Vakil exercise on sheaf associated to the divisor of rational section

Question

This is exercise 15.4.G. of Vakil's notes.

Let $\mathscr{L}$ be an invertible sheaf on an irreducible normal scheme $X$ with $s$ a rational section of $\mathscr{L}$. We want that $\mathscr{O}_X(\text{div}(s)) \cong \mathscr{L}$. The first part of the hint is to show that $\{U\subseteq X: \mathscr{O}_X(\text{div}(s))|_U\cong \mathscr{O}_X|_U\}$ forms a basis for the topology on $X$ which I think is just using that $\mathscr{O}_X(\text{div}(f_i))|_{U_i} \cong \mathscr{O}_X|_{U_i}$ with $U_i$ the trivializing open sets for $\mathscr{L}$ and $f_i$ the rational function corresponding to $s$ on $U_i$. What I am confused about is how to conclude the sheaves are isomorphic.

He says: on each $U$ where $\mathscr{O}_X(\text{div}(s))|_U\cong \mathscr{O}_X|_U$ let $\phi_U: \mathscr{O}_X(\text{div}(s))(U) \to \mathscr{L}(U)$ be given by $t\mapsto ts$ where $t$ is a "rational function with zeros and poles constrained by $s$." I cannot tell what he means by $t$ and $ts$ here. I am guessing that he means $t \in \mathscr{O}_X(\text{div}(s))(U)$ corresponds to some element $f$ of $\mathscr{O}_X(U)$ by our choice of open set and then we are really performing the map $t\mapsto fs$ which makes sense. But the dependence on the choice of isomorphism makes it very hard for me to think about gluing these together to some map on all of $X$ as the distinct isomorphisms for each $U$ don't have anything to do with each other.

Any clarification on what is going on here is greatly appreciated!

(Edit for detail since it isn't clear I've used the definition):

Here is what I have so far. Let $\{U_i\}$ be trivializing open subsets for $\mathscr{L}$ with $\psi_i: \mathscr{O}_X|_{U_i} \to \mathscr{L}|_{U_i}$ the isomorphisms. Say that $s\in \mathscr{L}(V)$ for a dense open $V \subseteq X$. Let $f_i = \psi_i^{-1}(s|_{U_i \cap V}) \in \mathscr{O}_X(U_i \cap V) \subseteq K(X)$ so that $\text{div}(s)|_{U_i} = \text{div}|_{U_i}(f_i)$ then we have $\mathscr{O}_X|_{U_i} \to \mathscr{O}_X(\text{div}(s))|_{U_i}$ given by $1\mapsto 1/f_i$ well defined since $\text{div}|_{U_i}(1/f_i) + \text{div(s)}|_{U_i}= 0 \geq 0$ and it is invertible (I think) since multiplication by $f_i$ takes $t\in \mathscr{O}_X(\text{div}(s))(U_i)$ to $t f_i$ which has $\text{div}|_{U_i}(t f_i) = \text{div}|_{U_i}(t) + \text{div}(s)|_{U_i} \geq 0$ since $t \in \mathscr{O}_X(\text{div}(s))(U_i)$ and normality ensures that this condition is enough that $t f_i \in \mathscr{O}_X(U_i)$. So the trivializing neighborhoods of $\mathscr{L}$ are also trivializing for $\mathscr{O}_X(\text{div}(s))$ and the former are a basis for $X$ since $\mathscr{L}$ is an invertible sheaf, so the first claim is fine. It is just not clear how to use these to get an isomorphism on all of $X$.

Answer 1

I think I have a solution that goes through some detail, if someone spots something that doesn't make sense please let me know!

