(This is the third post in a project started here to describe some of Grothendieck’s work on the fundamental group of curves in positive characteristic.)

Let {X_0 \rightarrow \mathrm{Spec} k} be a smooth curve over an algebraically closed field {k} of characteristic {p}. We are interested in determining a set of topological generators for this curve. To do this, we started by showing that if {A} is a complete DVR with residue field {k}, then one can “lift” (by using cohomological vanishing and formal-to-algebraic comparison theorems) {X_0 \rightarrow \mathrm{Spec} k} to a smooth, proper morphism {X \rightarrow \mathrm{Spec} A}.

Ideally, we will have chosen {A} to be characteristic zero itself. Now our plan is to compare the two geometric fibers of {X}: one is {X_0}, and the other is {X_{\overline{\xi}}} (where {\xi} is the generic point; the over-line indicates that one wishes an algebraic closure of {k(\xi) = K(A)} here) with each other. Ultimately, we are going to show two things:

 

  1. The natural map {\pi_1(X_0) \rightarrow \pi_1(X)} is an isomorphism.
  2. The natural map {\pi_1(X_{\overline{\xi}}) \rightarrow \pi_1(X)} is an epimorphism.

Here we have been loose with notation, as we have not indicated the relevant geometric points. The geometric point is, however, irrelevant for a connected scheme.

It will follow from this that there is an continuous epimorphism of profinite groups

\displaystyle \pi_1(X_{\overline{\xi}}) \rightarrow \pi_1(X_0).

However, {\pi_1(X_{\overline{\xi}})} will be seen to be topologically generated by {2g} generators (where {g} is the genus) by comparison with a curve over {\mathbb{C}}. For a smooth curve of genus {g} over {\mathbb{C}}, it is clear from the Riemann existence theorem (and the topological fundamental group) that {\pi_1} has {2g} topological generators.

Thus, it will follow, as stated earlier:

Theorem  If {X_0} is a smooth curve of genus {g} over an algebraically closed field of any characteristic, {\pi_1(X_0)} is topologically generated by {2g} generators.

 

One technical point will be, of course, that it is not entirely obvious that {\pi_1(X_{\overline{\xi}})} is the same as it would be for a curve over {\mathbb{C}}. This requires independent proof, but it will not be too hard.

1. The specific fiber

Let {(A, \mathfrak{m})} be a complete local noetherian ring, and let {X \rightarrow \mathrm{Spec} A} be a proper morphism. There is a “specific” fiber {X_0 \rightarrow \mathrm{Spec} k} of {X}, where {k = A/\mathfrak{m}} is the residue field of {A}.

Let {\overline{x} \rightarrow X_0} be a geometric point. Since there is a natural inclusion {X_0 \rightarrow X}, there is a natural map {\pi_1(X_0, \overline{x}) \rightarrow \pi_1(X, \overline{x})}. Recall the definition of this map. For a scheme {Y}, let {\mathrm{Et}(Y)} denote the category of schemes finite and étale over {Y} (i.e. étale covers). Then there is a functor (base-change)

\displaystyle \mathrm{Et}(X) \rightarrow \mathrm{Et}(X_0) .

Note that {\mathrm{Et}(X)} is identified with the category of finite continuous {\pi_1(X, \overline{x})}-sets, and {\mathrm{Et}(X_0)} with the category of finite continuous {\pi_1(X_0, \overline{x})}-sets. The map

\displaystyle \pi_1(X_0, \overline{x}) \rightarrow \pi_1(X, \overline{x})

is the one inducing the corresponding functor from {\pi_1(X, \overline{x})}-sets to {\pi_1(X_0,\overline{x})}-sets.

We want to show that this is an isomorphism. This will follow directly as a consequence of the next result.

Theorem 13 If {X \rightarrow \mathrm{Spec} A} is proper for {A} local, complete, and noetherian, and {X_0} is the specific fiber, then the functor of base-change induces an equivalence of categories between {\mathrm{Et}(X)} and {\mathrm{Et}(X_0)}.

 

This is a generalization of the more elementary fact that if {A} is a complete local noetherian ring with residue field {k}, then the functor { B \mapsto B \otimes_A k} induces an equivalence of categories between finite étale {A}-algebras and finite étale {k}-algebras.

Proof: Let us first start by showing that it is fully faithful. Let {Y, Y'} be étale {X}-schemes. Then a {X}-morphism {Y \rightarrow Y'} can be described as a certain subscheme of the product {Y \times_X Y'}, via its graph {\Gamma}. Namely, this subscheme {\Gamma} must be open and closed, and of degree one over {Y}.

To see this, note that there is a cartesian diagram

However, {Y'} is unramified over {X}, so the diagonal map {Y' \rightarrow Y' \times_X Y'} is an open immersion. It is also separated (being finite), so the diagonal map is a closed immersion. Thus {\Gamma} is a union of connected components of {Y \times_X Y'}. Moreover, any such open and closed {\Gamma \subset Y \times_X Y'} gives rise to a morphism {Y \rightarrow Y'} if the projection {\Gamma \rightarrow Y} is an isomorphism, which is equivalent to saying that it has degree one (because all these morphisms are étale). The same can be said for morphisms {Y_0 \rightarrow Y'_0}.

However, we will see below that open and closed subsets of {Y \times_X Y'} in natural bijection with open and closed subsets of {Y_0 \times_{X_0} Y'_{0}}; moreover, it is clear that the degree of such a subset {\Gamma \subset Y \times_X Y'} is one over {Y} if and only if the degree of {\Gamma_0} over {Y_0} is one. (This follows again because {Y} is connected if and only if {Y_0} is, which we will see below.)

