# Equations of a projective toric variety

Given complete fan $\Delta$ defining a projective toric variety (so that $\Delta$ is the normal fan of some polytope). How do one go on to find a defining ideal of the toric variety in projective space? Or, going the other way, given binomial equations defining a projective toric variety, is there some way to recover the fan?

In my case, I am trying to recover the fan of the hypersurface $Z(x_0x_3-x_1x_2)$ in $\mathbb{P}^n$ ($n>3$).

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To get the equations, one way to find this is to read David Cox's The Homogeneous Coordinate Ring of a Toric Variety, which you can find here.

If you are given the equations and you know the action of the torus on the variety, you can reconstruct the fan by looking at the way orbits are placed relative to each other. I think this is explained in Fulton's book, for example.

Everything should be explained in Cox's new book (together with Little and Schenk), aptly named Toric Varieties

Later. Let's do your example with $n=4$. There is a simply transitive action of the torus $T=(\mathbb C^\times)^3$ on the subset $\mathcal T$ of your variety $X$ of those points all of whose coordinates are non-zero such that $$(t_1,t_2,t_3)\cdot(x_0:x_1:x_2:x_3:x_4)=(x_0:t_1x_1:t_2x_2:t_1t_2x_3:t_3x_4).$$ The set of group homomorphisms $\mathbb C^\times\to T$ is parametrized by the abelian group $\mathbb Z^3$, with $(a,b,c)\in\mathbb Z^2$ corresponding to the map $$\chi_{a,b,c}:t\in\mathbb C^\times\longmapsto (t^a,t^b,t^c)\in T.$$ Now let $p=(1:1:1:1:1)$, a point in $%\mathcal T$, and the consider the limit $$\lim_{t\to0} \chi_{a,b,c}(t)\cdot p=\lim_{t\to0}(1:t^a:t^b:t^{a+b}:t^c)$$ Depending on the value if $(a,b,c)$ in $\mathbb Z^3$, this limit changes (and in some cases may not exists). This determies regions in $\mathbb R^3$ where the behaviour does not change: these regions are the relative interiors of the cones in the fan.

For example, in the regions

1. $a>0, b>0,c>0$
2. $a<0, a+b>0, c>0$
3. $b<0, a+b>0, c>0$

the limits exist (and are $(1:0:0:0:0)$, $(0:1:0:0:0)$ and $(0:0:1:0:0)$, respectively) so this gives us three of the cones. You'll surely have fun determining all the rest :)

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