Are extensions of linear algebraic groups (over a field) themselves linear algebraic? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T06:59:34Z http://mathoverflow.net/feeds/question/109456 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109456/are-extensions-of-linear-algebraic-groups-over-a-field-themselves-linear-algebr Are extensions of linear algebraic groups (over a field) themselves linear algebraic? Michael Thaddeus 2012-10-12T12:22:34Z 2012-10-12T16:26:19Z <p>The title says it all. </p> <p>A very similar question was asked and answered about linear groups, but none of the counterexamples are algebraic: <a href="http://mathoverflow.net/questions/22814/are-extensions-of-linear-groups-linear" rel="nofollow">http://mathoverflow.net/questions/22814/are-extensions-of-linear-groups-linear</a></p> <p>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?</p> http://mathoverflow.net/questions/109456/are-extensions-of-linear-algebraic-groups-over-a-field-themselves-linear-algebr/109462#109462 Answer by Angelo for Are extensions of linear algebraic groups (over a field) themselves linear algebraic? Angelo 2012-10-12T15:07:34Z 2012-10-12T16:26:19Z <p>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 $A$.</p> <p>: Sorry I had not noticed grp's comment, or I wouldn't have posted an answer.</p> <p>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 $C$.</p> <p>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.</p>