'Divides means contains' for Dedekind domains, by treating PIDs and localising

I'm trying to solve the following problem:

Suppose $R$ is a Dedekind domain which contains (nonzero) ideals $\mathfrak{a}$ and $\mathfrak{b}$. By first dealing with the case where $R$ is a PID and then localising, show that $\mathfrak{a} \supset \mathfrak{b} \Longleftrightarrow \mathfrak{a} \, \vert \, \mathfrak{b}$.

Now the PID case was pretty easy to do, but I don't really understand what the question means or intends by "treating the case where $R$ is a PID and then localising"; specifically I don't think I have a good understanding of what 'localising' means.

To my understanding, the localisation of $R$ by $S \subset R$ is $S^{-1}R = \{\frac{r}{s}: r \in R,\,s \in S\} \subset K$, where $K$ is the field of fractions of $R$. However, I don't have a clue what I'm meant to do to 'localise' for the case of a Dedekind domain here: could anyone help? Since I don't have any experience working with localisations, I'd be very grateful for any detailed but relatively simple explanations you could provide; thanks in advance.

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Hopefully $\mathfrak a | \mathfrak b \Rightarrow \mathfrak a \supset b$ was easy regardless of your strategy. Let's do the reverse. For a prime ideal $\mathfrak p$ of $R$ I'll use the standard notation of $R_\mathfrak p$ for $S^{-1}R$ with $S = R - \mathfrak p$. In the case that $R$ is a Dedekind domain and $\mathfrak p$ a non-zero prime ideal, $R_\mathfrak p$ will be a PID; moreover, it is what is called a discrete valuation ring.
Since $R$ is a Dedekind domain, our ideals have factorizations $$\mathfrak a = \prod_\mathfrak p \mathfrak p^{r_\mathfrak p} \quad \text{and} \quad \mathfrak b = \prod_\mathfrak p \mathfrak p^{s_\mathfrak p},$$ where the product runs over all non-zero prime ideals of $R$ and almost all of the exponents are zero. The goal is to show that $r_\mathfrak p \leq s_\mathfrak p$ for all $\mathfrak p$. What you need to use or prove is that for a given prime $\mathfrak p$, $$\mathfrak aR_\mathfrak{p} = \mathfrak p^{r_\mathfrak p}R_\mathfrak p \quad \text{and similarly} \quad \mathfrak bR_\mathfrak{p} = \mathfrak p^{s_\mathfrak p}R_\mathfrak p.$$ That $\mathfrak a \supset \mathfrak b$ will imply that $\mathfrak aR_\mathfrak{p} \supset \mathfrak bR_\mathfrak{p}$, and from there it shouldn't be hard to see the end.
 Apologies, my computer froze up and for some reason cleared my cookies - this is the OP. I can see why that statement is true, so many thanks for the help on that, it makes sense although I'm not sure I could have spotted the solution myself. Just to check; "by localising" here means "by considering the localisation at $R_\mathfrak{p}$"? One thing which isn't totally clear to me is why we use $S = R - \mathfrak{p}$ all the time: this choice of S seems to come up a lot. Is it for the nice properties it happens to have and the fact it makes $\mathfrak{p} maximal, or is there a deeper reason? – Spyam Jan 29 '12 at 23:56 @Ben I do think that the intent was for you to consider all of the localizations$R_\mathfrak p$. Using$S - \mathfrak p$is indeed common (the other common choice is$S = \{1, f, f^2, \ldots\}$for some element$f$, but this happens more in algebraic geometry). In this setting,$\mathfrak p$is already maximal in$R$, but I think the important part is that the ideal theory of$R_\mathfrak p$consists of what happens inside of$\mathfrak p$: we use here the fact that the other primes in the ideal factorization become irrelevant in the localization. – Dylan Moreland Jan 30 '12 at 0:06 ... so you can prove things about the ramification of prime ideals or the degree of the corresponding residue field extension (this is preserved as well) by working locally, and you can go even further by completing. Maybe the best motivation comes from algebraic geometry: you can view the set of prime ideals of$R$as a topological space$\operatorname{Spec} R$, with$R$being the ring of global functions on this space.$R_\mathfrak p$naturally corresponds to the germs of functions at the point$\mathfrak p\$. (Hence "localization") – Dylan Moreland Jan 30 '12 at 0:10 That makes sense. Originally I thought the question wanted a proof in both directions for PIDs, and then to somehow use localization to drop back down to the PID case when working in a Dedekind domain. However, I guess that was just poor phrasing of the question, since it immediately drops out in the way you suggested. – Spyam Jan 30 '12 at 0:21 @Ben It seems like overkill to use localization for the other direction, I agree. – Dylan Moreland Jan 30 '12 at 0:35