# Divisors over a projective curve

I have a few doubts about some algebraic geometry problems. This is the situation: X is a (smooth) curve on the projective plane $P^2(F)$, F being an algebraic closure of $F_2$ $X = Z(x_{1}^2x_{2}+x_1x^{2}_{2}+x^{3}_{0}+x^{3}_{2}$. $Y = X\cap U_2 = Z(y^2 + y + x^3 + 1)$ is affine. The point at infinity of $X$ is $(0:1:0)$. $X_0$ is the variety seen as defined over $F_2$ (this is not $X ∩ P^2(F_2)$). First I don't understand very well the distinction between X and $X_0$.. since $X$ is already given as zero-set of a polynomial with coefficient in $F_2$. The first question i can't answer is to list all the positive divisor of degree 2 on $X_0$. Two examples are: $2\infty, (0,z)+(0,z^2)$ where $z$ is such that $z^2+z+1=0$.

The second question is to find the divisor of the functions $x, y, x+1, y+1, x+y, x+y+1$. From theory i know that such divisors are supposed to have degree 0, but the divisor i've found have not. Isn't $(0:z:1) + (0:z^2:1) - \infty$ the divisor of $x$?

• The function $x$ has a pole of order two at the point of infinity $\infty=(0:1:0)$, so $(x)=(0:z:1)+(0:z^2:1)-2\infty$. – Jyrki Lahtonen Jan 2 '12 at 10:14
• Also apparently the question is asking you to list degree two divisors defined over $F_2$. Otherwise there would be quite a few :-) – Jyrki Lahtonen Jan 2 '12 at 10:15

First note that $\mathbb{P}^2(\mathbb{F}_2)$ contains only 7 points. $\mathbb{A}^3(\mathbb{F}_2)$ contains $2^3=8$, we remove $(0,0,0)$ and mod out by scalar multiplication, but there's only 1 nonzero scalar in $\mathbb F_2$.
Note that $\overline{\mathbb F_2}$ is an infinite field, so $X$ contains more points than $X_0$. You've already noted that it contains $[0:z:1]$ where $z^2+z+1=0$, but $z\not\in\mathbb{F}_2$, so $[0:z:1]\not\in X_0$.
Assuming I made my calculations correctly, $X_0$ contains only three points: $[0:1:0]$, $[1:0:1]$, and $[1:1:1]$. Therefore, the degree 2 divisors would be $2[0:1:0]$, $[0:1:0]+[1:0:1]$, $[0:1:0]+[1:1:1]$, $2[1:0:1]$, $[1:0:1]+[1:1:1]$, and $2[1:1:1]$.
Note that the function $x=x_0/x_2$, so $\operatorname{ord}_P x = \operatorname{ord}_P x_0 - \operatorname{ord}_P x_2$. The only point in $X_0$ where either $x_0$ or $x_2$ vanish is $[0:1:0]$ and $\operatorname{ord}_{[0:1:0]}x=1-1=0$. This gives $\operatorname{div}x=0$. You can play the same game for the other functions.
• The points with coordinates in $F_4$ (and at least one coordinate not in $F_2$) also give degree two divisors, won't they? – Jyrki Lahtonen Jan 2 '12 at 10:16