This is the fourth in a series of posts started here intended to describe Grothendieck’s work on the fundamental group of smooth curves in positive characteristic. Past posts were devoted to showing that a smooth proper curve $X_0$ could always be “lifted” to characteristic zero, via a proper smooth map $X \to \mathrm{Spec} A$ (with $A$ a complete discrete valuation ring) whose specific fiber is the initial curve $X_0$. The promise was to use various methods of algebraic geometry to compare the fundamental groups of the “generic” and the “special” fibers. In the previous post, we handled the case of the special fiber, where we argued that the fundamental group of the special fiber was the same as that of the whole thing. In this post, we want to develop the technology to handle the general fiber.

Our next goal is to obtain a small analog of the classical long exact sequence of a fibration in homotopy theory. Namely, the smooth proper morphism ${X \rightarrow \mathrm{Spec} A}$ constructed as earlier that lifts the curve ${X_0 \rightarrow \mathrm{Spec} k}$ will be the “fibration,” and we are going to take the (geometric) fiber over the generic point. Since the base ${\mathrm{Spec} A}$ has trivial ${\pi_1}$ (because ${A}$ is complete local and its residue field is algebraically closed), it will follow from this long exact sequence that

$\displaystyle \pi_1(X_{\overline{\xi}}) \rightarrow \pi_1(X)$

is a surjection.

1. An auxiliary result: the Stein factorization of proper, separable morphisms

We shall start by proving an auxiliary result. To motivate it, recall that a proper morphism ${f: X \rightarrow Y}$ has a Stein factorization via ${X \rightarrow \mathbf{Spec} f_*\mathcal{O}_X \rightarrow Y}$ (note that the second morphism is finite because ${f_* \mathcal{O}_X}$ is coherent). This result states that the Stein factorizations of certain morphisms give a means of constructing étale covers.

Theorem 16 Let ${f: X \rightarrow Y}$ be a proper, flat morphism of noetherian schemes with geometrically reduced fibers. Then the map ${\mathbf{Spec} f_*\mathcal{O}_X \rightarrow Y}$ is an étale cover.

Note that ${\mathbf{Spec} f_*\mathcal{O}_X \rightarrow Y}$ is the second half of the Stein factorization of ${f}$. Since ${f}$ is proper, it is automatic that this map is finite.

Proof:  We shall start by employing the standard Grothendieckian trick of reducing to the case where the base is the spectrum of a complete local noetherian ring (whose residue field is even algebraically closed).

Everything is local on ${Y}$. We may thus assume ${Y}$ is affine, ${Y = \mathrm{Spec} A}$. Moreover, the construction of ${f_* \mathcal{O}_X}$ commutes with flat base change (in general, the higher direct images by a separated morphism commute with flat base change). In particular, the {Stein factorization} commutes with flat base change. Since a morphism (say, of finite type) ${ Z_1 \rightarrow Z_2}$ is étale if and only if ${Z_1 \times_{Z_2} \mathrm{Spec} \mathcal{O}_{z, Z_2} \rightarrow \mathrm{Spec} \mathcal{O}_{z, Z_2}}$ is étale for each ${z \in Z_2}$, we may (in view of this observation) assume furthermore that ${A}$ is noetherian local.

Moreover, since étaleness descends under faithfully flat base change, we can make a faithfully flat base extension to ensure that ${A}$‘s residue field is algebraically closed. (In fact, it is a general lemma in EGA 0-III that one may produce flat local noetherian extensions of a local noetherian ring with any residue field extension whatsoever.) In view of the faithfully flat morphism ${A \rightarrow \hat{A}}$, we can even assume ${A}$ is complete local with an algebraically closed residue field.

So let ${A}$ be complete local with residue field ${k}$ (algebraically closed), and let ${X \rightarrow \mathrm{Spec} A}$ be a proper, flat morphism whose fibers are geometrically reduced. We shall show:

1. ${H^0(X, \mathcal{O}_{X} )}$ is a flat and finitely generated (i.e. free) ${A}$-module.
2. ${ H^0(X, \mathcal{O}_X) \otimes_A k}$ is a product of copies of ${k}$.

Together, these will imply that ${H^0(X, \mathcal{O}_X)}$ is an étale ${A}$-algebra (because it is flat and unramified). The techniques are essentially those in the proof of “cohomology and base change.” Indeed, this claim follows directly from that general result.

