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This question refers to the proof of Proposition 2.3/IV in Hartshorne, page 301. Let $R$ be the ramification divisor associated to a finite, separable morphism of curves $f:X \rightarrow Y$. Why is it true that $\Omega_{X/Y} \otimes \Omega_{X}^{-1} = O_R$?

Edited: More specifically the proof starts by tensoring the sequence $0 \rightarrow f^* \Omega_Y \rightarrow \Omega_X \rightarrow \Omega_{X/Y} \rightarrow 0$ with $\Omega_X^{-1}$ and getting $0 \rightarrow f^* \Omega_Y \otimes \Omega_X^{-1} \rightarrow O_X \rightarrow \Omega_{X/Y} \otimes \Omega_X^{-1} \rightarrow 0$. But instead of $\Omega_{X/Y} \otimes \Omega_X^{-1}$ he writes $O_R$. This is what i don't understand.

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I don't see where is he making such a claim. It looks that he claims $\mathcal{O}_R \simeq \Omega_{X/Y}$ and $f^*\Omega_Y \otimes \Omega_X^{-1} \simeq \mathcal{L}(-R)$. Am I missing something obvious? –  Giovanni De Gaetano Dec 5 '12 at 15:24
    
@GiovanniDeGaetano: He tensors the sequence $0 \rightarrow f^* \Omega_Y \rightarrow \Omega_X \rightarrow \Omega_{X/Y} \rightarrow 0$ with $\Omega_X^{-1}$ and he gets $0 \rightarrow f^* \Omega_Y \otimes \Omega_X^{-1} \rightarrow O_X \rightarrow \Omega_{X/Y} \otimes \Omega_X^{-1} \rightarrow 0$. But instead of $\Omega_{X/Y} \otimes \Omega_X^{-1}$ he writes $O_R$. This is what i don't understand. –  Manos Dec 5 '12 at 15:28
    
Ok, I see! Now I understand the question. :) It looks tough, because it seems to be in open contradiction with the claim he makes a few rows above and which I reported. I'm gonna think about it! –  Giovanni De Gaetano Dec 5 '12 at 15:31
    
@GiovanniDeGaetano: Where is he claiming what you said? I don't see it. –  Manos Dec 5 '12 at 15:32
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@GiovanniDeGaetano: Is there a possibility that $\Omega_{X/Y} \otimes \Omega_X^{-1} \cong \Omega_{X/Y}$? –  Manos Dec 5 '12 at 16:17

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