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

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.

For every $i\ge 0$, the inclusion $P^{i+1}\subseteq P^i$ is strict: equality, followed by multiplication by the inverse of $P^i$, would give $P=R$. Hence one can choosePut $J=(\pi )+P^{i+1}$. ThenThe classification just proved, applied modulo $P^{i+1}$, leaves no ideal strictly between consecutive powers. It follows thatFor $i=0$, the same argument says $(\pi )+P=R$, using $\pi \notin P$ and maximality.

The extension to every $r\ge 1$ follows by mathematical induction. Suppose $P^i=(\pi )+P^{i+r}$. Multiplication of the initial equality by $P^r$ givesSubstitution yieldsThusTaking $r=n-i$ for $0\le i<n$ givesThis is a principal ideal of the quotient ring $R/P^n$. Its zero ideal is generated by zero, so every ideal of this quotient ring is a principal ideal.

The printed domain assertion is false. For example, take $K=\mathbb{Q}$ and $I=(4)$. Then ${\mathcal{O}}_K/I=\mathbb{Z}/4\mathbb{Z}$ has the nonzero zero divisor $2+(4)$, so it is not an integral domain and cannot be a principal ideal domain. The valid conclusion is that ${\mathcal{O}}_K/I$ is a principal ideal ring. This error is present in the original PDF, not just its transcription.

To prove the corrected conclusion, factorinto powers of distinct nonzero prime ideals. These powers are pairwise comaximal ideals. Indeed, a maximal ideal containing two such powers would contain both underlying prime ideals, forcing them to be equal. The Chinese remainder theorem givesEvery factor is a principal ideal ring by the preceding part. Every ideal $H$ in a finite direct product of rings is a product $H_1\times \cdots \times H_s$ of ideals in the factors: multiplication by the coordinate idempotent elements separates the coordinates of every element of $H$. Write $H_j=(h_j)$. Thenso $R/I$ is a principal ideal ring. This proves the principal ideal property of quotients of a Dedekind domain in the present setting. In fact, this finite quotient ring is an integral domain precisely when $I$ is a prime ideal, in which case it is a field.

For the final deduction, choose any $0\ne a\in I$ and apply the result to the nonzero proper ideal $(a)$. The ideal $I/(a)$ of $R/(a)$ is generated by one residue class $b+(a)$, which can be represented by $b\in I$. For each $x\in I$, some $r\in R$ satisfies $x-rb\in (a)$. HenceThis proves two generators for an ideal in a Dedekind domain; moreover, its first generator can be any prescribed nonzero element of $I$.