The étale cohomology of curves over finite fields

When I was a graduate student, Zev Rosengarten (a former student of Brian Conrad) and I used to eat dinner at Stanford’s Arrillaga dining hall a lot. We’d talk about math for hours, but one thing that will forever be ingrained in my mind is how Zev was able to do all these complicated spectral sequence arguments off the top of his head – without ever writing anything down!

In today’s post, we’ll be calculating the étale cohomology of curves over finite fields. You guessed it: We’ll be using a goddamn spectral sequence. More specifically, for $k := \mathbf{F}_q$ and $X/k$ a smooth, proper, geometrically connected curve, we’ll calculate $H^n(X,\mathbf{G}_m)$ .

Caution to the reader: you cannot use Artin vanishing or Poincaré duality, because we are not over a separably closed field. In addition, $\mathbf{G}_m$ is not torsion/lcc/constructible.

Why would anyone want to calculate $H^n(X,\mathbf{G}_m)$ anyway? Well, it turns out that this is exactly what you need to prove the fundamental exact sequence of Brauer groups of global functions fields! When I learned class field theory, the proofs were a nightmare. God only knows how to construct the Artin map in the global situation, and in the local situation you fool around with Tate cohomology. The point is I don’t remember any of this stuff, and of course I don’t know how to prove the fundamental exact sequence for Brauer groups in general. But for function fields, this last fact is literally just pure algebraic geometry. For more details, see this mathoverflow post here.

Theorem 1: Define $k := \mathbf{F}_q$ and let $f: X \to \operatorname{Spec} k$ be a smooth, proper, geometrically connected curve. Then

$H^n(X,\mathbf{G}_m) = \begin{cases} \mathbf{G}_m, & \hspace{5mm} \text{if }n = 0 \\ \text{Pic}(X),& \hspace{5mm} \text{if }n = 1 \\ 0, & \hspace{5mm} \text{if }n = 2 \\ \mathbf{Q}/\mathbf{Z}, &\hspace{5mm }\text{if }n=3 \\ 0, &\hspace{5mm} \text{else.}\end{cases}$

To prove the theorem, we will record a couple of lemmas. Throughout, our cohomology will be étale cohomology (unless stated otherwise). In addition, $k$ will always denote a finite field of cardinality $q$ .

Lemma 1: Let $A/k$ be an abelian variety. Then $H^n(k,A) = 0$ for all $n > 0$ .

Proof: When $n = 1$ , this is Lang’s theorem. Now let $m$ be any positive integer (not necessarily prime to the characteristic of $k$ ), and consider the exact sequence of sheaves in the fppf topology

$0 \to A[m] \to A \stackrel{m \cdot}{\to} A \to 0.$

Since $k$ is perfect, we have

$H^n_{\text{fppf}}(k,A[m]) = H^n_{\text{ét}}(k,A[m]) = H^n(G_k, A[m](\bar{k}))$

which vanishes for $n > 1$ by cohomological dimension considerations. It follows that multiplication by $m$ is an isomorphism on $H^n_{\text{ét}}(k,A)$ for $n > 1$ . But $H^n_{\text{ét}}(k,A)$ is also torsion and hence the result.

Remark 1: A similar argument using the Kummer sequence shows that $H^n(k,\mathbf{G}_m) = 0$ for $n > 0$ , noting that in the $n= 1$ case, we use Hilbert’s Theorem 90 instead of Lang’s theorem.

Lemma 2: Let $f$ be as in the statement of the theorem. Then $R^n f_\ast \mathbf{G}_m = 0$ for all $n > 1$ .

Proof: It is enough to show that the stalk of $R^n f_\ast\mathbf{G}_m$ at the geometric point $\bar{x} : \operatorname{Spec} \bar{k} \to \operatorname{Spec} k$ is zero. We have

$(R^n f_\ast \mathbf{G}_m)_{\bar{x}} = H^n(X_{\bar{k}}, \mathbf{G}_m).$

An argument using the divisor exact sequence shows that this injects into $H^n\left(k(X_{\bar{k}}), \mathbf{G}_m\right)$ , which vanishes for $n > 1$ by Tsen’s theorem and we win.

Remark 2: In general, we need to be a little careful when we identify the stalk of the higher direct image as above, simply because the pullback of $\mathbf{G}_m$ is not always $\mathbf{G}_m$ in general. However, the identification above is still valid as the sheaf $\mathbf{G}_m$ is “compatible” with filtered colimits of étale maps.

Lemma 3: Consider the discrete group $\mathbf{Z}$ as the trivial $G_k$ -module. Then:

$H^n(k, \mathbf{Z}) = \begin{cases} 0, \hspace{5mm} & \text{if }n = 1 \\ \mathbf{Q}/\mathbf{Z}, \hspace{5mm} &\text{if }n = 2 \\ 0, \hspace{5mm} &\text{else.}$

Let us first deal with the $n = 1$ case. Since the Galois action is trivial,

$H^1(k,\mathbf{Z}) = \operatorname{Hom}_{\text{cts}}(G_k, \mathbf{Z})$

of which there are none since $G_k \simeq \widehat{\mathbf{Z}}$ is profinite while $\mathbf{Z}$ is discrete. Why? Let $\phi : G_k \to \mathbf{Z}$ be a continuous homomorphism. Since $\mathbf{Z}$ has the discrete topology, $\ker \phi$ is open in $G_k$ . On the other hand, an open subgroup of $G_k$ is of finite index. In summary, we have an injection from a finite group $G_k/\ker \phi \hookrightarrow \mathbf{Z}$ and therefore $\phi = 0$ .

