## BSD for a Large Class of Elliptic Curves

I’m giving up on the p-divisible group posts for awhile. I would have to be too technical and tedious to write anything interesting about enlarging the base. It is pretty fascinating stuff, but not blog material at the moment.

I’ve been playing around with counting fibration structures on K3 surfaces, and I just noticed something I probably should have been aware of for a long time. This is totally well-known, but I’ll give a slightly anachronistic presentation so that we can use results from 2013 to prove the Birch and Swinnerton-Dyer conjecture!! … Well, only in a case that has been known since 1973 when it was published by Artin and Swinnerton-Dyer.

Let’s recall the Tate conjecture for surfaces. Let ${k}$ be a finite field and ${X/k}$ a smooth, projective surface. We’ve written this down many times now, but the long exact sequence associate to the Kummer sequence

$\displaystyle 0\rightarrow \mu_{\ell}\rightarrow \mathbb{G}_m\rightarrow \mathbb{G}_m\rightarrow 0$

(for ${\ell\neq \text{char}(k)}$) gives us a cycle class map

$\displaystyle c_1: Pic(X_{\overline{k}})\otimes \mathbb{Q}_{\ell}\rightarrow H^2_{et}(X_{\overline{k}}, \mathbb{Q}_\ell(1))$

In fact, we could take Galois invariants to get our standard

$\displaystyle 0\rightarrow Pic(X)\otimes \mathbb{Q}_{\ell}\rightarrow H^2_{et}(X_{\overline{k}}, \mathbb{Q}_\ell(1))^G\rightarrow Br(X)[\ell^\infty]\rightarrow 0$

The Tate conjecture is in some sense the positive characteristic version of the Hodge conjecture. It conjectures that the first map is surjective. In other words, whenever an ${\ell}$-adic class “looks like” it could come from an honest geometric thing, then it does. But if the Tate conjecture is true, then this implies the ${\ell}$-primary part of ${Br(X)}$ is finite. We could spend some time worrying about independence of ${\ell}$, but it works, and hence the Tate conjecture is actually equivalent to finiteness of ${Br(X)}$.

Suppose now that ${X}$ is an elliptic K3 surface. This just means that there is a flat map ${X\rightarrow \mathbb{P}^1}$ where the fibers are elliptic curves (there are some degenerate fibers, but after some heavy machinery we could always put this into some nice form, we’re sketching an argument here so we won’t worry about the technical details of what we want “fibration” to mean). The generic fiber ${X_\eta}$ is a genus ${1}$ curve that does not necessarily have a rational point and hence is not necessarily an elliptic curve.

But we can just use a relative version of the Jacobian construction to produce a new fibration ${J\rightarrow \mathbb{P}^1}$ where ${J}$ is a K3 surface fiberwise isomorphic to ${X}$, but now ${J_\eta=Jac(X_\eta)}$ and hence is an elliptic curve. Suppose we want to classify elliptic fibrations that have ${J}$ as the relative Jacobian. We have two natural ideas to do this.

The first is that etale locally such a fibration is trivial, so you could consider all glueing data to piece such a thing together. The obstruction will be some Cech class that actually lives in ${H^2(X, \mathbb{G}_m)=Br(X)}$. In fancy language, you make these things as ${\mathbb{G}_m}$-gerbes which are just twisted relative moduli of sheaves. The class in ${Br(X)}$ is giving you the obstruction the existence of a universal sheaf.

A more number theoretic way to think about this is that rather than think about surfaces over ${k}$, we work with the generic fiber ${X_\eta/k(t)}$. It is well-known that the Weil-Chatelet group: ${H^1(Gal(k(t)^{sep}/k(t), J_\eta)}$ gives you the possible genus ${1}$ curves that could occur as generic fibers of such fibrations. This group is way too big though, because we only want ones that are locally trivial everywhere (otherwise it won’t be a fibration).

So it shouldn’t be surprising that the classification of such things is given by the Tate-Shafarevich group:

Ш $\displaystyle (J_\eta /k(t))=ker ( H^1(G, J_\eta)\rightarrow \prod H^1(G_v, (J_\eta)_v))$

Very roughly, I’ve now given a heuristic argument (namely that they both classify the same set of things) that ${Br(X)\simeq}$ Ш ${(J_\eta)}$, and it turns out that Grothendieck proved the natural map that comes form the Leray spectral sequence ${Br(X)\rightarrow}$ Ш${(J_\eta)}$ is an isomorphism (this rigorous argument might actually have been easier than the heuristic one because we’ve computed everything involved in previous posts, but it doesn’t give you any idea why one might think they are the same).

Theorem: If ${E/\mathbb{F}_q(t)}$ is an elliptic curve of height ${2}$ (occuring as the generic fiber of an elliptic K3 surface), then ${E}$ satisfies the Birch and Swinnerton-Dyer conjecture.

Idea: Using the machinery alluded to before, we spread out ${E}$ to an elliptic K3 surface ${X\rightarrow \mathbb{P}^1}$ over a finite field. As of this year, it seems the Tate conjecture is true for K3 surfaces (the proofs are all there, I’m not sure if they have been double checked and published yet). Thus ${Br(X)}$ is finite. Thus Ш${ (E)}$ is finite. But now it is well-known that if Ш${ (E)}$ being finite is equivalent to the Birch and Swinnerton-Dyer conjecture.

## 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!

## 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 ${\mathbf{Q}_p}$.