Proposition 5.13 of Atiyah and Macdonald

Question

Proposition 5.12. Let $A \subset B$ be rings, $C$ the integral closure of $A$ in $B$. Let $S$ be a multiplicatively closed subset of $A$. Then $S^{-1}C$ is the integral closure of $S^{-1}A$ in $S^{-1}B$.

Definition: An integral domain is said to be integrally closed if it is integrally closed in its field of fractions.

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Let $A$ be an integral domain. Then the following are equivalent:

(i) $A$ is integrally closed;

(ii) $A_{\mathfrak p}$ is integrally closed, for each prime ideal $\mathfrak p$;

(iii) $A_{\mathfrak m}$ is integrally closed, for each maximal ideal $\mathfrak m$.

Proof. Let $K$ be the field of fractions of $A$, let $C$ be the integral closure of $A$ in $K$, and let $f: A \to C$ be the identity mapping of $A$ into $C$. Then $A$ is integrally closed iff $f$ is surjective, and by (5.12) $A_{\mathfrak p}$ (resp. $A_{\mathfrak m}$) is integrally closed iff $f_{\mathfrak p}$ (resp.$f_{\mathfrak m}$) is surjective. Now use (3.9).

I can’t fill following detail in above proof: field of fraction of $A_\mathfrak{p}$ is $K_\mathfrak{p}$. We know $K_\mathfrak{p} = S^{-1}K=S^{-1}(\bar{S}^{-1} A)$ where $S=A-\mathfrak{p}$ and $\bar{S}=A-\{0\}$.

Answer 1

(3.9) is a theorem that a module morphism $f\colon M\to N$ is injective (resp.~surjective) if and only if every localization at a prime ideal $\mathfrak{p}$, $f_{\mathfrak{p}}\colon M_{\mathfrak{p}}\to N_{\mathfrak{p}}$, is injective (resp.~surjective), if and only if every localization at a maximal ideal $\mathfrak{m}$, $f_{\mathfrak{m}}\colon M_{\mathfrak{m}}\to N_{\mathfrak{m}}$ is injective (resp.~surjective).

So all you need to do is show that if $\mathfrak{p}$ is a prime ideal of $A$, and $K$ is the field of fractions of $A$, then $K_{\mathfrak{p}}$ is $Q(A_\mathfrak{p})$, the field of fractions of $A_{\mathfrak{p}}$.

Indeed, since $K$ is already a field, (the image of) every element of $A-\mathfrak{p}$ is already a unit in $K$, so $(A-\mathfrak{p})^{-1})K=K$. That is, $K_{\mathfrak{p}}=K$.

Now, since every element of $A-\mathfrak{p}$ maps to a unit under the canonical embedding $A\hookrightarrow K$, the universal property of localizations tells you that the map $A\hookrightarrow K$ induces a map $A_{\mathfrak{p}}\hookrightarrow K$. And because $K$ is a field, the universal property of the field of fractions tells that this map extends to a morphism $Q(A_{\mathfrak{p}})\to K$, which must be injective since they are both fields. Conversely, the composition $A\to A_{\mathfrak{p}}\to Q(A_{\mathfrak{p}})$ maps $A$ into a field, and hence induces a morphism $K\to Q(A_{\mathfrak{p}})$. It is now easy to verify that these two maps are inverses of each other, so that we have $K_{\mathfrak{p}}=K=Q(A_{\mathfrak{p}})$. Now (3.9) gives the desired equivalence.