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Let $X$ be a smooth projective variety and $\pi:Y\to X$ be a blowing up along the closed centre $Z$. Denote by $E$ the exceptional set of the blowing up. In general it seems not to be true that $\pi_*\mathcal{O}_Y(-E)$ coincides with the ideal sheaf of $Z$ in $X$. Am I right?

I am interested in the following specific example. Let $X=\mathbb{P}^1\times\mathbb{P}^1$ and $Y$ is the blowing up of $X$ at one point. Then $E$ is the line. Is this a counterexample? If yes, how one can compute $\pi_*\mathcal{O}_Y(-E)$ explicitly?

guest31
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  • I always find this issue confusing, so maybe I am missing something, but why do you say "it seems not to be true"? To me these two things seem equal. –  Feb 19 '15 at 09:15
  • I read the discussion at mathoverflow, there are counterexamples when $X$ is normal and singular. I do not see why this should be true in the smooth case too. – guest31 Feb 19 '15 at 16:54
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    Do you mind sharing the link to the post on mathoverflow? – Youngsu Feb 20 '15 at 08:20
  • http://mathoverflow.net/questions/66980/blowing-up-and-direct-image-sheaf – guest31 Feb 22 '15 at 08:19

1 Answers1

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If $X$ and $Z$ are smooth, there is no issue: by Zariski's Main Theorem, $\pi_*(\mathcal{O}_Y) = \mathcal{O}_X$ and $\pi_*(\mathcal{O}_E) = \mathcal{O}_Z$, so push down the short exact sequence

$$0 \to \mathcal{O}_Y(-E) \to \mathcal{O}_Y \to \mathcal{O}_E \to 0$$

to get

$$0 \to \pi_*\mathcal{O}_Y(-E) \to \mathcal{O}_X \to \mathcal{O}_Z \to \cdots$$

in fact this is right exact since $\mathcal{O}_X \to \mathcal{O}_Z$ is surjective. In any case, this shows that the pushforward is the ideal sheaf of $Z$.

Jake Levinson
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  • Thanx! I have one more question: how can I interpret $\pi_\mathcal{O}Y(E)$? If I twist the first exact sequence from your answer with $\mathcal{O}_Y(E)$ I will need to compute the direct image of $\mathcal{O}_E(E)$. If $(E.E)=m$ that will be actually $\pi\mathcal{O}_E(m)$. How to compute this? – guest31 Feb 23 '15 at 18:48
  • @guest31: for blowups in general I don't know the answer. But when $Z$ is a point on a smooth surface, $(E.E) = -1$, so the pushforward of $\mathcal{O}E(E)$ is zero, and so $\pi*\mathcal{O}_Y(E) = \mathcal{O}_X$. – Jake Levinson Mar 06 '15 at 16:08