Picard group and jacobian of a singular curve

Question

I found somewhere written that the jacobian of a reducible curve is the product of the jacobians, which seems quite reasonable to me. Where can I find some proof of it and a description of the Picard group? What can we say if we admit singularities in general?

Answer 1

This is a long comment.

Definition. A projective nodal curve $C$ is said to be of compact type if one of the following equivalent conditions holds:

• Its dual graph is a tree.
• Its Jacobian is compact.
• Every node is disconnecting, i.e. $C\setminus \{P\}$ is disconnected for every node $P$.
• Every connected component of $\textrm{Pic}^d(C)$ is proper.

To address one of the comments, I would like to say: life is "much easier" when the curve is of compact type (e.g. your union $C=E\cup F$ is).

The nice issue about curves $C=X_1\cup\dots\cup X_r$ of compact type is that, for all $d\geq 0$, $$\textrm{Pic}^d(C)=\coprod_{d_1+\dots+d_r=d}\textrm{Pic}^{d_1}(X_1)\times \dots\times \textrm{Pic}^{d_r}(X_r).\,\,\,\,\,\,\,\,\,\,\,\,(\star)$$ In particular, $$\textrm{Pic}^0(C)=\prod_{i=1}^r\textrm{Pic}^0(X_i).$$ In other words, specifying a line bundle of degree $d$ on a curve of compact type boils down to fixing a "multidegree" $(d_1,\dots,d_r)$ and to choosing a line bundle of degree $d_i$ on each component $X_i$. This is really special to curves of compact type (For instance, a union of two elliptic curves meeting in two points is not so well behaved).

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Two more words on the compact type condition: Why is this really a nice condition to ask when dealing with line bundles?

Suppose $C$ is a projective nodal curve, appearing as special fiber $f^{-1}(0)$ of a one-parameter family of curves $f:X\to B$, where $X$ is a smooth surface. Suppose that over the locus $U=B\setminus \{0\}$ the family is smooth, and we have a family of linear series $L_U$ on $X_U=X|_{f^{-1}(U)}$. $$Question. \textrm{Does $L_U$ extend to a linear series on the whole of }X\textrm{?}$$ It seems like there is no problem: as $X$ is smooth, divisors extend, so line bundle extend. No assumption on $C$ is needed! However, if for some reason we wanted to base change the family, the total space of the new family would not be smooth any longer, in general. Here is where an assumption on $C$ is needed: if $C$ is of compact type, then, even after base change, line bundles always extend. (Maybe not uniquely, as the Picard scheme $\pi:\textrm{Pic}(X/B)\to B$ is not proper, because of $(\star)$).