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.