Past exam of the algebraic number theory course of the University of Oxford / 2023 / 2 / b / i / Solution

By the correspondence theorem for ideals, an ideal of $R/P^n$ has the form $J/P^n$ with $P^n\subseteq J\subseteq R$. If $J\ne R$, apply unique factorization of ideals in a number field to $J$. Every prime ideal $Q$ dividing $J$ contains $J$ and hence $P^n$. Primality gives $P\subseteq Q$, and maximality gives $Q=P$. Thus $J=P^i$ for some $i\ge 1$.

The inclusion $P^n\subseteq P^i$ forces $i\le n$: if $i>n$, cancellation using the invertible fractional ideal $P^n$ would give $R\subseteq P^{i-n}$, which is impossible. The zero ideal corresponds to $i=n$, while the unit ideal corresponds to $i=0$. Consequently the nonzero proper ideals are exactlyWhen $n=1$, there are none, in agreement with $R/P$ being a field.