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Let $k$ be an algebraically closed field of characteristic $0$. A well-known result of Lazard's states that an algebraic group which is isomorphic as a variety to an affine space is unipotent (M. Lazard, Sur nilpotence de certains groupes algebriques, C.R.A.S. Paris 241 (1955), 1687-1689.).

In the language of Hopf algebras, this says that any Hopf algebra isomorphic as an algebra to $k[x_{1}, \ldots, x_{n}]$, for some $n$, is necessarily connected as a Hopf algebra (i.e. the sum of its simple subcoalgebras is just the base field, $k$).

Lazard's original proof of this result is completely in the language of algebraic groups, and involved passing to finite fields (see 'How to use finite fields for problems concerning infinite fields', Serre, https://arxiv.org/pdf/0903.0517.pdf Theorem 4.1, for a brief description). My question is as follows:

(a) Is there a known proof of Lazard's result purely in the language of Hopf algebras?

(b) Is there perhaps a simpler proof which does not involve passing to finite fields if one considers this Hopf-theoretic approach?

Edit: Here's one possible idea, but I'm not sure if it helps.

Let $H$ be a Hopf algebra, isomorphic as an algebra to $k[x_{1},\ldots, x_{n}]$ for some $n>0$. Let $C$ be a simple subcoalgebra of $H$. We aim to show that $C$ is one-dimensional.

We have a coalgebra morphism $C\rightarrow H$. Dualising, this corresponds to a surjective algebra morphism $$H^{\circ}\rightarrow C^{*}$$ where is the finite dual of $H$, and $C^{*}$ is a finite-dimensional simple algebra. Since $H$ is commutative, $H^{\circ}$ is cocommutative, so $H^{\circ}\cong kG(H^{\circ})\#U(P(H^{\circ}))$ by the Cartier-Gabriel-Konstant theorem. We know that elements of $G(H^{\circ})$ are precisely algebra homomorphisms $H\rightarrow k$, hence $G(H^{\circ})$ is the underlying algebraic group of $H$. This is about as far as I got.

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  • $\begingroup$ What do you take for granted? At the group level, Lazard/Serre assume for granted that once a connected affine algebraic group has no 1-dimensional subtorus (or even, no element of finite order), then it is unipotent, but relies on basic theory, which is usually not proved in the language of Hopf algebras. $\endgroup$
    – YCor
    Commented Dec 4, 2016 at 19:32
  • $\begingroup$ I'm not entirely sure what you mean, but I suppose the statement 'if an affine algebraic group has no 1-dimensional subtorus then it is connected' could be translated to a statement about Hopf algebras: 'if a commutative Hopf algebra satisfies property P, it is connected' (can't think off the top of my head what the Hopf analogue of no 1-d subtorus is). If this result is basic in algebraic group theory, the Hopf analogue is presumably basic as well. It is just Lazard's use of finite fields in his proof that I don't feel lends itself as well to using Hopf algebras, but I could easily be wrong. $\endgroup$
    – Paul Gilmartin
    Commented Dec 4, 2016 at 19:52
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    $\begingroup$ A 1-dim subtorus corresponds to a Hopf quotient map from $H$ to the Hopf algebra of the multiplicative group. $\endgroup$
    – YCor
    Commented Dec 4, 2016 at 19:54
  • $\begingroup$ If you take this for granted, then all you want is, for instance, to have a proof, avoiding both topology and finite fields, that every polynomial involution of the affine plane has a fixed point. That is, every $k$-algebra every involutive automorphism of $k[x_1,\dots,x_n]$ ($k$ algebraically closed of characteristic not 2) leaves some maximal ideal invariant. $\endgroup$
    – YCor
    Commented Dec 4, 2016 at 19:57
  • $\begingroup$ Could please you briefly explain how you arrived at that conclusion if we take that for granted, YCor ? $\endgroup$
    – Paul Gilmartin
    Commented Dec 5, 2016 at 21:17

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