Let us consider the functor ${T}$ from finitely generated ${A}$-modules to finitely generated ${A}$-modules,

$\displaystyle T(M) = H^0(X, \mathcal{O}_X \otimes_A M).$

Since ${X \rightarrow \mathrm{Spec} A}$ is flat, a short exact sequence of ${A}$-modules ${0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0}$ induces a short exact sequence ${0 \rightarrow \mathcal{O}_X \otimes_A M' \rightarrow \mathcal{O}_X \otimes_A M \rightarrow \mathcal{O}_X \otimes_A M'' \rightarrow 0}$. In particular, we see that ${T}$ is left exact.

We will show:

Lemma 17 ${T}$ is right-exact, and indeed, isomorphic to the functor ${M \mapsto T(A) \otimes_A M}$.

This will easily give the result that ${H^0(X, \mathcal{O}_X)}$ is flat, because ${T}$ is also left-exact.

Proof: There is a natural homomorphism

$\displaystyle T(A) \otimes_A M \rightarrow T(M).$

This is given by sending ${x \in T(A), m \in M}$ to the image of ${x}$ under the homomorphism ${A \stackrel{1 \mapsto m}{\rightarrow} M}$. We will show that this map is surjective. Denote the previous functor by ${U(M)}$; this is right-exact. The claim is that ${U(M) \rightarrow T(M)}$ is always surjective.

We shall first prove this assuming that ${A}$ is local artinian with maximal ideal ${\mathfrak{m} \subset A}$. Let ${X_0}$ be the specific fiber ${X \times_A k}$.

Then note that ${U(k) \rightarrow T(k)}$ is surjective. For ${U(k) = H^0(X, \mathcal{O}_X) \otimes_A k }$ and ${T(k) = H^0(X_0, \mathcal{O}_{X_0})}$. The latter is a free ${k}$-vector space on the connected components of ${X_0}$, since the geometric fibers of ${X \rightarrow \mathrm{Spec} A}$ are reduced. But the primitive idempotents are hit by elements of ${H^0(X, \mathcal{O}_X)}$ because the map ${\pi_0(X) \rightarrow \pi_0(X_0)}$ is an isomorphism.

In general, we note:

Lemma 18 Let ${A}$ be an artinian local ring with residue field ${k}$. Let ${T}$ be an additive, half-exact functor from the category of finitely generated ${A}$-modules to itself . Suppose ${T(A) \rightarrow T(k)}$ is surjective. Then ${T}$ is of the form ${M \mapsto T(A) \otimes_A M}$ and ${T(A)}$ is flat.

Proof: Indeed, if we let ${U(M) = T(A) \otimes_A M}$ as before, then we have a natural transformation ${U(M) \rightarrow T(M)}$. We have seen that this is a surjection for ${M = k}$. In general, induct on the length of ${M}$. Suppose ${M}$ is an ${A}$-module of length greater than one; there is then an exact sequence

$\displaystyle 0 \rightarrow M_1 \rightarrow M \rightarrow M_2 \rightarrow 0,$

where ${M_1, M_2}$ have smaller lengths. We can find an exact and commutative diagram:

The outside vertical maps are surjective. From this a diagram chase (or the snake lemma, if ${T(M_1)}$ is replaced by its image in ${T(M)}$) shows that ${U(M) \rightarrow T(M)}$ is surjective.

Finally, we want to see that ${U(M) \rightarrow T(M)}$ is bijective. We can see this, however, by noting now that both ${U}$ and ${T}$ are right-exact (for ${T}$ is now right-exact, as the surjective image of ${U}$) functors on the category of ${A}$-modules; there is a natural transformation ${U(M) \rightarrow T(M)}$ that is an isomorphism for free ${A}$-modules. By the “finite presentation trick,” (i.e. choosing a presentation ${F' \rightarrow F \rightarrow M \rightarrow 0}$ of any finitely generated ${A}$-module with ${F, F'}$ free) it now follows that the natural transformation is an isomorphism for any finitely generated ${A}$-module ${M}$.

