## Serre-Tate Theory 1

Today we’ll try to answer the question: What is Serre-Tate theory? It’s been a few years, but if you’re not comfortable with formal groups and ${p}$-divisible groups, I did a series of something like 10 posts on this topic back here: formal groups, p-divisible groups, and deforming p-divisible groups.

The idea is the following. Suppose you have an elliptic curve ${E/k}$ where ${k}$ is a perfect field of characteristic ${p>2}$. In most first courses on elliptic curves you learn how to attach a formal group to ${E}$ (chapter IV of Silverman). It is suggestively notated ${\widehat{E}}$, because if you unwind what is going on you are just completing the elliptic curve (as a group scheme) at the identity.

Since an elliptic curve is isomorphic to it’s Jacobian ${Pic_E^0}$ there is a conflation that happens. In general, if you have a variety ${X/k}$ you can make the same formal group by completing this group scheme and it is called the formal Picard group of ${X}$. Although, in general you’ll want to do this with the Brauer group or higher analogues to guarantee existence and smoothness. Then you prove a remarkable fact that the elliptic curve is ordinary if and only if the formal group has height ${1}$. In particular, since the ${p}$-divisible group is connected and ${1}$-dimensional it must be isomorphic to ${\mu_{p^\infty}}$.

It might seem silly to think in these terms, but there is another “enlarged” ${p}$-divisible group attached to ${E}$ which always has height ${2}$. This is the ${p}$-divisible group you get by taking the inductive limit of the finite group schemes that are the kernel of multiplication by ${p^n}$. It is important to note that these are non-trivial group schemes even if they are “geometrically trivial” (and is the reason I didn’t just call it the “${p^n}$-torsion”). We’ll denote this in the usual way by ${E[p^\infty]}$.

I don’t really know anyone that studies elliptic curves that phrases it this way, but since this theory must be generalized in a certain way to work for other varieties like K3 surfaces I’ll point out why this should be thought of as an enlarged ${p}$-divisible group. It is another standard fact that ${E}$ is ordinary if and only if ${E[p^\infty]\simeq \mu_{p^\infty}\oplus \mathbb{Q}_p/\mathbb{Z}_p}$. In fact, you can just read off the connected-etale decomposition:

$\displaystyle 0\rightarrow \mu_{p^\infty}\rightarrow E[p^\infty] \rightarrow \mathbb{Q}_p/\mathbb{Z}_p\rightarrow 0$

We already noted that ${\widehat{E}\simeq \mu_{p^\infty}}$, so the ${p}$-divisible group ${E[p^\infty]}$ is a ${1}$-dimensional, height ${2}$ formal group whose connected component is the first one we talked about, i.e. ${E[p^\infty]}$ is an enlargement of ${\widehat{E}}$. For a general variety, this enlarged formal group can be defined, but it is a highly technical construction and would take a lot of work to check that it even exists and satisfies this property. Anyway, this enlarged group is the one we need to work with otherwise our deformation space will be too small to make the theory work.

Here’s what Serre-Tate theory is all about. If you take a deformation of your elliptic curve ${E}$ say to ${E'}$, then it turns out that ${E'[p^\infty]}$ is a deformation of the ${p}$-divisible group ${E[p^\infty]}$. Thus we have a natural map ${\gamma: Def_E \rightarrow Def_{E[p^\infty]}}$. The point of the theory is that it turns out that this map is an isomorphism (I’m still assuming ${E}$ is ordinary here). This is great news, because the deformation theory of ${p}$-divisible groups is well-understood. We know that the versal deformation of ${E[p^\infty]}$ is just ${Spf(W[[t]])}$. The deformation problem is unobstructed and everything lives in a ${1}$-dimensional family.

Of course, let’s not be silly. I’m pointing all this out because of the way in which it generalizes. We already knew this was true for elliptic curves because for any smooth, projective curve the deformations are unobstructed since the obstruction lives in ${H^2}$. Moreover, the dimension of the space of deformations is given by the dimension of ${H^1(E, \mathcal{T})}$. But for an elliptic curve ${\mathcal{T}\simeq \mathcal{O}_X}$, so by Serre duality this is one-dimensional.

On the other hand, we do get some actual information from the Serre-Tate theory isomorphism because ${Def_{E[p^\infty]}}$ carries a natural group structure. Thus an ordinary elliptic curve has a “canonical lift” to characteristic ${0}$ which comes from the deformation corresponding to the identity.

