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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 questionthis question might apply here too. But it doesn't. Sorry for the mistake.

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.

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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Michael Zieve
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TheEdit: I made a mistake, my attempted answer is "no", forwrong. It was just the same reasonas @Felipe's comment, namely, that Jason Starr explained in hisStarr's answer to mathoverflow:118117this question.

Edit: in more detail, the point is that for $g\ge 23$, there is no nonconstant morphism from a rational variety to $M_g$ whose image contains the general point might apply here too. But if a general genus-$g$ curve $C$ admitted a morphism $C'\to C$ from a smooth plane curve $C'$, then we could deform $C'$ and the morphism so that the image still has genus $g$it doesn't. This gives a contradiction, since Sorry for the deformation space is a rational varietymistake.

The answer is "no", for the same reason that Jason Starr explained in his answer to mathoverflow:118117.

Edit: in more detail, the point is that for $g\ge 23$, there is no nonconstant morphism from a rational variety to $M_g$ whose image contains the general point. But if a general genus-$g$ curve $C$ admitted a morphism $C'\to C$ from a smooth plane curve $C'$, then we could deform $C'$ and the morphism so that the image still has genus $g$. This gives a contradiction, since the deformation space is a rational variety.

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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Michael Zieve
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The answer is "no", for the same reason that Jason Starr explained in his answer to mathoverflow:118117.

Edit: in more detail, the point is that for $g\ge 23$, there is no nonconstant morphism from a rational variety to $M_g$ whose image contains the general point. But if a general genus-$g$ curve $C$ admitted a morphism $C'\to C$ from a smooth plane curve $C'$, then we could deform $C'$ and the morphism so that the image still has genus $g$. This gives a contradiction, since the deformation space is a rational variety.

The answer is "no", for the same reason that Jason Starr explained in his answer to mathoverflow:118117.

The answer is "no", for the same reason that Jason Starr explained in his answer to mathoverflow:118117.

Edit: in more detail, the point is that for $g\ge 23$, there is no nonconstant morphism from a rational variety to $M_g$ whose image contains the general point. But if a general genus-$g$ curve $C$ admitted a morphism $C'\to C$ from a smooth plane curve $C'$, then we could deform $C'$ and the morphism so that the image still has genus $g$. This gives a contradiction, since the deformation space is a rational variety.

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Michael Zieve
  • 6.4k
  • 30
  • 43
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