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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?

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up vote 2 down vote accepted

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.

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