Riemann vs. Ihara's $\zeta$ Function Variable Question

The Euler product for the Riemann zeta function $\zeta(z)$ implies that $$\log\zeta_R(z)=\sum_{m>0}\frac{P(mz)}{m} \tag{R},$$ whereas the Ihara zeta function for a graph $G$--all can be expressed in the same basic form:

$$\log \zeta_I(z)= \sum_{m> 0} \frac{N_mz^m}{m} .\tag{I}$$

What confuses me is the different appearance of the variable $z$. Since $\displaystyle P(mz)=\sum_{p\,\in\mathrm{\,primes}} (p^{-z})^m$, we have $z$ in the exponent in $(R)$ , but polynomial in $(I)$.

Why is that?

EDIT: Further I found the following on The Riemann hypothesis for ($q+1$-regular) graphs: $$\displaystyle \zeta(s) = \prod_{p \ \rm prime} \frac { 1 } { 1 - p^{-s}} \ \ \ \ \ (2),$$ compared to $$\displaystyle z(s) = \prod_{p \ \rm prime\ cycle} \frac 1 { 1 - (\color{red}{q^{|p|}})^{-s }} \ \ \ \ \ (5)$$ which still looks confusing to me since we build a product over primes in Riemann's case, where the prime $p$ varies, while a product over a fixed variable $\color{red}q$ with the length of the prime cycle as an exponent in Ihara's case, but since $(5)$ would just use powers of $q$, like as I would restrict my set of primes to only one single prime in $(2)$ it might be OK...

• Why do you think they ought to be the same? – Greg Martin Dec 17 '14 at 22:03
• Why is it okay that $x^5$ and $5^x$ are different? Because there's no reason to think they should be the same. – Greg Martin Dec 21 '14 at 8:07
• Yes this is additive vs multiplicative norms. For a prime $\mathfrak{p}$ in one case the norm has the form $|\mathfrak{p}| = a^{deg(p)}$ in the other it is $|\mathfrak{p}| = a^{\ln p}$. The zeta function is a function of $a = e^{-s}$. In both case we are looking at multiplicative monoids (instead of graphs you should look at the polynomial ring $\mathbb{F}_p[x]$ where $deg(f)$ is the degree of a polynomial). The functional equation appears when the multiplicative monoid is inside a commutative ring with an acceptable Fourier theory (Poisson summation formula) – reuns Apr 6 '17 at 3:12
• I meant if $M$ is a free multiplicative monoid and we have a multiplicative function $h : M \to \mathbb{R}$ then $\zeta_M(s) = \sum_{x \in M} |h(x)|^{-s} = \prod_{p \in M} \frac{1}{1-h(p)^{-s}}$ where $p$ are the generators. For $M = \mathbb{Z}^*$ the natural choice is $h(n) = n$, for $M = \mathbb{F}_p[x]^*$ the natural choice is $h(f) = e^{deg(f)}$. In one case $h(n)$ is an integer, in the other case it is $\log h(f)$ that is an integer. – reuns Apr 6 '17 at 20:59
• If $M$ is inside a ring $R$ with an acceptable Fourier theory then there is a natural choice for $h$ (such that $\zeta_M$ has a functional equation), this is the local fields vs global fields theory. Both have a Riemann hypothesis, but the RH is non-trivial (involving prime numbers) only for global fields. For local fields, the $\zeta$ function is rational. – reuns Apr 6 '17 at 21:06