# Extension of line bundle defined over an open subscheme

Let $$X$$ be a normal projective (or, quasi-projective) variety over $$\mathbb{C}$$. Let $$U \subset X$$ be an open subscheme whose complement $$Z = X \setminus U$$ has codimension at least $$2$$ in $$X$$. Let $$L$$ be a line bundle on $$U$$. Is it possible to extend $$L$$ to a line bundle $$\widetilde{L}$$ on $$X$$ such that $$\widetilde{L}\vert_U \cong L$$?

Edit: If this is true, can we get $$\textrm{Pic}(X) \cong \textrm{Pic}(U)$$?

As explained here Extending vector bundles on a given open subscheme the only possible such extension is $$\tilde{L} = (i_*L)^{\vee\vee},$$ where $$i \colon U \to X$$ is the embedding. The sheaf $$\tilde{L}$$ is a reflexive sheaf of rank 1 on $$X$$. So, if $$X$$ is locally factorial (i.e., every Weil divisor on $$X$$ is Cartier) then $$\tilde{L}$$ is a line bundle. Otherwise, this is not necessarily true. For example, let $$X$$ be the 2-dimensonal quadratic cone, $$U$$ its smooth locus, and $$\tilde{L}$$ the ideal of a line on $$X$$. Then $$L = i^*\tilde{L}$$ is a line bundle, but its unique reflexive extension $$\tilde{L}$$ is not.
For any normal variety one has $$Cl(U) = Cl(X)$$, so you get $$Pic(U) = Pic(X)$$ if and only if $$X$$ is locally factorial along the complement $$X \setminus U$$.
• I trust your argument but don't understand one thing: why would not Pic remain unchanged if all (possible, arbitrarily bad) singularities of $X$ are inside $U$? – მამუკა ჯიბლაძე Nov 20 '18 at 5:17
• @მამუკაჯიბლაძე: you are completely right, my last sentence was incorrect (I presumed that $U$ is the smooth locus of $X$), but now I fixed that. – Sasha Nov 20 '18 at 19:03
• I see, thank you. Actually I only now noticed that the question is about normal $X$, so that all singularities would be of codimension $>1$ anyway... – მამუკა ჯიბლაძე Nov 20 '18 at 19:46