In general, of course, the complete local ring ${A}$ constructed earlier will not be artinian. However, the functor ${T}$ will be extra-friendly, though: it will have the property that, for any finitely generated ${A}$-module ${M}$,

$\displaystyle T(M) \rightarrow \varprojlim_{n} T(M/\mathfrak{m}^n M) \ \ \ \ \ (1)$

is an isomorphism. This is in fact a consequence of the formal function theorem and the completeness of ${A}$.

Now we use:

Lemma 19 Let ${A}$ be a complete local noetherian ring, ${T}$ a half-exact functor from finitely generated ${A}$-modules to finitely generated ${A}$-modules such that (1) is always an isomorphism. Then ${T}$ is right-exact, and so isomorphic to the functor ${M \mapsto T(A) \otimes_A M}$.

If ${T}$ is additionally left-exact, it follows that ${T(A)}$ must be flat.

Proof: It follows that the natural transformation

$\displaystyle T(A) \otimes_A M \rightarrow T(M)$

is an isomorphism for each finitely generated ${M}$ annihilated by a power of ${\mathfrak{m}}$, by the previous lemma for the artinian rings ${A/\mathfrak{m}^n}$.

Now fix a general ${M}$. We then have a natural isomorphism

$\displaystyle T(A) \otimes_A \varprojlim M/\mathfrak{m}^n M \rightarrow \varprojlim T(M/\mathfrak{m}^n M ) ;$

the left side is clearly isomorphic to ${T(A) \otimes_A M}$, and the right side to ${T(M)}$ (by the condition on (1)). The lemma is thus clear.

So it follows that in our case, for ${T(M) = H^0(X, \mathcal{O}_X \otimes_A M)}$, this half-exact functor is actually right-exact (since ${T(A) \rightarrow T(k)}$ is surjective). Moreover ${T(A)}$ is a flat ${A}$-module, since the functor is left-exact. Finally, ${T(k) = T(A) \otimes_A k}$ is the ring of regular functions on a reduced proper ${k}$-variety, so it is a product of copies of ${k}$. This proves the result.

Here is an important consequence of the general formalism developed above. If ${f: X \rightarrow Y}$ is a proper and flat morphism with geometrically reduced fibers, then the formation of ${f_*(\mathcal{O}_X)}$ commutes with base change. To be more precise, if one has a cartesian diagram

then ${f'_*(\mathcal{O}_{X'})}$ is the pull-back of ${f_*(\mathcal{O}_X)}$. This follows from the general formalism. Note if that ${f}$ satisfies these conditions, then so does the composite—in either direction—of ${f}$ with any finite étale morphism. One may see this by reducing to the case where ${Y}$ is complete local(!) by flat base change, and then using the above description of the functor ${T}$.

2. The exact sequence

Finally, we are able to construct the not-so-long exact sequence of a “fibration.”

Theorem 20 (Exact sequence) Let ${f: X \rightarrow Y}$ be a proper, flat morphism of connected noetherian schemes with geometrically reduced fibers. Suppose ${f_* \mathcal{O}_X \simeq \mathcal{O}_Y}$, and let ${\overline{y} \in Y}$ be a geometric point, lifting to a point ${\overline{x} \in X}$. Then there is an exact sequence

$\displaystyle \pi_1(X_{\overline{y}, }, \overline{x}) \rightarrow \pi_1(X, \overline{x}) \rightarrow \pi_1(Y, \overline{y}) \rightarrow 1.$

Note that by one form of Zariski’s Main Theorem, the fibers of ${f}$ are geometrically connected because ${f_* \mathcal{O}_X \simeq \mathcal{O}_Y}$.

To make sense of this result, we will have to think in terms of étale covers: after all, it is only in this sense that we even know what the morphisms of groups are. Namely, from the sequence of schemes ${X_{\overline{y}} \rightarrow X \rightarrow Y}$, we have functors ${\mathrm{Et}(Y) \rightarrow \mathrm{Et}(X) \rightarrow \mathrm{Et}(X_{\overline{y}})}$. If we let the three fundamental groups be ${G_{X_{\overline{y}}}, G_{X} , G_Y}$, then we have a collection of functors

$\displaystyle G_Y\mbox{--}\mathbf{Set} \rightarrow G_X\mbox{--}\mathbf{Set} \rightarrow G_{X_{\overline{y}}}\mbox{--}\mathbf{Set} .$

Here ${G\mbox{--}\mathbf{Set}}$ denotes the categories of finite continuous ${G}$-sets, for ${G}$ a profinite group.