Now observe for any $m \in \mathbf{N}$ that multiplication by $m$ on $H^n(k,\mathbf{Q})$ is an isomorphism. But $H^n(k,\mathbf{Q})$ is also torsion, and therefore $H^n(k,\mathbf{Q}) = 0$ for all $n > 0$ . We deduce that

$H^{n-1}(k,\mathbf{Q}/\mathbf{Z}) \simeq H^{n}(k,\mathbf{Z}).$

When $n= 2$ , by triviality of the Galois action the left hand side is again continuous homomorphisms from $G_k$ to $\mathbf{Q}/\mathbf{Z}$ . This is none other than $\mathbf{Q}/\mathbf{Z}$ . When $n> 2$ , the cohomology vanishes by cohomological dimension considerations since $\mathbf{Q}/\mathbf{Z}$ is torsion. This completes the proof of Lemma 3.

We can now prove Theorem 1.

Proof of Theorem 1: For $n = 0$ , this is simply the fact that the global sections of a proper, geometrically connected and geometrically reduced scheme over $k$ are just $k$ . For $n=1$ , this is usual descent theory for line bundles. Therefore, from now on we will concentrate on what happens for $n > 1$ . Consider the Leray spectral sequence

$E_2^{p,q} := H^p(k, R^q f_\ast \mathbf{G}_m) \implies H^{p+q}(X,\mathbf{G}_m).$

By construction of the Leray spectral sequence, the abutment $E_\infty^n$ admits a filtration

$0 = F^{-1} \subset F^0 \subset \ldots \subset F^n \subset F^{n+1} = E_\infty^n$

such that

$\begin{eqnarray*} F^0/F^{-1} &\simeq& E_\infty^{n,0} \\ F^1/F^0 &\simeq& E_\infty^{n-1,1} \\ &\vdots&\\ F^{n+1}/F^n &\simeq& E_\infty^{0,n}. \end{eqnarray*}$

Now first I claim that the only quotients we need to care about are $E_\infty^{n,0}$ and $E_\infty^{n-1, 1}$ . Indeed, by Lemma 2 above we have $E_2^{p,q} = 0$ for $q > 1$ , and hence the same is true of the abutment $E_\infty^{p,q} = 0$ .

We now turn to the $E_\infty^{n,0}$ term. You guessed it, we’ll look at the $E_2$ page first. By cohomology and base change, $f_\ast \mathbf{G}_m = \mathbf{G}_m$ and therefore $E_2^{n,0} = H^n(k, \mathbf{G}_m)$ which vanishes for $n > 0$ by Remark 1 above. Therefore, we also have $E_\infty^{n,0} = 0$ .

Now this is great! You’d think there would be lots of ugly crap with differentials, but boom! Lots of stuff is straight up zero.

Ok, so finally we must deal with the quotient $E_\infty^{n-1,1}$ . Unfortunately, we now have to look at differentials (sad face here). Let’s look at the $E_3$ page first. By definition,

$E_3^{n-1,1} = \frac{\ker(E_2^{n-1,1} \to E_2^{n+1,0} )}{\text{im}(E_2^{n-3,2} \to E_2^{n-1,1} )}.$

But that image in the denominator is zero by Lemma 2. On the other hand, the kernel in the numerator is just $E_2^{n-1,1}$ because $E_2^{n+1,0}$ is zero by Remark 1. In conclusion, $E_3^{n-1,1} = E_2^{n-1,1}$ . I’ll leave it to you to convince yourself that in fact

$E_2^{n-1,1} = E_3^{n-1,1} = E_4^{n-1,1} = \ldots = E_\infty^{n-1,1}.$

The essential point is that as we keep “turning the pages” of the spectral sequence, that image term comes from something “higher and higher up in the plane”, and hence is always zero by Lemma 2. In addition, the kernel term on the $r$ -th page is just $E_r^{n-1,1}$ , because now the differential out of it points to things in negative degree (in the $y$ -axis).

We can now concentrate fully on calculating $E_2^{n-1,1} = H^{n-1}(k, R^1 f_\ast \mathbf{G}_m)$ . To this end, let us recall that $R^1 f_\ast \mathbf{G}_m$ can be identified with the Picard functor $\operatorname{Pic}_{X/k}$ , with identity component $\operatorname{Pic}^0_{X/k}$ represented by an abelian variety of dimension $h^0(X, \Omega^1_X)$ over $k$ . Note that $\operatorname{Pic}^0_{X/k}$ and $\operatorname{Pic}_{X/k}$ fit into an exact sequence

$0 \to \operatorname{Pic}^0_{X/k} \to \operatorname{Pic}_{X/k} \to\mathbf{Z} \to 0$

where the last map sends a line bundle to its degree.

Now take the cohomology of this exact sequence. We get

$\ldots \to H^{n-1}(k, \operatorname{Pic}^0_{X/k}) \to H^{n-1}(k, \operatorname{Pic}_{X/k}) \to H^{n-1}(k, \mathbf{Z})\to \ldots$

The thing on the left is zero for all $n > 1$ by Lemma 1, and therefore

$H^{n-1}(k, \operatorname{Pic}_{X/k}) \simeq H^{n-1}(k, \mathbf{Z})$

for all $n > 1$ . It follows using Lemma 3 that

$E_\infty^{n-1,1} = \begin{cases} 0, & \hspace{5mm} \text{if }n = 2 \\ \mathbf{Q}/\mathbf{Z},& \hspace{5mm} \text{if }n = 3 \\ 0, &\hspace{5mm} \text{else.}\end{cases}$

This completes the proof of Theorem 1.