Why does $\mathbb{C}$ have transcendence degree $\mathfrak{c}$ over $\mathbb{Q}$? It's pretty well known that $\text{trdeg}(\mathbb{C}/\mathbb{Q})=\mathfrak{c}=|\mathbb{C}|$.
As a subset of $\mathbb{C}$, of course the degree cannot be any greater than $\mathfrak{c}$. I'm trying to understand the justification why it cannot be any smaller. The explanation in my book says that if $\mathbb{C}$ has an at most countable (i.e. finite or countable) transcendence basis $z_1,z_2,\dots$ over $\mathbb{Q}$, then $\mathbb{C}$ is algebraic over $\mathbb{Q}(z_1,z_2,\dots)$. Since a polynomial over $\mathbb{Q}$ can be identified as a finite sequence of rationals, it follows that $|\mathbb{C}|=|\mathbb{Q}|$, a contradiction. 
I don't see why the polynomial part comes in? I'm know things like a countable unions/products of countable sets is countable, but could someone please explain in more detail this part about the polynomial approach? Since $\mathbb{C}$ is algebraic over $\mathbb{Q}(z_1,z_2,\dots)$, does that just mean that any complex number can be written as a polynomial in the $z_i$ with coefficients in $\mathbb{Q}$? For example,
$$
\alpha=q_1z_1^3z_4z_6^5+q_2z_{11}+q_3z^{12}_{19}+\cdots+q_nz_6z_8z^4_{51}?
$$
Is the point just that the set of all such polynomials are countable?
Thanks,
 A: Note that, if $K$ is a countable field and $x$ is transcendental over $K$, then $K[x]$ is also countable (separate polynomials by degree), hence $K(x)$ is countable (the elements can be identified with pairs of elements of $K[x]$). Letting $K_0=\mathbb{Q}$ and $K_n=K_{n-1}(z_n)$, we have that
$$\mathbb{Q}(z_1,z_2,\ldots)=\bigcup_{n=0}^\infty K_n$$
is a countable union of countable sets, and hence is countable.
If a field $L$ is countable and $F$ is algebraic over $L$, then $F$ is countable, because $L[x]$ is countable and we can cover $F$ by a countable number of finite sets $S_f$, one for each $f\in L[x]$, where $S_f=\{a\in F\mid f(a)=0\}$.
Thus, if $\mathbb{C}$ had a countable transcendence basis $z_1,z_2,\ldots$ over $\mathbb{Q}$, then $\mathbb{C}$ is algebraic over $\mathbb{Q}(z_1,z_2,\ldots)$, and it would follow that $\mathbb{C}$ is countable, a contradiction.
(This does not explain why $\mathbb{C}$ has transcendence degree $\mathfrak{c}$, though.)
A: (Of course I assume the Axiom of Choice...) Choose a transcendence basis $X = \{x_i\}_{i \in I}$ for $\mathbb{C}$ over $\mathbb{Q}$.  Then $\mathbb{C}$ is an algebraic extension of $\mathbb{Q}(X)$.  Now here are two rather straightforward facts:
1: If $F$ is any infinite field and $K/F$ is an algebraic extension, then $\# K = \#F$.
2: For any infinite field $F$ and purely transcendental extension $F(X)$, we have $\# F(X) = \max (\#F, \# X)$.
Putting these together we find
$\mathfrak{c} = \# \mathbb{C} = \# \mathbb{Q}(X) = \max (\aleph_0, \# X)$.  
Since $\mathfrak{c} > \aleph_0$, we conclude $\mathfrak{c} = \# X$.  
A: The field $F=\mathbb Q(z_1,z_2,..)$ is countable as well as the polynomial ring $F[x]$. Since $\mathbb C$ is algebraic over $F$ you can define a surjective map $\phi:F[x]\to \mathbb C $ by sending $p(x)$ to one of its roots.
Edit: This just shows that the transcendence degree is uncountable but you can use the same argument for any base of size less than $\mathfrak c$.
