Dualizing sheaf of nodal genus two curve

I'm interested in extending Riemann-Roch and Serre duality to bundles or sheaves defined on nodal curves. The extension of Riemann-Roch I think is straightforward: simply replace the geometric genus by the arithmetic genus. Unless I'm mistaken, Serre duality can be extended to nodal curves by replacing the canonical bundle with the dualizing sheaf. I've never worked with a dualizing sheaf concretely, so I was hoping someone could help me in one very explicit example.

Let $C$ be the nodal curve given as the glueing of two elliptic curves at their origins. One can think of this as a particular kind of genus two curve. Obviously, the normalization is just the two disjoint elliptic curves. But what exactly is the dualizing sheaf $\omega_{C}$ and how does one carry out the computation?

First, simple nodal curves are locally planar and so the dualizing sheaf is a line bundle. So, if the elliptic curves are $E_1,E_2$, one has an exact sequence $0\to \mathcal{O}_C\to \mathcal{O}_{E_1}\oplus\mathcal{O}_{E_2}\to k(p)\to 0$, where $p$ is the origin. Dulaizing, you get, $0\to\mathcal{O}_{E_1}\oplus\mathcal{O}_{E_2}\to \omega_C\to k(p)\to 0$. Most calculations you need can be made using this sequence.
• Thanks, this helps a lot. For an example calculation, let's say I wanted to compute the first Chern class of $\omega_{C}$. Using your exact sequence, I seem to get $c_{1}(\omega_{C})=0$, by the additivity of $c_{1}$ on SES. Is this right? I'm a little unsure if by $\mathcal{O}_{E_{1}} \oplus \mathcal{O}_{E_{2}}$ you really mean some pushforward. So maybe my argument is flawed. – Benighted Sep 2 '17 at 19:22
• Chern class is computed where? For a good theory for non-irreducible varieties, these are not numbers anymore. For a baby case, think of two disjoint curves, then Chern class is a pair of numbers. $\mathcal{O}_{E_i}$ are quotients of $\mathcal{O}_X$ and thus sheaves on $X$. – Mohan Sep 2 '17 at 20:29
• I see, I guess I just naively tried to apply $c_{1}$ to the short exact sequence you gave. Is there a notion of the "degree of $\omega_{C}$" analogous to that on a smooth curve? In applying Serre duality to a Riemann-Roch computation, I would hope to argue something like $h^{0}(C, L^{\vee} \otimes \omega_{C})=0$ if the degree of $L^{\vee} \otimes \omega_{C}$ is negative, or something. – Benighted Sep 2 '17 at 21:10
• Degree makes perfect sense for line bundles, if you think of it as "total degree". In the above, $\deg \omega_C=2$. One way of defining it is of course (as Mumford does it) for a line bundle $L$ on a projective curve $X$, $\deg L=\chi(L)-\chi(\mathcal{O}_X)$. – Mohan Sep 2 '17 at 22:21
• Am I correct to say that if we added a third elliptic curve glued nodally to one of the previous two, we would get $\text{deg}\omega_{C}=4$? And if you added yet another you would get degree 6 I believe. – Benighted Sep 3 '17 at 1:41