## 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!

## Bayesianism in the Philosophy of Math

Today I’ll sketch an idea that I fist learned about from David Corfield’s excellent book Towards a Philosophy of Real Mathematics. I read it about six years ago while doing my undergraduate honors thesis and my copy is filled with notes in the margins. It has been interesting to revisit this book. What I’m going to talk about is done in much greater detail and thoroughness with tons of examples in that book. So check it out if this is interesting to you.

There are lots of ways we could use Bayesian analysis in the philosophy of math. I’ll just use a single example to show how we can use it to describe how confident we are in certain conjectures. In other words, we’ll come up with a probability for how plausible a conjecture is given the known evidence. As usual we’ll denote this P(C|E). Before doing this, let’s address the question of why would we want to do this.

To me, there are two main answers to this question. The first is that mathematicians already do this colloquially. When someone proposes something in an informal setting, you hear phrases like, “I don’t believe that at all,” or “How could that be true considering …” or “I buy that, it seems plausible.” If you think that the subject of philosophy of mathematics has any legitimacy, then certainly one of its main goals would be to take such statements and try to figure out what is meant by them and whether or not they seem justified. This is exactly what our analysis will do.

The second answer is much more practical in nature. Suppose you conjecture something as part of your research program. As we’ve been doing in these posts, you could use Baye’s theorem to give two estimates on the plausibility of your conjecture being true. One is giving the most generous probabilities given the evidence, and the other is giving the least generous. You’ll get some sort of Bayesian confidence interval of the probability of the conjecture being true. If the entire interval is low (say below 60% or something), then before spending several months trying prove it your time might be better spent gathering more evidence for or against it.

Again, mathematicians already do this at some subconscious level, so being aware of one way to analyze what it is you are actually doing could be very useful. Humans have tons of cognitive biases, so maybe you have greatly overestimated how likely something is and doing a quick Bayes’ theorem calculation can set you straight before wasting a ton of time. Or you could write all this off as nonsense. Whatever. It’s up to you.

If you’ve followed the posts up to now, you’ll probably find this calculation quite repetitive. You can probably guess what we’ll do. We want to figure out P(C|E), the probability that a conjecture is true given the evidence you’ve accumulated. What goes into Bayes’ theorem? Well, P(E|C) the probability that we would see the evidence we have supposing the conjecture is true; P(C) the prior probability that the conjecture is true; P(E|-C) the probability we would see the evidence we have supposing the conjecture is not true; and P(-C) the prior probability that the conjecture is not true.

Clearly the problem of assigning some exact probability to any of these is insanely subjective. But also, as before, it should be possible to find the most optimistic person about a conjecture to overestimate the probability and the most skeptical person to underestimate the probability. This type of interval forming should be a lot less subjective and fairly consistent. One should even have strong arguments to support the estimates which will convince someone who questions them.

Let’s use the Riemann hypothesis as an example. In our modern age, we have massive numerical evidence that the Riemann hypothesis is true. Recall that it just says that all the zeroes of the Riemann zeta function in the critical strip lie on the line with real part 1/2. Something like the first 10,000,000,000,000 zeroes have been checked by computer plus lots (billions?) have been checked in random other places after this.

Interestingly enough, if this were our “evidence” our estimation of P(E|C) may as well be 1, but P(E|-C) would have to contribute a significant non-trivial factor in the denominator of Bayes’ theorem. This is because we estimate this probability based on what we’ve seen in the past in similar situations. It turns out that in analytic number theory we have several prior instances of the phenomenon of a conjecture looking true for exceedingly large numbers before getting a counterexample. In fact, Merten’s Conjecture is explicitly connected to the Riemann hypothesis and the first counterexample could be around $10^{30}$ (no explicit counterexample is known, just that one exists, but we know by checking that it is exceedingly large).

It probably isn’t unreasonable to say that most mathematicians believe the Riemann hypothesis. Even giving generous prior probabilities, the above analysis would give a not too high level of confidence. So where does the confidence come from? Remember, that in Bayesian analysis it is often easy to accidentally not use all available evidence (subconscious bias may play a role in this process).

