I am reading Fulton's "Toric Varieties." In it, he explains that if $X$ is a toric variety and if $D_1, \ldots, D_d$ are the irreducible divisors invariant under the big torus action, then
$$ \Omega_X^n \cong \mathcal{O}_X \left( -\sum_{i=1}^d D_i \right). $$
He begins his proof by noting that
$$ \omega = \frac{dX_1}{X_1} \wedge \cdots \wedge \frac{dX_n}{X_n} $$
is a global section of $\Omega_T^n$, where $T$ is the big torus orbit and the $X_i$ are coordinates corresponding to a particular choice of basis of the lattice. Hence, $\omega$ is also a rational section of $\Omega^n_X$. He concludes by showing that on a special affine cover, the restriction of the divisor associated to $\omega$ agrees with the restriction of $-\sum_{i=1}^d D_i$.
My question is this: why does the divisor associated to $\omega$, and not that of any other rational section, give us our canonical divisor?
Best Answer
There is no such thing as the canonical divisor of a smooth variety, only the canonical class. If you choose a different rational section of $\Omega_X^n$, you'll get a different, but linearly equivalent, divisor. As an abuse of notation, people generally refer to any divisor in the canonical class as a canonical divisor.