Suppose I give you positive integers $g\geq 2$ and $N.$ Is it always possible to find an absolutely irreducible curve of genus $g$ over $\mathbb{Q}$ which has at least $N$ rational points? For that matter, what if $g=2?$ I assume that the answer is YES, but what do I know?
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asked Jan 23, 2017 at 20:30
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On the contrary, some conjectures suggest that the answer is NO! It follows from the Bombieri-Lang conjecture (sometimes known as Lang's conjectures) that a uniform bound should exist.
More precisely, given a number field $F$ and a genus $g\geq2$ BL implies that there's some bound $N(F,g)$ such that every smooth projective genus $g$ curve over $F$ has at most $N(F,g)$ points. Even better -- the bound depends only on the degree of $F$ over $\mathbb{Q}$, not $F$ itself.
The fact that Bombieri-Lang implies this was proved by Caporaso, (Joe) Harris and Mazur in 1997 and my recollection at the time was that some people regarded this as evidence against Bombieri-Lang rather than for the uniform bound. However, at this stage, the question is open, even for curves of genus 2 over $\mathbb{Q}$.
I once saw a talk of Elkies where he exhibited a curve of genus 2 over $\mathbb{Q}$ with something like 588 rational points, but this may not be the record any longer.
answered Jan 23, 2017 at 20:45
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2$\begingroup$ math.harvard.edu/~elkies/many_pts.pdf is what looks like a write-up of what Elkies knew a while back. It will contain all the references you need. I quite agree that it's an extraordinary situation! $\endgroup$ Commented Jan 23, 2017 at 20:47
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$\begingroup$ Wow, who would have thunk it! $\endgroup$ Commented Jan 23, 2017 at 20:52
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$\begingroup$ Armed with your words of wisdom, I found this: mathoverflow.net/questions/103327/… $\endgroup$ Commented Jan 23, 2017 at 20:59