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Let $K$ be a perfect field. In what follows, an algebraic group $G/K$ is by definition a group scheme of finite type over $K$.

The following seems to be well-known:

Theorem: Let $G/K$ be a connected smooth algebraic group. Then there is a connected smooth affine normal closed subgroup $N$ of $G$, an abelian variety $A/K$ and a homomorphism $G\to A$ with kernel $N$ such that the sequence $$0\to N\to G\to A\to 0$$ is exact for the fppf-topology (say).

Can someone give me a proper reference or a hint why this is true? (Checking the literature I find on the one hand plenty of references treating affine algebraic groups, and on the other hand references containing the theory of abelian varieties, but I was surprised not to find a reference containing a proof of this Theorem about the "mixed case".)

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    $\begingroup$ Maybe this will be helpful: Brian Conrad A modern proof of Chevalley's theorem on algebraic groups . $\endgroup$
    – Piotr Achinger
    Oct 25, 2010 at 15:33
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    $\begingroup$ Basti, here's something not well-known but deserves to be! First, Chevalley's thm is false over every imperfect field. Second, over fields of char. $> 0$ there is a better result (even for finite fields): there's an exact sequence $1 \rightarrow Z \rightarrow G \rightarrow H \rightarrow 1$ with smooth affine $H$ and central $Z$ that is semi-abelian (i.e., extension of abelian variety by a torus). Its buried in Demazure-Gabriel & godsend for cohomological purposes, since commutative term is on left rather than right. See Example A.3.8 and Thm. A.3.9 in "Pseudo-reductive groups" for more. $\endgroup$
    – BCnrd
    Oct 25, 2010 at 16:06
  • $\begingroup$ Thanks for pointing to the proper reference, Piotr. The paper of Brian is exactly what I wanted to have. @Brian: Thanks for the additional information! I was expecting that the Theorem in my post can become false for imperfect fields, but good to know that it fails for every imperfect field. The other exact sequence in positive characteristic is also very interesting. I was not aware of that. $\endgroup$
    – Sebastian Petersen
    Oct 26, 2010 at 9:54

1 Answer 1

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This question is answered in the Wikipedia page for algebraic groups. The article says that this is a difficult result of Chevalley, and it has a link to a modern write-up by Brian Conrad of Chevalley's result. So that is surely a proper reference. No particular hint leaps out at me for "why" it is true, but I can say something about what the proof is really saying. The subgroup $N$ appears as the common kernel of all algebraic homomorphisms from $G$ to all abelian varieties. So it is a functorial construction and $N$ is actually a characteristic subgroup, not just a normal subgroup. Relatively early on it is shown that there are no non-trivial algebraic homomorphisms from an affine group to an abelian variety, in fact not even any non-trivial algebraic morphisms that don't have to be homomorphisms.

It is also relatively quick to show that $G/N$ is an abelian variety. The really hard part is to show that the kernel is affine. You might as well let $G = N$ and you might as well let $K$ be algebraically closed. The hard theorem is that if $G$ does not have any non-trivial homomorphisms to an abelian variety, then it is affine. It is important to remember that $G$ is affine if and only if it is linear, i.e., an algebraic subgroup of $\text{GL}(V)$ for some finite-dimensional vector space $V$. This vector space $V$ is the most difficult construction of the paper.

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    $\begingroup$ Loosely speaking, the way affineness is obtained in the setup that Greg indicates at the end is to find "enough" divisors $D$ to make a projective representation of $G$ that is faithful on geometric points. That defines a homomorphism $f:G \rightarrow {\rm{PGL}}_n$ whose scheme-theoretic kernel is infinitesimal, whence $G$ is finite over its (necessarily smooth and closed) image $f(G)$ due to descent theory (since $G \rightarrow f(G)$ is flat, due to smoothness of $G$). But ${\rm{PGL}}_n$ is affine, whence $f(G)$ is affine and hence so is $G$. $\endgroup$
    – BCnrd
    Oct 25, 2010 at 15:59
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    $\begingroup$ As BCnrd suggests here and in his published note (the only helpful literature I know of), there is some intuition behind the theorem but it's not exactly transparent. My short "review" in Math Reviews doesn't address the intuition but does bring out a bit more of the history: MR1906417 (2003f:20078) 20G15 (11H99 14K05 14L15), Conrad, Brian (1-MI), A modern proof of Chevalley’s theorem on algebraic groups. J. Ramanujan Math. Soc. 17 (2002), no. 1, 1–18. $\endgroup$
    – Jim Humphreys
    Oct 25, 2010 at 16:16
  • $\begingroup$ @BCnrd Your paper says, "By choosing (non-canonically) a realization of [$\text{PGL}(V)$] as an algebraic subgroup of some $\text{GL}(W)$..." Actually there is a canonical realization: The adjoint representation of $\text{GL}(V)$ factors through $\text{PGL}(V)$. $\endgroup$
    – Greg Kuperberg
    Oct 26, 2010 at 6:51
  • $\begingroup$ Dear Greg: Right you are. Note that the projective representation as constructed there has an "interpretation" in terms of function theory on the group, but the adjoint repn built from it doesn't seem to have such an interpretation. More importantly, since the projective repn might have an infinitesimal kernel when in nonzero characteristic, even this adjoint repn may not provide a faithful repn for the original $G$ when ${\rm{char}}(k) > 0$. So in the end the proof doesn't directly construct a faithful realization inside some ${\rm{GL}}(V)$; it's indirect via descent theory, etc. $\endgroup$
    – BCnrd
    Oct 26, 2010 at 7:37
  • $\begingroup$ Thanks a lot for the proper reference and for the information on the strategy of the proof! $\endgroup$
    – Sebastian Petersen
    Oct 26, 2010 at 9:50

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