1
$\begingroup$

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

Improve this question
$\endgroup$
3
  • 7
    $\begingroup$ 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. $\endgroup$
    – Noam D. Elkies
    Commented Apr 12, 2012 at 16:07
  • $\begingroup$ That is great, though I feel that I should have seen that before now! Thank you very much. $\endgroup$
    – Tait
    Commented Apr 12, 2012 at 16:14
  • 2
    $\begingroup$ You're welcome. One correction: I should have written that any separable quadratic extension can be written in that Artin-Schreier form. $\endgroup$
    – Noam D. Elkies
    Commented Apr 13, 2012 at 0:56

1 Answer 1

Reset to default
1
$\begingroup$

If anyone is interested, a thorough treatment is given by the book referenced in the comments here: https://math.stackexchange.com/questions/137495/artin-schreier-extensions-over-characteristic-two-fields

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .