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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?

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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 $q=1$.)

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