# Are degrees and ramification degrees preserved upon passing to the smooth compactification?

Let $$\phi ：C_1\to C_2$$ be morphism of projective singular curve. Let $$\tilde{C}_1$$ and $$\tilde{C}_2$$ be their smooth compactification. Then $$\phi$$ extends to $$\tilde{\phi} : \tilde{C}_1\to \tilde{C}_2$$. Let $$\deg \phi$$ be degree of $$\phi$$ and $$e_\phi (P)$$ be ramification degree at $$P \in C_1$$.

Does $$\deg \phi=\deg \tilde\phi$$ and $$e_{\phi} (P)=e_{\tilde \phi} (P)$$ hold?

Background. If this is true, we can easily calculate the genus of $$C_1$$ from he genus of $$C_2$$ by applying the Riemann–Hurwitz formula to $$\tilde{\phi}$$.

N.B. Sorry, my link was wrong, the meaning of smooth compactification is in this paper of page $$2$$, https://arxiv.org/pdf/2112.02470.pdf

• I did some tidying, including trying to write a descriptive title rather than just the first two sentences of notation. I hope that was all right. I also edited $\tilde{C_1}$ \tilde{C_1} to $\tilde{C}_1$ \tilde{C}_1, which I think looks better; but, if you do prefer the tilde to be placed over everything, then perhaps $\widetilde{C_1}$ \widetilde{C_1} looks better. Finally, I had trouble understanding your "cf.", which I would usually take to mean "compare". Is that what is meant here? Apr 6 at 18:45
• @masquerade The definition in the Wikipedia link is only for a smooth curve. Apr 7 at 10:34

The first claim is true. The degree can be defined as the degree of the extension on function fields $$k(C_1)/k(C_2)$$, but $$k(\tilde{C}_1) = k(C_1)$$ and $$k(\tilde{C}_2)= k(C_2)$$ so $$k(C_1)/k(C_2) = k(\tilde{C}_1)/k(\tilde{C}_2)$$.
The second claim is true for smooth points lying over smooth points. This is because it is clear from the definition of ramification degree that it may be computed locally (say in the Zariski topology) and $$C_1$$ is isomorphic to $$\tilde{C}_1$$ locally in the Zariski topology around smooth points. I am not sure if one can define the ramification degree for singular points.
We would normally refer to what you call the "smooth compactification" as the "normalization" or maybe "resolution of singularities" since it's not the compactification of $$C$$ but rather of a smooth open subset of $$C$$.
• The meaning of Smooth completion is en.wikipedia.org/wiki/Smooth_completion. Smooth completion contains $C$ as a subset. Apr 7 at 2:58