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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 $\mathcal L$?

Johnny Apple
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    How do you define a rational section? For instance, is it "a section on a dense open subset"? – KReiser Nov 10 '20 at 21:24
  • Yes, or in terms of a colimit. Is there another useful way to define it? (Consider that a separate but possibly heavily overlapping question) – Johnny Apple Nov 10 '20 at 21:27
  • I'm just making sure we're on the same page. This means you don't have to bother with any gluing stuff, for instance. – KReiser Nov 10 '20 at 21:31

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The idea here is to find a dense affine open subset $\operatorname{Spec} A = U\subset X$ where

  1. $\mathcal{L}|_U\cong \mathcal{O}_X|_U$,
  2. $s$ is an honest section of $\mathcal{L}(U)$,
  3. $s$ 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.

KReiser
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  • Is the overall U you found actually guaranteed to be affine? I agree you've found one in each irreducible component. – Johnny Apple Nov 10 '20 at 21:51
  • Yes: $X$ is noetherian, so there are finitely many irreducible components, and all the affines in the proof are disjoint. The disjoint union of finitely many affine schemes is again affine. – KReiser Nov 10 '20 at 21:53
  • I discovered I have one more question: why does s^-1 being in A make it a section of L^-1(U)? – Johnny Apple Nov 11 '20 at 00:01
  • @JohnnyApple If $\mathcal{L}|_U\cong \mathcal{O}_X|_U$, then $\mathcal{L}^{-1}|_U\cong \mathcal{O}_X^{-1}|_U$. But $\mathcal{O}_X^{-1}\cong\mathcal{O}_X$. – KReiser Nov 11 '20 at 00:09
  • Excellent. That was my argument. – Johnny Apple Nov 11 '20 at 03:41
  • Sorry to bother, but why do we need the open set $U$ to be affine? – Mugenen Apr 21 '22 at 16:15
  • @Mugenen You don't, but you don't lose anything by doing it either and it can help with some related calculations (I copied parts of this out of some of my teaching notes where being able to use $U$ affine in the next proof was useful). – KReiser Apr 21 '22 at 18:37