Let ${A}$ be an abelian variety of dimension ${g}$ over a field ${k}$ of characteristic ${p}$. In the previous posts, we saw that to give a deformation ${R}$ of ${A}$ over a local artinian ring with residue field ${k}$ was the same as giving a continuous morphism of rings ${W(k)[[t_1, \dots, t_{g^2}]] \rightarrow A}$: that is, the local deformation space is smooth on ${g^2}$ parameters.

There is another description of the deformation space in terms of the ${p}$-divisible group, though. Given ${A}$, we can form the formal scheme

$\displaystyle A[p^\infty] = \varinjlim A[p^n],$

where each ${A[p^n]}$ is a finite group scheme of rank ${p^{2ng}}$ over ${k}$. As a formal scheme, this is smooth: that is, given a small extension ${R' \twoheadrightarrow R}$ in ${\mathrm{Art}_k}$ and a morphism ${\mathrm{Spec} R \rightarrow A[p^\infty]}$, there is an extension

In fact, we start by finding an extension ${\mathrm{Spec} R' \rightarrow A}$ (as ${A}$ is smooth), and then observe that this extension must land in some ${A[p^n]}$ if ${\mathrm{Spec} R}$ landed in ${A[p^\infty]}$.

As a result, we can talk about deformations of this formal (group) scheme.

Definition 16 deformation of a smooth, formal group scheme ${G}$ over ${k}$ over a ring ${R \in \mathrm{Art}_k}$ is a smooth, formal group scheme ${G'}$ over ${R}$ which reduces mod the maximal ideal (i.e., when restricted to ${k}$-algebras) to ${G}$.

Suppose ${G(k) = \ast}$. Using Schlessinger’s criterion, we find that (under appropriate finiteness hypotheses), ${G}$ must be prorepresentable by a power series ring ${R[[t_1, \dots, t_n]]}$ for some ${n}$. In other words, ${G}$ corresponds to a formal group over ${R}$.

Suppose for instance that ${A}$ was a supersingular elliptic curve. By definition, each of the ${A[p^n]}$ is a thickening of the zero point, and consequently ${A[p^\infty]}$ is the formal completion ${\hat{A}}$ of ${A}$ at zero. This data is equivalent to the formal group of ${A}$, and as we just saw a deformation of this formal group over a ring ${R \in \mathrm{Art}_k}$ is a formal group over ${R}$ which reduces to the formal group of ${A}$ mod the maximal ideal.

The main result is:

Theorem 17 (Serre-Tate) Let ${R \in \mathrm{Art}_k}$. There is an equivalence of categories between abelian schemes over ${R}$ and pairs ${(A, G)}$ where ${A }$ is an abelian variety over ${k}$ and ${G}$ a deformation of ${A[p^\infty]}$ over ${R}$.

So, in particular, deformations of ${A}$ over ${R}$ are equivalent to deformations of ${A[p^\infty]}$ over ${R}$.

In this post, I’ll describe an argument due to Drinfeld for this result (which I learned from Katz’s article).

1. Lifting

The key step in Drinfeld’s argument is the following idea. Let ${R \in \mathrm{Art}_k}$ with maximal ideal ${\mathfrak{m}}$, and let ${G}$ be a fppf sheaf of abelian groups on ${R}$-algebras, satisfying certain assumptions below. Let ${N = p^M }$ for ${M}$ sufficiently large (depending on ${R}$). The claim is that for any ${R}$-algebra ${R'}$, there exists a natural map of groups

$\displaystyle G(R'/\mathfrak{m}R') \stackrel{N''}{\rightarrow} G(R')$

which lifts the map ${G(R'/\mathfrak{m}R') \stackrel{N}{\rightarrow} G(R'/\mathfrak{m}R')}$. In other words, there is a functorial lift of multiplication by ${N}$ on the reduction mod ${\mathfrak{m}}$ to the thickened version. As we’ll see, this implies that any two deformations of an abelian variety are isogeneous (though not necessarily isomorphic).

We need the following assumptions:

1. ${G}$ is formally smooth (as a sheaf on ${R}$-algebras).
2. The “formal completion ${\hat{G}}$” defined via ${\hat{G}(A) = \ker( G(A) \rightarrow G(A^{red})}$ is a formal group scheme, represented by a formal group over ${k}$ (i.e., of the form ${\mathrm{Spf} k[[x_1, \dots, x_n]]}$).

This is satisfied by the sheaf of abelian groups associated to an abelian scheme over ${R}$, and ditto for a deformation of ${A_0[p^\infty]}$ if ${A_0}$ is an abelian variety over ${k}$.

