This is a continuation of the project outlined in this post yesterday of describing Grothendieck’s proof that the fundamental group of a smooth curve in characteristic $p$ has $2g$ topological generators (where $g$ is the genus). The first step, as I explained there, is to show that one may “lift” such smooth curves to characteristic zero, in order that a comparison may be made between the characteristic $p$ curve and something much more concrete in characteristic zero, that we can approach via topological methods. This post will be devoted to showing that such a lifting is always possible.

1. Introduction

It is a general question of when one can “lift” varieties in characteristic ${p}$ to characteristic zero. Doing so often allows one to bring in transcendental techniques (to the lift), as it will in this case of ${\pi_1}$. Let us thus be formal:

Definition 4 Let ${X_0}$ be a proper, smooth scheme of finite type over a field ${k}$ of characteristic ${p}$. We say that a lifting of ${X_0}$ is the data of a DVR ${A}$ of characteristic zero with residue field ${k}$, and a proper, smooth morphism ${X \rightarrow \mathrm{Spec} A}$ whose special fiber is isomorphic to ${X_0}$.

There are obstructions that can prevent one from making such a lifting. One example is given by étale cohomology. A combination of the so-called proper and smooth base change theorems implies that, in such a situation, the cohomology of the special fiber and the cohomology of the general fiber, with coefficients in any finite group without ${p}$-torsion, are isomorphic. As a result, if there is something funny in the étale cohomology of ${X_0}$, it might not be liftable. See this MO question.

In the case of curves, fortunately, it turns out there are no such problems, but still actually lifting one will take some work. We aim to prove:

Theorem 5 Let ${X_0 \rightarrow \mathrm{Spec} k}$ be a smooth, proper curve of finite type over the field ${k}$ of characteristic ${p}$. Then if ${A}$ is any complete DVR of characteristic zero with residue field ${k}$, there is a smooth lifting ${X \rightarrow \mathrm{Spec} A}$ of ${X_0}$.

One should, of course, actually check that such a complete DVR does exist. But this is a general piece of algebra, found for instance in Serre’s Local Fields.

The reason there won’t be any obstructions in the case of curves is that they are of dimension one, but we’ll see that the cohomological obstructions to lifting all live in ${H^2}$.

The strategy, in fact, will be to lift ${X_0 \rightarrow \mathrm{Spec} k}$ to a sequence of smooth schemes ${X_n \rightarrow \mathrm{Spec} A/\mathfrak{m}^n}$ (where ${\mathfrak{m} \subset A}$ is the maximal ideal) that each lift each other, using the local nilpotent lifting property of smooth morphisms.

This family ${\left\{X_n \rightarrow \mathrm{Spec} A/\mathfrak{m}^n\right\}}$ is an example of a so-called formal scheme, which for our purposes is just such a compatible sequence of liftings. Obviously any scheme ${X \rightarrow \mathrm{Spec} A}$ gives rise to a formal scheme (take the base-changes to ${\mathrm{Spec} A/\mathfrak{m}^n}$), but it is actually nontrivial (i.e., not always true) to show that a formal scheme is indeed of this form. But we will be able to do this as well in the case of curves.

2. Local lifting of smooth schemes

Let us start by lifting the smooth curve ${X_0}$ to a sequence of schemes ${X_n \rightarrow \mathrm{Spec} A/\mathfrak{m}^n}$, following the program outlined earlier. It will be convenient to do this in a more general setting. Let ${S}$ be a base scheme, and let ${S_0 \subset S}$ be a subscheme defined by an ideal ${\mathcal{I}}$ of square zero. Suppose ${X_0 \rightarrow S_0}$ is a smooth scheme. We want to know if there is a smooth scheme ${X \rightarrow S}$ whose restriction to ${S_0}$ is ${S}$. In general, this need not exist, but the next result states that the smooth lifting does locally.

Hereafter, all schemes are noetherian.

Proposition 6 (Local lifting of smooth morphisms) Lifting ${X_0 \rightarrow S_0}$ to a smooth ${S}$-scheme can always be done locally. If ${x_0 \in X_0}$, there is a neighborhood ${U_0 \subset X_0}$ of ${x_0}$, a smooth scheme ${U \rightarrow S}$ such that ${U \times_S S_0 \simeq U_0}$.Moreover, any two liftings are locally ${S}$-isomorphic. If ${S}$ is affine, and ${U, U'}$ lift the open affine ${U_0}$, there is an ${S}$-isomorphism ${U \simeq U'}$.

