# Depth of a contraction of a prime ideal of the integral closure of height one

I'm trying to prove Mori-Nagata theorem.

Let $$A$$ be a noetherian domain, $$K=\operatorname{Frac}(A)$$ and $$B$$ be the integral closure of $$A$$ in $$K$$.

I want to show that for every nonzero $$x\in B$$, there are finitely many primes of $$B$$ of height $$1$$ containing $$x$$.

Replacing $$A$$ by $$A[x]$$, we may assume that $$x\in A$$.

It suffices to show that for every prime ideal $$P$$ of $$B$$ of height $$1$$ containing $$x$$, $$P\cap A\in \operatorname{Ass}_A(A/xA)$$.

(I know that the corresponding morphism $$\operatorname{Spec}B\rightarrow \operatorname{Spec}A$$ has finite fibres.)

This claim is equivalent to the following:

Let $$P$$ be a prime ideal of $$B$$ of height $$1$$, and put $$p=P\cap A$$.

Then $$\operatorname{depth}A_p =1$$.