For central simple $F$-algebra $A$, where $F$ is a field, a field $E$ such that $F \subset E$ is called a splitting field if $A \otimes E$ is isomoprhic to $M_n (E)$ for some $n$. Why is algebraic closure of $F$ a splitting field for $A$?

  • $\begingroup$ What is this $n$? $\endgroup$ – Crostul Nov 16 '14 at 17:53
  • $\begingroup$ Yes, sorry for that, $A$ is central simple. I will fix that. $\endgroup$ – Elensil Nov 16 '14 at 18:36

Let $\bar F$ the algebraic closure of $F$ and let $A$ be a csa over $F$.

Then $A\otimes\bar F$ is a central simple algebra over $\bar F$. Wedderburn's theorem tells us that there is a finite dimensional $\bar F$-algebra and an $n\geq1$ such that $A\otimes\bar F\cong M_n(D)$. Since $\bar F$ is algebraically closed, there are no non-trivial finite dimensional division $\bar F$-algebras, so $D=\bar F$.

This means that $\bar F$ splits $A$.

  • $\begingroup$ Can you please explain why $A \otimes_F \bar{F}$ is a simple ring? $\endgroup$ – Dune Nov 17 '14 at 15:55
  • $\begingroup$ Well, this is a standard property of central simple algebras: when you extend scalars they remain central and simple. You should consult any textbook on the subject! $\endgroup$ – Mariano Suárez-Álvarez Nov 17 '14 at 15:57
  • $\begingroup$ Ok thanks. I just know the general situation when $A$ is semisimple, and then scalar extensions can fail to be semisimple. $\endgroup$ – Dune Nov 17 '14 at 15:59
  • $\begingroup$ @Dune The Tensor product of a central simple algebra with a simple algebra is always simple. $\endgroup$ – Brett Frankel Nov 17 '14 at 16:01

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