Quotient of an ideal in Dedekind Domain is principal ring

Question

The quotient ring $\mathcal{O}/\mathfrak{a}$ of a Dedekind domain by an ideal $\mathfrak{a}\ne 0$ is a principal ideal domain.

I am trying to show $\mathcal{O}/\mathfrak{p}^n$ is principal ring. Let $\mathfrak{p}^i/\mathfrak{p}^n$ be an ideal and choose $\pi\in\mathfrak{p}\setminus\mathfrak{p}^2$. Then how I show that $\mathfrak{p}^i = \mathcal{O}\pi^i + \mathfrak{p}^n$ ? Any help/hint in this regards would be highly appreciated. Thanks in advance!

Answer 1

You have $(\pi)^i \not\subset \mathfrak{p}^{i+1}$, but $(\pi)^i\subset \mathfrak{p}^i$.

Thus $(\pi)^i=J\cdot \mathfrak{p}^i$ for an ideal $J$ with $\mathfrak{p}\nmid J$ .

Now we have $(\pi)^i+\mathfrak{p}^{n}=\mathfrak{p}^{i}(J+\mathfrak{p}^{n-i})$. But $J$ and $\mathfrak{p}^{n-i}$ have no common prime ideal factors, thus $J+\mathfrak{p}^{n-i}=\mathcal{O}$ and we get the desired result.

Answer 2

Hint: $$\mathcal O/\mathfrak p^n\simeq\mathcal O_{\mathfrak p}/(\mathfrak pO_{\mathfrak p})^n.$$ What is the localisation of a Dedekind domain at a maximal ideal?