Exercise A0.V.2.09

Proposition

Let $R$ be an UFD and let $I$ be a prime ideal of height $1$ in $R.$ Then $I$ is a principal ideal. ⚠️

Proof

Let $a \in I$ be non-zero. We perform induction on the number of irreducible factors of $a$ to prove that there exists an irreducible $p$ such that $p \in I.$

If $a$ has one irreducible factor $p$, then $a = up$ for some unit $u.$ Since $I$ is prime, $u \in I$ or $p \in I.$ It cannot be that $u \in I,$ since that contradicts $I$ being prime. Thus $p \in I.$ This resolves the base case.

Suppose $a$ has more than one irreducible factor. Then we write $a = a'p$ for some irreducible factor $p$ of $a.$ Since $I$ is prime, either $a' \in I$ or $p \in I.$ If $p \in I$ we’re done. Otherwise, $a' \in I.$ Observe that $a'$ has one fewer prime factors than $a.$ Therefore, by the inductive hypothesis, there exists an irreducible $p' \in I$ as required.

Thus, there exists an irreducible $p \in I.$ In that case, there is a containment of prime ideals $(0) \subsetneq (p) \subseteq I$ and since $I$ has height of $1,$ we must conclude that $(p) = I,$ and hence $I$ is principal.

$\blacksquare$