'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.
 A: 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.
