Cohomology and normalization of a curve Suppose $C$ is a projective curve such that $H^1(C,\mathcal{O}_C)=0$ and let $\pi:\tilde{C}\longrightarrow C$ be the normalization morphism. How can I show that $H^1(\tilde{C},\pi^\ast\mathcal{O}_C)=0$? Can somebody please help with this?
Thanks a lot!
 A: Let me make up the details. As Alex said, $\pi$ is a finite morphism, hence it is affine morphism. Thus we have $H^1(\tilde{C},\pi^*\mathcal{O}_C)=H^1(C,\pi_*(\pi^*\mathcal{O}_C\otimes\mathcal{O}_\tilde{C}))=H^1(C,\pi_*\mathcal{O}_\tilde{C})$. This can be shown by using Leray spectral sequence $E_2^{pq}=H^p(C,R^q\pi_*(\pi^*\mathcal{O}_C))\Longrightarrow H^{p+q}(\tilde{C},\pi^*\mathcal{O}_C)$, since $\pi$ is finite, all the higher direct image vanish, thus the second page degenerates. We have $H^1(\tilde{C},\pi^*\mathcal{O}_C)=H^1(C,\pi_*(\pi^*\mathcal{O}_C\otimes\mathcal{O}_\tilde{C}))$.
Then we use the short exact sequence for normalization:$0\rightarrow \mathcal{O}_C\rightarrow \pi_*\mathcal{O}_\tilde{C}\rightarrow \oplus_{p\in C}\mathcal{\tilde{O}}_p/\mathcal{O}_p\rightarrow 0$. You know, after taking the long exact sequence, you have $H^1(C,\mathcal{O}_C)\rightarrow H^1(C,\pi_*\mathcal{O}_\tilde{C})\rightarrow 0$. Then you will get what you want. i.e $H^1(\tilde{C},\pi^*\mathcal{O}_C)=H^1(C,\pi_*\mathcal{O}_\tilde{C})=0$.
