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Let $a$ and $b$ be two relatively prime natural numbers. What is the largest number $c$ such that there is a curve with maps to $\mathbb P^1$ of degree $a$ and $b$ but no map to $\mathbb P^1$ of degree $c$?

It's easy to seee that a map of degree $a$ and $b$ implies a map of degree each nonnegative integer combination of $a$ and $b$. So $ c \leq ab-a-b$, the largest number that is not a nonegative integer combination of $a$ and $b$.

But I don't think that's optimal. Given a map of degree $a$ and a map of degree $b$, we get a map of degree $(a,b)$ to $\mathbb P^1 \times \mathbb P^1$ which has a Segre embedding into $\mathbb P^3$. By projecting from a line which intersects the curve at $0,1,$ or $2$ points, we can get maps of degree $a+b$, $a+b-1$, and $a+b-2$. By iterating this we can always get a much smaller upper bound. So what's the optimal bound?

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    $\begingroup$ If the curve is hyperelliptic of genus $g$ and $a=2$, then the lowest odd degree of a map is $b=2g+1$ and your bound $ab-a-b=2g-1$ for $c$ is optimal. This follows from Clifford's theorem, which might help in general too. $\endgroup$
    – Felipe Voloch
    Jul 7, 2014 at 17:55
  • $\begingroup$ @FelipeVoloch Can you expand on that argument? I don't see it. I think a generic $(2,2n+1)$ curve in $\mathbb P^1$ has genus $2n$, not $n$, and is hyperelliptic with a degree $2n+1$ map to $\mathbb P^1$. $\endgroup$
    – Will Sawin
    Jul 7, 2014 at 21:35
  • $\begingroup$ $y^2=f(x)$ with $f$ of degree $2n+1$ has genus $n$. Try $n=1,2$. The function $x$ has degree $2$ and $y$ has degree $2n+1$. $\endgroup$
    – Felipe Voloch
    Jul 7, 2014 at 22:28
  • $\begingroup$ @FelipeVoloch But how do you use Clifford's theorem to show that there is no map of degree $2n-1$? $\endgroup$
    – Will Sawin
    Jul 8, 2014 at 2:29
  • $\begingroup$ I am wrong. I was implicitly assuming that the two functions have poles at the same point. $\endgroup$
    – Felipe Voloch
    Jul 8, 2014 at 2:52

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