Proposition 11

— Suppose that K is of characteristic exponent p, and let E be an algebraic extension of K, generated by a set S. If E is separable over K, then $E = K(S^{p^n})$ for all integers $n \geq 0$; conversely if E is of finite degree over K and $E = K(S^p)$, then E is separable over K.

The case $p = 1$ is trivial by the Cor. of V, p. 37. Suppose from now on that $p \neq 1$.

By hypothesis E is algebraic over K and we have $E = K(S)$, hence

If E is of finite degree over K, it is a separable extension of K if and only if it is an etale algebra over K; the Cor. of V, p. 35 shows that this happens if and only if $E = K[E^p]$.

Suppose now that $E$ is separable and of infinite degree over $K$. Then $K[E^p]$ is the union of the subrings $K[{E'}^p]$ where $E'$ ranges over the set of subextensions of $E$ of finite degree over $K$; but such an extension $E'$ is separable over $K$ ($V$, p. 36, Prop. 1), whence $E' = K[{E'}^p] \subset K[E^p]$ by what has been said; finally we have $E = K[E^p]$. By induction on $n \geq 0$, the relation $E = K[E^p]$ implies that $E = K[E^{p^n}]$.