Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We define the variety $V$ as $\{ (x,y,z) \in \mathbb{A}^3\ |\ x = yz \}$. On this variety, I can make $(\mathbb{C}^*)^2$ act by $(\lambda, \mu) \star (x,y,z) = (\lambda \mu\ x , \lambda\ y,\mu\ z)$ on $V$.

How do you compute (=give an equation) the variety $V/(\mathbb{C}^*)^2$ ?

Thank you!

share|cite|improve this question
up vote 1 down vote accepted

First of all, the projection $(x,y,z) \mapsto (y,z)$ is an isomorphism between $V$ and $\mathbb A^2$, so you may as well just consider the case $V =\mathbb A^2$, with $(\mathbb C^{\times})^{2}$ acting in the obvious way on the coordinates.

Then the open subset $U = \{ (y,z) \, | \, y z \neq 0\} \subset \mathbb A^2$ (i.e. the complement of the two coordinate lines) is acted on simply transitively by $(\mathbb C^{\times})^2$. Thus $U/(\mathbb C^{\times})^2$ is a point, and since $U$ is open and dense in $V$, to the extent that $V/(\mathbb C^{\times})^2$ has any meaning as a variety, it will be a point too.

share|cite|improve this answer
You're right when you say that $V/(\mathbb{C}^*)^2$ must already has a meaning as a variety, I was trying to simplify the definition of $V$ so that I could explain the issue more easily... But,anyway, thank you for your answer, I'll try to work it out in my situation then. – ng_th Mar 15 '12 at 18:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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