# More on finite group schemes

In the last post, we discussed finite flat commutative group schemes (henceforth, ‘finite group schemes’ or even ‘finite groups’) and made some basic constructions. Before we get to the main objects of interest, ${p}$-divisible groups, I want to say a few more words about the structure of finite group schemes in characteristic ${p}$. The key point, which will be increasingly hammered into your brain as this series of posts continues, is that everything is controlled by the Frobenius and Verschiebung maps. Today, as a first step in this argument, I’ll discuss a few basic examples of finite group schemes, as well as what happens when you dualize the connected-étale exact sequence from last time.

By the way, I didn’t mention sources in my last post. I’ve found the following extremely helpful:

• These notes for material on finite group schemes and Witt vectors.
• Demazure’s Lectures on p-divisible groups, which has an interesting perspective on formal groups and proves the classification theorem.
• Tate’s paper on ${p}$-divisible groups, which proves the interesting result that ${p}$-divisible groups over integrally closed domains are controlled by their behavior over the general fiber (that is, the field of fractions). This is short but gives a good summary of things one should understand about ${p}$-divisible groups. I’d go through the result here if I understood the requisite class field theory.
• Messing’s The Crystals Associated to Barsotti-Tate groups; with Applications to Abelian Schemes. This is higher-level than the other sources. It proves deformation-theoretic results that might tie in to Lurie’s theorem — I haven’t looked at it in enough detail yet.

Let ${k}$ be a perfect field of characteristic ${p}$. Recall that ${\sigma}$ denotes the ‘absolute’ Frobenius map ${k \rightarrow k}$ (I’ll also write ${\sigma:\mathrm{Spec} k \rightarrow \mathrm{Spec} k}$), which in the case of ${k}$ perfect is an isomorphism. We defined ${X^{(p)}}$ to be the base change of a ${k}$-scheme ${X}$ along ${\sigma}$; the structure map ${X \rightarrow k}$ and ${\sigma_X:X \rightarrow X}$ then induce a relative Frobenius map ${F_X:X \rightarrow X^{(p)}}$ by the universal property of the pullback. This exists for any scheme. If ${X}$ is additionally a commutative group scheme, we get a Verschiebung map ${V_X:X^{(p)} \rightarrow X}$, which can define either by dualizing ${F}$, or by looking at the Hopf algebra ${A}$ of ${X}$, applying ${\Delta^p:A^{(p)} \rightarrow ((A^{(p)})^{\otimes p})^\Sigma_p}$, and mapping this to ${A}$ via ${(a \otimes \dotsb \otimes a) \mapsto a}$.

When ${k}$ is perfect, we can identify ${X^{(p)}}$ with ${X}$ via the isomorphism ${\sigma_X}$. This is not a map of ${k}$-schemes, though — the structure map has changed! To deal with this sort of situation, we define a map of ${k}$-modules ${f:M \rightarrow N}$ to be ${\sigma}$-linear if it satisfies

$\displaystyle f(\lambda m) = \sigma(\lambda)f(m),\, \lambda \in k,\, m \in M.$

There’s an obvious scheme version of this. Under this point of view, ${F}$ and ${V}$ are respectively ${\sigma}$-linear and ${\sigma^{-1}}$-linear maps ${X \rightarrow X}$. In particular, we can iterate them and get maps ${F^n, V^n:X \rightarrow X}$, which are respectively ${\sigma^n}$ and ${\sigma^{-n}}$-linear.

Okay, now let ${G = \mathrm{Spec} A}$ be a finite group scheme over a perfect field ${k}$ of characteristic ${p}$. Splitting ${G}$ into its connected components gives a splitting ${A = \prod A_i}$ as algebras, with each ${A_i}$ local and Artinian; the Frobenius morphism acting as ${x \mapsto x^p}$ (and depending only on the algebra structure), it’s easy to see that ${\bigcup \ker F^n_A = A^0}$, the connected component of the identity. So the Frobenius is nilpotent on ${G^0}$ and an isomorphism on ${G^{et}}$. Another way to think about this last clause is to note that ${G^{et} \otimes_k \overline{k} = \mathrm{Spec} \prod_\Gamma \overline{k}}$ for some abelian group ${\Gamma}$, on which the Frobenius map is evidently an isomorphism.

