TITLE: How to show $\bigcap_{m \textrm{: maximal ideal}} A_m=A$?
QUESTION [5 upvotes]: $A$ is an integral domain. For every maximal ideal $m$ in $A$, consider $A_m$ as a subring of the quotient field $K$ of $A$. Show $\bigcap A_m=A$, where the intersection is taken over all maximal ideals $m$ of $A$.
This is a homework problem. I thought I figured it out but soon I realized that my argument only works in the case that there are finitely many maximal ideals, which is much weaker.

REPLY [3 votes]: Hint: Let $\frac{a}{b} \in \bigcap_m A_m$. Consider the ideal $I = \{r \in A: r\frac{a}{b} \in A\}$. What can you say about $I$? You need to use the fact that any proper ideal is always contained in a maximal ideal.