I could do an entire series on the analogies and relations between the Riemann hypothesis for curves over finite fields and the standard Riemann hypothesis, so I won’t explain it here. The curves over finite fields case has been proven and provides quite good evidence in terms of making P(E|-C) small.

The Bayesian calculation becomes much, much more complicated in terms of modern mathematics because of all the analogies and more concretely the ways in which the RH is interrelated with theorems about number fields and Galois representations and cohomological techniques. We have conjectures equivalent to (or implying or implied by) the RH which allows us to transfer evidence for and against these other conjectures.

In some sense, essentially all this complication will only increase the Bayesian estimate, so we could simplify our lives and make some baseline estimate taking into account the clearest of these and then just say that our confidence is at least that much. That is one explanation of why many mathematicians beleive the RH even if they’ve never explicitly thought of it that way. Well, this has gone on too long, but I hope the idea has been elucidated.

## This Week’s Finds in Arithmetic Geometry

It has been going around the math blogosphere that in honor of John Baez’s 20 year anniversary of doing This Week’s Finds we all do one in our area. So here’s a brief This Week’s Finds in Arithmetic Geometry. Hopefully this will raise awareness of the blog that essentially pioneered math/physics blogging (and if you’re into arithmetic geometry some papers you might not have caught yet).

Since the content and style of Baez’s This Week’s Finds vary so much, I’ll just copy what Jordan Ellenberg did here and give some papers posted the last week that caught my attention.

The fields of definition of branched Galois covers of the projective line by Hilaf Hasson caught my attention because I just met him and saw him speak on this exact topic a few weeks ago at the Joint Meetings. Although the results are certainly interesting in themselves, the part that a blog audience might appreciate is the set of corollaries to the results.

Recall that a major open problem in number theory is the “Inverse Galois Problem” which asks which groups arise as Galois groups. I even posted an elementary proof that if you don’t care what your fields are, then any finite group arises as ${Gal(L/K)}$. In general (for example if you force ${K=\mathbb{Q}}$), then the problem is extremely hard and wide open.

If you haven’t seen this type of thing before, then it might be surprising, but you can actually use geometry to study this question. This is exactly the type of result that Hilaf gets.

Next is New derived autoequivalences of Hilbert schemes and generalised Kummer varieties by Andreas Krug. This topic is near and dear to me because I study derived categories in the arithmetic setting. I haven’t taken a look at this paper in any depth, but I’ll just point out why these types of things are important.

In the classification of varieties one often tries to study the problem up to some type of birational equivalence otherwise it would be too difficult. Often times birational varieties are derived equivalent, but not the other way around. So one could think of studying varieties up to derived equivalence as a slightly looser classification.

When trying to figure out what two varieties that are derived equivalent have in common, a typical sticking point is that you need to know certain automorphisms of the derived category (i.e. autoequivalence) exist to get nice cohomological properties or something. When papers constructing new autoequivalences come out it always catches my attention because I want to know if the method used transfers to situations I work in.

Lastly, we’ll do La conjecture de Tate entière pour les cubiques de dimension quatre sur un corps fini by François Charles and Alena Pirutka. If you don’t know what the Tate conjecture is, then lots of people refer to it as the Hodge conjecture in positive characteristic.

If you’ve ever seen any cohomology theory, then you should be at least passingly familiar with the idea that certain sub-objects (subvarieties or submanifolds etc) can be realized as classes in the cohomology. Sometimes this is due to construction and sometimes it is a major theorem.

The particular case of the Tate conjecture says the following. Consider the relatively easy to prove fact. If you take a cycle on your variety ${X/k}$, then the cohomology class it maps to (in ${\ell}$-adic cohomology) will be invariant under the natural Galois action ${Gal(\overline{k}/k)}$ (because it is defined over ${k}$!). The Tate conjecture is that any Galois invariant cohomology class actually comes from one of these cycles.

The fact that mathematicians can have honest arguments over whether or not the Tate conjecture or the Hodge conjecture (a million dollar problem!) is harder just gives credence to the fact that it is darned hard. If you weren’t convinced, then just consider that this paper is proving the Tate conjecture in the particular case of smooth hypersurfaces of degree ${3}$ in ${\mathbb{P}^5}$ just for the cohomology classes of degree ${4}$. People consider this progress, and they should.

