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Let $\pi:Z=Bl_{Y}(X)\rightarrow X$ be the blowing-up of a smooth projective variety X along a subvariety $Y$, $E$ the closed subscheme defined by $\pi^{−1}I_{Y,X}\cdot O_Z$. Is it true that (without assume $Y$ smooth) $\pi_{\ast}(O_Z(-E))=I_{Y,X}$?

Thanks for your help.

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  • $\begingroup$ gianna, what do you mean by the exceptional divisor? The exceptional set need not be a divisor in general. Even if it is, do you mean with the reduced scheme structure? Alternately, by $E$ do you mean $\pi^{-1} I_{Y,X} \cdot O_Z$? Also you should notice that $Z$ is not generally normal, so maybe you don't want to be talking about divisors on it in general? Finally, if you replace $Z$ by its normalization and choose E such that $O_Z(−E)= \pi^{−1} I_{Y,X}\cdot O_Z$, then the answer is no. However in that case $\pi_* O_Z(−E)$ is something called the integral closure of $I_{Y,X}$. $\endgroup$ Jun 5, 2011 at 19:03
  • $\begingroup$ Thank you, if $X=\mathbb{P}^N$, is true my statement? $\endgroup$
    – gio
    Jun 6, 2011 at 6:30

2 Answers 2

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This is not true. Actually, your question needs to be made a bit more precise. I suppose you mean that $E$ is the exceptional set and that $\mathscr O_Z(-E)$ denotes the ideal sheaf of $E$. (Notice that $E$ is not necessarily a divisor and then its ideal is not an invertible sheaf). However, even if you assume that $Z$ is smooth, $E$ is a smooth divisor, the statement is still not true. You can even assume that $Y$ is smooth and $X$ is normal.

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If $X$ is not normal, then $\pi_*\mathscr I_{E\subseteq Z}$ is not even necessarily in $\mathscr O_X$, let alone being equal to the ideal sheaf of $Y$. Just take $X$ to be a cuspidal (or nodal) cubic and $Y$ the singular point. The blow up is the map induced by $$k[t^2,t^3]\hookrightarrow k[t]$$ and $\pi_*\mathscr I_{E\subseteq Z}$ corresponds to the $k[t^2,t^3]$ module $k[t] \cdot t$.

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So, let's assume that $X$ is normal. Unfortunately it is still not true: Let $X$ be a quadric cone in $\mathbb A^3$ and $Y$ a line through the vertex of $X$. Then the blow up of $X$ along $Y$ is the same as the blow up of the vertex and $\pi_*\mathscr I_{E\subseteq Z}$ is the ideal sheaf of the vertex, not of the line.

Notice that in this example, $X$ is a normal Gorenstein variety and $Y$ is smooth, so you need quite a bit of assumptions to have a blanket statement like you wished for.

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Here is a criterion that implies what you want:

If $X$ is normal and the natural map $\mathscr O_Y\to \pi_*\mathscr O_E$ is an isomorphism, then $\mathscr I_{Y\subseteq X}\simeq \pi_*\mathscr I_{E\subseteq Z}$. The proof of this is very simple. Consider the diagram \begin{gather} 0 \quad \longrightarrow & \quad \mathscr I_{Y\subseteq X} \qquad \longrightarrow & \mathscr O_X \ \qquad \longrightarrow &\mathscr O_Y &\longrightarrow & 0\\ &\downarrow \qquad & \downarrow \qquad\qquad & \downarrow & \qquad \qquad \\ 0\quad \longrightarrow & {\pi_*\mathscr I_{E\subseteq Z}} \quad \longrightarrow &\pi_* \mathscr O_Z \qquad \longrightarrow &\pi_*\mathscr O_E & \end{gather}

The assumption that $X$ is normal implies that the second vertical arrow is an isomorphism and the other assumption is that so is the third. Then it follows easily that so is the first.

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  • $\begingroup$ Thanks for the interesting answers. With $E$ I mean the subscheme defined by $\pi^{-1}I_{Y, X}\cdot O_Z$. However, my statement is true if $X$ (not necessarily $Y$) is smooth ? I'm particularly interested in the case $X = P^N,$ a projective space over a field. $\endgroup$
    – gio
    Jun 5, 2011 at 22:16
  • $\begingroup$ @Sandor Kovacs, You have said that if $X$ is normal and the natural map $O_Y\rightarrow\pi_*O_E$ is an isomorphism, then, the required condition is satisfied. But by Zariski's main theorem, if the fibers are connected, then the isomorphism is there right. Are the fibers not connected in this case? $\endgroup$ Jun 13, 2016 at 9:03
  • $\begingroup$ @gradstudent: to use ZMT, you would need to know that $E\to Y$ is birational, but $E$ doesn't even have to have the same number of components as $Y$. It may not dominate to $Y$, or some of its components do not dominate $Y$, so this could fail in many ways. $\endgroup$ Jul 25, 2017 at 15:52
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As Sándor's nice answer shows, $\pi_{\ast}(O_Z(-E))=I_{Y}$ is not neccesarily true, even for $Y$ a normal subvariety of $X$. On the other hand, the following statement

$$\pi_{\ast}(O_Z(-mE))=I_{Y}^m \quad\mbox{ for $m\gg 0$}$$

always holds, even without any assumptions on the subscheme $Y$. Here is a short explaination why:

It suffices to deal with the case $X=\mbox{Spec} A$ is affine and $I=I_Y=(g_1,\ldots,g_n)\subset A$. The $g_i$'s determine a surjection $A^r \to I$, and hence a surjection of Rees algebras $\mbox{Sym}^*(O_X^r) \to \bigoplus_{m \ge 0} I^m$. Taking Proj this means that there is an embedding $Z=Bl_Y X \subset \mathbb{P}=\mathbb{P}(O_X^n)$ where the exceptional divisor $E$ corresponds to $O(-1)\big|_Z$. Now, if $p:\mathbb{P}(O_X^n)\to X$ is the projection, we have $$p_* O_{\mathbb{P}}(m) \to p_*O_Z(-mE)$$ is surjective for $m$ large by relative Serre vanishing. Moreover, since $p_*O_{\mathbb{P}}(m)=Sym^m(O_X^n)$ we can identify the image of this map with $I^m$ and hence we have $\pi_{\ast}(O_Z(-mE))=I^m$.

In particular, if $\pi_* O_E=O_Y$ holds (as in Sándor's answer) the above map will always be surjective and $\pi_{\ast}(O_Z(-mE))=I^m$ for all $m\ge 0$.

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