# Hyperelliptic curves over characteristic two fields

I have been looking at hyperelliptic curves over an algebraically closed field $k$ of characteristic two, with a view towards finding the basis for the vector space of holomorphic differentials. To do this I have viewed the curves as the function field $k(x,y)$, originally restricting to those defined by $$y^2 - y = f(x), \ f(x)\in k[x].$$

After this, I thought that to generalise to all hyperelliptic curves I should allow $f(x)$ to be any rational function. However, looking at the literature it seems like the definition is instead:

A hyperelliptic curve of genus $g$ ($g\geq 1$) is an equation of the form $$y^2 - h(x) y = f(x),\ f(x),h(x)\in k[x],$$ where the degree of $h(x)$ is at most $g$, and $f(u)$ is a monic polynomial of degree $2g +1$, with no elements of $k\times k$ satisfying the original equation and both of it's partial derivatives.

I don't see what was wrong with my initial intuition, so if anyone could tell me, or explain why the definition given is correct, I would be much obliged.

edit - to give fuller definition

-
You can write any quadratic extension of $k(x)$ as $y^2 - y = f(x)$ for some rational function $f$ (Artin-Schreier). If you're given the second form $\eta^2 - h(x) \eta = \phi(x)$, divide by $h^2$ to get $y^2 - y = \phi(x) / h^2(x)$ where $y = \eta/h(x)$. Note that the point at infinity is then a pole of odd order of $\phi/h^2$, and thus a rational Weierstrass point of the curve; indeed a hyperelliptic curve in characteristic $2$ can be put in that form if and only if it has a rational Weierstrass point. In your setting that's automatic because you assumed $k$ is algebraically closed. –  Noam D. Elkies Apr 12 '12 at 16:07
That is great, though I feel that I should have seen that before now! Thank you very much. –  Tait Apr 12 '12 at 16:14
You're welcome. One correction: I should have written that any separable quadratic extension can be written in that Artin-Schreier form. –  Noam D. Elkies Apr 13 '12 at 0:56