# 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 ${k=R/m}$.