 19:38
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If $A$ were always a sub-Hopf-algebra then its spectrum would be an algebraic group $H$ and there would be a non-constant map from $G$ to $H$. Furthermore, as an algebraic variety, $H$ would be isomorphic to affine $n$-space for some $n$ (equal to the dimension of $G$). Often this cannot happen. For example if $G$ is the multiplicative group, which would have been the first example I'd tried if I'd wanted to get my hands dirty and actually do the calculation, then the kernel of $G\to H$ would have to be finite, and the image would be an open dense subgroup and hence the whole thing. But the multiplicative group is not isomorphic to affine 1-space as an algebraic variety.

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