# 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.

# Galois Deformations 5: Flach’s Theorem

I’ll begin today by taking a poll. Usually I’d do this at the end, but I assume no one will be reading at that point. Since my research has a bit of an arithmetic flavor, one thing I’d like to do is post about some standard algebraic number theory where about a 10 week graduate class would leave off with an eye towards class field theory. I should probably be more comfortable with these things. My other idea is to blog about Neron models which I feel like I should understand better as well. If you strongly want to see one of these over another, then speak up soon in the comments (I’ll probably do both eventually).

The subject of today is to discuss the obstruction space a little. Recall that whenever we’ve discussed the universal deformation ring in the previous posts we had the caveat “if the deformations are unobstructed, then … “. It would be nice to have some theorems that tell us when this actually happens, or if it doesn’t happen at least get some good control on what the obstructions are.

Let’s quickly recall how this works. Fix ${S}$ a finite set of primes and some prime ${\ell}$ (not in ${S}$). Fix a an absolutely irreducible residual representation ${\overline{\rho}:G_S\rightarrow GL_2(\mathbb{F}_\ell)}$. In this case, we know the deformation functor is representable by some universal deformation ring denoted ${\mathcal{R}(\overline{\rho})}$. The structure of this ring is immensely important since it completely determines what all the deformations are. We said that the obstruction space for deforming is ${H^2(G_S, Ad(\overline{\rho}))}$, so if we suppose this is ${0}$, then we know that ${\mathcal{R}(\overline{\rho})\simeq \Lambda [[x_1, \ldots, x_d]]}$ where ${d=\dim H^1(G_S, Ad(\overline{\rho}))}$. This is why unobstructedness is so important. We can completely determine ${\mathcal{R}(\overline{\rho})}$ up to isomorphism.

A theorem of Flach in 1992 tells us one case when we get unobstructedness. Let ${E/\mathbb{Q}}$ be an elliptic curve having good reduction at ${\ell}$ and ${S}$ the primes of bad reduction together with ${\ell}$ and ${\infty}$. Let ${\rho=\rho_E:G_S\rightarrow GL_2(\mathbb{Z}_\ell)}$ and ${\overline{\rho}}$ the residual representation. Suppose we have the following three conditions satisfied as well: ${\rho}$ is surjective, for all ${p\in S\setminus \{\infty\}}$ we have ${E[\ell]\otimes E[\ell]}$ has no ${G_{\mathbb{Q}_p}}$-invariants, and ${\ell}$ does not divide ${L(Sym^2 T_\ell E, 0)/\Omega}$, then the deformation problem is unobstructed for ${\overline{\rho}}$.

We definitely won’t worry about what everything in that last condition is. You might be worried that this never happens, but in some sense that can be made precise these conditions are almost always satisfied. Now to finish the post off we’ll just very generally give an outline of the proof. Since we are trying to make some ${H^2}$ vanish, it isn’t surprising that we can reduce to checking that Ш${^1(G_s, E[\ell]\otimes E[\ell])\simeq}$
Ш${^1(G_s, \mu_\ell)\oplus}$ Ш${^1(G_S, Sym^2 E[\ell])}$ vanishes. We can already start to see the appearence of some of the conditions.

Using standard techniques in Galois cohomology like Hilbert 90 it can be checked directly that the first factor vanishes, so we only need to check that second one. Now define ${A=Sym^2 T_\ell E\otimes \mathbb{Q}_\ell/\mathbb{Z}_\ell}$. It turns out that we get an injection
Ш${^1(G_S, Sym^2 E[\ell])\hookrightarrow H_f^1(\mathbb{Q}, A)}$ into the Selmer group, so we can reduce to proving this Selmer group is ${0}$. Using a very hard, deep theorem and the fact that ${E}$ is modular we can conclude that ${\deg \phi H^1_f(\mathbb{Q}, A^*)=0}$ where ${\phi: X_0(N)\rightarrow E}$ is a modular parametrization. From this we can conclude the theorem of Flach.

