Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The title says it all.

A very similar question was asked and answered about linear groups, but none of the counterexamples are algebraic: Are extensions of linear groups linear?

If $A$, $B$ are affine and there is a rational section of $C \to A$ in $1 \to B \to C \to A \to 1$, then $C \to A$ is affine, so $C$ is affine. But if not?

share|improve this question
@Michael: Do you mean that in your exact sequence, $B$ and $A$ are affine algebraic groups, $C$ is merely algebraic and your are asking if $C$ is affine? Or you only mean that $C$ is an abstract group? Also, do you make any assumptions about the field? (If the field is perfect and $C$ is algebraic then the answer is positive; if $C$ is not required to be algebraic then the answer is negative in the case when the field is ${\mathbb R}$.) –  Misha Oct 12 '12 at 13:29
What you said first. An extension in the category of algebraic groups. –  Michael Thaddeus Oct 12 '12 at 13:33
Sure. If $1 \rightarrow G' \rightarrow G \rightarrow G'' \rightarrow 1$ is a short exact sequence of fppf group sheaves over a scheme $S$ with $G''$ representable and $G'$ is $S$-affine and fppf over $S$ then $G$ is representable and $G \rightarrow G''$ is affine and fppf (so $G$ is $S$-affine if $G''$ is, same for fppf). This is proved by identifying $G$ as a $G'$-torsor sheaf over $G''$ for the fppf topology (sheaf quotient maps have "local" sections!) and using effectivity of fppf descent for affine morphisms. It is explained in Oort's LNM book on commutative (!) group schemes. –  grp Oct 12 '12 at 13:50

1 Answer 1

up vote 2 down vote accepted

Yes. The point is that $C$ is a $B$-torsor over $A$. Since being affine is a local property in the fpqc topology, $C$ is affine over $A$.

[Edit]: Sorry I had not noticed grp's comment, or I wouldn't have posted an answer.

At to why there are local section, well, to me that's by the definition of an extension. Alternatively, assuming you are on a field, the injectivity of $A \to B$ means, I suppose, that $A$ is an embedding of algebraic groups. This defines a free action of $A$ on $C$; take the quotient $B/A$ (as an fppf sheaf, or étale, if $A$ is smooth); the projection $B \to B/A$ has local sections, by construction. It's a basic result that $B/A$ is represented by a group scheme. Then the exactness of the sequence should meant that $B \to C$ induces an isomorphism of $B/A$ with $C$.

If you are over an algebraically closed of characteristic 0, exactness of the sequence can be checked, in fact, at the level of closed points.

share|improve this answer
Very good, but that's basically the same argument I briefly indicated, only with the fppf or fpqc topology replacing the Zariski topology. My question for you, and for commenter grp above, is why do sections exist in the fppf or fpqc topologies? That is, why is a group extension a torsor for such topologies? I'll see if I can find this in Oort's book. –  Michael Thaddeus Oct 12 '12 at 15:18
PS: sorry, of course I meant local sections –  Michael Thaddeus Oct 12 '12 at 15:19
@Michael: I assumed (and probably Angelo did too) that you know the equivalence of several equivalent definitions of "group extension" (without which it is hard to work with this concept in a nice way). What definition are you using (especially if you aren't assuming smoothness of the groups)? –  grp Oct 12 '12 at 16:08
The map $C\to A$ is itself fppf (if your algebraic groups are reduced, flatness follows from generic flatness and homogeneity ; if they are not necessarily reduced, flatness should be somehow part of the definition of being surjective). Then you can base change $C\to A$ by the map $C\to A$ itself : here you have a tautological section. –  Olivier Benoist Oct 12 '12 at 16:14
@Olivier: Since Michael seems to be working over a general field, one should say "smooth" rather than "reduced". As you know, over any imperfect field there are reduced linear algebraic groups that are not smooth, and their relative Frobenius morphism is a finite surjective homomorphism which is not flat. Also, I see your intent by saying "flatness should somehow be part of the definition of being surjective", but this seems a bit risky since surjective has its own useful (ordinary) meaning for scheme maps. However,"fppf" requires fewer letters than "surjective", so using French solves it. :) –  grp Oct 12 '12 at 17:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.