Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 101 / 4

Suppose $xy\in J$ and $x\notin J$. Then the colon ideal $(J:x)$ properly contains $J$ because it contains $y$. If also $y\notin J$, then $J+(x)$ properly contains $J$. The maximality of $J$ in the family makes both larger ideals lie outside the family. Applying the stated closure condition with $I=J$ and $a=x$ would imply $J$ lies outside the family, a contradiction. Hence $x\in J$ or $y\in J$, and $J$ is a prime ideal. This argument is the Prime ideal principle for an Oka family.

Let $\mathcal{F}$ be the proper nonprincipal ideals. A chain in $\mathcal{F}$ has a nonprincipal union: if its union were $(a)$, then $a$ would belong to one member of the chain, forcing that member to equal the union and be principal. The union is also proper. Thus Zorn lemma gives a maximal member whenever $\mathcal{F}$ is nonempty.

The family satisfies the condition from part (a). Indeed, supposeWrite $a=bd$ and $b=i+ra$, where $i\in I$. Every $x=bs\in I$ has $as=dx\in I$, so $s\in (c)$ and $I\subseteq (bc)$. Conversely $ac\in I$ because $c\in (I:a)$, while $ic\in I$ because $i\in I$. Thus $bc=ic+rac\in I$, proving $(bc)\subseteq I$. Hence $I=(bc)$ is principal.

If a nonprincipal ideal existed, part (a) would therefore produce a nonprincipal prime ideal, contrary to the hypothesis. Every ideal is principal, so the integral domain is a principal ideal domain.

Suppose first that $A$ is a unique factorization domain, and let $\mathfrak{p}$ be a minimal nonzero prime ideal. Choose $0\ne a\in \mathfrak{p}$ and factor it into irreducibles. In a UFD each irreducible is a prime element, so one factor $\pi$ belongs to $\mathfrak{p}$. The nonzero prime ideal $(\pi )\subseteq \mathfrak{p}$ must equal $\mathfrak{p}$ by minimality.

Conversely, the ascending chain condition in the Noetherian domain implies that every nonzero nonunit factors into irreducibles. Let $\pi$ be irreducible and choose a prime $\mathfrak{p}$ minimal over $(\pi )$. The Krull principal ideal theorem gives $\mathrm{ht}\mathfrak{p}\le 1$. Since $A$ is a domain and $\mathfrak{p}\ne (0)$, it is a minimal nonzero prime and hence is principal, say $\mathfrak{p}=(q)$. The divisibility $q\mid \pi$ and irreducibility of $\pi$ force $q$ to be associate to $\pi$, so $(\pi )=\mathfrak{p}$ is prime. Thus every irreducible is prime, proving that $A$ is a UFD. This is the Minimal-prime criterion for a Noetherian unique factorization domain.

Let $A$ be a principal ideal domain that is not a field. A PID is Noetherian and is a unique factorization domain. Every nonzero prime ideal is generated by a prime element. Ifthen $a\mid p$, so primality of $p$ makes $a$ a unit or an associate of $p$. The proper alternative is $(a)=(p)$, and every nonzero prime is therefore maximal. Since $A$ has a nonzero prime ideal, its Krull dimension is exactly one.

Conversely, let $A$ be a Noetherian UFD of Krull dimension one. Every nonzero prime is minimal among nonzero primes, and the forward argument in part (c) makes it principal. The zero ideal is principal as well, so part (b) shows that $A$ is a PID. A field has Krull dimension zero, hence $A$ is not a field.