How can be possible $\cap R_M=R$? I'm doing this question in Hungerford's book:

I didn't understand how can be possible this intersection be equal to $R$, because $R_M$ is $S^{-1}R$ with $S=R-M$, maybe the author means they are isomorphic? I need a hint to begin to solve this question.
Thanks a lot
 A: Note that Hungerford says to consider $R_M$ as a subring of the quotient field of $R$.  So it consists of the elements $s^{-1}r$ for all $r\in R$ and all $s\in S=R-M$.  With this in mind, the intersection, over all maximal ideals $M$, of the rings $R_M$ is literally equal to $R$ (as Hungerford says), not just isomorphic.  
It is clear (by using $s=1$), that $R$ is included in every $R_M$ and therefore in the intersection.  The meat of the question is that, if an element of the quotient field can be expressed as $s^{-1}r$ in (possibly) many different ways, with all the $r$'s in $R$ and all the $s$'s outside various maximal ideals, then that element is in $R$, i.e., you don't actually need any $s^{-1}$ factors to express it.  HINT: The point is that your element has, for each maximal ideal $M$, an expression of the form $s^{-1}r$ with $s$ outside $M$.  The $s$'s that occur here, for all the various $M$'s, must generate the improper (i.e., unit) ideal of $R$, since no maximal ideal contains them all.  So write $1$ as a finite linear combination of some of these $s$'s and then use that to manipulate the corresponding expressions $s^{-1}r$.
