dualizing sheaf of a nodal curve I'm trying to understand the dualizing sheaf $\omega_C$ on a nodal curve $C$, in particular why is $H^1(C,\omega_C)=k$, where $k$ is the algebraically closed ground field.  I know this sheaf is defined as the push-forward of the sheaf of rational differentials on the normalization $\tilde{C}$ of $C$ with at most simple poles at the points lying over the nodal points of $C$ and such that the sum of residues at the two points lying over the node will be zero.  I can show that this is indeed an invertible sheaf on $C$, but I have no clue, despite my many attempts, how to show that $H^1(C,\omega_C)=k$.  I've been able to show it in some very simple cases using Cech cohomology, but can someone explain to me how to do it in general?
 A: If $\tilde{C}$ is the normalization, with two points $x$ and $y$ being identified under the map $\pi: \tilde{C} \to C$ to the node $z$ of $C$, then we have an exact sequence 
$$0 \to \Omega^1_{\tilde C} \to \Omega^1_{\tilde C}(x + y) \to k_x \oplus k_y \to 0,$$
where $k_x$ and $k_y$ are the skyscraper sheaves at the points $x$ and $y$.
Pushing forward (which is exact because the map $\pi$ is finite, and so in particular affine)
we get an exact sequence
$$0 \to \pi_* \Omega^1_{\tilde C} \to \pi_*\Omega^1_{\tilde C}(x+y) \to k_z^{\oplus 2} \to 0.$$
Now there is a short exact sequence $0 \to k_z \to k_z^{\oplus 2} \to k_z \to 0$,
where the third arrow is just given by adding the two components, and 
$\omega_C$ is the preimage of (the first copy of) $k_z$ under the surjection
$\pi_* \Omega^1_{\tilde C}(x+y) \to k_z^{\oplus 2}$.
In conclusion, we have an exact sequence
$$0 \to \pi_* \Omega^1_{\tilde C} \to \omega_{C} \to k_z \to 0.$$
Now taking cohomology (and recalling that $H^i(C,\pi_*\mathcal F) = H^i(\tilde{C},\mathcal F)$ for
a coherent sheaf on $\tilde{C}$), we obtain
$$0 \to H^0(\tilde{C},\Omega^1_{\tilde C}) \to H^0(C,\omega_C) \to
H^0(C,k_z) \to H^1(\tilde{C},\Omega^1_{\tilde C}) \to H^1(C,\omega_C) \to 0.$$
(The point here being that $H^1$ of a skyscraper sheaf such as $k_z$ vanishes.)
I claim that in this exact sequence the map $H^1(\tilde{C},\Omega^1_{\tilde C})
\to H^1(C,\omega_C)$ is an isomorphism, and hence that the latter is one-dimensional, since the former is.
For this, it is equivalent to show that the map
$H^0(C,\omega_C) \to H^0(C,k_Z) = k$ is surjective.
Now $H^0(C,\omega_C) \subset H^0(C,\pi_*\Omega^1_C(x+y)) = H^0(\tilde{C},\Omega^1(x+y)).$
The residue theorem shows that we may find a differential $\omega \in
H^0(\tilde{C},\Omega^1(x+y))$ whose residues at $x$ and $y$ are non-zero.  (These residues
are then negative to one another.)  Thought of as a section of
$H^0(C,\pi_*\Omega^1_C(x+y))$, this differential $\omega$ clearly lies in
$H^0(C,\omega_C)$.  Its image under the map $H^0(C,\omega_C)$ is non-zero (equal to
the residue at either $x$ or at $y$, depending on a choice that was implicitly made above),
and so indeed $H^0(C,\omega_C) \to k$ is surjective.
Summary: The residue theorem guarantees the existence of sections of $H^0(C,\omega_C)$
which have non-zero residues at $x$ and $y$ when pulled back to $\tilde{C}$, and
this in turn shows that $H^1(C,\omega_C)$ is isomorphic to $H^1(\tilde{C},\Omega^1_C)$,
and hence is one-dimensional.
