Suppose $C$ is an integral nodal curve with one node. It is claimed in the arxiv version of a paper by Bogomolov, Hassett, Tschinkel that the dualizing sheaf and the sheaf of differentials are related by the formula $$\Omega\_C \simeq \omega\_C \otimes I\_p,$$ where $I\_p$ is the ideal sheaf of a node: ([see p.10](http://arxiv.org/pdf/0907.3527.pdf)). Is this true? I was under the impression that $\Omega\_C$ has torsion at the point $p$, and that there was an exact sequence $$ 0 \to T \to \Omega\_C \to \mu\_\* \omega\_{C'} \to 0,$$
where $\mu: C' \to C$ is the normalization morphism and $T$ is a sky-scraper sheaf supported at $p$ (proof: argue locally analytically and assume $C$ is given by $st=0$). Yet we have $\mu\_\* \omega\_{C'} \simeq \omega\_C \otimes I\_p$, so this would appear to be a contradiction. 


What am I doing wrong?



Kaehler differentials on a nodal curve
As said the OP in the comments, in $\widehat{O}\_{C,p}$, the canonical map $\Omega^1\to \omega$ has a kernel $T$, isomorphic to the vector space generated by $sdt$. As $O\_{C,p}\to \widehat{O}\_{C,p}$ is flat, this implies that in $O\_C$, the kernel of the canonical map $\Omega^1\_C\to\omega\_{C}$ is a vector space of dimension $1$. 


On the other hands, the computations in the comments show that the image of $\Omega^1\_C\to\omega$ is $\omega\otimes I\_p\simeq I\_p\omega$. 


**Edit.** A concrete example with the integral nodal curve defined by $y^2=x^2(x+1)$. The differential form $(3x+2)ydx-2x(x+1)dy$ is killed by $x$ and is non-zero. 


