$A = \bigcap_{\mathfrak{p} \in \text{Spec(A)}} A_{\mathfrak{p}} = \bigcap_{\mathfrak{m} \in \text{MaxSpec(A)}} A_{\mathfrak{m}}$

Let $x$ be contained in the intersection and consider the ideal $I:=\{a \in A : ax \in A\}$. If $\mathfrak{m}$ is any maximal ideal, then there is some $b \in A \setminus \mathfrak{m}$ such that $bx \in A$ (since $x \in A_\mathfrak{m}$). This shows $I \not\subseteq \mathfrak{m}$. Hence, $I=A$, i.e. $x \in A$.


Key Idea the set of maximal ideals contain enough witnesses to show that every proper fraction has nontrivial denominator (ideal), $ $ i.e. if a fraction $\,f\not\in A\,$ then its denominator ideal $\, {\cal D}_f = \{ d\in A\ :\ d\:\!f\in A\} \ne (1),\:$ so $\, \cal D_f\,$ is contained in some maximal ideal $\,M,\,$ hence $\,f\not\in A_M.\:$ Therefore we conclude that $ \,f\not\in \bigcap A_{M}$ over all ${M \in \text{MaxSpec(A)}}.$