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?
Brauer group of a finite field is trivial
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$\begingroup$
abstract-algebra
number-theory
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0Given 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? =) – 2012-03-07
1 Answers
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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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0@Tom: That's correct. – 2012-03-07