Genus 2 - Points at Infinity

Question

I'm reading Cassels and Flynn's book on Genus 2 curves

In the background section, they have the following:

Still working over an algebraically closed field $\bar{k}$, we suppose that the characteristic is not 2 and take as an example the curve $\mathcal{C}$ given in the affine plane by $$ Y^2 = \prod_{j=1}^6 (X-\theta_j).$$ It is not complete: there are two points $\mathfrak{b}_1$, $\mathfrak{b}_2$ 'at infinity'.

I'm trying to see this. So, I homogenize to get $$Y^2Z^4 = \prod_{j=1}^6 (X-\theta_jZ).$$ A point at infinity is of the form $[a:b:0]$ and substituting implies that $a=0$ and so $[a:b:0]\sim [0:1:0]$. So, I only get one point at infinity. Do the authors mean that $\mathfrak{b}_1 = \mathfrak{b}_2$ occurs with multiplicity 2? In what sense do points at infinity have a multiplicity.

They go on to say

$X^{-1}$ is a local uniformizer at the $\mathfrak{b}_j$

Again, I'm having difficulty with points at infinity here. This implies that $X$ has a pole of order 1 at $\mathbf{b}_j$. In what sense does $X$ have a pole at $[0:1:0]$ and how do we determine it's multiplicity?

Answer 1

Turning my comment into an answer:

If you take the projective closure in $\DeclareMathOperator{\P}{\mathbb{P}} \P^2$, you get just one point at infinity, but it is singular, as can be checked by computing partial derivatives. If you resolve the singularity by blowing up, you'll get two nonsingular points.

Alternatively, people usually consider the closure of $C$ in the weighted projective space $\P(1,3,1)$, where it is given by the weighted homogeneous equation $Y^2 = \prod_{j=1}^6 (X-\theta_jZ)$ of degree $6$. Using this equation, if you look on the line $Z=0$, you will find 2 (smooth) points at infinity.

But why these weights? This can be explained by consider the orders of poles of the functions $x$ and $y$. The TL;DR is that $x$ has pole divisor $\renewcommand{\div}{\operatorname{div}} \div_\infty(x) = \infty_1 + \infty_2$, while $y$ has pole divisor $\div_\infty(y) = 3(\infty_1 + \infty_2)$. Thus it's natural to take $x$ and $y$ to have weights $1$ and $3$, respectively.

Suppose $C$ is a smooth projective genus $2$ curve, just abstractly without a defining equation. We will find an equation using Riemann-Roch. Let $K$ be a canonical divisor for $C$. Then $\deg(K) = 2g-2 = 2$. Since $h^0(K) = g = 2$, then there is a nonzero rational function $f \in H^0(K)$, so $\div(f) + K \geq 0$. Replacing $K$ by $\div(f) + K$, we may assume that $K$ is effective, so $K$ is the sum of $2$ points. Denoting these points by $\infty_1$ and $\infty_2$, then $K = \infty_1 + \infty_2$. (Generically these $2$ points will be distinct, but it is possible to take $K = 2P$ where $P$ is a Weierstrass point. In that case, the calculation below goes slightly differently.) Write $\newcommand{\iinfty}{\underline{\infty}} \iinfty = \infty_1 + \infty_2$, so $K = \iinfty$.

Let's compute $h^0(m \iinfty)$ for $m = 0, 1, 2, \ldots$. Since $C$ is projective, then there are no nonconstant regular functions, so $h^0(0) = 1$. Since $K = \iinfty$, then $$ h^0(\iinfty) = h^0(K) = g = 2 \, , $$ so there is some nonconstant function $x \in H^0(\iinfty)$. For divisors $D$ with $\deg(D) > 2g-2 = 2$, we have $$ h^0(D) = 1 - g + \deg(D) = \deg(D) - 1 $$ by Riemann-Roch. Thus for all $m \geq 2$, we have $h^0(m \iinfty) = 2m - 1$. Then $h^0(2 \iinfty) = 3$, but $1, x, x^2$ all belong to $H^0(2 \iinfty)$, so we don't find any new functions. For $m = 3$, however, we have $h^0(3 \iinfty) = 5$. Now $1, x, x^2, x^3$ are all in $H^0(3 \iinfty)$, but that means there's still one more linearly independent function: call this function $y$. Thus $\div_\infty(x) = \iinfty$ and $\div_\infty(y) = 3 \iinfty$, which explains the choice of grading.

If we continue on to $m = 6$, then we find that $h^0(6 \iinfty) = 11$ and $$ 1, x, x^2, x^3, y, x^4, xy, x^5, x^2 y, x^6, x^3 y $$ form a basis for $H^0(6 \iinfty)$. But $y^2$ also belongs to $H^0(6 \iinfty)$, so there is a linear dependence among these monomials. With a little more work, this yields a Weierstrass equation $$ y^2 + h(x) y = f(x) $$ where $\deg(h) \leq 3$ and $\deg(f) \leq 6$.

Here's a table summarizing these calculations: $$ \begin{array}{c|c|l} m & h^0(m \iinfty) & \text{basis for $H^0(m \iinfty)$}\\ \hline \hline 0 & 1 & 1\\ 1 & 2 & 1, x\\ 2 & 3 & 1, x, x^2\\ 3 & 5 & 1, x, x^2, x^3, y\\ 4 & 7 & 1, x, x^2, x^3, y, x^4, xy\\ 5 & 9 & 1, x, x^2, x^3, y, x^4, xy, x^5, x^2 y\\ 6 & 11 & 1, x, x^2, x^3, y, x^4, xy, x^5, x^2 y, x^6, x^3 y \end{array} $$

We also see that $1/x$ is a uniformizer at $\infty_1$ and $\infty_2$: $x$ has a simple pole at each of these points, so $1/x$ has a simple zero there.

For more on hyperelliptic curves, see these notes by Steven Galbraith.

Answer 2

The Riemann surface of the analytic function $Y=\sqrt{\prod_{j=1}^6 (X-\theta_j)}$ indeed has two points over infinity. It can be seen from the fact that the square root of a sixth degree polynomial doesn't change its sign when passing around origin along a circle of a big enough radius. The function $X^{-1}$ is an uniformizer for the Riemann sphere at infinity, and since the covering over infinity is unramified, the pullback of $X^{-1}$ is also an uniformizer for the points over infinity.

About the question in comments, if we put $Y=1$ and $X=0$, the equation of projective closure becomes $(\prod_{j=1}^6\theta_j)Z^6-Z^4=0$. This equation has a root $Z=0$ of multiplicity 4, and, in general, 2 other roots, so there are no contradiction with Bezout's Theorem.