Theorem 2

— Let K be a field of characteristic $p \neq 0$, $K^{p^{-\infty}}$ a perfect closure of K and A a commutative K-algebra. The following properties are equivalent: a) A is separable. b) There exists an extension K' of K such that K' is perfect and $K' \otimes_K A$ is reduced. c) The ring $K^{p^{-\infty}} \otimes_K A$ is reduced.

d) The ring $K' \otimes_K A$ is reduced for every extension $K'$ of $K$ which is of finite degree and $p$*-radical of height* $\leqslant 1$.

e) For every family $(a_i)_{i \in I}$, of elements of $A$ linearly free over $K$, the family $(a_i^p)_{i \in I}$, is linearly free over $K$.

f) There exists a basis $(a_i)_{i \in I}$, of the vector $K$*-space* $A$ suck that the family $(a_i^p)_{i \in I}$, is linearly free over $K$.

If an extension $K'$ of $K$ is a perfect field, it contains a subextension $K$-isomorphic to $K^{p^{-\infty}}$ ($V$, p. 6, Prop. 3); moreover, every $p$-radical extension of $K$ is isomorphic to a subextension of $K^{p^{-\infty}}$ ($V$, p. 26, Prop. 3). These remarks show the implications $a) \Rightarrow b) \Rightarrow c) \Rightarrow d)$.

Let us prove that d) implies e). Let $(a_i)_{i \in I}$, be a linearly free family in $A$ and let $(A_i)_{i \in I}$, be a family of finite support in $K$ such that $\sum \lambda_i a_i^p = 0$. Let $K'$ be the subextension of $K^{p^{-\infty}}$ generated by the elements $\lambda_i^{p^{-1}}$; it is of finite degree and height $\leqslant 1$. Put $x = \sum_{i \in I} \lambda_i^{p^{-1}} \otimes a_i$ in $K' \otimes_K A$; we have

By the hypothesis d) we have $x = 0$, whence $A_i = 0$ for all $i \in I$.

Clearly e) implies f) and it remains to show that f) implies a). Thus let $(a_i)_{i \in I}$ be a basis of $A$ over $K$ such that the family $(a_i^p)_{i \in I}$ is linearly free over $K$. Let $L$ be an extension of $K$ and let $x$ be an element of $L \otimes_K A$ such that $x^p = 0$. Write $x = \sum_{i \in I} \lambda_i \otimes a_i$ with $A_i \in L$ for all $i \in I$. We have $x^p = \sum_{i \in I} \lambda_i^p \otimes a_i^p = 0$ and since the family $(a_i^p)_{i \in I}$, is linearly free over $K$, we have $\lambda_i^p = 0$, whence $\lambda_i = 0$ for all $i \in I$; it follows that $x = 0$. So we have proved that $x^p = 0$ implies $x = 0$ in $L \otimes_K A$, from which it follows at once that $L \otimes_K A$ is reduced.