Ring equals max-localization intersection

Statement

Let $R$ be a commutative unital ring and $K (R)$ its total quotient ring. For each maximal ideal $M$ of $R$ , let $R_{M}$ denote the localization of $R$ at $M$ viewed as a subring of $K (R)$ . Then:

$R = ⋂ R_{M}$

Proof

Let $a \in ⋂ R_{M}$ . Let $I$ be the ideal comprising those $x \in R$ for which $a x \in R$ . $I$ is essentially the ideal of all possible denominators of fractions for $a$ in terms of elements of $R$ .

The claim is that $I = R$ . Suppose not. Then $I$ is a proper ideal of $R$ ,and since every proper ideal is contained in a maximal ideal, we can find a maximal ideal $M$ containing $I$ . But since $a \in R_{M}$ , $a$ can be written as $p / q$ where $q \notin M$ . Clearly $q \in I$ , and this contradicts $I \leq M$ .