Let $X$ be a nonsingular quasiprojective variety and $Z$ a closed subvariety of length $n$ i.e. $\dim H^0(Z, \mathcal{O}_Z) = n$. I found an identity in FGA Explained claiming that $\dim H^0(\mathcal{O}_Z \otimes K_X)^* = \dim H^0(\mathcal{O}_Z) = n$, where $K_X$ is the canonical bundle, but I can't see why that would be.
$\dim H^0(\mathcal{O}_Z \otimes K_X)^* = \dim H^0(\mathcal{O}_Z)$?
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algebraic-geometry
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1Length $n$, as in $Z$ is finite? Then $H^0$ of any line bundle restricted to $Z$ is just an $n$-dimensional vector space. – 2017-01-30