Wedderburn proved that any finite division algebra is a field. I saw a beautiful proof which uses class equation and some basic analysis in the book Proofs from the Book. However, I want to know whether we can prove the same result by using the cohomological argument and the Brauer group.
By using Hilbert's theorem 90 and the Herbrand quotient, we can show that $H^{2}(\mathrm{Gal}(l/k), l^{\times})=0$ for any finite extension of finite fields $l/k$. (In fact, $H^{r}=0$ in this case.) Since $\mathrm{Br}(l/k)\simeq H^{2}(\mathrm{Gal}(l/k), l^{\times})$, we have $\mathrm{Br}(l/k)=0$ for all $l/k$ and so $\mathrm{Br}(k)=0$ for all finite field $k$. I think this should prove the Wedderburn's theorem, but I don't know how to proceed from here. Could you fill the gaps? Thanks in advance.
I think awllower's arguement can be modified in this way:
Let $D$ be a finite division algebra over $k$ and let $l$ be its center. Then $D$ is a central simple algebra over $l$, so we can consider $[D]\in \mathrm{Br}(l)=0$. This implies $D$ splits over $l$, i.e. $D\simeq M_{n}(l)$ for some $n$. Since $M_{n}(l)$ can't be a division algebra for $n\geq 2$ (we can find explicit zero divisors easily), so $D\simeq M_{1}(l)\simeq l$.