Does there exist a group scheme of finite type over a field $k$ , which is neither affine, nor an abelian variety?

$\begingroup$ Related entry: mathoverflow.net/questions/43529/… $\endgroup$– Zoran SkodaMay 18, 2011 at 13:22

$\begingroup$ As my answer indicates, it's unclear precisely what you mean by your parenthetic statement. Being an affine group scheme doesn't imply having a realization as a matrix group. Certainly you want to consider the motivating examples mentioned in Brian Conrad's cited paper. $\endgroup$– Jim HumphreysMay 18, 2011 at 20:26
3 Answers
Yes, take a product of an abelian variety with an affine group scheme...

7$\begingroup$ But at least, we have something close to that. See en.wikipedia.org/wiki/Chevalley_theorem. $\endgroup$ May 18, 2011 at 10:18

1$\begingroup$ The ground field is required to be perfect in the Chevalley theorem. $\endgroup$ May 18, 2011 at 13:21
For me the most natural examples to think about are those coming from linear algebraic groups in prime characteristic, in the form of Frobenius kernels. These are extensively treated in Jantzen's book Representations of Algebraic Groups, with foundations laid in Part I. Concretely, you can start with the restricted universal enveloping algebra of the Lie algebra of such a group and associate to it naturally a finite group scheme. This is only the "first" Frobenius kernel.
Examples of this kind are "affine" but not linear algebraic groups.
Perhaps more in the spirit of the question, natural examples occur in connection with Chevalley's classical structure theorem: for a modern treatment see B. Conrad, "A modern proof of Chevalley's theorem on algebraic groups", J. Ramanujan Math. Soc. 17 (2002).
The type of Jacobian (generalised) discussed in Serre's book "Groupes algébriques et corps de classes" provides a large class of examples. They arise, in a sense, from the universality of the Albanese variety when extended to nonregular differentials.