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Apr 14, 2021 at 11:59 vote accept Mare
Apr 9, 2021 at 18:04 comment added David E Speyer So they get $\tfrac{q^{a-g+1}-1}{q-1} \geq \tfrac{q^a - 2g q^{a/2}+1}{a}$. It turns out that the optimal choice is $a = 2g-1$ (just big enough for RR to apply), and then this forces $g \leq 4$.
Apr 9, 2021 at 18:03 comment added David E Speyer But also, given any $\mathbb{F}_{q^a}$ point of $X$, summing up its $\mathrm{Gal}(\mathbb{F}_{q^a}/\mathbb{F}_q)$ conjugates gives a degree $a$ divisor defined over $\mathbb{F}_q$, which must be equivalent to $a x_{\infty}$ by the class number one hypothesis. Once $a$ is large enough for Riemann-Roch to kick in, the number of effective divisors equivalent to $a x^{\infty}$ is $\tfrac{q^{a-g+1}-1}{q-1}$.
Apr 9, 2021 at 18:02 comment added David E Speyer Oh, this is nice too! They get the same reduction to $q=2$ or $3$ that I do. Then they look at $\# X(\mathbb{F}_{q^a}$ in two ways. By Weil, this is $\geq q^a - 2 g q^{a/2} +1$.
Apr 9, 2021 at 16:10 comment added David E Speyer Huh, this paper takes a totally different approach. I used to like my approach better, but now I have discovered I miscomputed a term in the $q=2$, $g$ even case and my solution doesn't work. I'll go see how they handle it.
Apr 9, 2021 at 16:09 history edited David E Speyer CC BY-SA 4.0
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Apr 9, 2021 at 13:24 comment added pregunton I just found this paper which lists eight cases of function field with class number 1, including the four in your answer.
Apr 9, 2021 at 12:47 comment added David E Speyer Remarks: I did go back and check that the curves are unique; they are.
Apr 9, 2021 at 11:21 history bounty ended Mare
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Apr 8, 2021 at 13:23 history answered David E Speyer CC BY-SA 4.0