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:

$R = \bigcap R_M$

Proof

Let $a \in \bigcap R_M$. Let $I$ be the ideal comprising those $x \in R$ for which $ax \in R$. $I$ 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 $I \le M$.