## What’s up with the fppf site?

I’ve been thinking a lot about something called Serre-Tate theory lately. I want to do some posts on the “classical” case of elliptic curves. Before starting though we’ll go through some preliminaries on why one would ever want to use the fppf site and how to compute with it. It seems that today’s post is extremely well known, but not really spelled out anywhere.

Let’s say you’ve been reading stuff having to do with arithmetic geometry for awhile. Then without a doubt you’ve encountered étale cohomology. In fact, I’ve used it tons on this blog already. Here’s a standard way in which it comes up. Suppose you have some (smooth, projective) variety ${X/k}$. You want to understand the ${\ell^n}$-torsion in the Picard group or the (cohomological) Brauer group where ${\ell}$ is a prime not equal to the characteristic of the field.

What you do is take the Kummer sequence:

$\displaystyle 0\rightarrow \mu_{\ell^n}\rightarrow \mathbb{G}_m\stackrel{\ell^n}{\rightarrow} \mathbb{G}_m\rightarrow 0.$

This is an exact sequence of sheaves in the étale topology. Thus it gives you a long exact sequence of cohomology. But since ${H^1_{et}(X, \mathbb{G}_m)=Pic(X)}$ and ${H^2_{et}(X, \mathbb{G}_m)=Br(X)}$. Just writing down the long exact sequence you get that the image of ${H^1_{et}(X, \mu_{\ell^n})\rightarrow Pic(X)}$ is exactly ${Pic(X)[\ell^n]}$, and similarly with the Brauer group. In fact, people usually work with the truncated short exact sequence:

$\displaystyle 0\rightarrow Pic(X)/\ell^n Pic(X) \rightarrow H^2_{et}(X, \mu_{\ell^n})\rightarrow Br(X)[\ell^n]\rightarrow 0$

Fiddling around with other related things can help you figure out what is happening with the ${\ell^n}$-torsion. That isn’t the point of this post though. The point is what do you do when you want to figure out the ${p^n}$-torsion where ${p}$ is the characteristic of the ground field? It looks like you’re in big trouble, because the above Kummer sequence is not exact in the étale topology.

It turns out that you can switch to a finer topology called the fppf topology (or site). This is similar to the étale site, except instead of making your covering families using étale maps you make them with faithfully flat and locally of finite presentation maps (i.e. fppf for short when translated to french). When using this finer topology the sequence of sheaves actually becomes exact again.

A proof is here, and a quick read through will show you exactly why you can’t use the étale site. You need to extract ${p}$-th roots for the ${p}$-th power map to be surjective which will give you some sort of infinitesimal cover (for example if ${X=Spec(k)}$) that looks like ${Spec(k[t]/(t-a)^p)\rightarrow Spec(k)}$.

Thus you can try to figure out the ${p^n}$-torsion again now using “flat cohomology” which will be denoted ${H^i_{fl}(X, -)}$. We get the same long exact sequences to try to fiddle with:

$\displaystyle 0\rightarrow Pic(X)/p^n Pic(X) \rightarrow H^2_{fl}(X, \mu_{p^n})\rightarrow Br(X)[p^n]\rightarrow 0$

But what the heck is ${H^2_{fl}(X, \mu_{p^n})}$? I mean, how do you compute this? We have tons of books and things to compute with the étale topology. But this fppf thing is weird. So secretly we really want to translate this flat cohomology back to some étale cohomology. I saw the following claimed in several places without really explaining it, so we’ll prove it here:

$\displaystyle H^2_{fl}(X, \mu_p)=H^1_{et}(X, \mathbb{G}_m/\mathbb{G}_m^p).$

Actually, let’s just prove something much more general. We actually get that

$\displaystyle H^i_{fl}(X, \mu_p)=H^{i-1}_{et}(X, \mathbb{G}_m/\mathbb{G}_m^p).$

The proof is really just a silly “trick” once you see it. Since the Kummer sequence is exact on the fppf site, by definition this just means that the complex ${\mu_p}$ thought of as concentrated in degree ${0}$ is quasi-isomorphic to the complex ${\mathbb{G}_m\stackrel{p}{\rightarrow} \mathbb{G}_m}$. It looks like this is a useless and more complicated thing to say, but this means that the hypercohomology (still fppf) is isomorphic:

