Localization of a Dedekind domain.

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

This answer may be helpful for those more comfortable with ideal class groups than valuation rings.

Let $R$ be a Dedekind domain and let $\mathfrak{p}\subset R$ be a non-zero prime ideal. Let $r,s\in R$ both be non-zero. Let $k_r, k_s$ denote the exponents of $\mathfrak{p}$ in the factorisations of $(r),(s)$ respectively. Let $k=$min$(k_r,k_s)$. We know $k=k_r$ or $k=k_s$. Suppose $k=k_r$. We will show $\frac sr\in R_{\mathfrak{p}}$.

We have: \begin{align} (r)&=&J\mathfrak{p}^k,\\ (s)&=& I\mathfrak{p}^k,\\ \mathfrak{p}&\nsupseteq& J.\,\,\,\,\,\, \end{align}

Pick $a\in J \,\,\backslash\,\, \mathfrak{p}$. Pick an ideal $K$ such that $JK=(a)$. Also from the first two equations above, we know $I$ and $J$ represent the same element of the ideal class group, so we have $IK=(b)$, for some $b\in R$.

Note also $$(s)J\mathfrak{p}^k=(s)(r)=(r)(s)=(r)I\mathfrak{p}^k,$$ so $$(s)J=(r)I.$$

Putting this all together:$$(sa)=(s)(a)=(s)JK=(r)IK=(r)(b)=(rb),$$ so $sa=rbu$ for some unit $u\in R$. We selected $a\notin \mathfrak{p}$, so we may conclude: $$\frac sr=\frac{bu}a\in R_{\mathfrak{p}}.$$