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One of the classical uses of the existence of bases of vector spaces (which is equivalent to the axiom of choice) is the following theorem:

If $V$ is an infinite vector space over a field $F$, and $|V|>|F|$, then $|V|=\dim V$.

As a corollary, we immediately get that any two uncountable fields of the same characteristics are isomorphic as additive groups if and only if they have the same cardinality. Proof? They have the same dimension over their prime field. So for example $(\Bbb R,+)\cong(\Bbb C,+)$.

Without the axiom of choice, things are not that nice. If every set is Lebesgue measure, or every set has the Baire property, then $(\Bbb R,+)$ and $(\Bbb C,+)$ are not isomorphic. Therefore there are two $\Bbb Q$-vector spaces of the same uncountable cardinality, which are not isomorphic.

Question. Suppose that whenever $W,W'$ are uncountable $F$-vector spaces and $|W|=|W'|>|F|$, we have that $W\cong W'$. Does the axiom of choice holds? Or at least imply that $W$ has a basis over $F$?

(Looking through Rubin–Rubin II, there is a lot of focus on bases of vector spaces, but not so much on their cardinality. At most, there was a mention of the cardinality of a basis, but here I am not requiring that $W$ or $W'$ have a basis.)

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  • $\begingroup$ Usually, I'd ask myself that question. But my mind is elsewhere these days, and I feel that I'm missing something obvious. $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 20, 2018 at 23:50
  • $\begingroup$ I would say, first step: consult Rubin & Rubin, Equivalents of the Axiom of Choice $\endgroup$
    – Gerald Edgar
    Commented Sep 21, 2018 at 0:29
  • $\begingroup$ Yes, that's usually a good choice. But if my memory serves me right, this isn't there. I'll take a look tomorrow morning, see what I find there. Thanks. $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 21, 2018 at 0:30
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    $\begingroup$ Any two uncountable fields of the same characteristic are isomorphic as additive groups iff they have the same cardinality. In fact, you get something better in finite characteristic: any two infinite fields of characteristic $p\neq 0$ are isomorphic as additive groups iff they have the same cardinality. $\endgroup$
    – Alex Kruckman
    Commented Sep 21, 2018 at 17:41
  • $\begingroup$ @AlexKruckman: Boy was I tired when I posted this... :P $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 21, 2018 at 17:43

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