$\displaystyle \mathbf{H}^i_{fl}(X, \mu_p)=\mathbf{H}^i_{fl}(X, \mathbb{G}_m\stackrel{p}{\rightarrow} \mathbb{G}_m).$

Now here’s the trick. The left side is the group we want to compute. The right hand side only involves smooth group schemes, so a theorem of Grothendieck tells us that we can compute this hypercohomology using fpqc, fppf, étale, Zariski … it doesn’t matter. We’ll get the same answer. Thus we can switch to the étale site. But of course, just by definition we now extend the ${p}$-th power map (injective on the etale site) to an exact sequence

$\displaystyle 0\rightarrow \mathbb{G}_m \rightarrow \mathbb{G}_m\rightarrow \mathbb{G}_m/\mathbb{G}_m^p\rightarrow 0.$

Thus we get another quasi-isomorphism of complexes. This time to ${\mathbb{G}_m/\mathbb{G}_m^p[-1]}$. This is a complex concentrated in a single degree, so the hypercohomology is just the etale cohomology. The shift by ${-1}$ decreases the cohomology by one and we get the desired isomorphism ${H^i_{fl}(X, \mu_p)=H^{i-1}_{et}(X, \mathbb{G}_m/\mathbb{G}_m^p)}$. In particular, we were curious about ${H^2_{fl}(X, \mu_p)}$, so we want to figure out ${H^1_{et}(X, \mathbb{G}_m/\mathbb{G}_m^p)}$.

Alright. You’re now probably wondering what in the world to I do with the étale cohomology of ${\mathbb{G}_m/\mathbb{G}_m^p}$? It might be on the étale site, but it is a weird sheaf. Ah. But here’s something great, and not used all that much to my knowledge. There is something called the multiplicative de Rham complex. On the étale site we actually have an exact sequence of sheaves via the “dlog” map:

$\displaystyle 0\rightarrow \mathbb{G}_m/\mathbb{G}_m^p\stackrel{d\log}{\rightarrow} Z^1\stackrel{C-i}{\rightarrow} \Omega^1\rightarrow 0.$

This now gives us something nice because if we understand the Cartier operator (which is Serre dual to the Frobenius!) and know things how many global ${1}$-forms are on the variety (maybe none?) we have a hope of computing our original flat cohomology!

## Brauer Groups of Curves

Let ${C/k}$ be a smooth projective curve over an algebraically closed field. The main goal of today is to show that ${Br(C)=0}$. Both smooth and being over an algebraically closed field are crucial for this computation. The computation will run very similarly to the last post with basically one extra step.

We haven’t actually talked about the Brauer group for varieties, but there are again two definitions. One has to do with Azumaya algebras over ${\mathcal{O}_C}$ modulo Morita equivalence. The other is the cohomological Brauer group, ${Br'(C):=H^2(C, \mathbb{G}_m)}$. As already stated, it is a big open problem to determine when these are the same. We’ll continue to only consider situations where they are known to be the same and hence won’t cause any problems (or even require us to define rigorously the Azumaya algebra version).

First, note that if we look at the Leray spectral sequence with the inclusion of the generic point ${g:Spec(K)\hookrightarrow C}$ we get that ${R^1g_*\mathbb{G}_m=0}$ by Hilbert 90 again which tells us that ${0\rightarrow H^2(C, g_*\mathbb{G}_m)\hookrightarrow Br(K)}$. Now ${K}$ has transcendence degree ${1}$ over an algebraically closed field, so by Tsen’s theorem this is ${C_1}$. Thus the last post tells us that ${H^2(C, g_*\mathbb{G}_m)=0}$.

