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.
Best Answer
To make the field of fractions š¾, you used š“ 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.