Equality involving Hasse zeta function of commutative ring finitely generated over $\mathbb{Z}$ Let $\mathbb{F}_q$ be a finite field consisting of $q$ elements. Imitating Riemann's zeta function$$\zeta(s) = \sum_{n = 1}^\infty {1\over{n^s}},$$define$$\zeta_{\mathbb{F}_q[t]}(s) = \sum_f {1\over{\sharp(\mathbb{F}_q[T]/(f))^s}}$$where $f$ ranges over all monic polynomials in $\mathbb{F}_q[T]$ and $\sharp(\mathbb{F}_q[T]/(f))$ denotes the order of the finite set $\mathbb{F}_q[T]/(f)$. I know that$$\zeta_{\mathbb{F}_q[T]}(s) = {1\over{1 - q^{1-s}}}.\tag*{$(*)$}$$My question now is as follows. For a commutative ring $R$ which is finitely generated over $\mathbb{Z}$, is the following equality true:$$\zeta_{R[T_1, \ldots, T_n]}(s) = \zeta_R(s - n)?$$Here $\zeta_R(s)$ denotes the Hasse zeta function of $R$.
I've tried using induction on $n$ and the result $(*)$ on $\zeta_{\mathbb{F}_q[T]}(s)$, but not too much avail. Could anybody help?
 A: We want to prove that any maximal ideal of $R[T]$ contains a unique maximal ideal of $R$. Take any maximal ideal $\mathfrak{m}$ of $R[T]$. Consider the composition $\varphi := q \circ i$:$$R \overset{i}{\hookrightarrow} R[T] \overset{q}{\twoheadrightarrow} {{R[T]}\over\mathfrak{m}},$$where $i$ is the inclusion and $q$ is the natural projection. Then clearly$$\text{Ker}\,\varphi = R \cap \mathfrak{m}.$$So there is an injective homomorphism$${R\over{R \cap \mathfrak{m}}} \to {{R[T]}\over\mathfrak{m}}.$$We know that $R$ is a finitely generated commutative ring over $\mathbb{Z}$, hence so is $R[T]$. Also, if $\mathfrak{m}$ is a maximal ideal of $R$, then $R/\mathfrak{m}$ is a finite extension of a finite field, hence it is itself a finite field. Thus, we get that$${R\over{R \cap \mathfrak{m}}}$$is finite. But note that $R/\text{Ker}\,\varphi$ is an integral domain. Hence, it is a finite integral domain, so it is a field. Thus, $R \cap \mathfrak{m}$ is a maximal ideal of $R$. If $\mathfrak{n}$ is any other ideal of $R$ which is contained in $\mathfrak{m}$, then it is contained in $R \cap \mathfrak{m}$. By maximality, we have$$\mathfrak{n} = R \cap \mathfrak{m}.$$Thus, $R \cap \mathfrak{m}$ is the unique maximal ideal of $R$ which is contained in $\mathfrak{m}$. So$$\zeta_{R[T]}(s) = \prod_{\mathfrak{n} \in \text{max}(R)} \prod_{\substack{\mathfrak{m} \in \text{max}(R[T]), \\ \mathfrak{n} \subseteq \mathfrak{m}}} \left(1 - \#\left({{R[T]}\over{\mathfrak{m}}}\right)^{-s}\right)^{-1}.$$Next, note that the maximal ideals of $R[T]$ which contain a certain ideal $\mathfrak{n}$ are in bijective correspondence with maximal ideals of $(R/\mathfrak{n})[T]$. Thus, if $\mathfrak{m}$ is an ideal of $R[T]$, then$$\# {{R[T]}\over\mathfrak{m}} = \# {{(R/\mathfrak{n})[T]}\over{\overline{\mathfrak{m}}}},$$where $\overline{\mathfrak{m}}$ is the image of $\mathfrak{m}$ under the map $R \to (R/\mathfrak{n})$. So$$\zeta_{R[T]}(s) = \prod_{\mathfrak{n} \in \max(R)} \prod_{\overline{\mathfrak{m}} \in \max((R/\mathfrak{n})[T])} \left(1 - \#\left({{(R/\mathfrak{n})[T]}\over{\overline{\mathfrak{m}}}}\right)^{-s}\right)^{-1}.$$Since $(R/\mathfrak{n})[T]$ is a finite field, we have$$\zeta_{R[T]}(s) = \prod_{\mathfrak{n} \in \max(R)} \zeta_{(R/\mathfrak{n})[T]}(s) = \prod_{\mathfrak{n} \in \max(R)} \left(1 - \# \left({R\over \mathfrak{n}}\right)^{1 - s}\right)^{-1} = \zeta_R(s - 1).$$Thus,$$\zeta_{R[T_1, \ldots, T_n]}(s) = \zeta_{R[T_2, \ldots, T_n]}(s - 1) = \zeta_{R[T_3, \ldots, T_n]}(s - 2) = \ldots = \zeta_R(s - n).$$
