Question

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

Answer 1

[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.

Answer 2

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.