= Solution
Let $k$ be the full constant field of $K$. It is a finite extension of the given <finite field>, hence is itself finite. Since $k$ is perfect and algebraically closed in $K$, the one-variable extension $K/k$ is regular. In a common algebraic closure, put $E=K\overline k$. Then $E$ is the function field of a geometrically integral curve over the <algebraically closed field> $\overline k$.
The <Tsen theorem> makes $E$ a $C_1$ field, and the preceding argument gives $\operatorname{Br}(E)=0$. Thus $A\otimes_KE$ is a matrix algebra. A splitting isomorphism and its inverse involve only finitely many coefficients in $E$. All these coefficients lie in $Kk'$ for some finite extension $k'/k$, since $E$ is the union of these finite constant extensions. The two isomorphism identities consequently already hold over $Kk'$, so $A$ splits there.
Finite extensions of a <finite field> are cyclic, and regularity gives $K\cap k'=k$ and preserves their degree under this scalar extension. Therefore
$$
\operatorname{Gal}(Kk'/K)\cong\operatorname{Gal}(k'/k)
$$
is cyclic. \b[A finite cyclic extension of constants splits $A$.] This is <cyclic splitting by extension of constants>. The argument does not assert that the splitting extension has degree exactly the index of $A$; a larger cyclic constant extension is sufficient for the requested conclusion.
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