Let $f\in\mathbb{Z}\_p[X]$ and let $Z\_{f,p}(T)\in\mathbb{Z}\_{(p)}(T)$ be the $p$-adic Igusa zeta polynomial (i.e. $Z\_{f,p}(p^{-s})$ is the $p$-adic Igusa zeta function in the complex variable $s$, with $\Re(s)$ larger than the abscissa of convergence). 


It is not difficult to prove that $Z\_{f,p}(T)\in\mathbb{Z}\_{(p)}[T]$ if and only if $f$ has no roots in $\mathbb{Z}\_p$. Here is how the proof goes. Note first that, since the polynomial function $f:\mathbb{Z}\_p \to \mathbb{Z}\_p$ is continous, so is also the composition $|f|\_p:\mathbb{Z}\_p \to p^{\mathbb{Z}\_{\le 0}}\cup \{0\}$. Consequently, the image of $|f|\_p$ is compact. Now, if $f$ has no zeros in $\mathbb{Z}\_p$, then the image of $|f|\_p$ is contained in the discrete subset $p^{\mathbb{Z}\_{\le 0}}$, so it is finite. This implies that $Z\_{f,p}(T)\in \mathbb{Z}\_{(p)}[T]$. On the contrary, if $f$ has a zero in $\mathbb{Z}\_p$, then $0$ is an accumulation point for the image of $|f|\_p$. Therefore, there exist infinitely many $k\in \mathbb{Z}\_{\ge 0}$ such that the set
$\{x\in \mathbb{Z}\_p\,:\,|f(x)|\_p=p^{-k}\}$
 is non-empty. Since this is also open, it follows that it has positive measure and thus $Z\_{f,p}(T)\not\in \mathbb{Z}\_p[T]$.


Until here everything seems fine, but my problem comes when I try to get the same conclusion directly from Igusa's theorem. In the case of univariate polynomials, the exceptional divisors of a log-resolution are simply the roots of the polynomial, and if we work in the $\mathbb{Q}\_p$-analytic setting, then we are interested only in the $\mathbb{Q}\_p$-rational roots. From Igusa's theorem we get thus a pole for $Z\_{f,p}(T)$ as soon as $f$ has a $\mathbb{Q}\_p$-rational root. This implies that $Z\_{f,p}(T)\in \mathbb{Z}\_p[T]$ iff and only if $f$ has no roots in $\mathbb{Q}\_p$.


Comparing the two reasonings we see that, unless we assume that $\deg(f \mod p)=\deg f$, we seem to get a contradiction. On the other hand, this assumption is not made in the statement of Igusa's theorem, so I am wondering if I am actually not taking something into consideration. 



Igusa zeta functions of univariate polynomials: $\mathbb{Z}_p$ or $\mathbb{Q}_p$ in this statement
Going more attentively through the proof of Igusa's theorem ("An Introduction to he Theory of Local Zeta Functions", J.-I. Igusa, pp 122-123), I have actually realized what I was missing.


Let $h:Y\to \mathbb{Q}\_p$ be a log-resolution (in the $\mathbb{Q}\_p$-analytic setting). The exceptional divisors of $h$ are $(E\_{\alpha})\_{\alpha}$ for $\alpha$ ranging in the $\mathbb{Q}\_p$-rational roots of $f$, with corresponding numerical data $(r\_{\alpha},1)$, where $r\_{\alpha}$ denotes the multiplicity of the root $\alpha$. This means that we can write $h^{-1}(\mathbb{Z}\_p)$ as a disjoint union of compact open subsets $(B\_{\alpha})\_{\alpha}$, such that, for each $\alpha$, $B\_{\alpha}\supseteq E\_{\alpha}$, $\mathbf{1}\_{\mathbb{Z}\_p}\circ h|\_{B\_{\alpha}}=\mathbf{1}\_{\mathbb{Z}\_p}(\alpha)$ and there exists a chart $(B\_{\alpha},\phi\_{\alpha})$ on $Y$ with
$$ (f\circ h)(\phi\_{\alpha}(y))\sim\phi\_{\alpha}(y)^{r\_{\alpha}}\quad \text{and} \quad h^\*(dx)\sim dy.$$
Since $h\,:\,Y\setminus \bigcup\_{\alpha} E\_{\alpha}\to X\setminus \bigcup\_{\alpha}\{\alpha\}$ is $\mathbb{Q}\_p$-bianalytic, then we have 


$$ \begin{split} Z\_{f,p}(p^{-s})&=\sum\_{\alpha}\mathbf{1}\_{\mathbb{Z}\_p}(\alpha) C\_{\alpha} \int\_{\phi\_{\alpha}(B\_{\alpha})}|y|\_p^{-r\_{\alpha}s}dy \\ &=\sum\_{\alpha\,:\,\alpha\in\mathbb{Z}\_p} \frac{P\_{\alpha}(p^{-s})}{1-p^{1-r\_{\alpha}s}},\end{split}$$


where the $C\_{\alpha}$, $P\_{\alpha}$ are respectively constants and polynomials. 


In the reference I have mentioned, you can find the more details and the computation in a more general setting.


