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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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marked as duplicate by Gerald Edgar, Qiaochu Yuan, Allen Knutson, Kevin H. Lin, Jonas Meyer Jul 10 '10 at 23:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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. – Mariano Suárez-Alvarez Jul 10 '10 at 12:43
Let the record show that Mariano beat me by two minutes. – Gerry Myerson Jul 10 '10 at 12:59
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. – Martin Brandenburg Jul 10 '10 at 17:07
@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 $1$. – Arturo Magidin Jul 10 '10 at 17:10
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. – Pietro Majer Jul 10 '10 at 21:04
up vote 5 down vote accepted

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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This related question talks about the need for the axiom of choice:… – Chris Phan Jul 10 '10 at 13:16
... and thus this one should be closed ... – Gerald Edgar Jul 10 '10 at 16:09

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