Preimage of ideal in $R/p$ for Dedekind ring $R$ and prime ideal $p$.

Question

Let $R$ be a Dedekind ring, $p$ a prime ideal in $R$, $n>1$ some integer.

The question is about a fact that is stated in this answer: The ideals of $R/p^n$ are exactly the images of the ideals of $R$ containing $p^n$, i.e. $p^i$ for $1\le i \le n$.

I tried following: Consider a proper ideal in $R/p^n$. Then the preimage under the quotientmap must be a proper ideal, too. This preimage also contains $p^n$. Now this preimage could be $p^i$ for $1\le i \le n$. But i cannot exclude other ideals. I tried using that $p$ must be maximal, but still I have difficulties to understand why there are no other ideals.

Well, you say the preimage contains $p^n$. But $p^{n+1}$ does not contain $p^n$, so this should exclude other ideals right? — Sheel Stueber, May 16 '18 at 01:29

score 2 · Accepted Answer · answered May 16 '18 at 02:58

Let $p:R\rightarrow R/p^n$, and $I$ and ideal of $R/p^n$, write $J=p^{-1}(I)$. $J$ is contained in a maximal ideal $q$. For every $x\in p$, $x^n\in p^n\subset q$. Since $q$ is maximal thus prime, we deduce that $x\in q$ and $p=q$ since $p$ is maximal too. Since $J$ factors into primes (DD1 of the reference) and a prime factor $q$ of $J$ is equal to $p$ since $p^n\subset J\subset q$, like above, we deduce that $q=p$. This implies that $J=p^i$.

https://en.wikipedia.org/wiki/Dedekind_domain#Alternative_definitions

Preimage of ideal in $R/p$ for Dedekind ring $R$ and prime ideal $p$.

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