Blowing-up and direct image sheaf. 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.
 A: 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 $m\ge 0$.
A: 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.
1
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 $k[t] \cdot t$.
2
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.
3
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.
