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 $\textrm{Pic}(X) \cong \textrm{Pic}(U)$?


1 Answer 1


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 $X \setminus U$.

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    $\begingroup$ 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 $U$? $\endgroup$ Nov 20, 2018 at 5:17
  • $\begingroup$ @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. $\endgroup$
    – Sasha
    Nov 20, 2018 at 19:03
  • $\begingroup$ 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... $\endgroup$ Nov 20, 2018 at 19:46

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