Prove that if $\mathrm D$ is integral domain, then $\mathrm D$is UFD iff these conditions hold.

Question

Prove that id $\mathrm D$ is integral domain, the "$\mathrm D$ is UFD " iff

(1) $\mathrm D$ satisfies the ACC for principal ideals.

(2) every irreducible element is a prime element.

I've proven the right implication. The other direction is harder.

Please give me a proof or hint. Thanks!

Answer 1

You may follow these steps.

1. Let $a \in D$ be nonzero, and not a unit. Then $a$ is divisible by an irreducible element. (For this you use (1). The idea is that either $a$ is irreducible, and you're done, or it has a proper divisor $a_{1}$, that is, a divisor which is not a unit, nor associated to $a$. So you start constructing an AC of ideals $(a) \subset (a_{1}) \subset \dots$.)
2. Let $a \in D$ be nonzero, and not a unit. Then $a$ is a product of irreducible elements. (For this you use (1). The proof is similar to that of the previous point. If $a$ is irreducible, you're done. Otherwise $a = p_{1} a_{1}$, with $p_{1}$ irreducible. So you start constructing an AC of ideals $(a) \subset (a_{1}) \subset \dots$.)
3. Factorization is unique. (For this you use (2).)