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I have been looking at hyperelliptic curves in characteristic two, in particular using Algebraic Geometry and Arithmetic Curves by Qing Liu, which gives a description in all characteristics.

For the set up, we are given a hyperelliptic curve $X$ of genus $g$ over a field $k$ of characteristic two, with a separable morphism $f:X\rightarrow \mathbb P_k^1$ of degree 2. We then have (from Qing Liu) a corresponding Artin-Schreier extension $K$ of $k(x)$, where $K=k(t)[y]$, with the relation \begin{equation} y^2+Q(t)y=P(t), \end{equation} where $Q(t)$ and $P(t)$ have maximum degrees $g+1$ and $2g+2$ respectively.

I have been looking to compute the divisors of a number of functions in the function field $K$, and to this end have computed the ramification, which corresponds to the solutions of $Q(t)$ and a point at infinity if $\deg Q(t) < g+1$. In this case it is stated that $\deg P(t) = 2g+1$, else the point at infinity will be singular, and we want to consider smooth curves.

The point that I find confusing/counter-intuitive is that it seems to be implied that if $\deg Q(t) = g+1$, or equivalently there are two points in the pre-image of infinity, then the degree of $P(t)$ can be anything. Is this really true?

My confusion may just come from the form being different to that in characteristic not 2. But I also seem to have difficulties getting degree of the divisor associated to $y$ being zero, which it should if it is a meromorphic function in $K$. For example, if $P(t)=t$, and $\deg Q(t)=10$, then the orders of $y$ at the points at infinity are either both $\frac{1}{2}$ or 9, neither of which seems okay. I can only resolve these issues if $\deg P(t)=2g+2$, but the books doesn't seem to imply this should be the case, and I don't have a good justification.

To clarify, the main point of the question is: If $\deg Q(t)=g+1$, can the degree of $P(t)$ be anything (between zero and $2g+2$)?

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  • $\begingroup$ I don't know anything about hyperelliptic curves, but maybe the answer to your question is in this paper: math.uwaterloo.ca/~ajmeneze/publications/hyperelliptic.pdf. In particular, they don't put any restriction on the characteristic of the field of definition. $\endgroup$
    – M Turgeon
    Jul 12, 2012 at 13:37
  • $\begingroup$ Thanks, that is a good introductory paper, but they don't consider the case where $Q(t)$ (or $h(x)$ in their case) has degree $g+1$ at all. $\endgroup$
    – Joe Tait
    Jul 12, 2012 at 13:54

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Yes any degree between $0$ and $2g+2$ can happen. As you said, the case $\deg Q=g+1$ correspond to $X\to \mathbb P^1_k$ being unramified above $t=\infty$.

Let $y^2+Q(t)y=P(t)$ be an equation of $X$ with $\deg Q=g+1$ and $\deg P\le 2g+2$. Then the equation of $X$ above $t\ne 0$ is given by $$ (yt^{-g-1})^2+ (Q(t)/t^{g+1})(yt^{-g-1})=P(t)/t^{2g+2}\in k[1/t].$$ Or equivalently $$z^2+Q_1(1/t)z=P_1(1/t)$$ where $Q_1(s), P_1(s)\in k[s]$, $\deg Q_1(s)\le g+1$ and $\deg P_1(s)\le 2g+2$.

As examples, for any $1\le d\le 2g+1$, the equation $$ y^2+t^{g+1}y=1+ct+t^{d} $$ (with $c=0$ if $d=1$ and $c=1$ otherwise, to make the equation non-singular) defines a hyperelliptic curve of genus $g$ in characteristic $2$. The above change of variables leads to an equation $$z^2+z=s^{2g+2}+cs^{2g+1}+s^{2g+2-d}$$ and we are in the situation of $\deg Q(s)=0$ !

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  • $\begingroup$ Thank you for the clarification. I still have one query however. In your example you have $Q(t)=t^{g+1}$. I thought that repeated roots in $Q(t)$ would lead to a non-smooth curve. Namely, if say $t^2|Q(t)$, then $t^2|y^2+P(t)$. Hence $t^2$ divides the partial derivative with respect to $t$ (normally I could only say $t$ divides it, but since we are in characteristic two the stronger statement holds). This partial derivative is just $P'(t)$. But this contradicts $Q(t)$ and $Q'(t)^2P(t)+P'(t)^2$ being prime. Sorry if there is some stupid error, but I can't see what is wrong. $\endgroup$
    – Joe Tait
    Jul 14, 2012 at 10:06
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    $\begingroup$ @JoeTait: You are right in the sense that $P(t)$ must be choosen so that the equation defines a smooth curve. When $Q(t)=t^{g+1}$, the partial derivative with respect to $y$ is $t^{g+1}$ and vanishes only at $t=0$. We then look at the partial derivative w.r.t. $t$ at $t=0$: it is equal to $P'(0)$. So the equation defines a smooth (affine) curve if and only if $P'(0)\ne 0$. $\endgroup$
    – user18119
    Jul 14, 2012 at 13:03
  • $\begingroup$ Okay, I hadn't seen the edited answer somehow when I made the comment. I guess the error in my argument comes from going to two variables, but I will look at that. Thank you again for the clarification and the very nice example. $\endgroup$
    – Joe Tait
    Jul 14, 2012 at 18:03

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