# About direct image of ideal sheaves

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 ${(I_1)}_x$?

• I suspect that this isn't true. If I think of an easy example, I'll tell you. You may want to look at some papers of Kevin Tucker and also of Howard Thompson and Karen Smith, for computing multiplier ideals / jumping numbers on surfaces. They look at questions of jumping (ie, when you have strict containment). If your question is false for surfaces, you should be able to counter-examples based on the combinatorics worked out there. Dec 13, 2010 at 17:01
• Thanks a lot for the references. I think they may be very useful for me (at least to understand what can go wrong). You guessed right however, what I would need is exactly a jumping-like result for multiplier ideals. Dec 13, 2010 at 18:04
• By the way, Kevin doesn't really use mathoverflow, but he probably wouldn't mind a direct email on this. He's also much more likely than I to know the answer off the top of his head. Dec 13, 2010 at 18:57
• Why not? Thanks for the idea Dec 15, 2010 at 15:21

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 $(\mu_z \circ \mu_y)^*E_x=\tilde E_x + \tilde E_y +2 E_z$.

• Very simple..but it works! Thanks a lot! Dec 15, 2010 at 15:20