TITLE: Rank of a free module without the axiom of choice
QUESTION [9 upvotes]: Perhaps my question is really naive. I teach this semester in algebra. I am embarassed about the proof that a free module over an integral domain has a well-defined rank. It is based on the theorem that two bases have the same cardinal. But the proof I know of this statement uses a reduction modulo a maximal ideal (then we are brought back to the theory of vector spaces). It seems to me that the existence of a principal ideal needs in general the axiom of choice (not for concrete domains, of course). Whence my question:

Do we really need the axiom of choice to prove in full generality that the bases of a free module have the same cardinal ? 

One day later. I did not realize that the question even makes sense in the context of vector spaces. I'd like being able to accept a comment.

REPLY [6 votes]: Hans Läuchli [Auswahlaxiom in der Algebra, Comment. Math. Helv. 37, MR0143705, DOI:10.5169/seals-28602] has constructed a model of ZFA wherein a vector space has two bases of different size. (See also Jech, The Axiom of Choice, §10.3, problem 5.) The Jech–Sochor Embedding Theorem allows to transfer the result to ZF.