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Recall Deligne's theorem that for a finite flat commutative group scheme $G$ of order $n$, the multiplication by $n$ map $[n]: G \to G$ is the zero map.

I have seen the proof a few times but I can't really say I understand the idea behind it. In particular:

1) What is the general idea/strategy behind the proof?

2) Where does the proof break down in the non commutative case?

3) How much do we know about generalizations? Are there conditions weaker than non commutativity under which we can prove it?

4) Related to 2 - what is the conceptual reason for why the non commutative case is so hard? It is after all a very innocuous sounding statement but to be unproven after so many years, I guess there must be something interesting going on.

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    $\begingroup$ I think the commutativity is used for Cartier duality. Using this, one constructs over local rings, where $G$ is finite free, a trace map, see the exposition in [Tate-Oort], Group schemes of prime order. In [Schoof], Finite flat group schemes over local Artin rings, there are results over certain non-reduced Artinian rings. $\endgroup$
    – user19475
    Nov 10, 2018 at 6:01

1 Answer 1

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This is just expanding on user19475's comment, and is taken entirely from Schoof's notes, but as this question is the first thing to come up when googling, I thought it might be helpful to give my novice's interpretation of the key points of the proof.

First, by changing the base if needed, we are reduced to showing that the $m$th power map is trivial on the $R$ valued points of $G$, where $G=Spec(A)$ is a finite, free, affine group scheme of rank $m$ over $Spec(R)$.

Then, using Cartier duality, we may identify $G(R)\subset A^{\vee}$, since:$$G(R)=\text{Hom}_{R-Alg}(A,R)\subset Hom_{R-Mod}(A,R)=A^{\vee}$$ Further unravelling definitions, the points of $A^{\vee}$ that belong to $G(R)$ are exactly those $x$ which satisfy $$\Delta(x)=x\otimes x$$ For $\Delta$ the comultiplication on $A^{\vee}$.

The key tool now is the norm map, for any finite, free $R$ algebra $S$, we have the norm map $N:S^*\rightarrow R^*$ on units, given by $x\mapsto \det(\mu_x)$ where $\mu_x$ is the $R$ linear map of multiplication by $x$ on $S$ as an free $R$ module.

The core technical part of the proof is that for any $R\rightarrow S$, the norm map from $(A^{\vee}\otimes S)^*\rightarrow (A^{\vee})^*$ takes $G(S)$ into $G(R)$, and when we set $S=A$, that this norm map is invariant under the action of the translation automorphism $\tau(\alpha)$ on $G(A)$ for any $\alpha$ in $G(R)$.

Once we have these two compatibilities, we simply verify that for the element $\text{Id}_A$ in $G(A)$, we have $$N(\text{Id}_A)=N(\tau(\alpha)\text{Id}_A)=N(\alpha)N(\text{Id}_A)=\alpha^m N(\text{Id}_A)$$ Which gives the result.

It appears to me that in the proof, there doesn't seem to be any specific place where commutativity is required, as we are really only looking to prove the assertion at a single element of $G(R)$. The commutativity is important insofar as it provides the core tool used for the proof, the norm map, via the interpretion of points of $G(R)$ as elements of the Cartier dual.

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