## Brauer-Manin Obstruction

We will continue with the Brauer group after this post, but today let’s answer the question: Why should we care about the Brauer group? We gave one good reason back when talking about rational surfaces. It provided something that might be computable to detect whether or not our variety had good reduction.

Today let’s consider what is probably the most well-known use of the Brauer group. Suppose we have ${X/K}$ a (smooth, proper) variety over a number field. We want to know whether or not there are any ${K}$-rational points. Of course, classically if you hand me this variety using an equation, then the problem reduces to a Diophantine equation. So this problem is very old (3rd century … and one could argue we still don’t have a great handle on it).

Suppose ${Spec \ K\rightarrow X}$ is a rational point. Then we can use the embedding ${K\hookrightarrow K_v}$ into various completions of ${K}$ to get ${Spec \ K_v\rightarrow X}$ points over all the local fields. This isn’t saying much. It says that if we have a global point, then we have points locally everywhere. The interesting question is whether having points locally everywhere “glue” to give a global point. When a class of varieties always allows you to do this, then we say the varieties satisfy the Hasse principle.

This brings us to the content of today’s post. We will construct the Brauer-Manin obstruction to the Hasse principle. We’ve already considered the following pairing in a previous post. For any prime ${v}$, we get a pairing ${Br(X_{K_v})\times X(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}}$ which is just given by pullback on étale cohomology using the map defining the point ${Spec \ K_v\rightarrow X_{K_v}}$ followed by the canonical identification ${Br(K_v)\stackrel{\sim}{\rightarrow}\mathbb{Q}/\mathbb{Z}}$. We could write ${(\alpha, x_v)=x_v^*(\alpha)}$ or more typically ${\alpha(x_v)}$.

We can package this into something global as follows. By restriction we have ${\displaystyle Br(X)\hookrightarrow \prod_v Br(X_{K_v})}$ which we will write ${\alpha\mapsto (\alpha_v)}$. Now we just sum to get ${\displaystyle Br(X)\times \prod_v X(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}}$, so our global pairing is ${\displaystyle (\alpha, (x_v))\mapsto \sum_v \alpha_v(x_v)}$.

Recall that we have an exact sequence from the post on Brauer groups of fields ${0\rightarrow Br(K)\rightarrow \oplus_v Br(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}\rightarrow 0}$ by summing the local invariants. If our local points glue to give a global point ${x\mapsto (x_v)}$, then the pairing factors through this sequence and hence ${(\alpha, (x_v))=0}$ for any choice of ${\alpha}$.

This is our obstruction. The above argument says that if ${(\alpha, (x_v))\neq 0}$, then ${(x_v)}$ cannot possibly glue to give a rational point. We will give this set a name:

$\displaystyle X^{Br}=\{(x_v)\in \prod_v X(K_v) : (\alpha, x_v)=0 \ \text{for all} \ \alpha\in Br(K)\}$

We will call this the Brauer set of ${X}$ (earlier we called it being “Brauer equivalent to ${0}$” since we saw this type of condition was an equivalence relation on Chow groups). It is now immediate that ${X(K)\subset X^{Br}}$. We should think of this as the collection of local points that have some chance of coming from a global point. Now we have two obstructions to the existence of rational points. The first is that ${\prod_v X(K_v)\neq \emptyset}$ which is just the trivial condition that there are local points everywhere. The second is the Brauer-Manin obstruction which says that ${X^{Br}\neq \emptyset}$.

If the local points condition is necessary and sufficient for the existence of rational points, then of course that is exactly the Hasse principle, so we say the Hasse principle holds. If the Brauer-Manin condition (plus the local condition) is necessary and sufficient, then we say that the Brauer-Manin obstruction is the only obstruction to the existence of rational points. It would be fantastic if we could somehow figure out which varieties had this property.

Caution: It is not an open problem to determine whether or not the Brauer-Manin obstruction is the only obstruction. There are known examples where there is no B-M obstruction and yet there are still no rational points. As far as I can tell, it is conjectured, but still open that if the Tate-Shafarevich group of the Jacobian of a curve is finite, then the B-M obstruction is the only obstruction to the existence of rational points on curves over number fields. So even in such a special low-dimensional situation this is a very hard question.

## More Complicated Brauer Computations

Let’s wrap up some of our Brauer group loose ends today. We can push through the calculation of the Brauer groups of curves over some other fields using the same methods as the last post, but just a little more effort.