Let $X$ be normal and irreducible. Say $\mathscr{L}$ is invertible with $\{U \subseteq X: \psi_U: \mathscr{O}_X|_U \to \mathscr{L}|_U\}$ a basis for $X$ of trivializing opens and say that $s\in \mathscr{L}(V)$ is non-vanishing on some dense open $V\subseteq X$. Notice that for any such $U$ we have that $\mathscr{O}_X(\text{div}(s))|_U = \mathscr{O}_U(\text{div}|_{U}(\psi_U^{-1}(s|_{U\cap V})))$ by deifnition and $\mathscr{O}_X|_{U} \to \mathscr{O}_X(\text{div}(s))|_U$ given by $1\mapsto 1/\psi_U^{-1}(s|_{U\cap V})$ is well defined since we have that $\text{div}|_U(1/\psi_U^{-1}(s|_{U\cap V})) + \text{div}|_{U}(\psi_U^{-1}(s|_{U\cap V})) \geq 0$ and moreover it has an inverse where $t \mapsto t\psi_U^{-1}(s|_{U\cap V})$ which is an honest regular function in $\mathscr{O}_X(U)$ since $\text{div}|_U(t \psi_U^{-1}(s|_{U\cap V})) = \text{div}|_U(t) + \text{div}|_U(\psi_U^{-1}(s|_{U\cap V})) \geq 0$ since $t\in \mathscr{O}_X(\text{div(s)})(U)$ and since $X$ is normal and irreducible. Thus $\mathscr{O}_X(\text{div}(s))|_U \cong \mathscr{O}_X|_{U}$.

For each such $U$ we let $\phi|_U : \mathscr{O}_X(\text{div}(s))(U) \to \mathscr{L}(U)$ be given by $t\mapsto st$ where $st$ is the following. Say $t\in \mathscr{O}_X(W)$ with $W\subseteq U$ then $t|_{V\cap W} s|_{V\cap W} \in \mathscr{L}(V\cap W)$ and $\psi_U^{-1}(t|_{V\cap W} s|_{V\cap W}) = t|_{V\cap W} \psi_{U}^{-1}(s|_{V\cap W}) \in \mathscr{O}_X(V\cap W)$ and which has $\text{div}|_U(t|_{V\cap W} \psi_U^{-1}(s|_{V\cap W})) = \text{div}|_U(t \psi^{-1}_U(s|_{V\cap U}))$ since $t$ and $\psi^{-1}_U(s|_{V\cap U})$ both respectively agree with $t|_{V\cap W}$ and $\psi_U^{-1}(s|_{V\cap U})$ inside of $K(X)$ since $X$ is integral (so restrictions are inclusions). By the above we know $\text{div}|_U(t \psi_U^{-1}(s|_{V\cap U})) \geq 0$ so $st$ makes sense as a section of $\mathscr{L}(U)$. The use of $\psi_U$ is only to verify that $st$ is a good section of $\mathscr{L}(U)$. Thus, for each $U$ on which there exists a trivialization for $L$ the map $\phi_U: \mathscr{O}_X(\text{div}(s))(U) \to \mathscr{L}(U)$ taking $t\mapsto st$ makes sense, where we view $st \in \mathscr{L}_\eta$ with $\overline{\{\eta\}} = X$ and the existence of the trivializaiton over $U$ gives that $st \in \mathscr{L}(U)$.

The inverse to $\phi_U$ is $\phi_U^{-1}: \mathscr{L}(U) \to \mathscr{O}_X(\text{div}(s))(U)$ which takes $v$ to the rational function $\psi^{-1}_U(v)/\psi^{-1}_U(s) \in K(X)$ which is in $\mathscr{O}_U(\text{div}(s))(U)$ since the section $\psi_U^{-1}(v) \in \mathscr{O}_X(U)$ and moreover this element is independent of the trivialization we choose over $U$ since the quotient will remove the unit that the two trivializations differ by, and I think the compatibility with restriction is clear (at least for $\phi_U$ which is enough).

Answer 2

This always confused me - I don't remember the details now, but I remember there's a very good discussion on it in Griffiths and Harris principles of algebraic geometry when they introduce line bundles.

I think it's something along the lines of the $f_i$ that you use for local trivializations are the images of 1 in the sheaf considered as a subsheaf of $K(X)$, and these somehow correspond to $s$ so when you use the trivialization on some function $g$ to get $gf_i\in \mathcal{O}_X$, the $f_i$ is a local model for $s$ (?) so it's saying that $gs$ is integral ie that $g$ has poles bounded by $s^{-1}$.