Lemma 53.6.4. In Situation 53.6.2. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module which is globally generated and not isomorphic to $\mathcal{O}_ X$. Then $H^1(X, \omega _ X \otimes \mathcal{L}) = 0$.

Proof. By duality as discussed in Section 53.5 we have to show that $H^0(X, \mathcal{L}^{\otimes - 1}) = 0$. If not, then we can choose a global section $t$ of $\mathcal{L}^{\otimes - 1}$ and a global section $s$ of $\mathcal{L}$ such that $st \not= 0$. However, then $st$ is a constant multiple of $1$, by our assumption that $H^0(X, \mathcal{O}_ X) = k$. It follows that $\mathcal{L} \cong \mathcal{O}_ X$, which is a contradiction. $\square$

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