# 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

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.