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What's the dimension of $\mathbb C$ as a vector space over $\mathbb{R}$?

I think that the answer is $2$, because $\mathbb R^2 \cong \mathbb{C}$ (since all complex numbers can be mapped to $\mathbb R^2$ and vice versa).

Is this correct, or is it a trick question (in which case I'm wrong)?

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    $\begingroup$ You are correct, but a little non-rigorous. What do you mean by "all complex numbers can be mapped to $\Bbb R^2$ and visa versa"? Do you mean that there exists some bijection between them? $\endgroup$ – Omnomnomnom Oct 28 '14 at 14:28
  • $\begingroup$ @Omnomnomnom Yes. I mean that there's a 1-1 correspondence between $\mathbb R^2$ and $\mathbb{C}.$ Is this new phrase more rigorous? $\endgroup$ – user160325 Oct 28 '14 at 14:29
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    $\begingroup$ It's still not good enough. There is a 1-1 correspondence between $\Bbb R$ and $\Bbb R^2$ (i.e. they have the same cardinality), but these are not isomorphic vector spaces. What's special about $\Bbb C$ and $\Bbb R^2$? $\endgroup$ – Omnomnomnom Oct 28 '14 at 14:30
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Hint: $\lbrace 1,i\rbrace$ is a basis.

As for what you have said, there is an isomorphism $$\frac{\mathbb{R}[x]}{\langle x^2+1\rangle}\simeq \mathbb{C}$$ where $\mathbb{R}[x]$ is the polynomial ring.

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  • $\begingroup$ Therefore, the dimension is $2$, yes? $\endgroup$ – user160325 Oct 28 '14 at 14:26
  • $\begingroup$ That's correct. $\endgroup$ – Aaron Maroja Oct 28 '14 at 14:27
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Note that $\mathbf{R}^m$ is bijective with $\mathbf{R}^n$ for all $m,n > 0$, so the explicit argument you gave doesn't work.

However, there is more to the identification $\mathbf{R}^2 \cong \mathbf{C}$ than simply being a bijection... what you're trying to show is that it's actually a linear transformation between $\mathbf{R}$-vector spaces!

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