# 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.