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I have heard that from a given divisor $D \hookrightarrow X$ (where $X$ is a projective scheme and $D = \sum n_i Y_i$ such that each $Y_i$ is smooth and $Y_i \hookrightarrow X$ is a closed immersion), one can construct an exact sequence $0 \rightarrow \mathcal{O}(-D) \rightarrow \mathcal{O}_X \rightarrow i_{*} \mathcal{O}_D \rightarrow 0$, however what´s $\mathcal{O}_D$ in this case?

For an ordinary irreducible hypersurface, the sequence is clear, but, for a linear combination of such hypersurfaces, I don´t know exactly how to generalize this concept.

Thanks in advance.

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  • $\begingroup$ If $i:D\to X$ is the inclusion, the cokernel is just $i_\ast\mathcal O_D$. $\endgroup$
    – Brenin
    Commented May 6, 2014 at 13:42
  • $\begingroup$ @Brenin Yes, I know it if $D$ is just an irreducible hypersurface. But what´s $\mathcal{O}_D$ for a Weil divisor? $\endgroup$
    – user40276
    Commented May 6, 2014 at 13:45

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One requires that $D$ be an effective divisor, i.e. that $n_i \geq 0$ for all $i$. Then one can identify $D$ with a closed subscheme of $X$. Thus $\mathcal{O}_D$ makes sense. Note that each $Y_i$ need not be smooth, although perhaps in particular applications one might add this in.

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  • $\begingroup$ Thanks for the answer, but I don´t exactly understand how can you make this identification. If $D = \sum n_i Y_i$, how is this closed subscheme? $\endgroup$
    – user40276
    Commented May 6, 2014 at 14:11

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