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Let $X$ be a scheme and let $Y$ be a closed subscheme with ideal sheaf $I$. Let $F$ be a coherent sheaf on $X$. Is the sequence

$$ 0 \to I\otimes F \to F \to F \otimes O_Y \to 0 $$exact? This is obvious when $X$ is affine, but is it true for general schemes?

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Are you sure it's obvious when $X$ is affine? Try working out the example where $X=\mathbb{A}^1$, $Y=\{0\}$, and $F$ is $\mathcal{O}_Y$ (considered as a sheaf on $X$).

For a more obvious (to me) but less geometric counterexample, consider what happens to the exact sequence of $\mathbb{Z}$-modules $$ 0 \to 2\mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0 $$ when you take tensor products with $\mathbb{Z}/2\mathbb{Z}$.

The property you're interested in is flatness: an $R$-module $M$ is flat precisely when the operation of taking tensor products with $M$ preserves exactness. This turns out to be equivalent to your sequence being exact for all ideals $I$.

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  • $\begingroup$ @jacobi: Martin is right. The sequence in your comment is indeed exact, but is not what is obtained from your question if you take $F=\mathcal O_Y$. The second zero in the sequence of your comment should be replaced by $I/I^2$ , and the map emanating from there is zero. $\endgroup$ Apr 13 '11 at 19:34

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