The Stacks project

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 $x \in R$. $\square$

Comments (2)

Comment #8513 by on

I don't understand why we can assume that is a subring of (in the sentence “as is flat we see that ”). This is not the case for a non-domain : the preimage of the prime along is prime and thus different to .

Here's the proof I've thought of:

The total ring of fractions is not a functor: in general a ring homomorphism does not induce a ring homomorphism , let alone one that makes the corresponding square commute (this boils down to the observation that ring homos do not preserve nonzerodivisors in general). However, if is any multiplicative set, then the ring homo does preserve nonzerodivisors and it induces then a ring homomorphism that makes the diagram commute.

In particular, if , where is prime, the commutative diagram is Suppose satisfies an integral equation over . Then we can map this equation to to deduce that the image of in is integral over and hence over . Since we are assuming that is a normal domain, it follows that the image of in lies in .

Define , which is an ideal of . To see that is the unit ideal, it suffices to see that in not contained in . Write inside , where and is a nonzerodivisor. Then, inside , we can write , where and , i.e., in and therefore in . Hence, there is with in . Thus, inside , we can write , so . In particular, .

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

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