Intersection over all maximal ideals of localizations Let $A$ be an integral domain. Show that $$\bigcap \limits_{\mathfrak{m}}A_{\mathfrak{m}}=A,$$ where the intersection is taken over all maximal ideals $\mathfrak{m}$ in $A$.
P.S.  This is quite well-known problem in commutative algebra and I have found many topics in MSE which contains its solution. So please do not duplicate my question by the the following reason:
But  my question is not about the solution of this problem. How these object could be equal if  the LHS and RHS have elements of different structure. More prcisely, the elements of $A_{\mathfrak{m}}$ are equivalence classes denoted by $\dfrac{a}{s}$, where $a\in A$ and $s\notin \mathfrak{m}$.
Can anyone explain it in a rigorous way, please?
 A: To elaborate on Lord Shark's comment: if $A$ is an integral domain, then $A$ embeds into any localization $A\to S^{-1}A,$ provided the multiplicative set $S$ does not contain $0.$ Moreover, if $S\subseteq T\subseteq A$ are both multiplicative subsets of $A,$ then we have natural maps
$$
A\to S^{-1}A\to T^{-1}A,
$$
and the composition $A\to T^{-1}A$ is simply the localization map (the map $S^{-1}A\to T^{-1}A$ may either be thought of as a localization of $S^{-1}A$ or as the map induced by the universal property of localization).
An integral domain has a maximal multiplicative subset not containing $0$: namely, $A\setminus\{0\}.$ The localization of $A$ by this multiplicative subset is the field of fractions $K$ of $A.$ So, if $\mathfrak{m}$ is a maximal ideal of $A,$ we have natural embeddings
$$
A\to A_\mathfrak{m}\to K.
$$
Inside of $K,$ we may take the intersection of [the images of] all the $A_\mathfrak{m}$ and compare that with [the image of] $A,$ and one way to make the theorem precise is that these two subsets of $K$ are equal. Since $A\to K$ is an isomorphism of $A$ onto its image (and the same is true of each $A_\mathfrak{m}\to K$), it is reasonable to say that $\bigcap_\mathfrak{m} A_\mathfrak{m} = A$ (or at least that $\bigcap_\mathfrak{m} A_\mathfrak{m} \cong A$).
