Localisation of quotients of Dedekind Domains

Question

In a Dedekind Domain every ideal is the product of prime ideals. I was computing some localisations and got that: $$ (\mathbb{Z}/n\mathbb{Z})_{(p)}\cong\mathbb{Z}/p^\alpha\mathbb{Z}$$ where $p$ has multiplicity $\alpha$ in $n$. So, is it true that for any Dedekind domain R we have that: $$ (R/q)_{p}\cong R/p^\alpha$$ where $p$ has multiplicity $\alpha$ in $q$? Heuristically, in the localisation I am inverting everything but p so it seems convincing to me.

Answer 1

Your claim is true. Consider the following proposition:

Let $A$ be an artinian commutative ring and $S \subset A$ a multiplicative subset. Then the localization map $A \to S^{-1}A$ is surjective.

Proof. Let $\frac{a}{s} \in S^{-1}A$. Consider the chain $$(s) \supset (s^2) \supset \cdots.$$ The artinian property yields $(s^n)=(s^{n+1})$ for some $n$, i.e. $s^n=s^{n+1}b$ for some $b \in A$. We get $$s^n(a-sab)=0,$$ thus $\frac{a}{s}=\frac{ab}{1}$ in $S^{-1}A$ is contained in the image of the localization map. $\Tiny\square$

We can apply this to $R/q$ since this is a zero-dimensional noetherian ring, hence artinian.

Thus the map $$R \to R/q \to (R/q)_p$$ is surjective as a composition of surjective maps. Since localization and taking the quotient commute, this map can also be written as $$R \to R_p \to R_p/q_p \cong (R/q)_p.$$

It is straightforward to see that the kernel is precisely $p^a$, where $q$ can be factored as $q=p^ap_1^{a_1} \dotsb p_n^{a_n}$.