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AC in group isomorphism between R and R^2

Somewhere, I recall being told that there is an isomorphism between $\mathbb{R}$ and $\mathbb{C}$ under addition. However, despite a rather lengthy search, I have been unable to find anything to support this fact, although Paul Yale of Pomona College, in his paper, "Automorphisms of the Complex Numbers," showed that there are "wild" automorphisms of $\mathbb{C}$ that require the axiom of choice to construct. Given that rather surprising fact, it does not seem too unlikely that there could be an isomorphism between $\mathbb{R}$ and $\mathbb{C}$. So, the question is: Is it possible for there to be an isomorphism between $\mathbb{R}$ and $\mathbb{C}$, and if so, what is is?

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    $\begingroup$ If you are asking whether there exists an isomorphism $\mathbb C\to\mathbb R$ as abelian groups, then the answer is yes: both are in fact $\mathbb Q$-vector spaces of the same dimension, so they are isomorphic as such. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jul 10, 2010 at 12:43
  • $\begingroup$ Let the record show that Mariano beat me by two minutes. $\endgroup$
    – Gerry Myerson
    Commented Jul 10, 2010 at 12:59
  • $\begingroup$ If $K$ is a countable field and $V$ is a vector space, then its dimension equals its cardinality. In particular, such spaces are isomorphic as soon as they are equipotent. $\endgroup$
    – Martin Brandenburg
    Commented Jul 10, 2010 at 17:07
  • $\begingroup$ @Martin: that works if $|V|>|K|$; otherwise, note that $K$ is a vector space over itself, but the dimension is not $|K|=\aleph_0$, it's $1$. $\endgroup$
    – Arturo Magidin
    Commented Jul 10, 2010 at 17:10
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    $\begingroup$ Btw, talking about $\mathbb{R}$ as $\mathbb{Q}$ vector space, I'd like to mention that one can produce a Vitali set as a direct summand $V$ of the subspace $\mathbb{Q}$. This way $V$ can't have null Lebesgue measure since $\mathbb{R}=\cup_{q\in\mathbb{Q}} (V+q)$, and it can't have positive measure since $V−V\subset V$ is not a neighborhood of 0, as it should be were it a measurable set with positive measure. $\endgroup$
    – Pietro Majer
    Commented Jul 10, 2010 at 21:04

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As vector spaces over the rationals, they have the same dimension, so the only tricky part is the difficulty in finding a basis.

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    $\begingroup$ This related question talks about the need for the axiom of choice: mathoverflow.net/questions/25375/… $\endgroup$
    – PersonX
    Commented Jul 10, 2010 at 13:16
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    $\begingroup$ ... and thus this one should be closed ... $\endgroup$
    – Gerald Edgar
    Commented Jul 10, 2010 at 16:09

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