I'm supposed to prove that the map $\pi : (x,y) \mapsto x$ of $X = V(y^2 - g(x)) \subset \mathbb A^2$ where $g$ is cubic extends to a regular map of the projective closure $\overline \pi : \overline X \to \mathbb P^1$.

The projective closure of X (I think) is $V(y^2z - \overline g(x,z))$ where $\overline g(x,z)$ is the homogenization of g ($x^3 + $ stuff in terms of x and z). Now the problem is that the naive projection isn't defined at the point $\overline \pi(0,1,0) = (0,0)$ isn't defined. I'm given a hint that I should consider changing the map to $(x,y,z) \mapsto (x^3, x^2z)$, but I don't know how that is supposed to help.

  • 1
    $\begingroup$ Hint: for points on the curve, you can replace $x^3$ by $z(y^2-h(x,z))$ for some $h$ which is homogeneous of degree 2. Now you can simplify the expression for your map... $\endgroup$ – user64687 Mar 5 '14 at 8:56

Let $\infty =(0:1:0)\in \mathbb P^2$ and consider the projection from $\infty$ onto the projective line $L=\{y=0\}$, which is the map $$\Pi:\mathbb P^2\setminus \{\infty\}\to L:(u:v:w)\mapsto (u:0:w)$$ Its restriction to $X$ is precisely $\pi$.
So we have to show that $\Pi|X$ extends to $\overline X$.
For that we use the affine coordinates $\xi= \frac xy, \zeta=\frac zy$ in the neighbourhood of $\infty$.
The equation $y^2z=x^3+zq(x,z)$ ($q$ a quadratic form) for $\overline X$ becomes $\zeta= \xi^3+\zeta q(\xi,\zeta)$, so that $\xi$ is a uniformizing parameter at $\infty$ for $\bar X$ and we can write our equation as $\zeta= \xi^3+\xi^3f(\xi, \zeta)$ near $\infty$. The projection $\Pi|X$ becomes $$(\xi:1:\zeta)\mapsto (\xi:0:\zeta)=(\xi:0:\xi^3+\xi^3f(\xi, \zeta))=(1:0:\xi^2+\xi^2f(\xi, \zeta)) $$
This shows that we can extend $\pi$ to $\overline X$ by $\bar \pi(\infty)=(1:0:0)$ and obtain a regular map $\bar \pi:\overline X \to L$

Note carefully
The morphism $\Pi:\mathbb P^2\setminus \{\infty\}\to L$ cannot be extended regularly to $\mathbb P^2$, but its restriction to any smooth curve going through $\infty$ can be extended regularly through $\infty$.

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