Let $\mu:X'\rightarrow X$ be a birational morphism of normal complex projective varieties.


Consider the two ideal sheaves $I\_1= \mu\_\*\mathcal{O}\_{X'}(-\sum d(E)E)$, $I\_2=\mu\_\*\mathcal{O}\_{X'}(-\sum(d(E)+1)E)$, where the $d(E)$'s are non negative integers and the $E$'s are prime divisors. 


Suppose $x\in X$ is such that $x\in \mu(E\_0)$ for a prime Cartier divisor $E\_0$ such that $d(E\_0)>0$.
Can we say that the stalk ${(I\_2)}\_x$ is STRICTLY contained in ${(I\_1)}\_x$? 


Thanks for your help.



About direct image of ideal sheaves
The answer is no.


Consider $\mu=\mu\_z \circ \mu\_y \circ \mu\_x$ the blow up of a smooth surface at three points $x$, $y$, $z$, as follows: $x\in X$ is arbitrary, $y\in E\_x:=\mu\_x^{-1}(x)$, where $\mu\_x$ is the blowup of $X$ centered at $x$, and $z$ is the "satellite" point that appears after blowing up $y$, ie, $z=\tilde E\_x\cap E\_y$, where $E\_y:=\mu\_y^{-1}(y)$ is the exceptional of the second blowup $\mu\_y$ and $\tilde E\_x$ is the strict (birational) transform of $E\_x$.


Let $d(\tilde E\_x)=d(\tilde E\_y)=0$, $d(E\_z)=1$, where again $\tilde E\_x$ and $\tilde E\_y$ denote the strict transforms of the first and second exceptional divisors (but now on $X'$, ie, after blowing up the third). Then in your notation $I\_1=I\_2=\mathfrak{m}\_x$ is the maximal ideal of $x$.


An easy way to see it is that the pullback of $E\_x$ to $X'$ is precisely $(\mu\_z \circ \mu\_y)^\*E\_x=\tilde E\_x + \tilde E\_y +2 E\_z$.


