Ramification of Space Curves on Singular Surfaces

Question

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!

Answer 1

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 $x$.