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Let $g>1$ be an integer. Does there exist a (smooth projective) genus g curve $X$ which doesn`t dominate a curve of positive genus and genus smaller than $g$?

Surely such curves exist. Just take a curve with simple jacobian.

I`m wondering whether there are any other interesting examples, or is the above property equivalent to having simple jacobian?

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I bet there are curves with non-simple jacobians but such that the factors of the jacobian are not themselves isogenous to jacobians, so these curves don't dominate other curves. Maybe one can show these exist by dimension count. –  Felipe Voloch Jul 29 '13 at 18:21
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This is not quite what you are looking for, but: I have come across a genus 3 curve C that maps to an elliptic curve, but provably without a map to a genus 2 curve. (I.e., the Jacobian of C is, up to isogeny, E x J_H for some hyperelliptic curve H, but C does not map to H.) –  David Zureick-Brown Jul 29 '13 at 19:05
    
Is there some restriction on the polarization too? –  Ian Agol Jul 30 '13 at 22:56
    
I agree with Felipe. Take a genus 2 curve and ask whether it covers a genus one curve. It would seem to have 2 ramification points, hence exhaust only 2 dimensions (?) in the 3 dimensional moduli space of genus 2 curves. As g goes up it seems only to get worse. –  roy smith Aug 5 '13 at 2:23

1 Answer 1

I elaborate a bit on Roy Smith's comment.

One can show that the general curve of genus $g>1$ does not map onto a curve of genus $h>0$ by a dimension count, at least for complex curves. Let $f\colon C\to D$ be a map of degree $d$ from a curve of genus $g$ to a curve of genus $h>0$ and let $B$ the branch divisor of $f$ (a point $P\in D$ appears in $B$ with multiplicitiy equal to $\sum_{Q\mapsto P}(m_Q-1)$, where $m_Q$ is the order of ramification of $f$ at $Q$. The Hurwitz formula gives: $$2g-2=d(2h-2)+\deg B.$$ Let $S$ be the support of $B$ (i.e., $S$ is the set of critical values of $f$) and consider the restricted cover $f_0\colon C\setminus f^{-1}(S)\to D\setminus S$: this is a topological cover of degree $d$ and it determines $f$. There are finitely many such covers, hence the maps $f\colon C\to D$ as above depend on $3h-3+s$ parameters, where $s$ is the cardinality of $s$. Since $s\le \deg B$, the Hurwitz formula gives: $$(3g-3)-(3h-3+s)\ge 3(d-1)(h-1)+s/2>0.$$ So the general curve of genus $g>1$ does not have a map of degree $d$ onto a curve of genus $h>0$. Now it is enough to observe that, again by the Hurwitz formula, there are finitely many possibilities for $h$ and $d$.

I don't think that a curve without maps onto curves of positive genus must have a simple Jacobian: if one takes a curve $C$ inside an irregular surface $S$ such that $C$ is an ample divisor of $S$, then there is an injection $Pic^0(S)\to J(C)$, and I do not see why in general $C$ should have a map onto a curve of positive genus. To get an actual counterexample, one could try to look at an abelian surface $S$ with a polarization $L$ of type $(1,3)$ and a curve $C\in |L|$, but I don't know how to show that a general such $C$ does not map onto a curve of positive genus.

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