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By Riemann's existence theorem I mean this:

Let $X$ be some variety defined over $\mathbb{C}$, and let $Y$ be a topological covering space of $X$. Then $Y$ can be given the structure of a variety over $\mathbb{C}$, and furthermore this can be done so that the covering map will be algebraic.

It is frequently said that this theorem is not constructive. That is to say, that it is impossible to predict the polynomials that define $Y$ and the polynomial map that defines $Y\rightarrow X$. I want to read the proof, and understand for myself why this is true. Where can I find a good, preferably succinct, proof of this theorem in English?

Makhalan Duff
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    You need the covering to be finite, clearly. – Ben McKay Nov 12 '11 at 20:08
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    See Expose XII, Theoreme 5.1 on page 332 of SGA1 available at http://arxiv.org/abs/math/0206203 – Ariyan Javanpeykar Nov 12 '11 at 21:02
  • By the way, I'm not completely sure about what you're really looking for, but I think the proof of Theoreme 5.1 given in SGA1 of the essential surjectivity of $\psi$ is not constructive because the argument is partially a reduction via non-constructive reasoning to the case of an affine regular scheme. – Ariyan Javanpeykar Nov 12 '11 at 21:06
  • @Ariyan: I doubt that it is constructive even for affine regular schemes. Otherwise one would have an algebraic proof that $\pi_1(\mathbb{P}^1\smallsetminus a_1,...,a_r)\cong$ the profinite completion of $\langle \alpha_1,...,\alpha_r|\alpha_1 ...\alpha_r =1\rangle$. – James D. Taylor Nov 13 '11 at 00:32

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See Finite covers of complex varieties (all but two questions answered!)

Community
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Igor Rivin
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  • Aha! I imagine you mean Brian's comment. Do you have a reference of where that is in SGA1? Even for a projective variety it is not perfectly obvious to me how GAGA implies it... – Makhalan Duff Nov 12 '11 at 20:28