Lemma 10.164.3. Let $R \to S$ be a ring map. Assume that

1. $R \to S$ is faithfully flat, and

2. $S$ is a normal ring.

Then $R$ is a normal ring.

Proof. Since $S$ is reduced it follows that $R$ is reduced. Let $\mathfrak p$ be a prime of $R$. We have to show that $R_{\mathfrak p}$ is a normal domain. Since $S_{\mathfrak p}$ is faithfully over $R_{\mathfrak p}$ too we may assume that $R$ is local with maximal ideal $\mathfrak m$. Let $\mathfrak q$ be a prime of $S$ lying over $\mathfrak m$. Then we see that $R \to S_{\mathfrak q}$ is faithfully flat (Lemma 10.39.17). Hence we may assume $S$ is local as well. In particular $S$ is a normal domain. Since $R \to S$ is faithfully flat and $S$ is a normal domain we see that $R$ is a domain. Next, suppose that $a/b$ is integral over $R$ with $a, b \in R$. Then $a/b \in S$ as $S$ is normal. Hence $a \in bS$. This means that $a : R \to R/bR$ becomes the zero map after base change to $S$. By faithful flatness we see that $a \in bR$, so $a/b \in R$. Hence $R$ is normal. $\square$

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