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Question: Is it true that for every smooth compact complex curve $C$ there exists a smooth curve $C'$ in $\mathbb CP^2$ that admits a non-trivial morphism (i.e. holomorphic map) $C'\to C$?

Motivation. Unfortunately, I don't know yet any application for a positive answer to this question. But a negative answer to this question would solve in negative a great question of Francesco Polizzi: Surfaces in $\mathbb{P}^3$ with isolated singularities

Indeed, here is a simple exercise:

Exercise. Suppose that $C$ is a smooth curve that is not covered by any smooth plane curve. Then the surface $C\times \mathbb CP^1$ is not birational to any surface in $\mathbb CP^3$ with isolated singularities.

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  • $\begingroup$ Do you know if this is true for $C$ a hyperelliptic curve of genus 2 or 3? $\endgroup$
    – Piotr Achinger
    Commented Jan 6, 2013 at 22:08
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    $\begingroup$ @Piotr If $C: y^2=f(x)$ is hyperelliptic with $f$ of even degree $2n$, then $y^{2n}=f(x)$ is smooth and covers $C$. I suspect the answer to the general question is no, but I don't know how to do it. Maybe the answer that Jason Starr gave to a recent question of Mike Zieve will work here too. $\endgroup$
    – Felipe Voloch
    Commented Jan 6, 2013 at 22:25
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    $\begingroup$ I would be happy with any counter-example but after some thinking it seems to me that the answer to this question might be positive... $\endgroup$
    – aglearner
    Commented Jan 7, 2013 at 5:32
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    $\begingroup$ I don't know the answer, and this actually sounds difficult. Obviously, even when $C$ has general moduli among genus $g$ curves, a finite cover $C'$ does not need to have general moduli. This question does vaguely remind me of work of Bogomolov and Tschinkel: every curve over the algebraic closure of a finite field has a finite cover that is also an etale cover of a (fixed) hyperelliptic curve. $\endgroup$
    – Jason Starr
    Commented Jan 7, 2013 at 14:18

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Edit: I made a mistake, my attempted answer is wrong. It was just the same as @Felipe's comment, namely, that Jason Starr's answer to this question might apply here too. But it doesn't. Sorry for the mistake.

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  • $\begingroup$ Dear Zieve I would like to know why you can deform $C'$ in the way you descibe. I don't see this. $\endgroup$
    – aglearner
    Commented Jan 6, 2013 at 23:45
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    $\begingroup$ After some thinking I don't see any (non-superficial) relation between the question you cite and my question, so I would like to ask you give details on how you want to deform $C'$ with its morphism. $\endgroup$
    – aglearner
    Commented Jan 7, 2013 at 4:22
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    $\begingroup$ @Michael: In principle it is possible that each curve og genus $g$ can be covered by a plane curve, but these mappings do not "glue" into a dominant rational mapping from the space of plane curves of a given degree onto $M_g$. $\endgroup$
    – Serge Lvovski
    Commented Jan 7, 2013 at 6:47
  • $\begingroup$ Really? Then I take it back. I thought it would be straightforward to get that, if there were a morphism from a smooth plane curve of degree $d$ to a generic genus-$g$ curve, then there is a family of morphisms from a Zariski-open subset of degree-$d$ plane curves to genus-$g$ curves. If that's not true, then I was wrong. $\endgroup$
    – Michael Zieve
    Commented Jan 7, 2013 at 7:13
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    $\begingroup$ Michael, I don't see how you would use in your reasoning the fact that genus $g$ curve is generic. If you remove this assumption, it is very easy to construct a counterexample to that fact that you have assumed. Namely, the space of genus $3$ curves that admit a morphism to an elliptic curve is Zariski dense (in the moduli space $M_3$ of all genus $3$ curves), but it is by no means open. Since degree four curves in $\mathbb CP^3$ from a Zariski dense subset of $M_3$ we see a contradiction. $\endgroup$
    – aglearner
    Commented Jan 7, 2013 at 8:46

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