# Kaehler differentials on a nodal curve

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). 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?

• hey you should check the related questions! look at this: mathoverflow.net/questions/58559/… – IMeasy Jul 6 '13 at 12:22
• Admittedly the title is unfortunately the same. But as far as I can tell the question is completely different! I want to know if the sheaf of Kaehler differentials is torsion free for an integral curve, and if the formula $\Omega_C \simeq \omega_c \otimes I_p$ is really true. The related question was about how to define the line bundle $\omega_C$. My question is perhaps more about the Kaehler differentials than the dualizing sheaf (and I have now changed the title!). – mkemeny Jul 6 '13 at 17:10
• You are right that Kaehler differential sheaf has torsion. I also do not understand their formula. What is ideal sheaf here? – Alexander Chervov Jul 6 '13 at 17:43
• I think everything is explained by the fact that your exact sequence is backwards. There is an exact sequence $0\rightarrow I_p\rightarrow \mathcal{O}\rightarrow T\rightarrow 0$ which when you tensor with $\Omega_C$ gives the result. – Philip Engel Jul 7 '13 at 14:24
• @PhilipEngel: My exact sequence should be in the correct order. $T_p$ is a subsheaf of $\Omega_p$, generated by the torsion form $tds$. See the discussion in the comments below. – mkemeny Jul 7 '13 at 14:46

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.
• @AlexanderChervov: it is the $m_p$ in my answer above. The stalk $\omega_p$ is generated by $ds/s$, and $\Omega$ is generated by $ds=s.(ds/s)$ and $dt=-t.(ds/s)$, so $\Omega_p=(s,t)\omega_p=m_p \omega_p$. – user36560 Jul 7 '13 at 10:40
• Locally anaytically about $p$, $T_p$ is generated by the torsion differential $tds$. Since we have $st=0$ this cannot be a multiple of $ds/s$, so $\Omega_p \neq m_p\omega_p$, at least in the analytic case. – mkemeny Jul 7 '13 at 10:59