# Brauer group of a finite field is trivial

I know that all finite skew fields are field. How does it follow from this fact that the Brauer group of a finite field is trivial?

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Given a finite field, you can assume it is commutative, therefore isomorphic to the unique field of order $p^n$ for some $p$ prime and $n$ integer (more explicitly, the field extension generated by $\mathbb F_p$ and the polynomial $x^{p^n}-x$). Can you compute the Brauer group in those cases? =) – Patrick Da Silva Mar 7 '12 at 1:16

The Brauer group of a field $k$ is the group of (isomorphism classes of) division algebras whose center is $k$. Since any finite division algebra is a field, and hence is its own center, the only division algebra whose center is $\mathbb{F}_q$ is $\mathbb{F}_q$ itself. The statement that every finite division algebra is a field is known as Wedderburn's Theorem. (more detail here: http://en.wikipedia.org/wiki/Wedderburn%27s_little_theorem)
You may be more familiar with the definition of Brauer group as a group of classes of central simple algebras over $k$, but notice that in this definition each equivalence class contains exactly one division algebra.
Possibly, what is confusing Tom is that only finite rank division algebras show up in the Brauer group. A division algebra which is finite over $\mathbb{F}_q$ is clearly finite, and hence Wedderbrurn's little theorem is relevant. I imagine there are some infinite division algebras with center $\mathbb{F}_q$, but they don't matter. – David Speyer Mar 7 '12 at 2:02
@DavidSpeyer Arh I see. So you are saying that since D is finite dimensional over a finite field $K$, then D must be a finite field itself. And so it is its own center, $Z(D) = K$. And so there is only one element in the Brauer group. Is that right? – user Mar 7 '12 at 2:06