Showing that $\operatorname {Br}(\Bbb F_q)=0$

Question

I want to prove that $\operatorname {Br}(\Bbb F_q)=0$ using the cohomological description of the Brauer group.

We have:

$\operatorname {Br}(\Bbb F_q)=H^2(\operatorname {Gal}(\overline {\Bbb F_q}/\Bbb F_q), \overline {\Bbb F_q}^*)$.

Since $\operatorname {Gal}(\overline {\Bbb F_q}/\Bbb F_q)$ is just $\widehat {\Bbb Z}$, this is just

$H^2(\widehat{\Bbb Z}, \overline {\Bbb F_q}^*) = \operatorname{Ext}_{\Bbb Z[\widehat{\Bbb Z}]}^2(\Bbb Z, \overline {\Bbb F_q}^*)$.

So I just need to find a free $\Bbb Z[\widehat{\Bbb Z}]$-resolution of $\Bbb Z$:

$$...\to \Bbb Z[\widehat{\Bbb Z}] \to \Bbb Z\to 0$$

But already the kernel of the first map is complicated. Any suggestions?

EDIT: as suggested by @Mariano, I'll try another approach.

Assume $\operatorname {Br}(\Bbb F_q)\neq 0$ and let $D$ be division algebra over $\Bbb F_q$. Then $\dim_{\Bbb F_q} D = n^2$ for some integer $n$. Let $l=q^{n^2}$. Then it suffices to show that $\operatorname{Br}(\Bbb F_l / \Bbb F_q) =0$. $\operatorname {Gal}({\Bbb F_l}/\Bbb F_q)=\Bbb Z/n^2\Bbb Z$, so I need to compute $\operatorname{Ext}_{\Bbb Z[\Bbb Z/n^2\Bbb Z]}^2(\Bbb Z, \overline {\Bbb F_l^*})$

Answer 1

That cohomology group of the Galois group is not normal cohomology but continuous cohomology with respect to the Krull topology on the Galois group. In particular, your approach does not have many chances.

Now, the definition of both Krull topology and of continuous cohomology are made so that you can reduce everything to computing cohomology of finite quotients of the Galois group, which are cyclic. You should consult a textbook on Galois cohomology.

Answer 2

I'm not sure if your acquiescence to Mariano's suggestion of an alternate method means that you're open to different proofs, or if that is just another flavor of 'cohomological'. If so, then I apologize for not answering your actual question.

There is a much 'simpler' proof that finite fields have trivial Brauer group. If you are willing to move out of the cohomological point of view which, since we have the luxury of thinking about the Brauer group in several ways, seems like a good use of knowledge.

Now, finite fields are quasi-algebraically closed by an extremely famous theorem of Chevalley.

So, let $D$ be a division algebra over $\mathbb{F}_q$, of splitting degree $n$. We must show that $n=1$. But, if $e_1,\ldots,e_{n^2}$ is a $\mathbb{F}_q$-basis for $D$, then the reduced norm on $D$ can be interpreted as a homogenous polynomial in degree $n^2$.

More rigorously, there exists some homogenous $f(T_1,\ldots,T_{n^2})\in \mathbb{F}_q[T_1,\ldots,T_{n^2}]$ such that for any $\displaystyle x=\sum_{i=1}^{n^2}\alpha_i e_i\in D$, where $\alpha_i\in \mathbb{F}_q$, we have that the reduced norm of $x$ is $f(\alpha_1,\ldots,\alpha_{n^2})$. But, since the reduced norm of a non-zero element of $D$ is non-zero, we know that $f(T_1,\ldots,T_{n^2})$ has no non-trivial zeros. The quasi-algebraically closedness of $\mathbb{F}_q$ then implies that $n\geqslant n^2$, and so $n=1$ as desired.

I hope that was sufficiently cute, as to turn you away from a direct $H^2$ calculation :)