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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)\not\in \mathbb{Z}_p[T]$$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 $f\in\mathbb{Z}_p[X]$ 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 $Z_{f,p}(T)\not\in \mathbb{Z}_p[T]$ 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 $f\in\mathbb{Z}_p[X]$ 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.

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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 $Z_{f,p}(T)\not\in \mathbb{Z}_p[T]$ 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.