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,

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 $\mathfrak c$.
