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Is it possible to calculate the degree of a toric divisor directly from the fan of the toric variety? If so, how is this done? Or is there some alternative way to calculate the degree of these divisors?

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The set of Cartier T-divisors on a fan is exactly the set of piecewise linear integral functions on the fan. Given such a function, one can recover the divisor by finding the values of the function at the first lattice elements which are the coefficients of the irreducible divisors associated to rays ( 1- dimensional cones) – Ehsan M. Kermani Jul 22 '12 at 18:53

Let the toric variety be $X$; let $d = \mathrm{dim} \ X$. For any Cartier $T$-divisor $D$, we can describe $\dim H^0(X, \mathcal{O}(D))$ as the number of lattice points in a certain polytope $P_D$. See Section 5.3 of Fulton's Introduction to Toric Varieties. The degree of $D$ is $d! \mathrm{Vol}(P_D)$.

Proof: Let $h(n)$ be the Hilbert polynomial $h(n) = \dim H^0(X, \mathcal{O}(nD))$. We know that the leading term of $h(n)$ is $\mathrm{deg}(\mathcal{O}(D)) n^d/d!$. But also, the number of lattice points in $n \cdot P_D$ is $\mathrm{Vol}(P_D) n^d + O(n^{d-1})$. Equating these two, $\mathrm{deg}(\mathcal{O}(D)) = d! \mathrm{Vol}(P_D)$. $\square$

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I'm curious, wouldn't this imply that $\mathrm{deg}(\mathcal{O}(D))=D^d$ according to Fulton's book (p. 111)? (This lead to my own question…) – Mike Jul 24 '12 at 18:41

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