3
$\begingroup$

Suppose we are given the equation $$ y^2z = x(x - z)(x - 2z) $$ I would like to define a degree two map $g$ on this curve into complex projective space. I hate to say I am already lost here - how do I start this ?

So an element on complex projective space is an equivalence class of the form $[z,\omega]$ where $(z,\omega) \backsim (\lambda z,\lambda \omega)$ for any nonzero complex number $\lambda$. Also, I can see that the above equation gives a homogeneous polynomial of degree $3$, $$ f(x,y,z) = y^2z- x^3- 3x^2z +2z^2x $$ but how can I find the homogeneous coordinates ? My hunch is I should use the identification $\mathbb{C} P \cong S^2$ .. is that a good idea? Any help would be great!

$\endgroup$
3
$\begingroup$

Personally, I find the affine view much more intuitive, and your intuition may be different. I would dehomogenize with respect to $z$, to get the equation $Y^2=X(X-1)(X-2)$, whereupon your map to the complex line is clearly $(X,Y)\mapsto X$. Now rehomogenize to get $(x,y,z)\mapsto(x,z)$. Voilà!

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ +1; a minor comment is that one still needs to explain where the point (x,y,z)=(0,1,0) on the curve goes. (Or invoke the fact that a rational map on a smooth curve is regular, but that seems a little high-powered for this context...) $\endgroup$ – user64687 Apr 12 '13 at 17:13
  • $\begingroup$ @Asal, not so minor, I think! $\endgroup$ – Lubin Apr 13 '13 at 3:52

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.