Lemma 10.36.22. Let $R \to S$ be a ring map such that $S$ is integral over $R$. Let $\mathfrak p \subset \mathfrak p' \subset R$ be primes. Let $\mathfrak q$ be a prime of $S$ mapping to $\mathfrak p$. Then there exists a prime $\mathfrak q'$ with $\mathfrak q \subset \mathfrak q'$ mapping to $\mathfrak p'$.

**Proof.**
We may replace $R$ by $R/\mathfrak p$ and $S$ by $S/\mathfrak q$. This reduces us to the situation of having an integral extension of domains $R \subset S$ and a prime $\mathfrak p' \subset R$. By Lemma 10.36.17 we win.
$\square$

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