# conjugate prime ideals of integral extensions and relevance of the characteristic of the ground field

This question refers to the proof of theorem 9.3, p. 66 in Matsumura's Commutative Ring Theory: "if $A$ is an integrally closed domain, $K$ its field of fractions and $L/K$ a normal field extension, $B$ the integral closure of $A$ in $L$, then all prime ideals of $B$ lying over a prime ideal of $A$ are conjugates." In particular let's consider the case where $L/K$ is finite. By assuming that $P_1,P_2$ are prime ideals of $B$ lying over $p \in \operatorname{Spec} A$, that $\sigma_1, \cdots, \sigma_n$ are the elements of $Aut(L/K)$ and that $P_2 \notin \sigma_j^{-1}(P_1), \forall j$, then we can find some $x \in P_2$ such that $x \notin \sigma_j^{-1}(P_1), \forall j$. Then Matsumura defines the element $y=(\prod \sigma_j(x))^q$ where $q=1$ if $charK=1$ and $q=p^{\nu}$ for $\nu$ large enough if $char K=p>0$ and he proceeds to show that actually $y \in p$, since $y \in K$, $y$ is integral over $A$ and $y \in P_2$, thus yielding a contradiction.

Question: what is the relevance of $char K$ and why is it necessary to incorporate this $q=p^{\nu}$ when the characteristic is not zero?

The $y$ defined in the proof is defined to be a power of the norm of $x$, $$N_{L/K}(x)=\prod_{\sigma\in \text{Aut}(L/K)}\sigma(x).$$ The norm of any element of $L$ is obviously in the fixed field of $\text{Aut}(L/K)$, which you can easily show to be the purely inseperable closure of $K$ in $L$, i.e. the set of elements of $L$ which are purely inseperable over $K$. Now, an element is purely inseperable over a field $K$ if and only if it is a $p^{\nu}$-th root of some $a\in K$, where $p=\text{char}(K)$, hence some $p$-power of the Norm of $x$ lies in $K$. (Of course, if $K$ has characteristic $0$, then the norm is already in $K$, hence we take $q=1$.)