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 ?

  • $\begingroup$ This has come up before: mathoverflow.net/questions/64219 $\endgroup$ – Bruce Westbury Jun 27 '11 at 18:22
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    $\begingroup$ @Bruce: The two questions are clearly different. $\endgroup$ – François G. Dorais Jun 27 '11 at 18:31
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    $\begingroup$ 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. $\endgroup$ – Todd Trimble Jun 27 '11 at 18:53
  • $\begingroup$ I stand corrected. $\endgroup$ – Bruce Westbury Jun 27 '11 at 19:44
  • $\begingroup$ @Asaf: Thanks. Perhaps you can add references and post it as answer. $\endgroup$ – Ralph Jun 27 '11 at 19:55

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.

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    $\begingroup$ 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”. $\endgroup$ – Emil Jeřábek Aug 30 '11 at 15:12
  • $\begingroup$ Do you mean 'weaker'? $\endgroup$ – Jan Veselý Apr 10 '12 at 15:03
  • $\begingroup$ @JanVeselý, no, because every vector space has a generating set (itself!). $\endgroup$ – LSpice Apr 9 '18 at 17:05

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