First, note that with absolutely no extra effort we can run the same argument as yesterday in the following situation. Suppose ${X}$ is a regular, integral, quasi-compact scheme of dimension ${1}$ with the property that all closed points ${v\in X}$ have perfect residue fields ${k(v)}$. Let ${g: \text{Spec} K \hookrightarrow X}$ be the inclusion of the generic point.

Running the Leray spectral sequence a little further than last time still gives us an inclusion, but we will usually want more information because ${Br(K)}$ may not be ${0}$. The low degree terms (plus the argument from last time) gives us a sequence:

$\displaystyle 0\rightarrow Br'(X)\rightarrow Br(K)\rightarrow \bigoplus_v Hom_{cont}(G_{k(v)}, \mathbb{Q}/\mathbb{Z})\rightarrow H^3(X, \mathbb{G}_m)\rightarrow \cdots$

This allows us to recover a result we already proved. In the special case that ${X=\text{Spec} A}$ where ${A}$ is a Henselian DVR with perfect residue field ${k}$, then the uniformizing parameter defines a splitting to get a split exact sequence

$\displaystyle 0\rightarrow Br(A)\rightarrow Br(K)\rightarrow Hom_{cont}(G_k, \mathbb{Q}/\mathbb{Z})\rightarrow 0$

Thus when ${A}$ is a strict local ring (e.g. ${\mathbb{Z}_p}$) we get an isomorphism ${Br(K)\rightarrow \mathbb{Q}/\mathbb{Z}}$ since ${Br(A)\simeq Br(k)=0}$ (since ${k}$ is ${C_1}$). In fact, going back to Brauer groups of fields, we had a lot of trouble trying to figure anything out about number fields. Now we may have a tool (although without class field theory it isn’t very useful, so we’ll skip this for now).

The last computation we’ll do today is to consider a smooth (projective) curve over a finite field ${C/k}$. Fix a separable closure ${k^s}$ and ${K}$ the function field. First, we could attempt to use Leray on the generic point, since we can use that ${H^3(K, \mathbb{G}_m)=0}$ to get some more information. Unfortunately without something else this isn’t enough to recover ${Br(C)}$ up to isomorphism.

Instead, consider the base change map ${f: C^s=C\otimes_k k^s\rightarrow C}$. We use the Hochschild-Serre spectral sequence ${H^p(G_k, H^q(C^s, \mathbb{G}_m))\Rightarrow H^{p+q}(C, \mathbb{G}_m)}$. The low degree terms give us

$\displaystyle 0\rightarrow Br(k)\rightarrow \ker (Br(C)\rightarrow Br(C^s))\rightarrow H^1(G_k, Pic(C^s))\rightarrow \cdots$

First, ${\ker( Br(C)\rightarrow Br(C^s))=Br(C)}$ by the last post. Next ${H^1(G_k, Pic^0(C^s))=0}$ by Lang’s theorem as stated in Mumford’s Abelian Varieties, so ${H^1(G_k, Pic(C^s))=0}$ as well. That tells us that ${Br(C)\simeq Br(k)=0}$ since ${k}$ is ${C_1}$. So even over finite fields (finite was really used and not just ${C_1}$ for Lang’s theorem) we get that smooth, projective curves have trivial Brauer group.

## 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.

## Brauer Groups of Fields

Today we’ll talk about the basic theory of Brauer groups for certain types of fields. If the last post was too poorly written to comprehend, the only thing that will be used from it is that for fields we can refer to “the” Brauer group without any ambiguity because the cohomological definition and the Azumaya (central, simple) algebra definition are canonically isomorphic in this case.

Let’s just work our way from algebraically closed to furthest away from being algebraically closed. Thus, suppose ${K}$ is an algebraically closed field. The two ways to think about ${Br(K)}$ both tell us quickly that this is ${0}$. Cohomologically this is because ${G_K=1}$, so there are no non-trivial Galois cohomology classes. The slightly more interesting approach is that any central, simple algebra over ${K}$ is already split, i.e. a matrix algebra, so it is the zero class modulo the relation we defined last time.

I’m pretty sure I’ve blogged about this before, but there is a nice set of definitions that measures how “far away” from being algebraically closed you are. A field is called ${C_r}$ if for any ${d,n}$ such that ${n>d^r}$ any homogeneous polynomial (with ${K}$ coefficients) of degree ${d}$ in ${n}$ variables has a non-trivial solution.

