 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 $\mathrm{Br}(E)=0$. Thus $A{\otimes }_{K}E$ is a matrix algebra. A splitting isomorphism and its inverse involve only finitely many coefficients in $E$. All these coefficients lie in $K{k}^{\prime }$ for some finite extension $k^{\prime }/k$, since $E$ is the union of these finite constant extensions. The two isomorphism identities consequently already hold over $K{k}^{\prime }$, so $A$ splits there.

Finite extensions of a finite field are cyclic, and regularity gives $K\cap k^{\prime }=k$ and preserves their degree under this scalar extension. Thereforeis cyclic. 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.