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 $R \subset R_{\mathfrak p}$ 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 $x \in R$. $\square$

There are also:

• 3 comment(s) on Section 10.37: Normal rings

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).