Of course, it is not very deep to lift the schemes themselves: the composite ${X_0 \rightarrow S_0 \rightarrow S}$ would do. The point is to preserve essential properties (in this case, smoothness).

Proof: We are going to deduce it from the “équivalence remarkable de catégories” of Grothendieck, that states the following: if ${S_0 \subset S}$ is a closed subscheme defined by an ideal of square zero, then base-change gives an equivalence of categories between the collection of schemes étale over ${S}$ and the schemes étale over ${S_0}$. In other words, étale ${S_0}$-schemes can be lifted globally (and uniquely). For smooth morphisms the statement is weaker.

Indeed, we note that (by one characterization of smoothness) there is a neighborhood ${U_0 }$ of ${x_0}$ such that the map

$\displaystyle U_0 \rightarrow S_0$

factors as a composite

$\displaystyle U_0 \stackrel{g}{\rightarrow} \mathbb{A}^n_{S_0} \rightarrow S_0$

where ${g}$ is étale. Now ${U_0}$ then extends uniquely to a scheme ${U}$ étale over ${\mathbb{A}^n_S}$. This is the lifting we want.

Finally we need to show local uniqueness of the lifting. We will do this using the infinitesimal lifting property. Let ${U_0}$ be affine, ${U_0 = \mathrm{Spec} B_0}$. Suppose ${S}$ and ${S_0}$ are affine, without loss of generality, say ${S = \mathrm{Spec} A, S_0 = \mathrm{Spec} A_0 = \mathrm{Spec} A/I}$ where ${I \subset A}$ is an ideal of square zero. By hypothesis, we are given two smooth ${A}$-algebras ${B_a, B_b}$ (whose spectra are the two liftings ${U, U'}$), together with an isomorphism

$\displaystyle B_a/IB_a \simeq B_b/IB_b \simeq B_0.$

But here we use the infinitesimal lifting property of smooth morphisms. Namely, the map

$\displaystyle B_a \rightarrow B_b \otimes_A A/I$

can be lifted to an ${A}$-homomorphism ${B_a \rightarrow B_b}$ because ${B_a}$ is ${A}$-smooth. This homomorphism, moreover, induces the identity mod ${I}$ (when both are identified with ${B_0}$). The claim is that this is an isomorphism, which follows from the next lemma.

Lemma 7 Let ${A}$ be a noetherian ring. Let ${B, B'}$ be flat ${A}$-algebras, and let ${I \subset A}$ be an ideal of square zero. If a map ${f:B \rightarrow B'}$ is such that the reduction mod ${I}$, ${B \otimes_A A/I \rightarrow B' \otimes_A A/I}$, is an isomorphism, then ${f}$ is itself an isomorphism.

Proof: Indeed, we note that by flatness, ${(I/I^2) \otimes_A B = (I/I^2) \otimes_{A/I} (B/IB) = IB/I^2 B= IB}$. Similarly for ${B'}$. That is, flatness makes the associated graded behave nicely. But, again by flatness:

$\displaystyle I \otimes_A B = I B, \quad I \otimes_A B' = IB'$

and consequently the map ${IB\rightarrow IB'}$ is an isomorphism. In particular, the map ${B\rightarrow B'}$ induces an isomorphism on the associated gradeds of the ${I}$-adic filtration. Since ${I}$ is nilpotent, this gives the result. The point of the lemma is that ${B/IB, B'/IB'}$ determine the associated gradeds by flatness.

In the proof of the above result, we showed something more specific than just local unicity. When the base and the target are affine, then any two smooth liftings are isomorphic (noncanonically, in general).

3. Global lifting

So we can always lift smooth things locally. Of course, there will be lots of ways of doing that in general, and the question is whether we can patch them together. In the étale case, there are no nontrivial automorphisms of the lifting: that is, the functor of base-changing by ${S_0}$ is fully faithful (by the infinitesimal lifting property for étale morphisms). As a result, there is no problem patching local liftings.

For smooth morphisms, the problem is more delicate. We can always lift locally, but to patch the liftings one needs a “cocycle” condition, which is not automatic. As a result, there is a cohomological obstruction to lifting that comes from these automorphisms.

