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I am studying Cartier divisors, and I am confused about exactly how they correspond to rational sections of a line bundle, or what a rational section of a line bundle even is.

Let $X$ be an integral scheme with generic point $\eta$. Let $\mathscr{L}$ be an invertible sheaf. What is the precise definition of a rational section in this context? I have read the following three:

1) A section $s$ of $\mathscr{L}$ over a dense open set (since $X$ is integral, this is just any open set),

2) An equivalence class of sections of $\mathscr{L}$ over open sets with the obvious equivalence relation,

3) A global section $s \in \Gamma(X, \mathscr{L} \otimes_{\mathcal{O}_{X}} \mathcal{K})$, where $\mathcal{K}$ is the constant sheaf of rational functions on $X$.

I have been operating under the assumption of the second definition, which seems to me to just be saying that it's an element of the stalk $\mathscr{L}_{\eta}$ at the generic point. Is this correct? I also want to be able to say that 2) and 3) are equivalent, but I am not able to actually prove this. I was going to use the universal property for sheafifiation of the tensor product sheaf, but that argument seems to be broken because the line bundle may not have any global sections.

This is a precursor question to a question about Cartier divisors, but I wanted to clarify this first to make sure I can frame the question properly and iron out any confusion before I ask it.

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    $\begingroup$ In the affine case, say $X = Spec(R)$, then $\mathscr L$ corresponds to an invertible $R$-module, say $L$. Then $L \otimes_R K$ is exactly $L_{\eta}$ by property of localization so $2$ and $3$ coincide. It also coincide with $1$ since a section of $K$ is exactly a section defined on a non-empty open set. Also it's not true that $H^0(F) \otimes H^0(G) \cong H^0(F \otimes G)$, so even if $F$ has no global sections $F \otimes G$ might have a global section. $\endgroup$ Feb 12, 2019 at 9:32

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