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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 $trdeg_{\mathbb{C}}\mathbb{C}(X)= 2$?

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Such a surface admits a finite morphism onto ${\mathbb P}_{\mathbb C}^2$ (just subsequently project several times $X\subset{\mathbb P}_{\mathbb C}^n$ from a point $p\not\in X$ into ${\mathbb P}_{\mathbb C}^{n-1}$). No need to use that it is non-singular. It suffice to use that it is irreducible and reduced. –  Sasha Anan'in Oct 12 '13 at 13:32
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@SashaAnan'in You might as well turn this into a short answer. –  S. Carnahan Oct 13 '13 at 1:00
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2 Answers

up vote 3 down vote accepted

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 $2$.

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Thank you very much. Siegel prove that if $X$ is a compact complex connected manifold of dimension $n$ , then $trdeg_{\mathbb{C}}\mathbb{C}(X) \leq n$. You can find it in the book complex geometry write by Daniel Huybrechts(p54,proposition2.1.9). –  SWalker Oct 14 '13 at 1:55
    
Thank you, now I got your settings. You can easily modify the above sketch for compact analytic subsets in ${\mathbb P}_{\mathbb C}^n$ (i.e., locally given by finitely many analytic equations) and drop the assumption of irreducibility. As I hope, with the help of Weierstrass preparation theorem, everything should work. –  Sasha Anan'in Oct 14 '13 at 9:40
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What is your definition of "irreducible surface"? I thought it was an irreducible variety of dimension 2. An irreducible variety $X/k$ has a function field $k(X)$, and at least one of the standard definitions of the dimension of an irreducible variety (over a field $k$) is the transcendence degree of $k(X)$ over $k$. So this makes the answer to the question "by definition". Admittedly there are several other standard definitions of the dimension of an irreducible variety. But any standard text, e.g., Atiyah-McDonald, has a proof that they are all equivalent.

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Maybe, the OP meant $X$ as an analytic compact manifold analytically embedded into a projective space. In this case, it is worthwhile mentioning (to him/her, not to you, of course) that $X$ is known to be algebraic. –  Sasha Anan'in Oct 13 '13 at 12:15
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