Let ${X_0}$ be a smooth ${S_0}$-scheme. Suppose ${S_0 \hookrightarrow S}$ as a subscheme defined by an ideal of square zero. We are going to define a sheaf ${\mathcal{F}}$ on ${X_0}$ as follows: for each ${U_0 \subset X_0}$, ${\mathcal{F}(U_0)}$ will denote the set of ${S}$-automorphisms ${U \rightarrow U}$ (where ${U}$ is the pre-image of ${U_0}$ in ${X}$) inducing the identity on ${U_0}$.

We shall use:

Lemma 8 The sheaf ${\mathcal{F}}$ is canonically isomorphic to a quasi-coherent sheaf on ${X_0}$, which is independent of the lifting ${X}$.

This is a remarkable statement. It is clear that ${\mathcal{F}}$ can be made into a sheaf of groups (since ${S_0 \subset S}$ is defined by an ideal of square zero, it is easy to see that an automorphism of ${U}$ restricts to an automorphism of ${V}$). In fact it is not even obvious a priori that this sheaf of groups is a sheaf of abelian groups.

Proof: To see this, we will start by assuming ${X_0, S_0}$ (and consequently ${X}$) affine. Say ${S_0 = \mathrm{Spec} A_0 = \mathrm{Spec} A/I, S = \mathrm{Spec} A, X = \mathrm{Spec} B, X_0 = \mathrm{Spec} B_0 = \mathrm{Spec} B/IB}$. Then we are looking for the set of ${A}$-homomorphisms

$\displaystyle f: B \rightarrow B$

that induce the identity ${B/IB \rightarrow B/IB}$. Such a map is necessarily of the form ${1 + \phi}$, where ${\phi: B \rightarrow IB}$ is an ${A}$-homomorphism. One requires, of course:

$\displaystyle (b + \phi(b))(b' + \phi(b')) = bb' + \phi(b b').$

Since ${I^2 = 0}$, this is equivalent to saying that ${\phi}$ is an ${A}$-derivation ${B \rightarrow IB}$.

One can check that the composite ${(1 + \phi) \circ (1 + \phi') = 1 + \phi + \phi' + \phi \circ \phi' = 1 + \phi + \phi'}$, as the composite of two ${A}$-derivations ${B \rightarrow IB}$, ${B \rightarrow IB}$ is always zero. As a result, the group of such automorphisms is isomorphic to the group of derivations.

Such derivations are classified by maps of ${B}$-modules

$\displaystyle \Omega_{B/A} \rightarrow IB.$

However, these are the same as maps of ${B_0}$-modules ${\Omega_{B_0/A_0} \rightarrow IB}$ because ${IB}$ is a ${B_0}$-module, as ${I^2 = 0}$. It follows that the sheaf in question is the sheaf

$\displaystyle \mathcal{H}om_{\mathcal{O}_{X_0}}(\Omega_{X_0/S_0}, \mathcal{I}\mathcal{O}_X)$

where ${\mathcal{I}}$ is the ideal of square zero cut that cuts out ${S_0}$. Note that ${\mathcal{I} \mathcal{O}_X}$ is an ${\mathcal{O}_{X_0}}$-module. This is coherent and completely independent of ${X}$: indeed, it is also isomorphic to

$\displaystyle T_{X_0/S_0} \otimes_{\mathcal{O}_{S_0}} \mathcal{I}_{S_0}$

where ${T_{X_0/S_0}}$ is the dual of the (locally free) cotangent sheaf and ${\mathcal{I}_{S_0}}$ is the ideal of ${S_0}$ in ${S}$. This depends on the embedding ${S_0 \rightarrow S}$ but not the lifting ${X}$.

Now let us try to analyze the situation we are ultimately interested in. Let ${X_0 \rightarrow S_0}$ be a smooth separated scheme, where ${S_0 \hookrightarrow S}$ is a subscheme defined by an ideal of square zero. For simplicity, let us assume ${S_0, S}$ affine, since that is the only case we shall need. We know that there is an open affine cover ${\left\{U_0^\alpha\right\}}$ of ${X_0}$ consisting of schemes that lift to smooth affine ${S}$-schemes ${\{U^\alpha \}}$ such that

$\displaystyle U_0^\alpha = U^\alpha \times_S S_0.$

Now, for each ${\alpha, \beta}$, we have two liftings of the affine scheme ${U_0^\alpha \cap U_0^\beta}$: namely, ${U^{\alpha, \beta} \subset U^\alpha}$ and ${U^{\beta, \alpha} \subset U^\beta}$.

