TITLE: Does ZF prove that all PIDs are UFDs?
QUESTION [16 upvotes]: 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.)

REPLY [23 votes]: 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.)