Let $K$ be a finite extension of $\mathbb{F}_q(t)$ and define the curve $C$ by the equation $y^p=f(x)$ where $p=\mathbf{char} K$ and $f\in K[x]$. What is the genus of $C$? When does it have infinitely many $K$-points?
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It "changes genus". It has genus zero over the algebraic closure of $K$ but it behaves as if it had positive genus over $K$ unless there is a change of coordinates over $K$ that changes $f(x)$ into a polynomial defined over $K^p$. When it has positive genus over $K$ in this sense, then the set of $K$-rational points is finite. I proved this in Bull. SMF 119(1991), 121-126.
answered Jun 10, 2013 at 13:16
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$\begingroup$ Thanks, this is helpful. Perhaps you can also answer the following question: is there a good characterization of all polynomials $f\in K[x]$ s.t. $y^{p^k}=f(x)$ has infinitely many K-points for all k? $\endgroup$– AlexCommented Jun 10, 2013 at 19:57
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$\begingroup$ My guess would be that it's only for those polynomials for which there is a change of variables over $K$ taking the equation to one defined over $\mathbb{F}_q$. $\endgroup$ Commented Jun 10, 2013 at 20:28
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$\begingroup$ This doesn't seem to be correct as the example $f=tx^2+1$ over $K=\mathbb{F}_q(t)$ ($p>2$) shows. In this example the required change of variables is defined over a quadratic extension. Possibly the existence of such a change of variables over some separable extension is necessary, but it is not clear whether it is sufficient. $\endgroup$– AlexCommented Jun 11, 2013 at 8:38
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$\begingroup$ I think you are right. The existence of a change of variables over a separable extension should be equivalent to your family of curves having infinitely many points over a separable extension. I don't know the answer to the question in your first comment. $\endgroup$ Commented Jun 11, 2013 at 11:37
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$\begingroup$ I think I know how to prove this. Thanks for the helpful discussion. $\endgroup$– AlexCommented Jun 11, 2013 at 11:48