Thus the condition ${C_0}$ just says that all polynomials have non-trivial solutions, i.e. ${K}$ is algebraically closed. The condition ${C_1}$ is usually called being quasi-algebraically closed. Examples include, but are not limited to finite fields and function fields of curves over algebraically closed fields. A more complicated example that may come up later is that the maximal, unramified extension of a complete, discretely valued field with perfect residue field is ${C_1}$.

A beautiful result is that if ${K}$ is ${C_1}$, then we still get that ${Br(K)=0}$. One could consider this result “classical” if done properly. First, by Artin-Wedderburn any finite dimensional, central, simple algebra has the form ${M_n(D)}$ where ${D}$ is a finite dimensional division algebra with center ${K}$. If you play around with norms (I swear I did this in a previous post somewhere that I can’t find!) you produce the right degree homogeneous polynomial and use the ${C_1}$ condition to conclude that ${D=K}$. Thus any central, simple algebra is already split giving ${Br(K)=0}$.

We might give up and think the Brauer group of any field is ${0}$, but this is not the case (exercise to test understanding: think of ${\mathbb{R}}$). Let’s move on to the easiest example we can think of for a non-${C_1}$ field: ${\mathbb{Q}_p}$ for some prime ${p}$. The computation we do will be totally general and will actually work to show what ${Br(K)}$ is for any ${K}$ that is complete with respect to some non-archimedean discrete valuation, and hence for ${K}$ a local field.

The trick is to use the valuation ring, ${R=\mathbb{Z}_p}$ to interpolate between the Brauer group of ${K}$ and the Brauer group of ${R/m=\mathbb{F}_p}$, a ${C_1}$ field! Since ${K}$ is the fraction field of ${R}$, the first thing we should check is the Leray spectral sequence at the generic point ${i:Spec(K)\hookrightarrow Spec(R)}$. This is given by ${E_2^{p,q}=H^p(Spec(R), R^qi_*\mathbb{G}_m)\Rightarrow H^{p+q}(G_K, (K^s)^\times)}$.

By Hilbert’s Theorem 90, we have ${R^1i_*\mathbb{G}_m=0}$. Recall that last time we said there is a canonical isomorphism ${Br(R)\rightarrow Br(\mathbb{F}_p)}$ given by specialization. This gives us a short exact sequence from the long exact sequence of low degree terms:

$\displaystyle 0\rightarrow Br(\mathbb{F}_p)\rightarrow Br(\mathbb{Q}_p)\rightarrow Hom(G_{\mathbb{F}_p}, \mathbb{Q}/\mathbb{Z})\rightarrow 0$

Now we use that ${Br(\mathbb{F}_p)=0}$ and ${G_{\mathbb{F}_p}\simeq \widehat{\mathbb{Z}}}$ to get that ${Br(\mathbb{Q}_p)\simeq \mathbb{Q}/\mathbb{Z}}$. As already mentioned, nothing in the above argument was specific to ${\mathbb{Q}_p}$. The same argument shows that any (strict) non-archimedean local field also has Brauer group ${\mathbb{Q}/\mathbb{Z}}$.

To get away from local fields, I’ll just end by pointing out that if you start with some global field ${K}$ you can try to use a local-to-global idea to get information about the global field. From class field theory we get an exact sequence

$\displaystyle 0\rightarrow Br(K)\rightarrow \bigoplus_v Br(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}\rightarrow 0,$

which eventually we may talk about. We know what all the maps are already from this and the previous post. The first is specialization (or corestriction from a few posts ago, or most usually this is called taking invariants). Then the second map is just summing since each term of the direct sum is a ${\mathbb{Q}/\mathbb{Z}}$.

Next time we’ll move on to Brauer groups of curves even though so much more can still be said about fields.

## 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 ${k=R/m}$.

## Finiteness of X(k)/B for Rational Surfaces

Recall our setup. We start with a projective surface ${X/k}$ that becomes rational after some finite extension of scalars ${k'/k}$. Let ${C_0(X)}$ be the group of ${0}$-cycles of degree ${0}$. Last time we defined the Manin pairing ${(-,-): C_0(X)\times (Br(X)/Br(k))\rightarrow Br(k)}$ using the corestriction map ${(\sum n_ix_i, a)=\prod_i cor_{k(x_i)/k}(a(x_i))^{n_i}}$. Two rational points are called Brauer equivalent if ${(x-y, a)=1}$ for all ${a\in Br(X)}$, and denote the set of rational points up to Brauer equivalence by ${X(k)/B}$.

