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Let $X$ be a normal projective variety, $Z\subseteq X$ a closed subvariety, and suppose that $\mathfrak{a}\subseteq \mathcal{O}_X$ is an ideal sheaf. I tried to prove the following:

There is a positive number $C>0$ such that $$ \mathrm{ord}_E\mathfrak{a}\le C\mathrm{ord}_EZ$$ holds for any prime divisor $E$ over $X$ whose center on $X$ is in $Z$. ($C$ does not depend on $E$.)

Can we prove this one? If $Z$ contains both the singular locus and the locus of $\mathfrak{a}$, then it is Proposition 2.4 in [BdFFU15]. (But I don't follow the proof) Can we drop the condition?

[BdFFU15]: S. Boucksom, T. de Fernex, C. Favre, and S. Urbinati: Valuation spaces and multiplier ideals on singular varieties, Recent advances in algebraic geometry, London Math. Soc. Lecture Note Ser. 417 (2015), 29–51, Cambridge Univ. Press, Cambridge.

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  • $\begingroup$ maybe you need to show that $I_Z^C\subset a$ for sufficiently large $C$? $\endgroup$
    – Chen Jiang
    Commented Sep 7 at 13:09
  • $\begingroup$ No. In general, $\mathcal{I}^C_Z\subseteq\mathfrak{a}$ is false. (Just take $X=\mathbb{P}^2, Z=\{\mathrm{pt}\},$ and $\mathfrak{a}$ an ideal sheaf of an ample divisor on $X$ passing $Z$.) $\endgroup$
    – nariri
    Commented Sep 7 at 13:11
  • $\begingroup$ in your example Z does not contain the locus of a $\endgroup$
    – Chen Jiang
    Commented Sep 7 at 13:12
  • $\begingroup$ Sorry for inconvenience, there is no assumption that $Z$ contains the locus of $\mathfrak{a}$... Maybe I added the assumption because I obessed in the condition in [BdFFU15]. $\endgroup$
    – nariri
    Commented Sep 7 at 13:19

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