Let $k$ be a field, $A$ a finite-dimensional semisimple $k$-algebra. If $k$ is a perfect field (every finite field extension of $k$ is seperable), then $A$ is separable.
I know a proof that uses the classification of seperable $k$-algebras:
Seperable $k$-algebras are the same as finite products of matrix algebras over finite-dimensional division algebras whose centers are finite-dimensional separable field extensions of the field $k$. Thus, if $k$ is perfect, then seperable $k$-algebras are the same as finite products of matrix algebras over finite-dimensional division algebras over the field $k$. By the Wedderburn–Artin theorem these are precisely semisimple finite-dimensional semisimple $k$-algebras.
Is there another (maybe slick, maybe modern) proof? Does anyone have a reference for a proof of the classification of separable algebras?
Consider the answer here. A semisimple $k$-algebra has trivial Jacobson radical, so is obviously a semiprimary $k$-algebra. So if you can believe that a tensor product over $k$ of semisimple $k$-algebras is semisimple when $k$ is perfect, we can proceed. This amounts to the claim that a tensor product over $k$ of two (f.d. assoc.) division algebras over $k$ is a direct sum of division algebras over $k$ when $k$ is perfect. I admittedly don't know a proof of this fact, although it is easy if $k$ is algebraically closed, and probably false if $k$ is an arbitrary field.
Given that unproved claim, Auslander's paper applies, and we find that $gl.dim(A \otimes A^{op}) = 2gl.dim(A)$. Since $A$ is semisimple, every $A$-module is projective, hence, $gl.dim(A) = 0$. So $gl.dim(A \otimes A^{op}) = 0$, and hence every $A \otimes A^{op}$-module is projective. This includes $A$ as the identity bimodule, hence $A$ is projective as an $A \otimes A^{op}$-module, which is an equivalent characterization of separability.