Let $A$ be a Dedekind domain and $p$ be a prime ideal in $A$ then how to prove that $A/p^n \equiv A_p/p^nA_p$.

Question

Let $A$ be a Dedekind domain and $p$ be a prime ideal in $A$ then how to prove that $A/p^n \equiv A_p/p^nA_p$.

Clearly, $A_p$ is a D.V.R. Considering the map $\phi : A/p^n \to A_p/p^nA_p$ s.t $\phi(a)= \frac a1+p^nA_p$.

This well defined and ring homomorphism is fine.

But how to show that $\frac a1+p^nA_p=0$ then $\frac a1= \frac xs$ where $x \in p^n$ this implies $a \in p^n$?? That is how to show $\phi$ is injective?

Then how to prove $\phi$ is surjective?

Answer 1

If $A$ is Dedekind, it has Krull dimension one, i.e. non-zero prime ideals are maximal. There results that $\mathfrak p^n$ is contained in the sole maximal ideal $\mathfrak p$, and thence $A/\mathfrak p^n$ is a local ring of dimension $0$. It is therefore equal to its localisation at $\mathfrak p/\mathfrak p^n$, and $$A/\mathfrak p^n=\bigl(A/\mathfrak p^n\bigr)_{\mathfrak p/\mathfrak p^n}\simeq A_{\mathfrak p}/\mathfrak p^nA_{\mathfrak p}.$$