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Let $\mu:Y\to X$ be a birational morhism between normal projective (complex) varieties. Suppose, furthermore, that $Y$ is smooth. Let $D$ be a Weil divisor on $X$ and let $\mathcal{O}_X(D)$ be its corresponding reflexive sheaf. If we denote by $\widetilde{D}$ the strict transform of $D$, is it true that the push-forward of the invetible sheaf $O_Y(\widetilde{D})$ is equal to $\mathcal{O}_X(D)$?

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up vote 6 down vote accepted

No, the is definitely not true. Let me give you an example, suppose that $X = \mathbb{A}^2$ and that $\pi : Y \to X$ is the blowup of $X$ at the origin with exceptional divisor $E$. Suppose that $D$ is the Cartier divisor corresponding made up of $n$ lines through the origin, $n \geq 2$. Then $\widetilde{D}$ is just a disjoint union of $n$ lines on the blown up space, and $\pi^* D = \widetilde{D} + nE$.

Suppose now that $\pi_* O_Y(\widetilde D) = O_X(D)$. Then by the projection formula, $$ \pi_* O_Y(-nE) = \pi_* O_Y(\widetilde D - \pi^* D) = O_X $$ However, it is obvious that $\pi_* O_Y(-nE) = \mathfrak{m}^n$ is the maximal ideal of the origin raised to the $n$th power. This is a contradiction.

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Thanks a lot Karl! Very clear example. – Gianni Bello Nov 10 '12 at 13:01

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