To say that ${\varphi: G_X \rightarrow G_Y}$ is surjective is to say that any connected ${G_Y}$-set, when considered as a ${G_X}$-set via ${\varphi}$, is also connected. Via the Galois correspondence, this means that the base-change to ${X}$ of a connected étale cover of ${Y}$ is a connected étale cover of ${X}$. This is easy to see.

Next, to say that the kernel of ${G_X \rightarrow G_Y}$ is contained in the image of ${G_{X_{\overline{y}}}}$ is to say that if a connected ${G_X}$-set (say, of the form ${G_X/H}$ for ${H}$ an open subgroup) is such that its restriction to a ${G_{X_{\overline{y}}}}$-set is trivial (i.e. ${H}$ contains the image of ${G_{X_{\overline{y}}}}$), then ${G_X}$ arises as a ${G_Y}$-set.

That is, if ${X' \rightarrow X}$ is a connected étale cover such that ${X' \times_X X_{\overline{y}}}$ is trivial, then ${X'}$ is the base-change of an étale cover of ${Y}$.

Proof: We shall first start by establishing the surjectivity of ${\pi_1(X, \overline{x}) \rightarrow \pi_1(Y, \overline{y})}$. Let ${Y' \rightarrow Y}$ be a connected étale cover. We want to show that ${X \times_Y Y'}$ is connected; this is equivalent to surjectivity, by the above discussion. But ${f': X \times_Y Y' \rightarrow Y'}$ satisfies the same conditions as ${f}$ (because push-forward commutes with flat base change!). Now ${f'}$ is proper and surjective, so it is a quotient map. Also, its fibers are connected by Zariski’s Main Theorem. It follows formally that ${X \times_Y Y'}$ is connected if ${Y'}$ is.

Next, it is clear that the composite ${\pi_1(X_{\overline{y}}, \overline{x}) \rightarrow \pi_1(Y, \overline{y})}$ is trivial, because the map ${X_{\overline{y}} \rightarrow Y}$ itself is a constant map.

The hardest part will be to show exactness at the middle. As we discussed above, this amounts to saying that if ${X' \rightarrow X}$ is a connected étale cover such that ${X' \times_X X_{\overline{y}} = X' \times_Y \overline{y}}$ is trivial, then ${X'}$ is the base-change of an étale cover of ${Y}$. Here, we consider the Stein factorization of ${ g: X' \rightarrow Y}$; we get a composite ${X' \rightarrow Y' = \mathbf{Spec} g_*(\mathcal{O}_{X'}) \rightarrow Y}$, where as we have seen, ${Y' \rightarrow Y}$ is an étale cover. The claim is that the diagram

is cartesian. This will imply what we want.

To see this, note that ${X \times_Y Y' \rightarrow X}$ is an étale cover, and there is a natural map ${X' \rightarrow X \times_Y Y' }$. This is necessarily an étale cover itself, and it is in fact surjective: this follows because ${X' \rightarrow Y'}$ is surjective (as ${X' \rightarrow Y}$ is surjective and ${Y' \rightarrow Y}$ is finite), so we need to see that ${X' \rightarrow X \times_Y Y'}$ has degree one.

To check this, however, we can base-change to ${X_{\overline{y}}}$ (as both schemes are connected, we just need to check the degree at one point). This is the same as base-changing both ${X'}$ and ${X \times_Y Y'}$ to ${\overline{y}}$ over ${Y}$. To do this, we will show that

$\displaystyle X' \times_Y \overline{y} \rightarrow X \times_Y Y' \times_Y \overline{y} \ \ \ \ \ (2)$

is an isomorphism. To do this, note first that since ${g: X' \rightarrow Y}$ is proper and flat with geometrically reduced fibers as above, the formation of ${g_* (\mathcal{O}_X)}$ commutes with arbitrary base-change. Moreover, these properties are preserved under base change themselves. The upshot is that we can just assume ${Y}$ is the spectrum of an algebraically closed field.

Now ${X}$ is connected, and ${X'}$ splits as a sum of copies of ${X}$, so it is clear (because ${X}$ is reduced) that ${Y'}$ is a sum of copies of the appropriate number of copies of ${k}$, and that the map ${X' \rightarrow X \times_Y Y'}$ is an isomorphism.