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I have a question on localizations of Dedekind rings which I am learning about in an undergraduate class. Let $R$ be a Dedekind ring with quotient field $K$, $\mathfrak p$ a nonzero prime ideal in $R$. Let $R_\mathfrak p$ be the localization of $R$ at $\mathfrak p$. One assignment question is to show that if $x\in K-R_\mathfrak p$ then $x^{-1}\in R_\mathfrak p$.

If $x=r/s\in K-R_\mathfrak p$ then $r\in R$ and $s\in \mathfrak p$. I don't see how we can deduce $r\notin \mathfrak p$ since we want $s/r\in R_\mathfrak p$.

A hint will be very useful. Thanking you in advance.

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This exercise is asking you to prove that $R_\mathfrak{p}$ is a valuation ring (see the first definition in the Wikipedia entry: http://en.wikipedia.org/wiki/Valuation_ring#Definitions).

My answer to the following question explains this:

Discrete valuation ring associated with a prime ideal of a Dedekind domain

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  • $\begingroup$ So the way to solve the question is to show that $R_\mathfrak p$ is a valuation ring through a valuation $v_\mathfrak p$ on $K$? Is $v_\mathfrak p(x)$ the power of $\mathfrak p$ present in the factorization of $xR_\mathfrak p$? Then do we know that $K$ is the quotient field of $R_\mathfrak p$? Sorry for my basic question, I have only started learning this and it seems quite hard. Thank you. $\endgroup$ – Young Lee Apr 13 '14 at 1:30
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    $\begingroup$ I can't think of a more direct way. Yes, the valuation is the $\mathfrak{p}$-adic one, meaning that $v_\mathfrak{p}(x)$ is the power of $\mathfrak{p}$ in the fractional ideal $Rx$ (or equivalently $xR_\mathfrak{p}$). If $R$ is a domain with fraction field $K$ and $S\subseteq R\setminus\{0\}$ is a multiplicative set, then there is always a unique $R$-algebra injection $S^{-1}R\hookrightarrow K$, and this realizes $K$ as the quotient field of $S^{-1}R$. $\endgroup$ – Keenan Kidwell Apr 13 '14 at 1:34
  • $\begingroup$ Do you show that $K$ is the quotient field of $S^{-1}R$ by showing that every element of $K$ lies in $S^{-1}R$? Also in your answer to the linked question, you mention the uniformizer $\pi$ in $R_\mathfrak p$. Is $\pi$ the generator of the unique maximal ideal of $R_\mathfrak p$ and is it a generator of $\mathfrak p$ in $R$ also? $\endgroup$ – Young Lee Apr 13 '14 at 2:23
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    $\begingroup$ If every element of $K$ were in $S^{-1}R$ then the map $S^{-1}R\to K$ would be an isomorphism. This is not generally the case (think about $R=\mathbf{Z}$). Every element of $K$ is a quotient of elements of $R$, and in particular, is also a quotient of elements of $S^{-1}R$. Regarding $\pi$, it's a generator of the maximal ideal $\mathfrak{p}R_\mathfrak{p}$, but not necessarily of $\mathfrak{p}$, and there is no reason for $\mathfrak{p}$ to be principal in general. $\endgroup$ – Keenan Kidwell Apr 13 '14 at 13:24
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Let $R = \mathbb{Z}$ and $\mathfrak{p} = (p)$, with $p$ a rational prime. See what happens in this case, and that should indicate how to proceed in the general case.

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