Corollary 2

— Let K be a field of characteristic exponent p, $K^{p^{-\infty}}$ a perfect closure of K and L a separable extension of K. Then the ring $L \otimes_K K^{p^{-\infty}}$ is a field. If moreover L is algebraic over K, then $L \otimes_K K^{p^{-\infty}}$ is a perfect closure of L.

The case $p = 1$ is trivial, so let us suppose $p \neq 1$. Let $\Omega$ be a perfect closure of L. By Cor. 1 there exists a K-algebra homomorphism of $L \otimes_K K^{p^{-\infty}}$ onto $L[K^{p^{-\infty}}]$ which maps $x \otimes y$ to $xy$ for $x \in L$ and $y \in K^{p^{-\infty}}$. Since $K^{p^{-\infty}}$ is algebraic over K, the ring $L[K^{p^{-\infty}}]$ is a subfield of $\Omega$ (V, p. 18, Cor. 1). Suppose further that L is algebraic over K, then $L[K^p]$ is an algebraic extension of the perfect field $K^{p^{-\infty}}$, hence it is a perfect field (V, p. 43, Cor. 1); finally, since the field $L[K^{p^{-\infty}}]$ is ap-radical extension of L (V, p. 25, Cor.), it is a perfect closure of L.