Say that $K \subseteq L$ are two algebraically closed fields of characteristic $0$. Let $C$ be a curve (not nec. proj., but maybe assume smooth) over $K$. Supposedly $\pi_1(C_L)$ maps isomorphically to $\pi_1(C)$. Is it true that any etale cover of $C_L$ is already defined over $C$? Why should that be true?
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asked Oct 22, 2010 at 20:53
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1$\begingroup$ Yes, it's true, and at least morally I think it's equivalent to the statement about the natural map on $\pi_1$'s being an isomorphism. Someone else will probably come along with a more detailed answer, but in the meantime have you tried looking in Szamuely's book Galois groups and fundamental groups? $\endgroup$– Pete L. ClarkCommented Oct 22, 2010 at 23:21
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2$\begingroup$ Ramification at the missing points is tame, and the discrepancy between the $\pi_1$'s of the given curves and their compactifications is governed by the (tame!) inertia groups at the missing points. Since tame inertia doesn't notice $L/K$, one can make covers sufficiently ramified at missing points and etale over $C$ that "eat up" ramification for the covering of $C_L$, and so by the Galois correspondence between group theory and intermediate curves reduce to everywhere unramified covers of certain curves over $C$. Then SGA1 results on specialization of $\pi_1$ in proper case can be applied. $\endgroup$– BCnrdCommented Oct 23, 2010 at 0:06
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$\begingroup$ In case you know, where can I find a proof (preferably in English, but French is fine) of this map between fundamental groups being an isomorphism? I've seen similar proofs, but not quite that. $\endgroup$– James D. TaylorCommented Oct 23, 2010 at 0:57
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