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Let $X$ be a singular curve over an algebraic closed field $k$ with characteristic zero. Let $Z$ be the closed subset of singular points on $X$ and $U=X-Z$ be the smooth part, which is an open subset of $X$.

Let $\mathcal{L}$ be a line bundle on $U$. Could we always extend $\mathcal{L}$ to a line bundle on $X$, i.e. could we find a line bundle $\widetilde{\mathcal{L}}$ on $X$ such that $\widetilde{\mathcal{L}}|_U\cong \mathcal{L}$? If not, do we have counter examples?

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The answer is 'yes'. One way to argue this is to first find a Cartier divisor $D$ on $U$ whose associated line bundle is $\mathcal{L}$ (the existence of such a divisor is ensured, for instance, by [EGA IV$_4$, 21.3.4 a)]), extend $D$ to a Cartier divisor $\widetilde{D}$ on the whole $X$ (e.g., by applying [EGA IV$_4$, 21.9.4]), and then let $\widetilde{\mathcal{L}}$ be the line bundle associated to $\widetilde{D}$.

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  • $\begingroup$ This may be a long shot since it's an old question, but I was wondering if you have an alternative source for the EGA IV 21.9.4 result? My French isn't as good as I'd like it to be and I'm really struggling to find the result actually written anywhere else. $\endgroup$
    – Luke
    Commented May 2, 2018 at 4:54
  • $\begingroup$ I don't know another reference, but I think the existence of an extension may be argued directly. Namely, a Cartier divisor is a quasi-coherent ideal sheaf $\mathscr{I} \subset \mathscr{O}_U$ locally generated by a nonzero divisor, so $\mathscr{O}_U/\mathscr{I}$ vanishes at the generic points and hence is supported at a finite set of closed points $D \subset U$. Now, $D$ is also closed in $X$ (it contains no generic point), so we may extend the ideal $\mathscr{I}$ by glueing it with $\mathscr{O}_{X - D}$ over $X - D$. The extension is still locally generated by a nonzero divisor. $\endgroup$
    – Kestutis Cesnavicius
    Commented May 2, 2018 at 15:16
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A more direct approach is the following. Let $U=U_0\cup \dots \cup U_r$ be an open cover of $U$ such that $\mathscr L\left|_{U_i}\right.\simeq \mathscr O_{U_i}$ for all $i=0,\dots,r$. Define $X_0:=U_0\cup Z$, $X_i=U_i$ for $i>0$ and let $\overline{\mathscr L_i}:= \mathscr O_{X_i}$. Now glue $\overline{\mathscr L_i}:= \mathscr O_{X_i}$ together by the gluing data of $\mathscr L$ on $U_i\cap U_j=X_i\cap X_j$ (assume that $i\neq j$).

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    $\begingroup$ In this case is it necessary that $X_0$ is an open subset of $X$? $\endgroup$
    – Zhaoting Wei
    Commented May 5, 2015 at 1:18
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    $\begingroup$ On a curve a non-empty subset is open if and only if its complement is finite. $X_0$ contains the open subset $U_0$ and hence the complement of $X_0$ is contained in the complement of $U_0$, which is finite. OK? $\endgroup$
    – Sándor Kovács
    Commented May 6, 2015 at 18:24

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