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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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    Yes, that's fine.2012-05-14
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    Looks good to me.2012-05-14
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    Amazing community, answer after 15 second. ArturoMagidin, Brian M. Scott big thanks!2012-05-14
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    The proof if OK. But NOTE you use the Axiom of Choice in the proof. This result CANNOT be proved in ZF set theory (provided ZF is consistent).2012-05-14
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    @GEdgar, I don't care about ZF and completely accept Axiom of Choice with all its "strange" consequnces.2012-05-14
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    @GEdgar in which part he used the Axiom of Choice?2012-11-08
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    Existence of a Hamel basis for $\mathbb R$ over $\mathbb Q$ uses the Axiom of Choice.2012-11-08
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    @BrianM.Scott: Hi, I have 1 question, though, what's the theoretic bijection $i: \Lambda \to \Lambda$? I haven't come across this one before. How is it defined?2013-06-16
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    @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 $\Lambda\times\Lambda$.2013-06-16
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    I have a question: if B is Hamel basis for R over Q, why BxB is Hamel basis for R^2 over Q?2013-12-02
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    @GAJO Because given $x=\sum_i \alpha_i e_i$ and $y=\sum_j \beta_j e_j$ you can say that $(x,y)=\sum_{i,j} \alpha_i\beta_i (e_i,e_j)$2013-12-02
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    @Norbert I am confused, assume $x = \alpha_{1} e_{1}$ and $y = \alpha_{2} e_{2}$ your saying that $(x,y)= \alpha_{1}\alpha_{2}(e_{1},e_{2})$ = $(\alpha_{1}\alpha_{2}e_{1},\alpha_{1}\alpha_{2}e_{2})$2013-12-02
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    @GAJO nevermind, that was wrong, let's delete this comments2013-12-02
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    @BrianM.Scott please see the last question in comments by GAJO. I think it makes sense and my soltion in the original answer is wrong. The correct one basis is $\{(e_\lambda,0):\lambda\in\Lambda\}\cup \{(0,e_\lambda):\lambda\in\Lambda\}$. Isn't it?2013-12-02
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    @Norbert: Yes, you’re right. (I’m surprised that so many of us missed it. We must all have been seeing what we expected to see!)2013-12-02
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    @GAJO You are the best - you found the mistake that all others failed to see!2013-12-02
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    Very verrrry late to this party, but since the question I noted it on may get closed dupe, here's a discussion about this question in the absence of choice: http://journals.cambridge.org/download.php?file=%2FJAZ%2FJAZ1_19_03%2FS1446788700031505a.pdf&code=3275bc464c70c444d990e75cf703ee45 )2015-01-21
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    Related: https://math.stackexchange.com/questions/302514/2018-11-28

1 Answers 1

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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

$$\text{Frac}(R)\otimes_R\text{Frac}(R)\cong\text{Frac}(R)\otimes_R\text{Frac}(R)^n\cong(\text{Frac}(R)\otimes_R\text{Frac}(R))^n$$

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 $m\ne n$.

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    Sorry, I'm a newbie in algebra. What is $\mathrm{Frac}(R)$?2012-05-14
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    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).2012-05-14
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    Thanks, Alex. I've not suspected that this can be generalized to such extent.2012-05-14
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    @ Alex , can you tell me why is $F$ isomorphic to $F^n$ and only for finite $n$ ?2012-05-14
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    @Ananda It's actually isomorphic for any cardinal with cardinality less then $\dim_\mathbb{F_0}F$. This is because $\lambda\cdot\mu=\min\{\lambda,\mu\}$.2012-05-14