Corollary 1 (Mac Lane)

— Let $K$ be a field of characteristic exponent $p$, $\Omega$ a perfect extension of $K$ and $L$ a subextension of $\Omega$. Then the following conditions are equivalent:

a) $L$ is separable over $K$.

b) $L$ is linearly disjoint from $K^{p^{-\infty}}$ over $K$.

c) $L$ is linearly disjoint over $K$ from every $p$*-radical extension of* $K$ contained in $\Omega$, of finite degree and height $\leqslant 1$.

The case where $K$ is of characteristic 0 is trivial because then $L$ is separable over $K$ (Th. 1) and $K^{p^{-\infty}} = K$ by convention. Suppose then that $p \neq 1$, and let us first show that a) implies b). Assume that $L$ is separable over $K$ and let $(a_i)_{i \in I}$, be a basis of $L$ over $K$. Let $(\lambda_i)_{i \in I}$ be a family of finite support of elements of $K^{p^{-\infty}}$ such that $\sum_{i \in I} \lambda_i a_i = 0$; there exists an integer $f \geqslant 0$ and elements μ_i of K such that λ_i = μ_i^{p^{-f}}. We have

and Th. 2 implies, by induction on f, that the family $(a_i^{p^f})_{i \in I}$ is linearly free over K. We thus have $\mu_i = 0$, whence $\lambda_i = 0$ for all $i \in I$. Finally L is linearly disjoint from $K^{p^{-\infty}}$ over K.

It is clear that b) implies c). Lastly suppose that c) holds and let K' be an extension of K of finite degree and p-radical of height $\leq 1$. The ring $K' \otimes_K L$ is isomorphic to a subring of $\Omega$, hence is reduced. Now it follows from Th. 2 that L is separable over K.