The new step is that we need to relate ${H^2(C, g_*\mathbb{G}_m)}$ to ${Br(C)}$. On the étale site of ${C}$ we have an exact sequence of sheaves

$\displaystyle 0\rightarrow \mathbb{G}_m\rightarrow g_*\mathbb{G}_m\rightarrow Div_C\rightarrow 0$

where ${\displaystyle Div_C=\bigoplus_{v \ \text{closed}}(i_v)_*\mathbb{Z}}$.
Taking the long exact sequence on cohomology we get

$\displaystyle \cdots \rightarrow H^1(C, Div_C)\rightarrow Br(C)\rightarrow H^2(C, g_*\mathbb{G}_m)\rightarrow \cdots .$

Thus it will complete the proof to show that ${H^1(C, Div_C)=0}$, since then ${Br(C)}$ will inject into ${0}$. Writing ${\displaystyle Div_C=\bigoplus_{v \ \text{closed}}(i_v)_*\mathbb{Z}}$ and using that cohomology commutes with direct sums we need only show that for some fixed closed point ${(i_v): Spec(k(v))\hookrightarrow C}$ that ${H^1(C, (i_v)_*\mathbb{Z})=0}$.

We use Leray again, but this time on ${i_v}$. For notational convenience, we’ll abuse notation and call both the map and the point ${v\in C}$. The low degree terms give us ${H^1(C, v_*\mathbb{Z})\hookrightarrow H^1(v, \mathbb{Z})}$. Using the Galois cohomology interpretation of étale cohomology of a point ${H^1(v,\mathbb{Z})\simeq Hom_{cont}(G_{k(v)}, \mathbb{Z})}$ (the homomorphisms are not twisted since the Galois action is trivial). Since ${G_{k(v)}}$ is profinite, the continuous image is compact and hence a finite subgroup of ${\mathbb{Z}}$. Thus ${H^1(C, v_*\mathbb{Z})=0}$ which implies ${H^1(C, Div_C)=0}$ which gives the result that ${Br(C)=0}$.

So again we see that even for a full curve being over an algebraically closed field is just too strong a condition to give anything interesting. This suggests that the Brauer group really is measuring some arithmetic properties of the curve. For example, we could ask whether or not good/bad reduction of the curve is related to the Brauer group, but this would require us to move into Brauer groups of surfaces (since the model will be a relative curve over a one-dimensional base).

Already for local fields or ${C_1}$ fields the question of determining ${Br(C)}$ is really interesting. The above argument merely tells us that ${Br(C)\hookrightarrow Br(K)}$ where ${K}$ is the function field, but this is true of all smooth, proper varieties and often doesn’t help much if the group is non-zero.

## Intro to Brauer Groups

I want to do a series on the basics of Brauer groups since they came up in the past few posts. Since I haven’t really talked about Galois cohomology anywhere, we’ll take a slightly nonstandard approach and view everything “geometrically” in terms of étale cohomology. Everything should be equivalent to the Galois cohomology approach, but this way will allow us to use the theory that is already developed elsewhere on the blog.

I apologize in advance for the sporadic nature of this post. I just need to get a few random things out there before really starting the series. There will be one or two posts on the Brauer group of a “point” which will just mean the usual Brauer group of a field (to be defined shortly). Then we’ll move on to the Brauer group of a curve, and maybe if I still feel like continuing the series of a surface.

Let ${K}$ be a field and ${K^s}$ a fixed separable closure. We will define ${Br(K)=H^2_{et}(Spec(K), \mathbb{G}_m)=H^2(Gal(K^s/K), (K^s)^\times)}$. This isn’t the usual definition and is often called the cohomological Brauer group. The usual definition is as follows. Let ${R}$ be a commutative, local, (unital) ring. An algebra ${A}$ over ${R}$ is called an Azumaya algebra if it is a free of finite rank ${R}$-module and ${A\otimes_R A^{op}\rightarrow End_{R-mod}(A)}$ sending ${a\otimes a'}$ to ${(x\mapsto axa')}$ is an isomorphism.

