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According to several sources, it is conjectured (or at least believed) that the rational points of curves over the rationals of genus $g > 1$ are uniformly bounded by $g$. E.g. here p. 1.

Assuming the curve is irreducible, the singular points (which are also rational) can easily be made unbounded for fixed $g > 1$.

For natural $n$, define $j(n)=\prod_{i=1}^n(x-i)$.

Consider the curve $C_n : j(n)^2(x^5+13)=y^2$.

It birationally equivalent to $x'^5+13=y'^2$ which is genus $2$.

$C_n$ has the rational (singular) points $(1,0),(2,0),\ldots(n,0)$ which are unbounded, since $n$ is unbounded.

The same applies for $2$ replaced by integer $d>1$.

For $d > 2$ the curve need not be hyperelliptic.

Q. What is the exact statement about uniform boundedness?

According to both sage and magma, $C_n$ is irreducible for small values of $n$.

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    $\begingroup$ Yes, you can take any number of Galois orbits of closed points and form a coproduct to turn those Galois orbits into highly singular rational points. So there is no bound on rational points on a singular curve of fixed geometric genus. However, the arithmetic genus of that curve grows as the number of singular points. So if you want a boundedness result for singular curves, you should work with the arithmetic genus. $\endgroup$ – Jason Starr Nov 18 '15 at 12:14
  • $\begingroup$ @JasonStarr What is a reference for this? I will give my references. Believe they definitely mean geometric genus by "genus". $\endgroup$ – joro Nov 18 '15 at 12:29
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    $\begingroup$ The geometric genus equals the arithmetic genus for smooth curves. For singular curves, as your example illustrates, it is absurd to ask for a bound only in terms of the geometric genus. Since the arithmetic genus is at least as large as the geometric genus, and also it is at least as large as the length of the singular locus, any uniform bound on rational points of smooth curves of geometric genus $\leq g$ implies a uniform bound on rational points of singular curves of arithmetic genus $\leq g$. $\endgroup$ – Jason Starr Nov 18 '15 at 12:39
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The precise statement of the conjecture is:

Uniformity Conjecture. Let $K$ be a number field and $g\geq2$ an integer. There exists a number $B(K,g)$ such that for any smooth curve $X$ of genus $g$ defined over $K$

$$|X(K)|\leq B(K,g)$$

and the original reference for it is:

The conjecture refers indeed to geometric genus, following Faltings early results on finiteness of rational points.

Also note (here of example) that this can be proved conditionally on Lang's conjecture.

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  • $\begingroup$ Thanks. Does your second reference misses smooth by mistake (on the first page)? $\endgroup$ – joro Nov 18 '15 at 13:27
  • $\begingroup$ In Corollary 1.2 they write "any curve $C$ of genus $g$", assuming Lang's conjecture. $\endgroup$ – joro Nov 18 '15 at 13:33
  • $\begingroup$ @joro Yes, I agree it is a bit confusing. In paragraph 3 they make clear that their result holds for "all smooth curves", but they don't mention it on the statement of theorem 1.1 and corollary 1.2. But they definitely mean smooth. $\endgroup$ – Myshkin Nov 18 '15 at 13:39
  • $\begingroup$ Indeed, I believe theorem/corollary takes precedence over comments. $\endgroup$ – joro Nov 18 '15 at 14:35
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    $\begingroup$ Patricia L. Pacelli, Uniform boundedness for rational points, Duke Math. J. 88 (1997), no. 1, 77–102, shows (under the same assumption of Lang's conjecture) that the bound would depend on the field $K$ only through its degree $[K : {\mathbb Q}]$. $\endgroup$ – Michael Stoll Nov 18 '15 at 17:52

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