Lemma 10.39.19. Let $R \to S$ be flat. Let $\mathfrak p \subset \mathfrak p'$ be primes of $R$. Let $\mathfrak q' \subset S$ be a prime of $S$ mapping to $\mathfrak p'$. Then there exists a prime $\mathfrak q \subset \mathfrak q'$ mapping to $\mathfrak p$.

**Proof.**
By Lemma 10.39.18 the local ring map $R_{\mathfrak p'} \to S_{\mathfrak q'}$ is flat. By Lemma 10.39.17 this local ring map is faithfully flat. By Lemma 10.39.16 there is a prime mapping to $\mathfrak p R_{\mathfrak p'}$. The inverse image of this prime in $S$ does the job.
$\square$

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