Define an equivalence relation on the collection of Azumaya algebras over ${R}$ by saying ${A}$ and ${A'}$ are similar if ${A\otimes_R M_n(R)\simeq A'\otimes_R M_{n'}(R)}$ for some ${n}$ and ${n'}$. The set of Azumaya algebras over ${R}$ modulo similarity form a group with multiplication given by tensor product. This is called the Brauer group of ${R}$ denoted ${Br(R)}$. Often times, when an author is being careful to distinguish, the cohomological Brauer group will be denoted with a prime: ${Br'(R)}$. It turns out that there is always an injection ${Br(R)\hookrightarrow Br'(R)}$.

One way to see this is that on the étale site of ${Spec(R)}$, the sequence of sheaves ${1\rightarrow \mathbb{G}_m\rightarrow GL_n\rightarrow PGL_n\rightarrow 1}$ is exact. It is a little tedious to check, but using a Čech cocycle argument (caution: a priori the cohomology “groups” are merely pointed sets) one can check that the injection from the associated long exact sequence ${H^1(Spec(R), PGL_n)/H^1(Spec(R), GL_n)\hookrightarrow Br'(R)}$ is the desired injection.

If we make the extra assumption that ${R}$ has dimension ${0}$ or ${1}$, then the natural map ${Br(R)\rightarrow Br'(R)}$ is an isomorphism. I’ll probably regret this later, but I’ll only prove the case of dimension ${0}$, since the point is to get to facts about Brauer groups of fields. If ${R}$ has dimension ${0}$, then it is a local Artin ring and hence Henselian.

One standard lemma to prove is that for local rings a cohomological Brauer class ${\gamma\in Br'(R)}$ comes from an Azumaya algebra if and only if there is a finite étale surjective map ${Y\rightarrow Spec(R)}$ such that ${\gamma}$ pulls back to ${0}$ in ${Br'(Y)}$. The easy direction is that if it comes from an Azumaya algebra, then any maximal étale subalgebra splits it (becomes the zero class after tensoring), so that is our finite étale surjective map. The other direction is harder.

Going back to the proof, since ${R}$ is Henselian, given any class ${\gamma\in H^2(Spec(R), \mathbb{G}_m)}$ a standard Čech cocycle argument shows that there is an étale covering ${(U_i\rightarrow Spec(R))}$ such that ${\gamma|_{U_i}=0}$. Choosing any ${U_i\rightarrow Spec(R)}$ we have a finite étale surjection that kills the class and hence it lifts by the previous lemma.

It is a major open question to find conditions to make ${Br(X)\rightarrow Br'(X)}$ surjective, so don’t jump to the conclusion that we only did the easy case, but it is always true. Now that we have that the Brauer group is the cohomological Brauer group we can convert the computation of ${Br(R)}$ for a Henselian local ring to a cohomological computation using the specialization map (pulling back to the closed point) ${Br(R)\rightarrow Br(k)}$ where ${k=R/m}$.

## Taniyama-Shimura 2: Galois Representations

Fix some proper variety ${X/\mathbb{Q}}$. Our goal today will seem very strange, but it is to explain how to get a continuous representation of the absolute Galois group of ${\mathbb{Q}}$ from this data. I’m going to assume familiarity with etale cohomology, since describing Taniyama-Shimura is already going to take a bit of work. To avoid excessive notation, all cohomology in this post (including the higher direct image functors) are done on the etale site.

For those that are intimately familiar with etale cohomology, we’ll do the quick way first. I’ll describe a more hands on approach afterwards. Let ${\pi: X\rightarrow \mathrm{Spec} \mathbb{Q}}$ be the structure morphism. Fix an algebraic closure ${v: \mathrm{Spec} \overline{\mathbb{Q}}\rightarrow \mathrm{Spec}\mathbb{Q}}$ (i.e. a geometric point of the base). We’ll denote the base change of ${X}$ with respect to this morphism ${\overline{X}}$. Suppose the dimension of ${X}$ is ${n}$.

