Corollary 3

— Suppose that K has characteristic $p \neq 0$. For an element x of E to belong to K it is necessary and sufficient for it to be both separable algebraic and p-radical over K. The stated condition is clearly necessary. Conversely suppose that x is separable algebraic over K and p-radical of height e over K. Since x is separable over K, the minimal polynomial f of x over K does not belong to K[X^p] (Prop. 4 and 5); since x is p-radical of height e over K, we have $f(X) = X^{p^e} - x^{p^e}$ (V, p. 24, Prop. 1); we conclude that $e = 0$, so $x \in K$.