Lemma 4

Let A be an integrally closed domain, K its field of fractions, p the characteristic exponent of K, K' a radicial extension of K and A' a subring of K' containing A and integral over A. For every prime ideal p of A, there exists a unique prim ideal p' of $A'$ lying above $p$ and $p'$ is the set of $x \in A'$ such that there exists an integer $m \geq 0$ for which $x^{p^m} \in p$.

The existence of $p'$ follows from no. 1, Theorem 1. If $x \in p'$, there exists $m \geq 0$ such that $x^{p^m} \in K$, whence $x^{p^m} \in A$ since $A$ is integrally closed, hence $x^{p^m} \in p' \cap A = p$. Conversely, if $x \in A'$ is such that $x^{p^m} \in p \subset p'$, then $x \in p$ since $p'$ is prime.