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I know that from ZF + the Axiom of Choice (AC) follows that every vector space has a basis.

And, conversely, Blass proved that in ZF set theory, the assumption that every vector space has a basis leads to AC.

Suppose we fix a field $k$, and we assume in ZF set theory that every $k$-vector space has a basis.

I understand that the answer to this question is not known, at least not for all fields.

My question is: does Blass's proof work for prime fields (finite fields of prime number order and the rationals), and if not, why not ?

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    $\begingroup$ It's an open question. Essentially, Blass' construction, taken "as is", requires you to have fairly complicated fields to encode for the choice function. So it's not clear if you can reduce those to their prime field. $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 28, 2021 at 19:56
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    $\begingroup$ Blass's beautiful and ingenious proof is quite short (the paper is three pages long, and can be found on his web page), so I think the best way for you to answer your question is to read it. But briefly, the answer to the question "why not?" is that, starting with a counterexample to AC (strictly speaking, to the Axiom of Multiple Choice), the vector space without a basis that he constructs is over a field that depends on that counterexample. $\endgroup$
    – Jeremy Rickard
    Commented Jul 28, 2021 at 20:30

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