# An integral domain $A$ is exactly the intersection of the localisations of $A$ at each maximal ideal

This result appears to be ubiquitous as an algebra exercise. How do you prove this result?

Let $A$ be an integral domain with field of fractions $K$, and let $A_{\mathfrak{m}}$ denote the localisation of $A$ at a maximal ideal $\mathfrak{m}$ considered as a subring of $K$. Prove that $$A = \bigcap_{\mathfrak{m}} A_{\mathfrak{m}}\,,$$ where the intersection is taken over all maximal ideals $\mathfrak{m}$ of $A$.

Or equivalently, suppose $z$ is not in $A$. Also consider the ideal $I=A:z=\{x\in A: xz \in A\}$. Then $I$ is a proper ideal since $1\not\in I$. Then there exists a maximal ideal $\mathfrak{m}$ containing $I$. The element $z$ must be not in $A_{\mathfrak{m}}$, otherwise, there exists some $s\not\in \mathfrak{m}$ such that $sz \in A$, hence $s\in I$ which contradicts the choice of the maximal ideal $\mathfrak{m}$.

• Thank you for this great concise proof! Jun 26, 2019 at 19:15

Since $A$ is a domain, the localization map $A \to A_{\mathfrak m}$ is injective for every maximal ideal $\mathfrak m$, so $A \subseteq \cap_{\mathfrak m \in \text{mSpec} A} A_{\mathfrak m}$. The other inclusion is more interesting: suppose $z \in K$, with $z \in A_{\mathfrak m}$ for every maximal ideal $\mathfrak m$. Consider the $A$-ideal $I := A :_A Az = \{x \in A \mid xz \in A\}$. We want $1 \in I$, i.e. $I = A$. But to show this we can localize: $I_{\mathfrak m} = (A :_A Az)_{\mathfrak m} = A_{\mathfrak m} :_{A_{\mathfrak m}} A_{\mathfrak m}z = A_{\mathfrak m}$ (since $Az$ is a finitely generated $A$-module) for every maximal ideal ${\mathfrak m}$, so indeed $I = A$.

We may think of $$\bigcap_{m}A_m$$ as an $$A$$-module. It is clear that $$A\subset \bigcap_{m}A_m$$ is an $$A$$-submodule of $$\bigcap_{m}A_m$$. Our aim is to show that the $$A$$-module $$M=\frac{(\bigcap_mA_m)}{A}$$ is zero. This would establish what we want to prove.

From local-to-global principles, it suffices to show that $$M_n$$ is zero for every maximal ideal $$n\subset A$$.

So, lets compute, $$M_n=\left(\frac{\bigcap_mA_m}{A}\right)_n=\frac{(\cap_mA_m)_n}{A_n}.$$

Notice that the last equality follows from the fact that localization commutes with quotients (which is true as localization is exact functor).

Now, it is clear that $$(\bigcap_mA_m)_n\subset\bigcap_m (A_m)_n$$, Hence, $$\frac{(\bigcap_mA_m)_n}{A_n}\subset \frac{\bigcap_m (A_m)_n}{A_n}= \frac{\bigcap_m (A_n)_m}{A_n}$$ (we have changed the order of localization). Now, $$\bigcap_m (A_n)_m\subset (A_n)_n=A_n$$, and so $$\frac{\bigcap_m (A_n)_m}{A_n}\subset \frac{A_n}{A_n}=0$$.

Combining all these containement, we conclude that $$M_n\subset \frac{(\cap_mA_m)_n}{A_n} \subset \frac{A_n}{A_n}=0$$. That is, $$M_n=0$$ for every maximal ideal $$n$$, hence it is the zero $$A$$-module and so $$A=\bigcap_m A_m$$.