By formal smoothness (namely, the infinitesimal lifting property), we have isomorphisms

$\displaystyle t^{\alpha \beta}: U^{\alpha, \beta} \simeq U^{\beta, \alpha}$

that induce the identity ${U_0^\alpha \cap U^0_\beta \rightarrow U_0^\alpha \rightarrow U_0^\beta}$. This is where we have used the affineness and separatedness hypotheses. The hope is that these would satisfy the cocycle condition, and that we could glue all the ${\left\{U^\alpha\right\}}$ together. Unfortunately, they needn’t.

So let’s recall the cocycle condition: for any three indices ${\alpha, \beta, \gamma}$, one must have

$\displaystyle t^{\beta \gamma} \circ t^{\alpha \beta} = t^{\alpha \gamma}: U^{\alpha \gamma} \cap U^{\alpha \beta} \rightarrow U^{\gamma \alpha} \cap U^{\gamma \beta} .$

If we had this, then we could just glue the ${U^{\alpha}}$ and get a smooth lifting of ${X_0}$ to ${S}$.

What we can do is to consider the differences ${ (t^{\alpha \gamma})^{-1} \circ t^{\beta \gamma} \circ t^{\alpha \beta}}$, which are automorphisms of ${U^{\alpha \gamma} \cap U^{\alpha \beta}}$. By the previous lemma, this is a 2-Cech cocycle with values in the sheaf ${\mathcal{F}}$ defined as above, over the open cover ${\{U_0^\alpha\}}$. (One should check that it is actually a 2- cocycle, but this is formal.)

Now, if it were a boundary, then we would be done and we could make the lifting. For if we have a 1-cochain in ${\mathcal{F}}$, this would be a collection of automorphisms ${q^{\alpha, \beta}}$ of each ${U^{\alpha \beta}}$ such that, for each triple ${(\alpha, \beta, \gamma)}$, one has:

$\displaystyle (t^{\alpha \gamma})^{-1} \circ t^{\beta \gamma} \circ t^{\alpha \beta} = q^{\alpha \beta} q^{\beta \gamma} ( q^{\alpha \gamma})^{-1}.$

The point is now that if this 2-cocycle is a coboundary, then we can use the ${q^{\alpha \beta}}$ to modify the transition maps ${t^{\alpha \beta}}$ (by, say, precomposition) so as to have satisfied the cocycle condition. In particular, if the Cech ${H^2}$ of this sheaf vanishes with respect to the open cover ${U_0^\alpha}$, the lifting exists.

Note that this is a good Cech cover of ${X_0}$ because ${X_0}$ is separated and eaech ${U_0^\alpha}$ is affine. It follows that if ${X_0}$ is of dimension one and ${H^2(X_0, \mathcal{F}) =0}$, then the lifting exists.

Corollary 9 Let ${X_0 \rightarrow k}$ be a smooth curve. Let ${A}$ be a DVR with residue field ${k}$ and maximal ideal ${\mathfrak{m}}$. Then there is a compatible system of smooth schemes ${X_n \rightarrow \mathrm{Spec} A/\mathfrak{m}^n}$.

4. From formal to actual

Let ${X_0 \rightarrow k}$ be a smooth curve. If ${A}$ is a complete DVR with residue field ${k}$, then we have seen how to lift ${X_0 \rightarrow k}$ to a sequence ${X_n \rightarrow \mathrm{Spec} A/\mathfrak{m}^n}$ of compatible smooth schemes by the previous section. Namely, we first lift to ${\mathrm{Spec} A/\mathfrak{m}^2}$, lift that to ${\mathrm{Spec} A/\mathfrak{m}^3}$, and continue repeatedly. This is still rather far from our ultimate goal, which is a smooth, proper scheme ${X \rightarrow \mathrm{Spec} A}$.

Now, with the mechanics of the lifting procedure behind us, we want to turn the system of schemes ${X_n \rightarrow \mathrm{Spec} A/\mathfrak{m}^n}$ into an actual scheme. We have already stated that this is a so-called formal scheme, which we will denote by the symbol ${\mathfrak{X}}$, and write formally

$\displaystyle \mathfrak{X} \rightarrow \mathrm{Sppf} A,$

to indicate the system of maps ${X_n \rightarrow \mathrm{Spec} A_n}$. We shall think of formal schemes very naively; we do not need to worry about their general theory, so shall treat them as a black box here. We shall write ${\mathfrak{X} \times_A k = X_0}$, and similarly for ${\mathfrak{X} \times_A A/\mathfrak{m}^n}$ for ${X_n}$.

