Are extensions of linear algebraic groups (over a field) themselves linear algebraic?

2012-10-12T12:22:34Z

The title says it all.

A very similar question was asked and answered about linear groups, but none of the counterexamples are algebraic: http://mathoverflow.net/questions/22814/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?

Answer by Angelo for Are extensions of linear algebraic groups (over a field) themselves linear algebraic?

2012-10-12T15:07:34Z

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.

Are extensions of linear algebraic groups (over a field) themselves linear algebraic? - MathOverflow

Are extensions of linear algebraic groups (over a field) themselves linear algebraic?

Answer by Angelo for Are extensions of linear algebraic groups (over a field) themselves linear algebraic?