Lemma 10.120.3. Let $R$ be a domain. Consider the following conditions:

1. The ring $R$ satisfies the ascending chain condition for principal ideals.

2. Every nonzero, nonunit element $a \in R$ has a factorization $a = b_1 \ldots b_ k$ with each $b_ i$ an irreducible element of $R$.

Then (1) implies (2).

Proof. Let $x$ be a nonzero element, not a unit, which does not have a factorization into irreducibles. Set $x_1 = x$. We can write $x = yz$ where neither $y$ nor $z$ is irreducible or a unit. Then either $y$ does not have a factorization into irreducibles, in which case we set $x_2 = y$, or $z$ does not have a factorization into irreducibles, in which case we set $x_2 = z$. Continuing in this fashion we find a sequence

$x_1 | x_2 | x_3 | \ldots$

of elements of $R$ with $x_ n/x_{n + 1}$ not a unit. This gives a strictly increasing sequence of principal ideals $(x_1) \subset (x_2) \subset (x_3) \subset \ldots$ thereby finishing the proof. $\square$

Comment #4339 by David Speyer on

I don't know if you have a standard convention about marking uses of Choice. If so, it might be worth pointing out that this argument requires Choice; see https://mathscinet.ams.org/mathscinet-getitem?mr=422022 . I ran across this page while trying to find out if Choice was needed for this Lemma, so maybe this comment will help the next person.

Comment #4489 by on

Thanks David! To everybody: you should always assume everything in the Stacks project uses choice because this is one of the starting axioms of the set theory we use.

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