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This question is immediately related to Discriminant ideal in a member of Barsotti-Tate Group dealing with Barsotti–Tate groups and here I would like to clarify a proof presented by Anonymous in the comments from linked thread which I do not completely understand. Although meanwhile I found another proofs of the claim below I have a big interest on understanding this proof below.

Assume $G$ is a finite flat commutative group scheme over a field $k$ of order $p^N$. Assume $p$ prime and $p \in k \setminus \{0\}$, equivalently invertible on the base.

Claim: Any finite flat commutative group scheme of $p$-power order is etale if $p$ is invertible on the base.

Anonymous' proof works as follows: Firstly we reduce to case over a field (because a finitely presented flat map is etale if it is fibrewise etale). Since we assumed $G$ commutative the multiplication by $p^N$ map $f_{p^N}: G \to G$ is well defined and by Deligne's theorem $p^N$ kills $G$ since it's the order of $G$. That means that $f_{p^N}$ is the zero map: equivalently it factorizes over $\operatorname{Spec}(k)$.

What now comes I do not understand:

It is claimed that $f_{p^N}$ is unramified "as the map on the tangent spaces is given by $p^N$, which is invertible".

Question I: why is the induced by $f_{p^N}$ maps on tangent spaces given by $p^N$?

Question II: assume we understand Question I. Why does this imply $G$ is unramified?

when we can answer these two question we are done because unramified finite type schemes over a field are etale.

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    $\begingroup$ You don't need Deligne's theorem to prove this. For the standard proof, see, for example, 11.31 of Milne, Algebraic Groups, CUP, 2017. $\endgroup$
    – anon
    Jun 24, 2020 at 2:57
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    $\begingroup$ Nor do you need to assume it's commutative. Another reference is Waterhouse Introduction to affine group schemes, §11.4, §14.4. $\endgroup$ Jun 24, 2020 at 3:29
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    $\begingroup$ But if you insist on this argument, I follows since the derivative of $\mu \colon G \times G \to G$ is addition $T_e G \times T_e G \to T_e G$ (use $\mu \circ (\operatorname{id} \times e) = \operatorname{id} = \mu \circ (e \times \operatorname{id})$ where $e \colon \operatorname{Spec} k \to G$ is the unit). II is a near immediate consequence of the definitions (see e.g. Tags 00UT and 00RS). $\endgroup$ Jun 24, 2020 at 3:39
  • $\begingroup$ I finished the title (which ended mid-sentence), and added a link to the comment I think that you meant. If I got the wrong comment, please fix it or let me know. $\endgroup$
    – LSpice
    Jun 24, 2020 at 3:58
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    $\begingroup$ @LSpice: you linked the correct comment $\endgroup$
    – user267839
    Jun 24, 2020 at 12:43

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