Let ${\ell}$ be a prime. We consider the constructible sheaf ${R^n\pi_*(\mathbb{Z}/\ell^m)}$. Now we have an equivalence of categories between these sheaves and continuous ${G=Gal(\overline{\mathbb{Q}}/\mathbb{Q})}$-modules by taking the stalk at our geometric point. Thus ${R^n\pi_*(\mathbb{Z}/\ell^m)_v\simeq H^n(\overline{X}, \mathbb{Z}/\ell^m)}$ has a continuous action of ${G}$ on it, and hence we get a continuous representation ${\rho_{X,m}: G\rightarrow Aut(H^n(\overline{X}, \mathbb{Z}/\ell^m)\simeq GL_d(\mathbb{Z}/\ell^m)}$. These all form a compatible family and hence we can take the inverse limit and tensor with ${\mathbb{Q}_\ell}$ to get what is known as an ${\ell}$-adic Galois representation ${\rho_X: G\rightarrow GL_d(\mathbb{Q}_\ell)}$. For a technicality that will come up later, we will abuse notation and now relabel ${\rho_X}$ to be the dual (or contragredient) representation.

If you aren’t comfortable with etale cohomology, then you can just use it as a black box cohomology theory to get the same thing as follows. First take the base change ${\overline{X}\rightarrow \mathrm{Spec} \overline{\mathbb{Q}}}$. Given any element of the Galois group ${\sigma \in G}$ we get an automorphism of ${\overline{\mathbb{Q}}}$. Thus we can fill in the diagram:

${\begin{matrix} \overline{X} & \stackrel{\sigma}{\rightarrow} & \overline{X} \\ \downarrow & & \downarrow \\ \mathrm{Spec} \overline{\mathbb{Q}} & \stackrel{\sigma}{\rightarrow} & \mathrm{Spec} \overline{\mathbb{Q}} \end{matrix}}$

Since ${\sigma}$ was an automorphism, then only thing you have to believe about cohomology is that you then get an isomorphism via pullback ${H^n(\overline{X}, \mathbb{Q}_\ell)\stackrel{\sigma^*}{\rightarrow} H^n(\overline{X}, \mathbb{Q}_\ell)}$. Thus we get a continuous group homomorphism ${G\rightarrow Aut(H^n(\overline{X}, \mathbb{Q}_\ell))}$ as before. Again, we’ll actually use the dual of this in the future.

To return to an elliptic curve ${E}$ over ${\mathbb{Q}}$, we know that these are just tori, and hence the first Betti number is ${2}$. In this case we get that our Galois representation ${\rho_E: G\rightarrow GL_2(\mathbb{Q}_\ell)}$. If you’ve seen Taniyama-Shimura explained before, this should look familiar. This turns out to be exactly the same representation as the one you get from the Galois action on the Tate module. But the definition of the Tate module requires a group law, and hence the ability to get such a representation doesn’t generalize to all varieties in the way that using middle $\ell$-adic cohomology does. This is the standard modern approach to defining modularity for other types of varieties.

## Problems with de Rham and l-adic

Let’s begin by reviewing the fact that when we work over ${\mathbb{C}}$ we have a nice comparison theorem. Last time it was briefly mentioned that if ${X}$ is smooth over ${\mathbb{C}}$, then we can consider the analytic topology on the ${\mathbb{C}}$-valued points which we’ll call ${X^{an}}$. Using GAGA and the degeneration of the Hodge-de Rham spectral sequence from last time we immediately get that ${\mathbf{H}^*(X, \Omega_{X/\mathbb{C}}^\cdot)\stackrel{\sim}{\rightarrow} \mathbf{H}^*(X^{an}, \Omega_{X^{an}}^\cdot)\simeq H^*(X, \mathbb{C})}$.

Thus we have a comparison theorem that allows us to use purely algebraic methods to recover the topological information of the singular cohomology. But what happens if we work over a field that is not ${\mathbb{C}}$? Well let’s assume that ${X}$ is smooth and of finite type over ${k}$. If ${k}$ has characteristic ${0}$, then the standard idea of the Lefschetz principle gives us that everything works out again and we can recover the singular cohomology.

