Let $C$ be a connected, reduced but reducible projective curve over an algebraically closed field. Denote by $\pi:\tilde{C}\to C$ its normalisation. There are two standard references for computing $J(C)=\rm{Pic}^0(C)$, namely Harris-Morrison's 'Moduli of Curves' section 5.B and Bosch-Lütkebohmert-Raynaud's 'Néron Models' section 9.2.
I am happy with the computations of the Jacobian for the node, the cusp and for compact type curves, but am experiencing some confusion applying the occasionally unclear text of the above references to more complicated cases.
Take for example the case of three distinct smooth rational curves meeting at one point - i.e. a star-shaped cubic in $\mathbb{P}^2$. The arithmetic genus is one (as it's a plane cubic) whereas the normalisation consists of three disjoint $\mathbb{P}^1$'s. For any $C$ as in the first paragraph we have an induced exact sequence $$0\to\mathcal{O}^*_C\to\pi_*\mathcal{O}^*_{\tilde{C}}\to F\to0$$ where $F$ is a torsion sheaf supported on the singular locus of $C$. Harris-Morrison in exercise 5.11(2) give a formula for computing how many $\mathbb{G}_m$ and $\mathbb{G}_a$ factors the sheaf $F$ comprises of. In their notation, if $C$ is the star-shaped cubic with singularity $p$, one finds $b=2$ as the number of branches is 3, whereas $\dim(\pi_*\mathcal{O}^*_{\tilde{C}}/\mathcal{O}^*_C)_p$ should be ''the contribution at $p$ to the difference in arithmetic genus between $C$ and $\tilde{C}$''. This would imply $a+b=1$ meaning $a=-1$ which is already confusing and my first point in need of clarification.
Moving on under the assumption $a=0$ though, the cohomology of the above sequence gives $$0\to\mathbb{G}_m\to\mathbb{G}_m^3\to\rm{H}^0(F)\to J(C)\to J(\tilde{C})\to 0.$$ $J(\tilde{C})=0$ as it consists of disjoint rational curves whereas the above computation gives $\rm{H}^0(F)=\mathbb{G}_m^2$. This leads to $J(C)=0$ by a dimension count from the sequence. This is at odds with my expectation that the Jacobian of $C$ should be one dimensional.
Some further comments:
A further way to see that $\rm{H}^0(F)=\mathbb{G}_m^2$ in the above example is the following. The stalk at $p$ of the first short exact sequence above is as follows: the middle group is generated by three functions $f_1,f_2,f_3$, each defined on a neighbourhood of a preimage of $p$ on each of the $\mathbb{P}^1$'s of the normalisation, and it contains $\mathcal{O}^*_{C,p}$ those functions so that $f_1(p)=f_2(p)=f_3(p)$. The quotient $F_p$ should then be given by $(\frac{f_1(p)}{f_2(p)}, \frac{f_2(p)}{f_3(p)})$ ie $\mathbb{G}_m^2$ - although there may be a mistake here.
Unraveling Propositions 9 and 10 (and the definition of the intermediate curve $X'$) in the 'Néron Models' section, one expects copies of $\mathbb{G}_a$ to appear in $F$ whenever the singularity admits a resolution which is a homeomorphism (eg a cusp) in the sense that there's only one preimage. Whenever this is not the case, we expect a copy of $\mathbb{G}_m$ in $F$ for the number of branches at the singularity, minus 1 (the long sequence in the middle of p249).
