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For instance say that $X$ is a degree-$d$ hypersurface in $\mathbb{P}^n$. To be precise, put $R = k[x_0, \ldots, x_n]$, $d > 0$, $f \in R_d$ squarefree, and $X = V(f) \subset \mathbb{P}^n$. Then we have an exact sequence of graded $R$-modules

$0 \to R(-d) \xrightarrow{-\cdot f} R \to R/(f) \to 0$;

when we promote this to sheaves, apparently we have

$0 \to \mathcal{O}_{\mathbb{P}^n}(-d) \to \mathcal{O}_{\mathbb{P}^n} \to \mathcal{O}_X \to 0$.

Now normally I'd think of "$\mathcal{O}_X$" as representing a sheaf on $X$, but this construction seems to have produced a sheaf on $\mathbb{P}^n$ instead.

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See this question. –  Zhen Lin Dec 28 '12 at 11:03

1 Answer 1

up vote 1 down vote accepted

Just for the sake of resolving this question, the sequence should read

$0 \to \mathcal{O}_{\mathbb{P}^n}(-d) \to \mathcal{O}_{\mathbb{P}^n} \to i_\ast \mathcal{O}_X \to 0$,

where $i : X \to \mathbb{P}^n$ is the inclusion. Using $\mathcal{O}_X$ in place of $i_\ast \mathcal{O}_X$ is apparently an accepted abuse of notation in this context.

Thanks to Zhen Lin for clearing this up in the comments.

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