MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the absence of the Axiom of Choice, it is not necessarily true that all vector spaces over a field have bases.

What are the possible global dimensions of fields in a model of ZF in which AoC does not hold?

For comparison, a related result of Osofsky is that the global dimension of a countable product of fields is $k+1$ if $2^{\aleph_0}$ is $\aleph_k$, so that dimension depends on how badly the Continuum Hypothesis fails in one's model of set theory.

share|cite|improve this question
Which definition of global dimension do you use? I think that the equivalence between the usual definitions/characterizations might need AC. – Martin Brandenburg Jul 10 '13 at 8:56
The least upper bound on the length of projective resolutions of modules! – Mariano Suárez-Alvarez Jul 10 '13 at 8:57
(Fields are von Neumann regular and noetherian. Normally, such a thing is semisimple, but I don't know if the proof of this carries over to a non-AoC situation) – Mariano Suárez-Alvarez Jul 10 '13 at 9:11
Need there even be enough projectives without Choice? I don't see why an infinitely generated free module need be projective, in particular. – Jeremy Rickard Jul 10 '13 at 10:57
Presumably the reduction to finitely generated modules uses choice? – Benjamin Steinberg Jul 10 '13 at 11:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.