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A homework problem is divided into 2 parts:

I managed to solve the first part, which states:

Prove Koethe's theorem: If $D$ is a finite dimensional central division $k$-algebra and $K_0 \subset D$ is a separable extension of $k$ then $D$ has a maximal subfield $K$ such that $K_0 \subset K$ and $K/k$ is separable.

I am having difficulties with the second part:

Deduce that if $S$ is a finite dimensional central simple $k$-algebra then there exists a Galois extension $K/k$ such that $K$ splits $S$.

My attempt at the second part:

The algebra $S$ is not necessarily a division algebra, but even if it was, Koethe's theorem would give me a maximal subfield separable over $k$. This subfield would split $S$ because it is maximal, but to show that it is Galois I would still need to prove that it is normal over $k$.

I think that one of the things I might be missing is the relation between a subfield that splits an algebra and the usual notion of a splitting field.

1 Answers 1

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1) Recall that any finite-dimensional central simple $k$-algebra $S$ is isomorphic to a $k$-algebra of the form $M_r(D)\simeq M_r(k) \otimes_k D$, where $D$ is a division algebra with center $k$ (the integer $r$ and the isomorphism class of $D$ being uniquely determined by $S$) .
As a consequence, any extension $K/k$ which splits $D$ also splits $S$.
Indeed, if $D \otimes_k K\simeq M_t(K)$, we'll have
$S \otimes_k K\simeq (M_r(k) \otimes_k D)\otimes_k K\simeq M_r(K)\otimes_K M_t(K)\simeq M_{rt}(K)$ Hence in order to split $S$ it suffices to split the division ring $D$.

2) Suppose that $K$ is a separable extension of $k$ inside $D$ which splits $D$.
Since $K/k$ is separable, it has a Galois closure $K\subset K_{gal}$, which is (as the name says!) Galois over $k$.
But since already $K$ splits $D$, so will a fortiori $K_{gal}$ and you are home.

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    Dear @Amy, I have strongly expanded my previous version of 1) in order to make it completely (I hope !) explicit.2012-01-28