Let $X \subset \mathbb{P}^3$ be a singular surface. For simplicity take $X$ to be a cone over a smooth conic. And let $C \subset X$ be a curve passing through the singular point, with normalization morphism
\begin{equation\*}
\nu: \tilde{C} \rightarrow C.
\end{equation\*}


My question is: how can we determine the degree of the ramification divisor of the morphism $\nu$?


And specifically, in the case of a quadric, let's just consider the case where $\tilde{C} \subset \tilde{X}$, where $\pi: \tilde{X} \rightarrow X$ is the minimal resolution of $X$ and $\tilde{C}$ intersects the exceptional $(-2)$-curve transversely. Can we calculate the degree of ramification in this case?


My thought was to look at the exact sequence
\begin{equation\*}
0 \rightarrow \Omega\_X \rightarrow \Omega\_X^{\vee \vee} \rightarrow Q \rightarrow 0,
\end{equation\*}
where the cokernel $Q$ is supported at the singular point and then pull this back via $\pi$
\begin{equation\*}
\pi^\* \Omega\_X \rightarrow \pi^\*(\Omega\_X^{\vee \vee}) \rightarrow \pi^\* Q \rightarrow 0.
\end{equation\*}


and then restrict this to $\tilde{C}$. So to make this work, I need to calculate $Q$ and $\pi^\* Q$. Is there a good reference for this calculation? Or a better way to understand the ramification more directly?


Thanks!



Ramification of Space Curves on Singular Surfaces
I think there is no ramification in the case you are interested in. 


The surface $\tilde X$ is the minimal ruled surface $\mathbb F\_2$. The map $\tilde X \to X$ contracts the $-2$-curve $E$ to a point and maps the fibers of the ruling of $\mathbb F\_2$ to lines in $\mathbb P^3$. So at every point of $E$ the map has differential of rank 1 and the kernel of the differential is the tangent space to $E$, because $E$ is contracted.


If $E$ intersects $\tilde C$ transversally, 
by the previous remarks, $\tilde C \to C$ has non zero differential at $x$. 


