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Suppose $X$ is a separated, projective curve, and $Z = V(\mathcal{J})$ (where $\mathcal{J}$ is a nilpotent sheaf of ideals) is a closed subscheme of $X$ with closed immersion $i : Z \to X$.

Is it true that $H^{1}(Z,\mathcal{O}^*_Z) \cong H^{1}(X, i_*\mathcal{O}^*_Z)$ (in a canonical way?).

I stumbled across this in the proof of Liu, Lemma 7.5.11, where he seems to use this to get from the exact sequence $$1 \to 1 + \mathcal{J} \to \mathcal{O}_X^* \to \mathcal{O}_Z^* \to 1$$ the exact sequence of cohomology groups $$\mathcal{O}_X(X)^* \to \mathcal{O}_Z(Z)^* \to H^{1}(X, 1 + \mathcal{J}) \to \text{Pic}(X) \to \text{Pic}(Z) \to 0.$$ (He assumes here that $\mathcal{J}^2 = 0$.)

Now, for a quasi-coherent sheaf $\mathcal{F}$ on $Z$, we have $H^{1}(Z,\mathcal{F}) \cong H^{1}(X, i_*\mathcal{F})$ by Liu, Ex 5.2.3, but to my understanding, $\mathcal{O}_Z^*$ is not quasi-coherent.

Verroq
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1 Answers1

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No need to play with the result from exercise 5.2.3 - this is just a consequence of $X$ and $Z$ having the same underlying topological space. As $X$ is noetherian, all the potential ways to calculate cohomology (as the derived functor of global sections from abelian sheaves on a space or from $\mathcal{O}$-modules or from quasi-coherent $\mathcal{O}$-modules) are the same, see here, so we can just think of the "sheaf of abelian groups on a topological space" description and note that $\Gamma(X,-)$ and $\Gamma(Z,-)$ are literally the same functors.

KReiser
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