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Assume $S$ is a normal surface (so 2-dimensional $k$-scheme). Let $\mathcal{L}, \mathcal{N}$ be zwo invertible sheaves on $S$ such that they coinside on a dense open subset $U \subset S$; so $\mathcal{L} \vert_U= \mathcal{N} \vert _U$.

Why does this already imply that $\mathcal{L}=\mathcal{N}$on whole $S$?

My considerations: Since $\mathcal{L} \otimes \mathcal{N}^{\vee}$ is also invertible and therefore wlog $\mathcal{L}= O_S$.

Assume that $\mathcal{N} \vert_U= O_U$. How does this equality extend to whole $S$?

My ideas: for every $U \subset V$ we have diagram

$$ \require{AMScd} \begin{CD} O_S(V) @>{a_V???} >> \mathcal{N} \vert_V \\ @VVcanV @VVcanV \\ O_S(U) @>{\cong}>> \mathcal{N} \vert_U; \end{CD} $$

and we want that this diagram extends to $a_V$ and futhermore it is an isomorphism. From here I'm stuck. How to use here the normality condition to obtain the desired result?

user267839
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    Any such $N$ is isomorphic to the trivial bundle on a suitable dense open set, so unless you put restrictions on $U$, this will not be true. – Mohan May 28 '19 at 02:22
  • Are you sure this is true? What about $\mathbb{P}^2$ and $\mathcal{O}$ and $\mathcal{O}(1)$? They are isomorphic over $U=\mathbb{A}^2$. I think you need complement of $U$ to have codimension $2$. This may be what you are thinking of: https://math.stackexchange.com/questions/1258398/extending-regular-function-on-normal-variety-from-a-subvariety-of-codimension-2 – usr0192 May 28 '19 at 02:23

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