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What is information about the existence of rational points on hyperelliptic curves over finite fields available?

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One fact about hyperelliptic curves (which distinguishes them from arbitrary curves) is that they have quadratic twists, so that a hyperelliptic curve over F_q has at most 2q+2 rational points. There are several constructions of hyperelliptic curves with no points, as well as results improving the `obvious' bound on the genus of a pointless hyperelliptic curve which follows from the Weil bound. Could you make your question more precise, to clarify what type of situation you have in mind? –  Michael Zieve Feb 12 '13 at 23:29
    
I want criterion (or something like it) that curve has rational point –  Alexey Feb 12 '13 at 23:39
    
Not trivial constructions of hyperelliptic curves with no points are interesting too. –  Alexey Feb 12 '13 at 23:42
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2 Answers

[Edited to remove material subsumed and improved by Felipe's answer.]

Here is some historical info. Dickson studied this question in his 1909 paper "Definite forms in a finite field". For Dickson, a "definite form" is a homogeneous $f(x,z)\in\mathbb{F}_q[x,z]$ which takes nonzero square values for all $(x,z)$ in $\mathbb{F}_q\times\mathbb{F}_q$ except $(0,0)$. If $q$ is odd and $f(x,z)$ is not a square then Dickson's condition is equivalent to saying that the hyperelliptic curve $y^2=f(x,1)$ has $2q+2$ points over $\mathbb{F}_q$, or equivalently, its quadratic twist $y^2=nf(x,1)$ has no points (where $n$ is any nonsquare in $\mathbb{F}_q$).

In modern language, Dickson showed that there are no pointless genus-$2$ curves over $\mathbb{F}_q$ if $q$ is odd and $q\ge 13$. Carlitz took up this topic in a series of papers, and among other things made the connection with Weil's bound, which implies that a pointless hyperelliptic curve over $\mathbb{F}_q$ has genus at least $(q+1)/(2\sqrt{q})$, or roughly $\sqrt{q}/2$. As Felipe's answer indicates, this bound is essentially best possible when $q$ is an odd square. It can be improved by a factor of roughly $\sqrt{2}$ (and possibly much more) when $q$ is prime.

It is known that genus-$2$ pointless hyperelliptic curves exist over $\mathbb{F}_q$ precisely when $q\le 11$, and in genus-$3$ the analogous result is $q\le 25$ (the latter is due to Howe, Lauter, and Top). Further experimental results over small prime fields appear in papers by Glazunov.

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As Mike says, there isn't much beyond the Weil bound. For prime fields, there is a slight improvement due to Stark. If the field has $q=r^2$ elements with $q$ odd, then one can find $a,b$ such that $y^2=ax^{r+1}+b$ has no points, which shows that you cannot improve the Weil bound. For arbitrary $q$ odd, there is this beautiful idea of I. Shparlinski:

For a monic irreducible polynomial $f$ of degree $d$, the vector of values of $f$ modulo squares is one of $2^q$ possibilities. There are about $q^d/d$ choices for $f$ and so if $q^d/d > 2^q$ or thereabouts, you get two monic irreducibles $f,g$ such that $fg$ takes only square values and so, if $c$ is a non-square, $y^2=cf(x)g(x)$ has no points and you can take $d$ about $q/\log q$. I don't know how to do better than this.

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Felipe, maybe this is what Stark did, but for prime fields you can improve the estimate provided by the Weil bound (for the genus of a pointless hyperelliptic curve) by a factor of $\sqrt{2}$. This was done by Mitkin 1975 "Existence of rational points...", El Baghdadi 1995, and Zannier 1998 "Polynomials modulo $p$...". –  Michael Zieve Feb 13 '13 at 1:00
    
@Mike: Thanks. I guess I misconstrued what you said. BTW, the answer you deleted had some additional useful information. –  Felipe Voloch Feb 13 '13 at 1:17
    
A peripheral remark: all the known improvements of Weil's bound in the case of hyperelliptic curves over prime fields are done via variants of Stepanov's method, which is based on the construction of an auxiliary rational function of fairly small degree which vanishes to high order at the rational points. I can't find a statement of Stark's result online, but I do see on page 15 of Stohr-Voloch "Weierstrass points..." that Stark's bound can be deduced via their geometric approach to constructing these auxiliary functions. –  Michael Zieve Feb 13 '13 at 2:35
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