The $x$-coordinate map of $X:y^2+y=x^3\to\mathbb{P}^1$

Question

Let $X$ be the smooth projective model of $y^2+y=x^3$ over a field $k$ of characteristic $2$, and let $\pi:X\to\mathbb{P}^1$ be the $x$-coordinate map. We get the projective model by homogenizing $y^2+y=x^3$, which gives us $y^2z+z^2y=x^3$. Setting $z=0$ then tells us that there is one points at infinity, $[0:1:0]$. Thus, $\pi$ is given by $(x,y)\to (x:1)$ for affine points $(x,y)$ of $X$. But what does infinity map to and how can I explicitly see this?

Answer 1

This is an instance of the curve-to-projective extension theorem. The way to solve this is to find a uniformizer $t$ in the local ring of at $[0:1:0]$, then write $x=ut^n$ and $z=vt^m$ in terms of the uniformizer for $u,v$ units so that our map becomes $[ut^n:vt^m]$. Then if $n>m$, we can write our map as $[ut^{n-m}:v]$ and evaluate to get $[0:1]$, or if $m>n$ we can write our map as $[u:vt^{m-n}]$ to get $[1:0]$, and if $n=m$ we can write our map as $[u:v]$ and we'll have to work a little more to see what we get.

Let's look for a uniformizer. To check whether an element $f\in\mathcal{O}_{X,[0:1:0]}$ is a uniformizer, we can compute the dimension of $\mathcal{O}_{X,[0:1:0]}/(f)$ as a vector space - this ring is a DVR, so $\nu(f)=\dim_k\mathcal{O}_{X,[0:1:0]}/(f)$. We'll check $x$ and $z$ first to see if we get lucky with an easy choice.

$\mathcal{O}_{X,[0:1:0]}/(x) \cong k[x,z]_{(x,z)}/(z+z^2+x^3,x) \cong k[z]_{(z)}/(z+z^2)\cong k[z]_{(z)}/z\cong k$, as $z+z^2=z(1+z)$ and $(1+z)$ is a unit in this ring. So $x$ is a uniformizer, and we note that $z=\frac{1}{1+z}x^3$ has valuation 3. Therefore our map can be written as $[x:\frac{1}{1+z}x^3]=[1:\frac{1}{1+z}x^2]$ near $[0:1:0]$, and so $[0:1:0]\mapsto [1:0]$ under the projection to the $y$-axis.