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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

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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.

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Ok, so how do I know that the only division algebra whose center is Fq is Fq itself ? (I opened a new question with this before I got back on this account) –  user Mar 7 '12 at 1:51
    
@Tom: See the link that I edited into the post. –  Brett Frankel Mar 7 '12 at 1:57
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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
    
@Tom: That's correct. –  Brett Frankel Mar 7 '12 at 2:11

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