If $X$ is a surface, projective and non-singular. Let $\mathbb{C}(X)$ be the function field of $X$. By a theorem of Siegel, we know that $trdeg\_{\mathbb{C}}\mathbb{C}(X)\leq 2$. But how to argue that $trdeg\_{\mathbb{C}}\mathbb{C}(X)= 2$？



How to determine the transcendence degree of function field
OK, make it an answer (which is just a piece of basic algebraic geometry). We do not assume that $X$ is smooth, just projective and irreducible. So, $X\subset{\mathbb P}\_{\mathbb C}^n$. If $n=2$, we a done as
$X={\mathbb P}\_{\mathbb C}^2$ and
$\text{tr}\,\text{deg}{\mathbb C}({\mathbb P}\_{\mathbb C}^2)=2$. Otherwise, we can project $X$ from a point $p\in{\mathbb P}\_{\mathbb C}^n\setminus X$ into ${\mathbb P}\_{\mathbb C}^{n-1}$, $\pi:X\to{\mathbb P}\_{\mathbb C}^{n-1}$. The image $\pi(X)$ is irreducible because $X$ is irreducible and projective (i.e., Zariski closed in ${\mathbb P}\_{\mathbb C}^{n-1}$). Moreover, the map $\pi:X\to\pi(X)$ is a finite morphism and, in particular, it is finite-to-one. So, whatever you mean by "surface", $\pi(X)$ is a one. Keeping projecting, we arrive at a finite (hence, surjective) morphism
$\varphi:X\to{\mathbb P}\_{\mathbb C}^2$. In this situation, the field extension
${\mathbb C}({\mathbb P}\_{\mathbb C}^2)\subset{\mathbb C}(X)$ is finite.


I should confess I do not know any Siegel theorem applicable to this case.


**Late edit.** Using the Weierstrass preparation theorem, one can show (following the above sketch) that there is a finite (hence, surjective) morphism
$X\to{\mathbb P}\_{\mathbb C}^2$ for any compact analytic subset $X\subset{\mathbb P}\_{\mathbb C}^n$ of pure dimension $2$.


