In a course I attended at university, we calculated the Brauer group of $\mathbb{F}_q((t))$ with $q=p^n$ , $p$ prime number and we proved it was $\dfrac {\mathbb{Q}}{\mathbb{Z}}=Br(\bar{\mathbb{F}_q}((t))/ \mathbb{F}_q((t)))$.

To do this,we had to show that every central division algebra over $\mathbb{F}_q((t))$ splits over $\bar{\mathbb{F}_q}((t))$.

This is the part I did not understand. The aim of the proof was to show that given a central finite dimensional division algebra $D$, there was $x \in D$ such that $x \in \bar{\mathbb{F}_q}((t))$, so that $D \otimes \mathbb{F}_q((t))[x]$ is not a domain, and one can apply induction.

To demonstrate this statement, our professor used some number theory facts about discrete valuation over these division algebras, but I did not really understand what he did.

  • $\begingroup$ "In particular, we showed...." what you wanted to do! I can't follow what the central division algebra stuff is relevant for... $\endgroup$ – DonAntonio Apr 24 '18 at 9:31
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    $\begingroup$ This is because of how Brauer group is done:if every centrale simple algebra splits over some extension the Brauer group relative to that extension and to the separable closure coincide $\endgroup$ – Tommaso Scognamiglio Apr 24 '18 at 9:39

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