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This is proposition 5.13 from Atiyah-Macdonald:
Let $A$ be an integral domain. Then the following are equivalent:

  • $A$ is integrally closed
  • $A_{\mathfrak p}$ is integrally closed, for each prime ideal $\mathfrak p$
  • $A_{\mathfrak m}$ is integrally closed, for each maximal ideal $\mathfrak m$

This is the proof given in the text:
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). $\blacksquare$

Thm 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 $S^{-1}B$.

Thm 3.9) Let $\phi: M \to N$ be an $A$-module homomorphism. Then the following are equivalent:

  • $\phi$ is surjective
  • $\phi_{\mathfrak p}:M_{\mathfrak p} \to N_{\mathfrak p}$ is surjective for each prime ideal $\mathfrak p$
  • $\phi_{\mathfrak m}:M_{\mathfrak m} \to N_{\mathfrak m}$ is surjective for each maximal ideal $\mathfrak m$

I am not able to see where we have used the hypothesis that $A$ is an integral domain. I'm pretty sure that I'm missing something. Please can somebody guide me.

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    $\begingroup$ @K02: Atiyah and MacDonald use a generalised definition of the ring of fractions that applies to any commutative ring. However, the definition they give for integrally closed is restricted to integral domains (and uses the term "field of fractions" because for integral domains, you know the ring of fractions is a field.) $\endgroup$
    – Rob Arthan
    Nov 27, 2023 at 20:52
  • $\begingroup$ Again, answers as comments... sigh $\endgroup$ Nov 27, 2023 at 21:40
  • $\begingroup$ @MartinBrandenburg My bad! MSE complains if I post too many answers that are short, but now I've moved it according. $\endgroup$
    – Shrugs
    Nov 27, 2023 at 22:13
  • $\begingroup$ You could also add what Rob wrote ^^ $\endgroup$ Nov 27, 2023 at 22:14
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    $\begingroup$ @MartinBrandenburg: tentative or partial answers seem to me to be quite appropriate as comments on MSE. So what are you objecting to? "Sigh" is utterly unhelpful as a criticism. If there are MSE guidelines that you believe I have transgressed, then please tell me what they are. $\endgroup$
    – Rob Arthan
    Dec 1, 2023 at 22:31

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To make the field of fractions 𝐾, you used 𝐴 is an integral domain.

Also, @Rob Arthan points out that the book uses the term "ring of fractions" as well as a generalized definition which applies a similar construction to any commutative ring. However, the definition they give for "integrally closed" is only restricted to integral domains and hence, the term field of fractions. The ring of fractions of an integral domain is always a field.

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