In Vakil's chapter 14 of "The Rising Sea", he talks about a map from invertible sheaves to divisors. Specifically, for a Noetherian irreducible (can be relaxed apparently) scheme regular in codimension one, it makes sense to talk about the divisor of a rational section $s$ of an invertible sheaf $\mathcal{L}$. He then has a map:
$$\{(\mathcal{L}, s)\}/ \text{isomorphism} \rightarrow \operatorname{Weil} X$$
I am not sure what this isomorphism of a pair $(\mathcal{L}, s)$ is supposed to mean. One way to interpret it is that two such pairs are isomorphic if there is sheaf isomorphism which takes one section to another. But I don't think that's all there is to it, because we also have the case when we can extend a rational section to a bigger open set and in that case it is still supposed to be from the same isomorphism class. So what is it?
Another way to think of it was along the lines of rational functions. And I haven't studied the non-irreducible case but in case of irreducible scheme rational functions have a natural notion of "isomorphism" in the sense that they are same if they map to the same element in the function field. Is there a similar notion for invertible sheaves? Do we similarly have an injective map from regular sections on open sets to the stalk of the invertible sheaf at the generic point?
Edit: In the last paragraph, I mean function field of an integral scheme, not just irreducible.
Best Answer
You're close - the correct notion of isomorphism of a pair $(\mathcal{L},s)$ here is that there's an isomorphism $\varphi:\mathcal{L}_1\to\mathcal{L}_2$ under which $\varphi(s_1)$ and $s_2$ agree on a dense open subset (sometimes we say that "they're equal as rational sections"), which is equivalent to agreeing at the generic point.
For your final paragraph, you're not too far away either. Suppose $\eta$ is the generic point of $X$ and $\mathcal{L}$ is an invertible sheaf. Then $\mathcal{L}_\eta \cong \mathcal{O}_{X,\eta}$: since stalks can be calculated from any open neighborhood, simply pick an open set where $\mathcal{L}\cong\mathcal{O}_X$ as it's locally free. On the other hand, it's not true that $\mathcal{O}_X(U)\to\mathcal{O}_{X,\eta}$ is an injection without the condition that $X$ is reduced - think about adding some nilpotent fuzz on a closed subset. For instance, $\operatorname{Spec} k[x,y]/(x^2,xy)$ is $\operatorname{Spec} k[y]$ with fuzz at the origin, and indeed $x\neq 0$ is in the kernel of the map from the global sections to the stalk at the generic point.