# Does ZF prove that all PIDs are UFDs?

Main Question:

Does ZF (no axiom of choice) prove that every Principal Ideal Domain is a Unique Factorization Domain?

The proofs I've seen all use dependent choice.

Minor Questions:

Does ZF + Countable Choice prove all PIDs are UFDs?

Does ZF prove "If all PIDs are UFDs, then [some choice principle]"?

(If anyone knows how I could force line breaks to put the questions on their own lines, please tell me.)

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Nope, it doesn't. I'm also pretty sure this is a duplicate question. – Harry Gindi Jul 12 '10 at 6:57
Gjergji, that's a more nicely put together proof than the others I've seen, but it still uses dependant choice. – Ricky Demer Jul 12 '10 at 7:15
While this is very similar to the question cited, I don't see it as identical. This question amounts roughly to is "PID => UFD" equivalent in ZF to some choice principle. It's maybe more a question for logicians than algebraists. – Robin Chapman Jul 12 '10 at 8:41
Re line breaks, <br /> will force a line break. – Chris Phan Jul 12 '10 at 14:41

## 1 Answer

ZF alone does not prove that every PID is a UFD, according to this paper: Hodges, Wilfrid. Läuchli's algebraic closure of $Q$. Math. Proc. Cambridge Philos. Soc. 79 (1976), no. 2, 289--297. MR 422022.

One result in this paper is the following:

COROLLARY 10. Neither (a) nor (b) is provable from ZF alone:
(a) Every principal ideal domain is a unique factorization domain.
(b) Every principal ideal domain has a maximal ideal.

By the way, I didn't know the answer to this question until today. To find the answer, I consulted Howard and Rubin's book Consequences of the Axiom of Choice. (Actually, I did a search for "principal ideal domain" of their book using Google Books.)

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