# Igusa zeta functions of univariate polynomials: $\mathbb{Z}_p$ or $\mathbb{Q}_p$ in this statement

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.

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.