Proposition 18 Let ${G: \mathrm{Alg}_R \rightarrow \mathbf{AbGrp}}$ be a functor satisfying the above two conditions (where ${R \in \mathrm{Art}_k}$ has maximal ideal ${\mathfrak{m}}$). Then, there exists an ${N = p^m}$ (depending on ${R}$) with the following property: for any ${R}$-algebra ${R'}$, there is a map of abelian groups

$\displaystyle G(R'/\mathfrak{m}R') \stackrel{N''}{\rightarrow} G(R')$

lifting multiplication by ${N}$ on ${G(R'/\mathfrak{m}R')}$. This is natural in homomorphisms of such functors.

Proof: The idea of the proof is simple: given ${x \in G(R'/\mathfrak{m}R')}$, we use formal smoothness to lift this to an element ${X \in G(R')}$, and then take ${NX \in G(R')}$ to be the required image. The whole point is that the choice of ${NX}$ doesn’t depend on the choice of ${X}$. In fact, given two choices ${X, X' \in G(R')}$ lifting ${x}$, we find that

$\displaystyle X - X' \in \hat{G}(R').$

But now ${\hat{G}}$ is a formal Lie group, so ${X - X'}$ is represented by an element in coordinates ${(Y_1, \dots, Y_n)}$ where the ${Y_i \in \mathfrak{m} R'}$. Multiplication by ${N}$ now annihilates these if ${N}$ is appropriate, because ${R}$ is artinian and ${N \in \mathfrak{m}}$ (e.g. we could take ${N = p^{\mathrm{length} R}}$). $\Box$

Let’s now use this to see that any two deformations of an abelian variety ${A}$ over ${R \in \mathrm{Art}_k}$ are isogeneous. Let ${A_0, A_1}$ be two such. The claim is that there is a morphism of abelian schemes

$\displaystyle A_0 \rightarrow A_1$

which lifts multiplication by ${N}$ on ${A}$. For this we can work at the level of abelian sheaves for the flat topology, and then we define, for an ${R}$-algebra ${R'}$, the map

$\displaystyle A_0(R') \rightarrow A_0(R'/\mathfrak{m}R') \simeq A_1(R'/\mathfrak{m}R') \stackrel{N''}{\rightarrow} A_1(R')$

where ${N''}$ is the morphism of the previous lemma, and we note that ${A_0, A_1}$ are the same on ${R/\mathfrak{m}R}$-algebras (they are ${A}$). By construction, this is a lifting of multiplication by ${N}$ on ${A}$, and is consequently an isogeny.

Note that this morphism is necessarily unique among lifts of multiplication by ${N}$. We could see this by the rigidity lemma, or we could note that two lifts ${f, g: A_0 \rightarrow A_1}$ would have their difference ${f-g: A_0 \rightarrow \hat{A_1}}$. Since multiplication by ${N}$ is a surjection on ${A_0}$ (when considered as a fppf sheaf on ${R}$-algebras) while multiplication by ${N}$ is zero on ${\hat{A_1}}$, it follows that ${f-g}$ must be zero.

In fact, we can state:

Lemma 19 Let ${A, B: \mathrm{Alg}_R \rightarrow \mathbf{AbGrp}}$ be two fppf abelian sheaves satisfying the two conditions above (i.e., formally smooth with formal completions represented by a formal Lie group). Suppose multiplication by ${N}$ is a surjection on each, moreover. Then reduction mod ${\mathfrak{m}}$ gives an injection

$\displaystyle \hom(A, B) \rightarrow \hom(A \times_{\mathrm{Spec} R} \mathrm{Spec} k, B \times_{\mathrm{Spec} R} \mathrm{Spec} k)$

and, moreover, there is a map ${N'': \hom(A \times_{\mathrm{Spec} R} \mathrm{Spec} k, B \times_{\mathrm{Spec} R} \mathrm{Spec} k) \rightarrow \hom(A, B)}$ lifting multiplication by ${N}$: if ${f: A \times_{\mathrm{Spec} R} \mathrm{Spec} k \rightarrow B \times_{\mathrm{Spec} R} \mathrm{Spec} k}$, ${N''f}$ lifts ${Nf}$.

In other words, we cannot necessarily lift a morphism ${A \times_{\mathrm{Spec} R} \mathrm{Spec} k \rightarrow B \times_{\mathrm{Spec} R} \mathrm{Spec} k}$ to a morphism ${A \rightarrow B}$, but we can always lift ${N}$ times that. This lift is necessarily unique, because multiplication by ${N}$ is a surjection and reduction mod the maximal ideal is injective.

