Corollary

If an extension E of a field K is generated by a set of $p\text{-radical}$ elements over K, then it is p-radical over K.

For $E_r$ is a subextension of E and by hypothesis $E = K(E_\infty)$ whence $E = E_r$; so E is a p-radical extension of K.

Under the conditions of Proposition 2, $E_r$ is said to be the relative $p\text{-radical}$ closure of K in E.

We shall apply Prop. 2 particularly to the case where E is an algebraically closed extension of K ; then E is perfect and we have $E_n = K^{p^{-n}}$ for all $n \geq 0$. In this case we denote by $K^{p^{-\infty}}$ the set of elements of E that are p-radical over K ; it is the subfield of E which is the union of the ascending sequence $(K^{p^{-n}})_{n > 0}$ of subfields of E. By Prop. 2, $K^{p^{-\infty}}$ is the largest subextension of E which is p-radical over K ; by Prop. 3 of V, p. 6, $K^{p^{-\infty}}$ is a perfect closure of K and it is also the smallest perfect subfield of E containing K. When K is perfect we clearly have $K = K^{p^{-n}} = K^{p^{-\infty}}$ for all n. If K is imperfect, we have $K \neq K^P$, whence

$^1$ A p-radical extension is also called a purely inseparable extension, or a radical extension.

for $n \geq 1$; the subfields $K^{p^{-n}}$ of E are then pairwise distinct and $K^{p^{-\infty}}$ is an algebraic extension of infinite degree of K.