Are $(\mathbb{R},+)$ and $(\mathbb{C},+)$ isomorphic as additive groups under the axiom of choice? 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.  
 A: 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.
A: 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.
A: 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 $1+1=2\ne 0$.
