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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

user75086
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  • If you're copying an exercise verbatim, it is good to say where it is from, i.e. author and book. – Pedro Jun 07 '13 at 01:18
  • @PeterTamaroff yes, thank you for the remark – user75086 Jun 07 '13 at 01:20
  • The localization maps $R\to R_M$ and $R_M \to F$ ($F$ the quotient field) are all injective (since $R$ is an integral domain), so you want to identify $R$ and $R_M$ with their isomorphic copy inside $F$. – dc2814 Jun 07 '13 at 01:25
  • Incidentally, this means that the difficult inclusion is $\bigcap R_M \subset R$. – dc2814 Jun 07 '13 at 01:29

1 Answers1

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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 $s^{-1}r$.

Andreas Blass
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  • I have the following questions: 1)In the second paragraph $s^{-1}$ is the inverse of $s$?2)Why $R_M$ a subring of the quotient field of $R$ implies that it consists of the elements $s^{-1}r$ for all $r\in R$ and all $s\in S=R-M$? thank you very much for your answer! – user75086 Jun 07 '13 at 02:03
  • For your first question, yes, $s^{-1}$ is the inverse of $s$. For the second, many people define $R_M$ to consist of these fractions $s^{-1}r$, and in fact it seems you said this in the question: "$R_M$ is $S^{-1}R$ with $S=R-M$." If you want to use some other definition, you should say which definition you want (perhaps the universal property?). – Andreas Blass Jun 07 '13 at 22:05
  • I didn't understand, for me $r/s$ is different than $s^{-1}r$. The former is an equivalence class in $S^{-1}R$ and the latter is the multiplication of the inverse of $s$ with r. – user75086 Jun 08 '13 at 03:27
  • The distinction you mention between $r/s$ and $s^{-1}r$ is reasonable in the abstract, but obliterating this distinction is precisely how one considers $R_M$ as a subring of the quotient field of $R$; one identifies the element $r/s$ of the abstract localization with the element $s^{-1}r$ of the quotient field. – Andreas Blass Jun 09 '13 at 22:58