Going down for integral extensions of normal domains

Statement
Suppose $$A$$ is a normal domain i.e. an integral domain that is integrally closed inside its fraction field. Suppose $$B$$ is an integral extension of $$A$$, and $$B$$ is also an integral domain.

Let $$P_1 \supset P_2$$ be primes of $$A$$ and let $$Q_1$$ be a prime of $$B$$ contracting to $$P_1$$. Then, there exists a prime $$Q_2$$ of $$B$$ such that $$Q_1 \supset Q_2$$, and such that $$Q_2$$ contracts to $$P_2$$.

Proof outline
The proof has several steps:
 * We first reduce to the case where the extension is a finite extension. Going from finite extensions to arbitrary integral extensions is an application of a Zorn's lemma argument.
 * We next reduce to the case where the field of fractions of $$B$$ is a normal field extension of the field of fractions of $$A$$, and $$B$$ is normal i.e. $$B$$ is integrally closed in its field of fractions.
 * We then split the extension into its separable part and purely inseparable part. For the purely inseparable part, we use the fact the every element has power in subring implies bijective on spectra. For the separable part, we use going down for fixed-point subring under finite automorphism group