I know that there was also a similar question asked before but I don't really understand the solution.

There was said:

"Assuming the axiom of choice, yes.

Observe that both these abelian groups are actually $\mathbb Q$-vector spaces, and they have the same dimension, so they must be isomorphic as vector spaces, and such isomorphism is also a group isomorphism. This is in fact a stronger requirement than just group isomorphism, but nevermind that.

It is consistent with the failure of the axiom of choice that these two are not isomorphic, though. So one cannot give an explicit isomorphism between them."

My question is: $\dim_{\mathbb{Q}}(\mathbb{R})=\dim_{\mathbb{Q}}(\mathbb{C})=\infty$. Why can we say that these $\mathbb Q$-vector spaces are isomorphic because I only know that vector spaces are isomorphic if they have the same finite dimension. Does this always works for vector spaces with infinite dimension? Then why can't we take for example $\mathbb F_2$ as a field, so isn't: $\dim_{\mathbb F_2}(\mathbb{R})=\dim_{\mathbb F_2}(\mathbb{Q})=\infty$. But $\mathbb{Q}$ and $\mathbb{R}$ are not isomorphic.

Also I have another question: As I first tried to answer the original question, I made up a proof where I couldn't find the mistake yet.

Assume: there exists an isomophism $\phi :(\mathbb{R},+) \rightarrow (\mathbb{C},+)$ $\Rightarrow \phi(0)=(0,0)$

Now let $\phi(1)=(a,b)$ with $a, b \in \mathbb{R}$

$\Rightarrow \phi(x)=(xa,xb)$ for all $x \in \mathbb{R}$

So this function has to be proportional. As a result $\phi$ is not surjective because it doesn't exists a $x \in \mathbb{R}$ so that $\phi(x)=(a,b/2)$.

I know that my way of thinking is false and I would be very glad if you can help me. Thanks for taking your time.

  • 2
    $\begingroup$ Why is it true that $\phi(x)=(xa,xb)$? $\endgroup$ – leibnewtz Oct 29 '16 at 1:11
  • $\begingroup$ now i see that it doesn't work $\endgroup$ – milui Oct 29 '16 at 7:40

You're right, it can't just be two arbitrary infinite dimensions. However, being more rigorous, the dimension is a cardinal number, in this case $2^{\aleph_0}$ (by cardinality arguments).

Selecting any two linear bases for $\mathbb{R}$ and $\mathbb{C}$ as $\mathbb{Q}$-vector spaces, the fact that they have the same cardinality means that there is a bijection $\varphi$ between them. Extending this linearly gives a linear isomorphism, in particular a group isomorphism.

Also, as far as your solution is concerned, you don't actually have $\varphi(x)=(xa,xb)$ for any $x\in \mathbb{R}$, only for $x\in\mathbb{Q}$. There is no reason to think that this must also then hold true for arbitrary $x\in\mathbb{R}$, since there is no reason to suspect that $\varphi$ is continuous.

  • $\begingroup$ To clarify the last sentence for the OP: what you have shown is that there is no continuous isomorphism. So while they are isomorphic as abelian groups, they are not isomorphic as topological groups. $\endgroup$ – Noah Schweber Oct 29 '16 at 2:27

If ${\cal B}$ is a $\Bbb Q$-basis for $\Bbb R$, then $|{\cal B}| = 2^{\aleph_0}$ and ${\cal B}\cup i{\cal B}$ is a $\Bbb Q$-basis for $\Bbb C$. So $$\dim_{\Bbb Q}\Bbb C = |{\cal B}\cup i{\cal B}| = 2|{\cal B}| = 22^{\aleph_0} = 2^{\aleph_0+1}=2^{\aleph_0} = \dim_{\Bbb Q}\Bbb R.$$

If one cardinal was greater then the other, the argument would fail. In this case it works.

  • $\begingroup$ The dimension of $\mathbb{R}$ as a $\mathbb{Q}$-vector space is $2^{\aleph_0}$. $\endgroup$ – Hayden Oct 29 '16 at 1:02
  • $\begingroup$ @Hayden oops, I slipped. Thanks for pointing, fixed. $\endgroup$ – Ivo Terek Oct 29 '16 at 1:31

The other answers already told ypur main error, but I want to point out also another error:

Neither $\mathbb Q$ nor $\mathbb R$ is an $\mathbb F_2$ vector space. If they were, you'd have for any $x\in\mathbb Q$ resp. $x\in\mathbb R$, $$x+x = 1_{\mathbb F_2}x + 1_{\mathbb F_2}x = (1_{\mathbb F_2} + 1_{\mathbb F_2})x = 0_{\mathbb F_2}x = 0$$ Clearly that is not correct, as e.g. in $\mathbb R$ we have $1+1=2\ne 0$.

  • $\begingroup$ thank you that was a main mistake of me $\endgroup$ – milui Oct 29 '16 at 7:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.