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missed $y$ in the differential form.
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user36560
user36560

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)dx-2x(x+1)dy$$(3x+2)ydx-2x(x+1)dy$ is killed by $x$ and is non-zero.

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 $(3x+2)dx-2x(x+1)dy$ is killed by $x$ and is non-zero.

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.

added 170 characters in body
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user36560
user36560

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 $(3x+2)dx-2x(x+1)dy$ is killed by $x$ and is non-zero.

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 $\omega\otimes I_p\simeq I_p\omega$.

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 $(3x+2)dx-2x(x+1)dy$ is killed by $x$ and is non-zero.

added 52 characters in body
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user36560
user36560

There are several possible definitions of torsion forAs said the OP in the comments, in $\widehat{O}_{C,p}$, the canonical map $\Omega^1\to \omega$ has a module. Inkernel $\Omega_C$$T$, there can be torsion elementsisomorphic to the vector space generated by (i.e$sdt$. killed by some non-zero elements), but, as yourAs $T$$O_{C,p}\to \widehat{O}_{C,p}$ is skyscraperflat, any differential formthis implies that in your $T$ is necessarily zero because it must be killed by some power$O_C$, the kernel of the maximal ideal $m_p$ defining $p$canonical map (go to the completion$\Omega^1_C\to\omega_{C}$ is a vector space of dimension $O_{C,p}$$1$.

On the other hands, and restrict such a differential form tothe computations in the comments show that the image of $s\ne 0$ and to$\Omega^1_C\to\omega$ is $t\ne 0$)$\omega\otimes I_p\simeq I_p\omega$.

There are several possible definitions of torsion for a module. In $\Omega_C$, there can be torsion elements (i.e. killed by some non-zero elements), but, as your $T$ is skyscraper, any differential form in your $T$ is necessarily zero because it must be killed by some power of the maximal ideal $m_p$ defining $p$ (go to the completion of $O_{C,p}$, and restrict such a differential form to $s\ne 0$ and to $t\ne 0$).

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 $\omega\otimes I_p\simeq I_p\omega$.

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user36560
user36560