We now that one way of obtaining a formal scheme is to start with an actual scheme ${X \rightarrow \mathrm{Spec} A}$ and consider each of the reductions mod ${\mathfrak{m}^n}$. Such formal schemes are said to be algebrizable. This is the formal analog of complex analytic spaces that come from algebraic varieties. There are formal schemes, even proper ones, that are not algebrizable, so we are going to need a special tool in the case of curves.

That tool is:

Theorem 10 Let ${A}$ be a complete local ring, ${\mathfrak{X} \rightarrow \mathrm{Sppf} A}$ a formal scheme. Suppose ${\mathfrak{X} \times_A k}$ is proper. Suppose moreover there is a compatible system of line bundles ${\mathcal{L}_n}$ on each ${X_n = \mathfrak{X} \times_A A/\mathfrak{m}^n}$ such that ${\mathcal{L}_0}$ is very ample on ${X_0 \rightarrow k}$.Then there is a projective morphism ${X \rightarrow \mathrm{Spec} A}$, such that the “formal scheme” ${\mathfrak{X}}$ is obtained from ${X}$: that is, ${\mathfrak{X}}$ is algebrizable.

This is a consequence of Grothendieck’s “formal GAGA” and appears in EGA III.5. We shall not prove this.

But we shall use it. Consider a formal scheme ${\mathfrak{X} \rightarrow \mathrm{Sppf} A}$ obtained by successively lifting a smooth proper curve ${X_0}$ over the residue field ${A/\mathfrak{m}}$. Now ${X_0}$ is projective, so it has a very ample line bundle on it. If we can lift this to each ${X_n \rightarrow \mathrm{Spec} A/\mathfrak{m}^n}$, then the above result will imply that the formal scheme is algebrizable, and then we will have lifted ${X_0}$ to characteristic zero.

Here the fact that we are in dimension one saves us (again!), because we can successively lift the ample line bundle step by step. We use:

Lemma 11 Let ${X}$ be a scheme of dimension one, ${X_0 \subset X}$ a closed subscheme defined by an ideal of square zero. Then the map ${\mathrm{Pic}(X) \rightarrow \mathrm{Pic}(X_0)}$ is surjective.

Proof: Suppose ${X_0}$ is defined by the ideal ${\mathcal{I}}$ of square zero. There is an exact sequence

$\displaystyle 0 \rightarrow \mathcal{I} \rightarrow \mathcal{O}_{X}^* \rightarrow \mathcal{O}_{X_0}^* \rightarrow 0,$

where the first map sends ${x \mapsto 1+x}$. This is a general fact about rings, even. Now since ${\mathrm{Pic}(X) = H^1(X, \mathcal{O}_X^*)}$ (and similarly for ${X_0}$), the long exact sequence in cohomology and ${H^2(X, \mathcal{I}) = 0}$ gives the result.

It follows that we can lift the sequence of smooth schemes ${\left\{X_n \rightarrow \mathrm{Spec} A/\mathfrak{m}^n\right\}}$ to a projective scheme ${X \rightarrow \mathrm{Spec} A}$. The only thing left to see is that ${X}$ is smooth over ${\mathrm{Spec} A}$. This follows because it is smooth on the special fiber: indeed, one checks flatness by the infinitesimal criterion. It is a general fact that if ${f: X \rightarrow Y}$ is a finite-type morphism of noetherian schemes, and ${x \in X}$ is such that the fiber ${X_{f(x)}}$ is smooth over ${k(f(x))}$ and ${f}$ is flat at ${x}$, then ${f}$ is smooth at ${x}$. With this in mind, it is clear that ${X \rightarrow \mathrm{Spec} A}$ is smooth on the specific fiber.

But this means in particular that ${X}$ is smooth everywhere! Indeed, the smooth locus of a morphism is always open. Let ${T \subset X}$ be the collection of points where ${f}$ is not smooth. Then the image of ${T}$ is closed because ${X}$ is proper over ${\mathrm{Spec} A}$, but this image does not contain the closed point; as a result, it is empty. So ${X}$ is ${A}$-smooth.

This completes the proof that smooth curves can be lifted to characteristic zero.