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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.

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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

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