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.
Localisation of quotients of Dedekind Domains
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commutative-algebra
localization
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1Your generalization isn't a direct parallel, in the example you look at a quotient of a localized ring, in your proposed new idea you localize **after** taking a quotient. And it cannot be true, take $R=\Bbb Z$, $q=(3)$ and $p=2$ with $\alpha>0$. Then $R/q$ has characteristic $3$ and $R/2^{\alpha}$ has no element of [additive] order $3$. – 2017-01-30
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0How localization at $p \subset S$ affects a product of two rings $R \times S$ ? – 2017-01-30
1 Answers
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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 $q=p^ap_1^{a_1} \dotsb p_n^{a_n}$.