The Inverse of a Rational Section is a Rational Section of the Dual This question comes after reading the last paragraph of Vakil's FOAG, p. 400.
We consider the set of $\{(\mathcal L, s) \}$, where $\mathcal L$ is an invertible sheaf on a Noetherian, reduced, regular in codimension 1 (in case any of that matters) scheme $X$, and $s$ is a nonzero rational section of $\mathcal L$.
The claim is that once you mod out by isomorphism, this set is an abelian group under $\otimes$, with inverse $\{(\mathcal L^*, 1/s) \}$. Thematically, this all works out well given that we know $\mathcal L \otimes \mathcal L^* \simeq \mathcal O_X$, but why is $1/s$ a nonzero rational section of the dual?
The best I can say is that both sheaves are locally isomorphic to $\mathcal O_X$, so perhaps we mean to consider $s$ and $1/s$ as rational sections of $\mathcal O$, and then we glue to produce $s$ and $1/s$ as rational sections of $\mathcal L$?
 A: The idea here is to find a dense affine open subset $\operatorname{Spec} A = U\subset X$ where

*

*$\mathcal{L}|_U\cong \mathcal{O}_X|_U$,

*$s$ is an honest section of $\mathcal{L}(U)$,

*$s$ is invertible.

Then the image of $s$ under the isomorphism $\mathcal{L}|_U\cong \mathcal{O}_X|_U$ is a unit in $A$, so $s^{-1}$ is an element of $A$ and therefore a section of $\mathcal{L}^{-1}(U)$ and thus a rational section of $\mathcal{L}^{-1}$.
To find such a $U$, it suffices to go irreducible component by irreducible component: since $X$ is noetherian, it has finitely many irreducible components, so for any irreducible component the set of points belonging to only that irreducible component is open. Now, for irreducible $X$, take the intersection of the open sets where $s$ is an honest section and an open subset where $\mathcal{L}|_U\cong\mathcal{O}_X|_U$: this is again a nonempty open subset by irreducibility, and since the affine opens form a basis for the topology we can find an affine open subset $\operatorname{Spec} A$ inside it. We're done.
