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.