# Galois Deformations 3: Tangent and Obstruction

Recall last time that if we start with some finite field ${k}$ and some absolutely irreducible residual Galois representation ${\overline{\rho}: G_S\rightarrow GL_n(k)}$, the deformation functor is (pro)representable by some universal deformation ring which we will denote ${\mathcal{R}(\overline{\rho})}$ (again, this was known even in the original papers to be true in a much more general situation than this one). Today we’ll try to figure out some properties of this ring.

First we note something that better be true. We really only consider representations up to equivalence, so if ${\overline{\rho}}$ and ${\overline{\rho}'}$ are equivalent then there is a matrix that conjugates one to the other. By functoriality of everything involved we can consider the natural transformation of deformation functors induced by the group scheme homomorphism given by this conjugation action. By representability the natural transformation gives a homomorphism of the universal rings ${\mathcal{R}(\overline{\rho})\rightarrow \mathcal{R}(\overline{\rho}')}$. It can be checked that this is an isomorphism. Geometrically this means that the “local space” of deformations is the same for equivalent representations.

By similar arguments to the last time I talked about deformation theory, we can compute that the tangent space to the functor given by ${Def_{\overline{\rho}}(k[\epsilon])}$ is isomorphic to ${H^1(G_S, Ad(\overline{\rho}))}$ where ${Ad(\overline{\rho})}$ just means the adjoint representation. So as in the geometric situation, if ${\mathcal{R}}$ is smooth, we can compute the dimension of the space of deformations by computing the dimension of some first cohomology. Thus if the deformations are completely unobstructed, then we know that ${\mathcal{R}\simeq \Lambda[[x_1, \ldots, x_d]]}$ where ${d=\dim H^1(G_S, Ad(\overline{\rho}))}$.

The standard next question is to ask what is the obstruction space. Very standard arguments that can be made with almost any deformation functor show us the following. I’ll leave out the details, but tell you what to fill in. By an obstruction, we mean take a map in our deformation ring category ${R_1\rightarrow R_0}$ that has kernel ${I}$ with the property that ${I\cdot m_{R_1}=0}$ so that it is naturally a ${k}$-vector space. Take some representation ${\rho: G_S\rightarrow GL_n(R_0)}$. We want to know if there is some ${\rho': G_S\rightarrow GL_n(R_1)}$ so that upon composing ${GL_n(R_1)\rightarrow GL_n(R_0)}$ we get ${\rho}$ back. If you’ve seen this geometrically, then your intuition should be to show that “locally” you can do it, and then the ability to glue to get a global object is some class in ${H^2}$ causing the obstruction. This is again the idea here. You formally make the lift set-theoretically. The obstruction to being a homomorphism is a class ${Ob(\rho)\in H^2(G_S, Ad(\overline{\rho})\otimes_k I)}$. If the class is 0, then we can lift the representation, otherwise it is obstructed. We call ${H^2(G_S, Ad(\overline{\rho}))}$ the “obstruction space” to the deformation functor.

This gives us an expected dimension for the deformation space. In some sense, we’d expect that each dimension of the obstruction space would cut the dimension of the space by one. If ${d_1=\dim H^1(G_S, Ad(\overline{\rho}))}$ and ${d_2=\dim H^2(G_S, Ad(\overline{\rho}))}$, then we’d expect ${Kdim(\mathcal{R}/m_{\Lambda}\mathcal{R})=d_1-d_2}$, but this is still an open conjecture and is probably hard since it generalizes Leopoldt’s conjecture. What we do get is that ${Kdim(\mathcal{R}/m_{\Lambda}\mathcal{R})\geq d_1-d_2}$. The rough idea behind having an inequality is that each dimension of the obstruction space is introducing a relation, and so the relation at most cuts the dimension down by one. But maybe we get to the fifth relation and it is redundant, then the dimension won’t go down.

Using some hard algebraic number theory we can actually convert ${d_1-d_2}$ to something only involving ${H^0}$, but it is beyond the scope of these posts. We’ll end here for today. Next time we’ll specialize to Galois representations that are modular, since that was our motivation and see what more information we can get in that case (this means if you are following along with the Gouvea article, we will be skipping lectures 5 and 6).