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Let $k$ be a field. In 1984 Andreas Blass proved that the axiom "for every extension $K|k$, every vector space over $K$ has a basis" implies the axiom of choice. He also raised the question

Does the axiom "every vector space over $k$ has a basis" imply the axiom of choice ?

What's the current status of the question ? Has there been progress ?

edited Jun 27 2011 at 19:21

asked Jun 27 2011 at 18:11

Ralph
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This has come up before: mathoverflow.net/questions/64219 – Bruce Westbury Jun 27 2011 at 18:22

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@Bruce: The two questions are clearly different. – François G. Dorais♦ Jun 27 2011 at 18:31

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Right -- Andreas even says toward the end of the note that the question is open for specific fields like the rationals. I hope that anyone else who votes to close this question leaves a note as to why it should be closed. – Todd Trimble Jun 27 2011 at 18:53

I stand corrected. – Bruce Westbury Jun 27 2011 at 19:44

@Asaf: Thanks. Perhaps you can add references and post it as answer. – Ralph Jun 27 2011 at 19:55

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Asaf Karagila · Accepted Answer · 2011-06-27 20:23:59Z UTC

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It has been shown for $K=\mathbb F_2$ (the field with two elements) by Keremedis (Available here)

In the dictionary of AC equivalences it shows that not a lot is known on the connection between the existence of a basis over a fixed field and the axiom of choice.

answered Jun 27 2011 at 20:23

Asaf Karagila
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	As far as I can see, Keremedis shows that AC is equivalent to “every generating set of a vector space over $\mathbb F_2$ contains a basis”, which is a stronger statement than “every vector space over $\mathbb F_2$ has a basis”. – Emil Jeřábek Aug 30 2011 at 15:12
	Do you mean 'weaker'? – Jan Veselý Apr 10 2012 at 15:03

Axiom of choice and bases of vector spaces over a fixed field

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