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Reading a book I met the following claim, and I don't understand how to justify it. [Actually I misunderstood the claim, below is the corrected version of it]

Let $X$ be a variety and $E\subset X$ a divisor. Suppose we have a global section $$ s: \mathcal{O}_X\to \mathcal{O}_X(E) $$ such that the vanishing locus of $s$ is a divisor $D\subset X$, and that $s$ vanishes of order $1$ there. Under these assumptions, it follows that $L(D)$ and $L(E)$ are isomorphic.

Why is this the case?

Abramo
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1 Answers1

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Ok, my problem was I misunderstood the claim.

Now it's quite clear: the conclusion follows from the well-known fact that a section $s$ belongs to $H^0(X, \mathcal{O}(D))$ if and only if $(s) = D$, i.e. if and only if $s$ vanishes of order $1$ over $D$ (and nowhere else).

Abramo
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