To say the same thing about the Verschiebung map, it’s useful to have a few new words.

Definition 1 A finite group scheme is multiplicative if it’s the Cartier dual of an étale group scheme. It is unipotent if it’s the Cartier dual of a connected group scheme.

Then by dualizing the above, we get the following.

Theorem 2 Let ${G}$ be a finite group scheme of ${p}$-power order over a perfect field ${k}$ of characteristic ${p}$, ${F}$ and ${V}$ its Frobenius and Verschiebung maps. The following equivalences hold:

• ${G}$ is connected iff ${F}$ is nilpotent.
• ${G}$ is étale iff ${F}$ is an isomorphism.
• ${G}$ is unipotent iff ${\bigcap_n \mathrm{Im}\, V^n = 0}$.
• ${G}$ is multiplicative iff ${V}$ is an isomorphism.

Moreover, there is a natural decomposition

$\displaystyle G \cong G_{cu} \times G_{eu} \times G_{cm}$

with the subscripts ${c}$, ${e}$, ${u}$, ${m}$ respectively indicating that the group is connected, étale, unipotent, or multiplicative.

Note that if a group is étale and multiplicative, then ${F}$ and ${V}$ are both isomorphisms, so ${[p]}$ is an isomorphism — thus, this can’t happen for finite group schemes of ${p}$-power order. (On the other hand, over a field of characteristic zero, it’s easily checked that every group is étale and multiplicative.)

Let’s take a look at the three basic group schemes of order ${p}$:

$\displaystyle \mu_p = \mathrm{Spec} k[X]/(X^p - 1) = \ker ([p]:\mathbb{G}_m \rightarrow \mathbb{G}_m)$

$\displaystyle \alpha_p = \mathrm{Spec} k[X]/(X^p) = \ker (F:\mathbb{G}_a \rightarrow \mathbb{G}_a)$

$\displaystyle \underline{{\mathbb Z}/p{\mathbb Z}} = \mathrm{Spec} \prod_{{\mathbb Z}/p{\mathbb Z}} k.$

Looking at the algebras, it’s clear that ${F = 0}$ on the first two and ${F = 1}$ on the last one. It follows that ${V = 0}$ on ${\underline{{\mathbb Z}/p{\mathbb Z}}}$. For ${\mu_p}$, we have

$\displaystyle \Delta^p(X) = X \otimes \dotsb \otimes X,$

which is sent to ${X}$ under the map ${(A^\otimes p)^{\Sigma_p} \rightarrow A}$ discussed last time; thus, ${V = 1}$ here. For ${\alpha_p}$,

$\displaystyle \Delta^p(X) = X \otimes 1 \otimes \dotsb \otimes 1 + \dotsb + 1 \otimes \dotsb \otimes 1 \otimes X,$

which I leave it to you to show is sent to ${0}$ under the analogous map, showing that ${V = 0}$. Thus, ${\underline{{\mathbb Z}/p{\mathbb Z}}}$ is étale and unipotent, ${\mu_p}$ is connected and multiplicative, and ${\alpha_p}$ is connected and unipotent. In fact, ${\mu_p}$ and ${\underline{{\mathbb Z}/p{\mathbb Z}}}$ are Cartier dual to each other, and ${\alpha_p}$ is self-dual.

This goes some way towards explaining the word ‘multiplicative’. In fact, it’s more generally true that multiplicative group schemes (over any base) Zariski-locally embed into a group that étale-locally looks like ${\mathbb{G}_m^n}$ (a ‘torus’). This is Proposition 1.3.1 in Messing’s book. Another good way to characterize these groups is to note that the dual of the constant Hopf algebra ${k^\Gamma}$ is the group algebra ${k[\Gamma]}$; thus, every multiplicative group is étale-locally of the form ${\mathrm{Spec} k[\Gamma]}$ (and again, this is true for arbitrary base schemes).

