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Does anybody know of a good way to figure out if the variety defined by the ideal $I=(a_1y_2-a_2y_1,b_1y_2-b_2y_1)$ in $\mathbb A^2_{y_1,y_2}\times\mathbb P^1_{a_1,a_2}\times \mathbb P^1_{b_1,b_2}$ is Toric?

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Your scheme has two irreducible components: On one of them, the equation $y_1=y_2=0$ holds, on the other one $a_1 b_2 = a_2 b_1$ holds. Each component is toric. So this isn't a variety (at least, according to Hartshorne) and isn't a toric variety (at least, according to Fulton). –  David Speyer Feb 17 '11 at 18:07
    
I actually hadn't though about it being irreducible or not! Do you know how can I find the components? The in the second component you mention, the equation $a_1b_2-a_2b_1$ is not enough, since this imposes no conditions on the $y$'s and you must have $[y_1,y_2]=[a_1,a_2]$ on $X$ as long as $(y_1,y_2)\neq (0,0)$. Thanks! –  Enrique Feb 17 '11 at 18:51

1 Answer 1

The torus $(\mathbb C^\times)^2$ acts on your variety $X$, with $(s,t)\in (\mathbb C^\times)^2$ sending the point $(y_1,y_2,a_1:a_2,b_1:b_2)\in X$ to $(sy_1,ty_2,t^{-1}a_1:s^{-1}a_2,t^{-1}b_1:s^{-1}b_2)$. There is an open faithful and dense orbit. That's enough.

Of course, "enough" depends on what definition of toric varieties you have in mind...

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But note that if $(y_1,y_2)=(0,0)$ then the $a$'s and the $b$'s could be anything, so $(0,0)\times \mathbb P^1\times \mathbb P^1$ lies in the variety. I'm no sure the action you are giving has a dense orbit. –  Enrique Feb 17 '11 at 17:38
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The projection $X\to\mathbb A ^2$ is an isomorphism in the complement of the origin. The image of the section you are referring to is going to give you points with orbits that don't intersect $(0,0)\times\mathbb P^1\times\mathbb P^1$, so the orbits can't be dense since $X$ is two dimensional. –  Enrique Feb 17 '11 at 17:50

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