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In Zermelo-Fraenkel set theory $ZF$ consider the following statement defined for every field $K$:

$B_K$ : Every vector space over $K$ has a basis.

It is well-known that $AC \Rightarrow \forall K (B_K)$. A. Blass proved the converse. In his paper from 1984 he said that for a fixed field $K$, the problem of proving $B_K \Rightarrow AC$ is open. What is the current status? Specifically, I would like to know:

Does $B_{\mathbb{Q}}$ imply $AC$?

Does $B_{\mathbb{C}}$ imply $AC$?

Martin Brandenburg
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In their paper, Howard and Tachtsis discuss these questions. The paper was published just last year, so I suspect that there hasn't been any significant progress since then.

Paul Howard and Eleftherios Tachtsis, On vector spaces over specific fields without choice, MLQ Math. Log. Q. 59 (2013), no. 3, 128--146.

In general, it seems that the axiom of regularity might play an important part in such proof. Which means that there might be more than just the structure of the vector spaces involved (although this might be mitigated by going at it as Blass did, by showing that you can prove $\sf MC$ rather than $\sf AC$).

If we slightly extend the result to "Every spanning set includes a basis", which in particular means that every vector space has a basis, then Keremedis showed in his paper

Kyriakos Keremedis, Extending independent sets to bases and the axiom of choice, Math. Logic Quart. 44 (1998), no. 1, 92--98.

That over $\Bbb Q$ if every generating set includes a basis, then the axiom of choice holds. I don't know the proof, but it might be possible to extend it to $\Bbb R$ or $\Bbb C$.

All in all, it seems that the questions you particularly address are still very much open, and perhaps new techniques are needed before we can find answers.

Cameron Buie
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Asaf Karagila
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  • "Howard and Tachtsis discuss these questions."- Yes, I had already found that paper, but don't know its contents since I don't have access to it. Can you add in how far the paper answers my questions? Are you saying that my two questions are open? – Martin Brandenburg Oct 28 '14 at 08:43
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    Well, there are no real results. Just a few results which are relatively simple about partial choice from finite subsets under several additional conditions in some cases where $F$ is a field satisfying some conditions, and assuming $B(F)$. There is no additional actual discussion about $B(\Bbb Q)$. The question whether or not there is even $F$ such that $B(F)$ implies choice, or that it doesn't remains open. $B(\Bbb R)$ specifically mentioned in a question whether or not it implies $\Bbb R$ can be well-ordered. But again, no actual discussion. So yes, both questions are quite open. – Asaf Karagila Oct 28 '14 at 10:26