Now let us show that the functor is essentially surjective. That is, given an étale cover {Y_0 \rightarrow X_0}, we need to find an étale cover {Y \rightarrow X} whose fiber is {Y_0 \rightarrow X_0}. To do this, we start by considering the family of infinitesimal thickenings {X_n = X \times_A A/\mathfrak{m}^{n+1}}. We know that {Y_0} can be lifted uniquely to a series of étale covers {Y_n \rightarrow X_n}, by the “équivalence remarquable de catégories” that any scheme étale over another scheme lifts uniquely to a nilpotent thickening (and in a fully faithful way). Note that said equivalence is not restricted to étale covers.

So given {Y_0 \rightarrow X_0}, we get a family of finite étale covers {Y_n \rightarrow X_n}. In other words, we get a formal scheme {\mathfrak{Y} \rightarrow \mathfrak{X}}, where {\mathfrak{X}} is identified with the family {\left\{X_n\right\}}. We want to know now that {\mathfrak{Y}} is algebrizable, or that it comes from a proper scheme {Y \rightarrow X}.

To see this, we will use Grothendieck’s existence theorem in formal geometry. This theorem states:

 

Theorem 14 (Formal GAGA) If {X \rightarrow \mathrm{Spec} A} is proper and {A} complete, local, and noetherian, then there is an equivalence of categories between coherent sheaves on {X} and coherent sheaves on the formal completion {\mathfrak{X}}.

We shall not prove this. In fact, we have not even properly defined a formal completion, let alone what a coherent sheaf on a formal scheme is. For our purposes, it suffices to say that the formal completion is the family of schemes {\left\{X_n \rightarrow \mathrm{Spec} A/\mathfrak{m}^{n+1}\right\}} (note that each {X_n} is the reduction of {X_{n+1}}) and a coherent sheaf on the formal completion is a compatible family of coherent sheaves on the {\left\{X_n\right\}}. The point is that this compatible family has to come from a well-defined coherent sheaf on {X}.

The analogous ordinary GAGA states that, on a proper variety over {\mathbb{C}}, analytification establishes an equivalence of categories between the category of coherent sheaves on {X} and coherent analytic sheaves on the analytification. As with formal GAGA, the hard part is not the full faithfulness of the functor, but the fact that any analytic sheaf comes from an algebraic one. The proof of ordinary GAGA is strikingly similar to that of formal GAGA, however. (It is perhaps more correct to say that sentence in the reverse order: formal GAGA came after Serre’s paper!)

Anyway, let’s return to the main goal: if one has a formal étale cover {\mathfrak{Y} \rightarrow \mathfrak{X}}—that is, a compatible family of étale covers {\left\{Y_n \rightarrow X_n\right\}}—then the one compatible family of coherent sheaves on the {\left\{X_n\right\}}, because each {Y_n} is a {\mathbf{Spec}} of a coherent {\mathcal{O}_{X_n}}-algebra, say {\mathcal{F}_n}. This family {\left\{\mathcal{F}_n\right\}} must come from a coherent sheaf {\mathcal{F}} on {X}, and since each {\mathcal{F}_n} is an algebra, it follows again formally that we can lift the algebra structure to {\mathcal{F}}. We get a finite map

\displaystyle Y = \mathbf{Spec} \mathcal{F} \rightarrow X.

If we can show that this is étale, then we will be done.

But étaleness now follows formally, as before: namely, by the infinitesimal criterion of flatness and étaleness of {Y_0 \rightarrow X_0}, we find that {Y \rightarrow X} is étale at all points in the special fiber. By properness, as before, it follows that {Y \rightarrow X} is étale everywhere, since the étale locus is open. This proves the essential surjectivity.

 

Finally, to complete the proof of the above theorem, we need to prove:

Lemma 15 Let {X} be a proper scheme over {\mathrm{Spec} A}, where {A} is a complete local noetherian ring, and let {X_0 = X \times_A k}. Then the map {\pi_0(X_0) \rightarrow \pi_0(X)} is an isomorphism.

Here {\pi_0} denotes the functor of connected components.

Proof: Indeed, this follows from the formal function theorem, since open and closed subsets of any locally ringed space {(X, \mathcal{O}_X)} are in bijection with idempotents in the ring {\Gamma(X, \mathcal{O}_X)}.

Note that we can consider the infinitesimal liftings {X_n = X \times_A A/\mathfrak{m}^{n+1}} for each {n}. Each of these has the same topological space as {X_0}, and in particular the same {\pi_0}.

However, by the formal function theorem we have

\displaystyle \Gamma(X, \mathcal{O}_X) = \varprojlim \Gamma(X_n, \mathcal{O}_{X_n}).

In particular, the idempotents in {\Gamma(X, \mathcal{O}_X)} are in bijection with {\varprojlim \mathrm{Idem} \Gamma(X_n, \mathcal{O}_{X_n})} where {\mathrm{Idem}} is the functor that assigns to each ring its set of idempotents. ({\mathrm{Idem}} is a corepresentable functor, so it commutes with projective limits.) Anyway, since {\mathrm{Idem} \Gamma( X_n, \mathcal{O}_{X_n}) = \Gamma(X_0, \mathcal{O}_{X_0})}, it follows that

\displaystyle \mathrm{Idem} \Gamma(X, \mathcal{O}_X) = \mathrm{Idem} \Gamma(X_0, \mathcal{O}_{X_0}).

By the connection between idempotents and {\pi_0}, this completes the proof.

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