Lemma 10.37.12. A normal ring is integrally closed in its total ring of fractions.
Proof. Let R be a normal ring. Let x \in Q(R) be an element of the total ring of fractions of R integral over R. Set I = \{ f \in R, fx \in R\} . Let \mathfrak p \subset R be a prime. As R \to R_{\mathfrak p} is flat we see that R_{\mathfrak p} \subset Q(R) \otimes _ R R_{\mathfrak p}. As R_{\mathfrak p} is a normal domain we see that x \otimes 1 is an element of R_{\mathfrak p}. Hence we can find a, f \in R, f \not\in \mathfrak p such that x \otimes 1 = a \otimes 1/f. This means that fx - a maps to zero in Q(R) \otimes _ R R_{\mathfrak p} = Q(R)_{\mathfrak p}, which in turn means that there exists an f' \in R, f' \not\in \mathfrak p such that f'fx = f'a in R. In other words, ff' \in I. Thus I is an ideal which isn't contained in any of the prime ideals of R, i.e., I = R and x \in R. \square
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