For a variety $V$ of dimension $n$, let $Irr(V)$ denote the minimal degree of a dominant rational map $V\to \mathbb{P}^n$.
Suppose that a curve $X$ admits a dominant map from a variety $V$ with $Irr(V)=2$. Does it follow that $X$ is hyperelliptic?
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$\begingroup$ Probably you want $Irr(V)$ to be the minimal degree of a dominant rational maps $V\to\mathbb P^n$. $\endgroup$– Joe SilvermanCommented Oct 20, 2016 at 1:05
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$\begingroup$ @JoeSilverman: yes. $\endgroup$– Nico BellicCommented Oct 20, 2016 at 1:18
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2$\begingroup$ Degree of irrationality? I thought that was called the gonality. $\endgroup$– Jason StarrCommented Oct 20, 2016 at 12:37
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1 Answer
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Yes, because then there is a nonconstant map from projective space to the symmetric square of $C$.
answered Oct 20, 2016 at 2:34
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$\begingroup$ I see the map from projective space to the symmetric square of $C$ but how does it imply that $C$ is hyperelliptic? $\endgroup$ Commented Oct 20, 2016 at 3:28
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3$\begingroup$ One way to see it is to compose with the map from $C$ to its Jacobian. That map does have to be constant, because that's true of every map from projective space to an abelian variety. But then there are two distinct points in the symmetric square that map to the same point in the Jacobian. So the difference between them is the divisor of a rational function. That function has degree $2$. So we have a degree-2 rational function on $C$, QED. $\endgroup$ Commented Oct 20, 2016 at 3:38