Obviously lots can go wrong in positive characteristic. Say ${\mathrm{char}(k)=p}$, then we’ll throw out the case of affine things because for instance if ${X}$ is an affine curve the de Rham cohomology is not even finitely generated. But if we assume ${X/k}$ is proper and have finite generatedness, we still have problems with de Rham. It doesn’t recover the correct topological information. For instance, we’d like a cohomology that gives us “correct” Betti numbers.

If we define the Hodge Betti numbers to be ${\displaystyle b^i_H=\sum_{p+q=i}\dim_k H^q(X, \Omega^p)}$ and the de Rham Betti numbers ${b^i_{dR}=\dim_k H^i_{dR}(X/k)}$ we can just look at the HdR SS and see that ${b^i_H\leq b^i_{dR}}$ and we have equality if and only if the HdR SS degenerates at ${E_1}$. Degeneration of this spectral sequence is related to having some sort of analogue of ${X}$ in characteristic ${0}$, also known as a “lift” of ${X}$. Even if this spectral sequence degenerates, it merely relates these two Betti numbers, but doesn’t fully relate back to the topological Betti numbers.

Also, despite the fact that there is still a Lefschetz fixed point formula for de Rham cohomology in positive characteristic, it only gives the right answer mod ${p}$. We’ve seen that de Rham cohomology is a nice attempt at coming up with an algebraic way of defining singular cohomology (and is quite useful for lots and lots of things), but it seems very tied to smoothness and characteristic ${0}$ issues.

Anyway, it was good to point that out, but many of you might be frustrated because we already knew that de Rham was going to be a failure in characteristic ${p}$. If you’ve been introduced to ${\ell}$-adic cohomology, then you probably already realized that this was basically invented to solve some of these problems. Now we’ll impose the extra condition that ${k}$ be algebraically closed. Recall that we can just pick a prime ${\ell\neq p}$ and define ${H^i_\ell (X)=H^i(X_{et}, \mathbb{Z}_\ell)=\lim H^i(X_{et}, \mathbb{Z}/\ell^n)}$. This is a ${\mathbb{Z}_\ell}$-module.

Using the comparison theorem we see that if ${k=\mathbb{C}}$, then ${H^i_\ell(X)\simeq H^i(X^{an}, \mathbb{Z})\otimes \mathbb{Z}_\ell}$. This is quite nice. ${H^i_\ell(X)}$ tells us exactly the topological information we sought, since we recover the rank of ${H^i(X, \mathbb{Z})}$ and the ${\ell}$-primary torsion. Thus if we run through all primes ${\ell}$ we can completely recover ${H^i(X, \mathbb{Z})}$ up to isomorphism. Also, we see that the rank of ${H^i_\ell(X)}$ is finite and independent of ${\ell}$.

Here is the first of many problems. Although for a smooth projective scheme in positive characteristic we do get that the rank of ${H^i_\ell(X)}$ is finite, it is unknown if this is a number independent of ${\ell}$. The second much more severe problem is that we only have the reasonable properties listed in the previous paragraph for ${\ell\neq p}$. So even though we can get a bunch of topological information about ${H^i(X, \mathbb{Z})}$, we are unfortunately left clueless about the ${p}$-torsion. We also lose a bunch of classical theorems (even when ${\ell\neq p}$) such as the Lefschetz hyperplane theorem. Or if we consider the induced map from ${X\rightarrow X}$ on ${H^i_\ell(X)\rightarrow H^i_\ell(X)}$ there may be powers of ${p}$ in the denominator of the characteristic polynomial that we won’t ever be able to get information about.

Clearly, ${\ell}$-adic cohomology isn’t going to quite do the trick. All I seem to be doing is complaining about how these cohomology theories are failing what I want, but I haven’t ever told you exactly what it is that I do want from a cohomology theory. So next time we’ll start working on stating those properties and seeing if anything we’ve come across satisfies them and if not how we’re going to construct something that does.