What is the cardinality of a transcendence basis of $\mathbb{C}$ over $\mathbb{Q}$? What is the cardinality of a transcendence basis of $\mathbb{C}$ over $\mathbb{Q}$? Is it that of the continuum? Proof?
 A: If $S$ has infinite cardinality $\kappa$, then $|\mathbb Q(S)|=\kappa$.
And if $|F|=\kappa$, then $|F[X]|=\kappa $ and finally $|\bar F|=\kappa$. Therefore we need exactly $\kappa=2^{\aleph_0}$ if we want $\overline{\mathbb Q(S)}=\mathbb C$.
A: (Assume the axiom of choice)
If $V$ is a vector space over $F$, and $B$ is a basis for $V$, and at least $B$ or $F$ are infinite, then every $v\in V$ is identified with a unique finite function which is a subset of $F\times B$. Namely, the unique decomposition of $v$ to non-zero scalars and basis elements.
We therefore have that $|V|\leq |F\times B|^{<\omega}=|F\times B|=\max\{|F|,|B|\}$, and on the other hand $|F|\leq|V|$ and $|B|\leq|V|$, therefore: $$\max\{|F|,|B|\}\leq|V|\leq\max\{|F|,|B|\}\implies |V|=\max\{|F|,|B|\}$$
Apply this to the case of $\mathbb Q$ and $\mathbb C$, we have that $|\mathbb C|=2^{\aleph_0}$ and therefore the basis has to have size $2^{\aleph_0}$.

Without assuming the axiom of choice it is consistent that there is no such basis, so we cannot talk too much about that. It is also open whether the existence of a basis implies the well-orderability of $\mathbb C$. If it does, then the above theorem holds and it has to have size continuum; if it doesn't then it's hard to say too much, although it is likely to be provable that its size is again $2^{\aleph_0}$.
