Let $Ω\_p$ denote the set of the real parts of the poles of the Igusa zeta function of a polynomial $f∈\mathbb{Z}[X\_1,…,X\_m]$ (assume $f(0)=0$ so that $\Omega\_p\ne \emptyset$) at the prime p. From Denef's formula for the motivic zeta function, we know that if $\{(N\_i,\nu\_i)\}\_{i\in I}$ are the numerical data of a log resolution $h$ of the ideal $(f)$ over $\mathbb{Q}$, then for any "good" prime $p$ with respect to $h$ (in the sense of Denef) one has


$$ \Omega\_p\subseteq \Omega:=\Big\{ -\frac{\nu\_i}{N\_i} \,:\,i\in I\Big\} $$


and the maxima of the two sets agree (and it is independent of $h$).


However, $\Omega\_p$ is in general much smaller than $\Omega$, because of the many cancellations that take place in Denef's formula. 


**I was wondering whether these cancellations are actually uniform in $p$ (for $p$ good with respect to $h$), or, in other words, whether $\Omega\_p$ is independent of $p$ for $p$ big enough. If it is the case, do this cancellation have a geometric interpretation at the level of the motivic zeta function?**


Thank you.



Uniformity of the set of poles of Igusa local zeta functions
The answer is negative. One can pick for example $f(X)=X(X^2+1)^2$. Igusa's formula from "An Introduction to the Theory of Local Zeta Functions" (p.123) tells us that $\Omega\_p=\{1\}$ for $p\equiv 3 \mod 4$, but $\Omega\_p=\{1,1/2\}$ for $p\equiv 1 \mod 4$.


