# 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.

# The Grothendieck Spectral Sequence

Well, I meant to do lots more examples building up some more motivation for how powerful spectral sequences can be in some simple cases. But I’m just running out of steam on posting about them. Since we’ve done spectral sequences associated to a double complex, we may as well do the Grothendieck Spectral Sequence, then I might move on to another topic for a bit (I admit it is sort of sad to not prove the Kunneth formula using a SS).

I haven’t scoured the blogs to see whether these topics have been done yet, but I’m thinking about either basics on abelian varieties a la Mumford, or some curve theory, possibly building slowly to and culminating in Riemann-Roch.

In any case, we have the tools to do the Grothendieck Spectral Sequence (GSS) quite easily. Let $\mathcal{A}, \mathcal{B}, \mathcal{C}$ be abelian categories with enough injectives. Let $\mathcal{A}\stackrel{G}{\to}\mathcal{B}\stackrel{F}{\to}\mathcal{C}$ be functors (and $FG:\mathcal{A}\to\mathcal{C}$ the composition). Suppose that $F$ and $G$ are left exact and for every injective $J\in\mathcal{A}$ we have $G(J)$ is acyclic. This just means that $R^iF(J)=0$ for all positive $i$.

Then there exists a spectral sequence (the GSS) with $E_2^{pq}\simeq (R^pF)(R^qG)(X)\Rightarrow R^{p+q}(FG)(X)$ with differential $d_{r}:E_r^{pq}\to E_r^{p+r, q-r+q}$.

The proof of this is just to resolve $X$ using the injectives that we know exist. This gives us a double complex. From a double complex, the way to get the $E_2$ term is to take vertical then horizontal homology, or horizontal and then vertical. Both of these will converge to the same thing. One way completed collapses to the “0 row” due to the fact that the exact sequence remained exact after applying the functor except at the 0 spot. Thus it stabilizes at this term and writing it out, you see that it is exactly $R^{p+q}(FG)(X)$. Taking homology in the other order gives us exactly $(R^pF)(R^qG)(X)$ by definition of a derived functor. This completes the proof.

I probably should write the diagram out for clarity, but really they are quite a pain to make and import into wordpress. The entire outline of the proof is here, so if you’re curious about the details, just carefully fill in what everything is from the previous posts.

This is quite a neat spectral sequence. It is saying that just by knowing the derived functors of $F$ and $G$ you can get to the derived functors of the composition of them. There are two important spectral sequence consequences of this one. They are the Leray SS and the Lyndon-Hochschild-Serre SS. The later computes group cohomolgy.

I promised early on to do the Leray SS for all the algebraic geometers out there. The Leray SS gives a way to compute sheaf cohomology. Let $\mathcal{A}=Sh(X)$ and $\mathcal{B}=Sh(Y)$ be the category of sheaves of abelian groups on X and Y. Let $\mathcal{C}=Ab$ the category of abelian groups. Let $f:X\to Y$ be a continuous map, then we have the functor $F=f_*$ and the two global section functors $\Gamma_X$ and $\Gamma_Y$.

Applying the GSS to these functors, we get that $H^p(Y, R^qf_*\mathcal{F})\Rightarrow H^{p+q}(X, \mathcal{F})$.

There are a few things to verify to make sure that the GSS applies, and we need the fact that $\Gamma_Y\circ f_*=\Gamma_X$. It would also be nice to have an example to see that this is useful. So maybe I’ll do those two things next time.