# Could we extend any line bundle on the smooth part of a singular curve to a line bundle on the whole curve?

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?

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 $\widetilde{D}$.
• 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. – Kestutis Cesnavicius May 2 '18 at 15:16
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 $i\neq j$).
• In this case is it necessary that $X_0$ is an open subset of $X$? – Zhaoting Wei May 5 '15 at 1:18
• 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? – Sándor Kovács May 6 '15 at 18:24