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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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    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.2012-07-12

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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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    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.2012-07-14