In the category of groups, it is elementary that all central extensions of a cyclic group are abelian. Is the same true, in the category of (finite?) group schemes over a field $k$, for central extensions of the group $\mu_n$ of $n$th roots of unity?

$\begingroup$ Unless I'm missing some subtlety, this should be true by using the functor of points. That is, you will get an exact sequence $1\rightarrow A(S) \rightarrow B(S) \rightarrow C(S)$ for every $k$scheme $S$. Now, any subgroup of a cyclic group is cyclic, so what you want should follow from the usual fact. $\endgroup$– Minhyong KimAug 23, 2011 at 1:12

$\begingroup$ @Minhyong: It seems that $\mu_n(S)$ is not cyclic for general $S$. I don't quite follow your argument, so I don't know if this is a problem. $\endgroup$– S. Carnahan ♦Aug 23, 2011 at 2:51

$\begingroup$ Scott: You're probably right. The argument is OK when the characteristic of $k$ doesn't divide $n$, but I suppose all the fun is when it does. I'll think about it some more. $\endgroup$– Minhyong KimAug 23, 2011 at 3:10

1$\begingroup$ In the additive case there are interesting central extensions of $G_a$ over a finite field, the "fake Heisenberg groups" whose representation theory is a motivating example in the works of Boyarchenko and Drinfeld. $\endgroup$– David BenZviAug 23, 2011 at 3:40

1$\begingroup$ Heh. I see my kneejerk response was very naive. I will leave the comment up anyways, so others can learn from my silliness. $\endgroup$– Minhyong KimAug 23, 2011 at 6:06
1 Answer
If we have a central extension of group schemes $1\rightarrow B \rightarrow C\rightarrow A\rightarrow1$ with $A$ abelian, then we get a commutator mapping $\Lambda^2A\rightarrow B$ (of sheaves as $\Lambda^2A$ in general is not a group scheme) and the extension is abelian precisely when this map is zero. Hence for an nonabelian extension to exist there must be a nonzero map $\Lambda^2A\rightarrow B$. Let us now assume that $A=\mu_n$ and consider first the case when $n=p$, the characteristic of the field $k$ (which we may assume is algebraically closed). A nonzero map $\Lambda^2A\rightarrow B$ would give a nonzero map $A\rightarrow\mathrm{Hom}(A,B)$, where the right hand side is the sheaf of group homomorphisms. As the Frobenius map is zero on $\mu_p$ we may replace $B$ by its Frobenius kernel so we may assume that $B$ is either $\mu_p$ or the Cartier dual of $\alpha_{p^m}$. Now, as sheaves $\mathrm{Hom}(A,B)$ is isomorphic to $\mathrm{Hom}(D(B),D(A))$, where $D()$ denotes the Cartier dual. However $D(\mu_p)=\mathbb Z/p$ so when $A=\mu_p$ we get that $\mathrm{Hom}(D(B),D(A))=\mathbb Z/p$ and there is only the zero map from $A=\mu_p$ into it. In the other case $D(A)=\alpha_{p^m}$ and $\mathrm{Hom}(\alpha_{p^m},\mathbb Z/p)$ is zero. If instead $n=p^k$, the argument is the same. The case when $n=\ell^k$ is even simpler so in all cases all possible commutator maps are zero and the extension is commutative.
(When $A=B=\mathbb G_a$ then there are candidates for commutator maps and in fact $(a,b)(a',b')=(a+a',b+b'+a^pa')$ gives a noncommutative central extension which I imagine is the fake Heisenberg group.)
Addendum: A general comment is that it is more convenient to work with sheaves (in the fppf topology say) as that means that we essentially can pretend that we work with settheoretic groups. It is however also necessary if we want to see the commutator map as a map $\Lambda^2A\to B$ as the sheaf $\Lambda^2A$ (of $A$ considered as an abelian sheaf) is in general not reprsentable. The $\langle,\rangle\colon\Lambda^2A\to B$ view point is convenient as it allows us to do what one usually does when having a pairing: We get for instance a map $A\to\mathrm{Hom}(A,B)$ given by $a\mapsto (a'\mapsto \langle a,a'\rangle)$ just from the fact that $\langle,\rangle$ is biadditive.
I have implicitly assumed that $B$ is of finite type (as I claim that its Frobenius kernel is finite) even though it may not be necessary (a limit argument anyone?).

$\begingroup$ "Now, as sheaves Hom(A,B) is isomorphic to Hom(D(A),D(B))". Shouldn't that be Hom(D(B),D(A))? $\endgroup$ Aug 23, 2011 at 8:05


$\begingroup$ [There's still one Hom(D(A),D(B)) left.] $\endgroup$ Aug 23, 2011 at 20:11

$\begingroup$ Hmmm... I don't know what $\Lambda^2$ of a group scheme is, but I suppose I don't have to, as I certainly agree that the commutator will determine a morphism $A \to \mbox{Hom}(A,B)$. Also, this proof seems to assume $B$ is finite? For otherwise, even after Frobenius, couldn't $B$ be something humongous? Anyway, that's the only case I care about. Many thanks!! $\endgroup$ Aug 23, 2011 at 23:23

$\begingroup$ @Kevin: Sorry for the sloppy first editing. @Michael: I have added some comments which hopefully clarifies. $\endgroup$ Aug 24, 2011 at 4:29