Proof: This is now proved in exactly the same way as for abelian schemes. Given an ${R}$-algebra ${R'}$ and ${f: A \times_{\mathrm{Spec} R} \mathrm{Spec} k \rightarrow B \times_{\mathrm{Spec} R} \mathrm{Spec} k}$, we define

$\displaystyle A(R') \rightarrow A(R'/\mathfrak{m}R') \stackrel{f}{\rightarrow} B(R'/\mathfrak{m}R') \stackrel{N''}{\rightarrow} B(R'),$

where we use the multiplication by ${N}$ lift ${B(R'/\mathfrak{m}R') \rightarrow B(R')}$ mentioned earlier. $\Box$

2. Full faithfulness

Now, we can prove the Serre-Tate theorem. The theorem states that a certain forgetful functor is an equivalence of categories between abelian schemes over ${R}$ and abelian schemes over ${k}$ together with a deformation of their ${p}$-divisible group. In order to see this, we’ll have to show that it is fully faithful and essentially surjective.

Let’s start by showing that it is fully faithful.

Lemma 20 Given abelian schemes ${A, B}$ over ${R \in \mathrm{Art}_k}$ with ${k}$-reductions ${A_0, B_0}$, a morphism ${f_0: A_0 \rightarrow B_0}$ of abelian varieties, and a lift ${f': A[p^\infty] \rightarrow B[p^\infty]}$ lifting ${f_0: A_0[p^\infty] \rightarrow B_0[p^\infty]}$, then there is a unique morphism ${f: A \rightarrow B}$ of abelian ${R}$-schemes lifting these.

Proof: By the previous analysis, we know that there is a morphism of abelian schemes

$\displaystyle N''f : A \rightarrow B$

which lifts ${Nf_0: A_0 \rightarrow B_0}$, and which is unique with this property. We need to show that ${N''f}$ can be “divided by ${N}$,” which amounts (by descent theory) to saying that it annihilates the kernel ${A[N] \subset A}$ of multiplication by ${N}$.

To see this, note that this morphism ${N''f}$ induces a map ${A[p^\infty] \rightarrow B[p^\infty]}$ which lifts ${N f_0 : A_0[p^\infty] \rightarrow B_0[p^\infty]}$. There is another morphism which has this property: namely, ${N f'}$ (here this is honest multiplication by ${N}$!). By the uniqueness assertions of the previous analysis, we find that

$\displaystyle N'' f = N f': A[p^\infty] \rightarrow B[p^\infty]$

because both lift the same morphism ${A_0[p^\infty] \rightarrow B_0[p^\infty]}$. Anyway, the conclusion is that ${N''f}$ is actually (when restricted to the ${p}$-divisible group) a multiple of ${N}$, which means that it has to annihilate ${A[N]}$. It now follows that ${N''f}$ itself can be divided by ${N}$, which means that we can lift ${f_0}$ and ${f'}$. $\Box$

3. Essential surjectivity

The last step in the proof of the Serre-Tate theorem is the essential surjectivity. That is, given an abelian variety ${A_0/k}$, and a deformation ${G}$ of ${A_0[p^\infty]}$ over ${R}$, we need to produce an abelian scheme over ${R}$ lifting all this data. The first step is to lift ${A_0}$ itself over ${R}$ to an abelian scheme: we’ve seen that deformations of abelian varieties are unobstructed, so this is not an issue. So, we get a lifting ${A/R}$, which induces a new lifting ${A[p^\infty]}$ of the ${p}$-divisible group ${A_0[p^\infty]}$. This is not the same as ${G}$ most likely, but we can use the same analysis to see that it is isogeneous to ${G}$.

In fact, we have a map

$\displaystyle \phi_0: A_0[p^\infty] \simeq G_0[p^\infty]$

and this induces a lift of ${N\phi_0}$,

$\displaystyle N''\phi : A[p^\infty] \rightarrow G .$

(Technically, the ${\phi}$ should be in quotes as well, since it does not make sense.) Similarly, we can lift ${\phi_0^{-1}}$ to a map ${N'' \phi^{-1}: G \rightarrow A[p^\infty]}$, and the composite either way must be ${N^{2}}$ (by uniqueness). It follows now that the map ${N''\phi}$ is an isogeny, and the kernel ${\ker N \phi''}$ is a finite (indeed, flat) subgroup scheme ${K \subset A[p^\infty]}$.

Now we consider the quotient abelian scheme ${A/K}$. This comes with a map ${(A/K)[p^\infty] \stackrel{N''\phi}{\simeq}G}$, by construction, and the claim is that it also lifts ${A_0}$; in fact, we have a map

$\displaystyle (A/K)_0 \stackrel{N}{\rightarrow} A,$

which is an isomorphism, as ${K}$ lifts the subgroup scheme ${A_0[N] \subset A_0}$. In particular, we find that the quotient ${A/K}$ is a deformation of ${A_0}$ with the appropriate ${p}$-divisible group ${G}$.