A third way to think about multiplicative groups is to note that the Hopf algebra of a multiplicative group has no primitives. To elaborate, recall that an element ${x}$ of a Hopf algebra is called primitive if ${\Delta(x) = 1 \otimes x + x \otimes 1}$ and ${\epsilon(x) = 0}$. Such elements correspond to maps of group schemes ${G \rightarrow \mathbb{G}_a}$, or dually, maps of ${k}$-schemes ${\mathrm{Spec} k[T]/T^2 \rightarrow G^\vee}$. But ${G^\vee}$ is étale if and only if every such map is trivial. Thus, ${G}$ is multiplicative if and only if it has no nontrivial homomorphisms to ${\mathbb{G}_a}$.

I’m not really sure about the etymology of the word ‘unipotent’. A better word might be ‘additive’, as any unipotent group of finite type over ${k}$ is in fact an extension of subgroups of ${\mathbb{G}_a}$ (of which both ${\alpha_p}$ and ${\underline{{\mathbb Z}/p{\mathbb Z}} = \mathrm{Spec} k[X]/X(X-1)\dotsm (X-p+1)}$ are examples). To see this, observe that ${G}$ is connected iff it has no nontrivial étale quotients; dualizing this, ${G}$ is unipotent iff it has no nontrivial multiplicative subgroups. By the previous paragraph, this means every subgroup has a nontrivial map to ${\mathbb{G}_a}$. The result follows by induction on the order of ${G}$.

By way of conclusion, here’s why I call ${\alpha_p}$, ${\mu_p}$, and ${\underline{{\mathbb Z}/p{\mathbb Z}}}$ the ‘three basic group schemes of order ${p}$.’

Proposition 3 Over an algebraically closed field ${k}$ of chracteristic ${p}$, every finite group scheme of ${p}$-power order is an extension of copies of ${\alpha_p}$, ${\mu_p}$, and ${\underline{{\mathbb Z}/p{\mathbb Z}}}$.

Proof: The field ${k}$ is perfect, so we can consider the separate cases of connected-unipotent, étale-unipotent, and connected-multiplicative groups. Any étale group is constant, and thus corresponds to a finite abelian group of ${p}$-power order, which is an extension of cyclic groups of order ${p}$; the constant group scheme functor is exact, so we can automatically express this constant group scheme as an extension of copies of ${\underline{{\mathbb Z}/p{\mathbb Z}}}$.

Any connected multiplicative group is dual to an étale unipotent group. The duality functor is exact and the dual of ${\underline{{\mathbb Z}/p{\mathbb Z}}}$ is ${\mu_p}$, so any connected multiplicative group is an extension of copies of ${\mu_p}$.

Any connected unipotent group is a subgroup of ${\mathbb{G}_a}$, which is Spec of a Hopf algebra quotient of the additive Hopf algebra ${k[T]}$. We didn’t really talk Hopf algebra quotients in a whole lot of detail, but they’re not terribly hard to do: if ${A' \subseteq A}$ is an augmented subalgebra that’s closed under comultiplication, the Hopf algebra quotient is ${A \otimes_{A'} k}$, with its induced comultiplication and augmentation. Since we’re talking connected subgroups, the only subalgebras allowed are those of the form ${k \oplus (T^n)}$ for some ${n}$, giving a quotient ${k[T]/(T^n)}$. So the question is which ideals ${(T^n)}$ are closed under comultiplication. Well,

$\displaystyle \Delta(T^n) = T^n \otimes 1 + nT^{n-1} \otimes T + \dotsb + 1 \otimes T^n,$

so for this to be in ${(T^n \otimes 1, 1 \otimes T^n)}$, all terms but the first and last must vanish, giving ${n = p^r}$ and ${G = \alpha_{p^r}}$. Obviously, this has ${\alpha_p}$ as a subgroup, and by induction, it’s an extension of copies of ${\alpha_p}$. $\Box$

In the next post, I’ll define ${p}$-divisible groups.