3
$\begingroup$

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.

$\endgroup$
  • 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
1
$\begingroup$

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 $\infty$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.