I am trying to prove that a finite group scheme over a field is killed by the order. I think I can do the case where the field $k$ is perfect, here's a poorly written synopsis:
First, by definition a finite group scheme over a field is affine. For a finite group scheme $G = \text{Spec}A$ where $A$ is a $k$-algebra, which is finitely generated over $k$ as a module. It's clear to see that $G$ is just a disjoint union of points. Since $k$ is perfect, $G_\text{red}$ is actually a closed subgroup, and so we can assume WLOG that $G$ is reduced. Further, by base extension with $\overline{k}$, we can assume $k$ is algebraically closed. Now we can just use the fact that for any finite extension $K$ of $k$, $K\otimes_k \overline{k}$ is isomorphic to the sum of injections of $K$ into $\overline{k}$ to see that the orders of the group schemes are comparable, and so since $G$ is a finite group scheme over an algebraically closed field, it's a constant group scheme so the result follows from group theory.
Here we used twice that $k$ is perfect:
- Using $G_{red}$ as a subgroup
- In the base extension with $\overline{k}$
which is why I don't think this proof generalizes to the case where $k$ is not perfect.
I'm stuck trying to prove the case where $k$ is not perfect. I know that somehow I am meant to use Frobenius morphisms, but I am not quite sure how to do this. The only material that I can find online is regarding Deligne's theorem which as far as I can tell only applies to commutative group schemes. Any help or hints would be appreciated.
Best Answer
[Posting my comment as an answer, following OP's request.]
On the webpage of Ben Moonen, there are drafts about a book on Abelian Varieties. It also covers some general theory of group schemes. I found your problem as a guided exercise, though it does not give a written solution. The exercise may be found in Chapter 4, at page 22 of the .pdf, exercise 4.4. In fact, in proves the result over any reduced irreducible base scheme, under the assumption that the group scheme is finite locally free. It's equivalent to it being finite flat, with the further assumption that the base is locally noetherian (which of course it is, in the case of a field).
Here is the link to Ben Moonen's webpage
There must be a published reference with a written proof somewhere else, but I wasn't able to find it. If somebody knows about such a paper, please let us know in the comments.