Now let ${N=NS(X_{k'})}$ be the Néron-Severi group of ${X_{k'}}$. It turns out we can factor the Manin pairing as follows:

$\displaystyle \begin{matrix} C_0(X)\times (Br(X)/Br(k)) & \longrightarrow & Br(k) \\ \downarrow & & \uparrow \\ A_0(X)\times H^1(G, N) & \longrightarrow & H^1(G, N\otimes \overline{k}^\times)\times H^1(G, N)\end{matrix}$

The goal of today is to say something about this factoring. Last time we wrote down the Hochschild-Serre spectral sequence and said the map ${Br(k)\rightarrow Br(X)}$ was just the quotient map ${E_2^{2,0}\rightarrow E_\infty^{2,0}}$ followed by the inclusion. Note that since all differentials are ${0}$ for all ${E_n^{1,1}}$ we get that it equals ${H^1(G, N)}$ and sits inside ${Br(X)}$. Thus we have a sequence ${Br(k)\rightarrow Br(X)\rightarrow H^1(G,N)}$ whose composition is ${0}$ and hence gives a map

$\displaystyle Br(X)/Br(k)\rightarrow H^1(G, N).$

This defines for us the left vertical map, since the left factor is just projection from all ${0}$-cycles of degree ${0}$ to ${0}$-cycles modulo rational equivalence of degree ${0}$. The right vertical map is just the one induced on group cohomology via the standard intersection pairing on the surface ${N\otimes \overline{k}^\times \times N\rightarrow \overline{k}^\times}$.

This leaves us with the bottom map. Call it ${\Phi \times id}$ where ${\Phi:A_0(X)\rightarrow H^1(G, N\otimes \overline{k}^\times )}$. It turns out that the majority of Bloch’s paper is merely defining this map and checking that the above diagram commutes, so we won’t get into that. It involves lots of K-theory which I’m not going to get into.

Supposing the above, the main theorem of the paper is that ${Im \Phi}$ is finite in the case of our hypotheses. We can check the nice corollary that ${X(k)/B}$ is finite. If ${X(k)}$ is empty we’re done, so fix some ${x_0\in X(k)}$. The proof is that we can make ${\Psi: X(k)\rightarrow H^1(G, N\otimes \overline{k}^\times)}$ by ${\Psi(x)=\Phi([x]-[x_0])}$. Since ${Im \Psi\subset Im \Phi}$, it must have finite cardinality. We need only check that distinct Brauer classes stay distinct to get the result, but this follows from commutativity of the diagram and the fact that Brauer classes are by definition distinguished under the Manin pairing.

It turns out that Manin had already proved that ${X(k)/B}$ is finite for cubic surfaces, so Bloch’s result extends this to any rational surface. As a consequence of the construction of ${\Phi}$, Bloch also gets the strange result that if ${X}$ is a conic bundle, i.e. ${X\rightarrow \mathbf{P}^1}$ has generic fiber a conic, and ${k}$ is local, then if ${X}$ has good reduction then ${|Im\Phi |=1}$. Thus at places of good reduction ${A_0(X)}$ is trivial.

Note how useful this is. For example, take some conic bundle over ${\mathbf{Q}_p}$. Good reduction means that there exists some proper, regular model ${\frak{X}/\mathbf{Z}_p}$ whose generic fiber is ${X}$ and whose special fiber is non-singular. It is hard to tell whether or not ${X}$ has good reduction, because you might accidentally be picking the wrong model. With this condition of Bloch, one can sometimes explicitly calculate some non-trivial element of ${A_0(X)}$ (Manin actually does this using the defining equation of a class of Châtalet surfaces!) which tells you ${X}$ has bad reduction.

To phrase this a different way, to test for honest bad reduction without some criterion requires a choice of model over ${\mathbf{Z}_p}$. There could be infinitely many distinct choices here, so it could be hard to tell if you’ve exhausted all possibilities. This criterion of Bloch says that no choice needs to be made. Bad reduction can be tested inherently from the variety over ${\mathbf{Q}_p}$.