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 $x \in R$. $\square$

Comment #8513 by on

I don't understand why we can assume that $R$ is a subring of $R_\mathfrak{p}$ (in the sentence “as $R \subset R_{\mathfrak p}$ is flat we see that $R_{\mathfrak p} \subset Q(R) \otimes_R R_{\mathfrak p}$”). This is not the case for a non-domain $R$: the preimage of the prime $(0)\subset R_\mathfrak{p}$ along $R\to R_\mathfrak{p}$ is prime and thus different to $(0)\subset R$.

Here's the proof I've thought of:

The total ring of fractions is not a functor: in general a ring homomorphism $A\to B$ does not induce a ring homomorphism $Q(R)\to Q(R')$, 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 $S\subset R$ is any multiplicative set, then the ring homo $\varphi:R\to S^{-1}R$ does preserve nonzerodivisors and it induces then a ring homomorphism $Q(\varphi):Q(R)\to Q(S^{-1}R)$ that makes the diagram commute.

In particular, if $S=R\setminus\mathfrak{p}$, where $\mathfrak{p}\subset R$ is prime, the commutative diagram is Suppose $x\in Q(R)$ satisfies an integral equation over $R$. Then we can map this equation to $Q(R_\mathfrak{p})$ to deduce that the image of $x$ in $Q(R_\mathfrak{p})$ is integral over $R$ and hence over $R_\mathfrak{p}$. Since we are assuming that $R_\mathfrak{p}$ is a normal domain, it follows that the image of $x$ in $Q(R_\mathfrak{p})$ lies in $R_\mathfrak{p}$.

Define $I=\{f\in R\mid fx\in R\}$, which is an ideal of $R$. To see that $I$ is the unit ideal, it suffices to see that $I$ in not contained in $\mathfrak{p}$. Write $x=g/s$ inside $Q(R)$, where $g,s\in R$ and $s$ is a nonzerodivisor. Then, inside $Q(R_\mathfrak{p})$, we can write $g/s=h/t$, where $h\in R$ and $t\in R\setminus\mathfrak{p}$, i.e., $gt=hs$ in $Q(R_\mathfrak{p})$ and therefore in $R_\mathfrak{p}$. Hence, there is $u\in R\setminus\mathfrak{p}$ with $gtu=hsu$ in $R$. Thus, inside $Q(R)$, we can write $\frac{g}{s}tu=hu\in R$, so $tu\in I\cap (R\setminus\mathfrak{p})$. In particular, $I\not\subset\mathfrak{p}$.

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