I think the answer is yes.

Sketch of the proof Consider $\mathbb{R}$ as a vector space over $\mathbb{Q}$. Let $\{e_\lambda:\lambda\in\Lambda\}\subset\mathbb{R}$ be its Hamel basis. Then $\{(e_{\lambda_1},e_{\lambda_2}):\lambda_1,\lambda_2\in\Lambda\}\subset\mathbb{R}^2$ is a Hamel basis of $\mathbb{R}^2$. Consider set theoretic bijection $i:\Lambda\to\Lambda\times\Lambda$ define map $\varphi:\mathbb{R}\to\mathbb{R}^2$ by its action on element of Hamel basis of $\mathbb{R}$ by equality $\varphi(e_\lambda)=(e_{i(\lambda)_1},e_{i(\lambda)_2})$. This is an isomorphism between $\mathbb{Q}$-vector spaces $\mathbb{R}$ and $\mathbb{R}^2$.

Question Could you tell me is this proof correct, and if it is not, where is the mistake?

EDIT: Thanks to GAJO we found that this proof is wrong. The correct one basis is $$ \{(e_\lambda,0):\lambda\in \Lambda\}\cup\{(0,e_\lambda):\lambda\in\Lambda\} $$ So we had to look at bijections of the form $i:\Lambda\to\Lambda\coprod\Lambda$ and the desired linear operator is defined by equalities $\varphi(e_\lambda)=(e_{i(\lambda)},0)$ if $i(\lambda)$ lies in the first copy $\Lambda$ and $\varphi(e_\lambda)=(0,e_{i(\lambda)})$ otherwise.

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    $\begingroup$ Yes, that's fine. $\endgroup$ May 14, 2012 at 19:37
  • $\begingroup$ Looks good to me. $\endgroup$ May 14, 2012 at 19:37
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    $\begingroup$ Amazing community, answer after 15 second. ArturoMagidin, Brian M. Scott big thanks! $\endgroup$
    – Norbert
    May 14, 2012 at 19:39
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    $\begingroup$ @GEdgar, I don't care about ZF and completely accept Axiom of Choice with all its "strange" consequnces. $\endgroup$
    – Norbert
    May 14, 2012 at 21:22
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    $\begingroup$ @user49685: Assuming the axiom of choice, $|S\times S|=|S|$ for any infinite set $S$, so there is a bijection from $S$ to $S\times S$. One lets $\kappa=|S|$, and it’s not too hard to define an explicit bijection between the initial ordinal $\kappa$ and $\kappa\times\kappa$. Here $\Lambda$ is infinite, so there is a bijection from $\Lambda$ to $\Lambda\times\Lambda$. $\endgroup$ Jun 16, 2013 at 17:41

1 Answer 1


This was too long for a comment.

This obviously generalizes. Indeed, if $F$ is an uncountable abelian group which can be given the structure of a characteristic zero field then $F\cong F^{n}$ for any finite $n$. This is because, as you have noticed, $\dim_\mathbb{Q} F$ is necessarily infinite and thus $\dim_\mathbb{Q} F=\dim_\mathbb{Q} F^n$ so that they are isomorphic as $\mathbb{Q}$-vector spaces and thus as abelian groups.

The same argument shows that if $F$ is an infinite abelian group which can be given the structure of a characteristic $p$ field then $F\cong F^2$ as abelian groups. The same trick applies since necessarily $\dim_{\mathbb{F}_p} F$ is infinite.

Of course, both of these are really saying that if you can give the abelian group $F$ the structure of a field which is infinite dimensional over its prime subfield then it's isomorphically idempotent (as groups).

Of course, ALL OF THIS is just saying that if $F$ is a group which can be given the structure of a free $R$-module of infinite rank then it is isomorphically idempotent. So, this applies equally well to, say, $\mathbb{Q}[x]$.

That said, it's definitely not true for things like, say, $\mathbb{Q}$. Indeed, if $R$ is any integral domain then $\text{Frac}(R)\not\cong\text{Frac}(R)^n$ as $R$-modules for any finite $n$. This is because if this were true then


as $\text{Frac}(R)$-modules. But, I'll leave it to you to show that $\text{Frac}(R)\otimes_R\text{Frac}(R)\cong\text{Frac}(R)$ as $\text{Frac}(R)$ modules (just show it's nonzero and singly generated) from where you get (since fields have the IBN property) that $n=1$.

In fact, I think it's pretty obvious that the above generalizes to show that $\mathbb{Q}^n\not\cong\mathbb{Q}^m$ (or the more general $\text{Frac}(R)$ version) as groups for any finite $m\ne n$.

  • $\begingroup$ Sorry, I'm a newbie in algebra. What is $\mathrm{Frac}(R)$? $\endgroup$
    – Norbert
    May 14, 2012 at 19:47
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    $\begingroup$ I'm sorry, I should have said. It means field of fractions. So, for example, $\text{Frac}(\mathbb{Z})=\mathbb{Q}$. So the above says that $\mathbb{Q}\not\cong\mathbb{Q}^n$ as groups for any finite $n$ (or obviously infinite). $\endgroup$ May 14, 2012 at 19:47
  • $\begingroup$ Thanks, Alex. I've not suspected that this can be generalized to such extent. $\endgroup$
    – Norbert
    May 14, 2012 at 19:58
  • $\begingroup$ @ Alex , can you tell me why is $F$ isomorphic to $F^n$ and only for finite $n$ ? $\endgroup$
    – Theorem
    May 14, 2012 at 20:21
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    $\begingroup$ @Ananda It's actually isomorphic for any cardinal with cardinality less then $\dim_\mathbb{F_0}F$. This is because $\lambda\cdot\mu=\min\{\lambda,\mu\}$. $\endgroup$ May 14, 2012 at 20:59

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