For the arithmetic zeta function of (say) a nonsingular projective variety $X$, one has the following Euler product
\begin{equation} \zeta_X(s) = \prod_{p\ \mbox{prime}}\zeta_{X\vert\mathbb{F}_p}(s), \end{equation} where $\zeta_{X \vert \mathbb{F}_p}(s)$ is the local zeta function given by $\exp(\sum_{m = 1}^\infty \frac{\vert X_{p^m} \vert}{m}p^{-sm})$ (here, $X_{p^m}$ is the set of $p^m$-rational points of $X$). For any field $\mathbb{F}_q$, one also has a local zeta function for $X$, being $\zeta_{X \vert \mathbb{F}_q}(s) = \exp(\sum_{m = 1}^\infty \frac{\vert X_{q^m} \vert}{m}q^{-sm})$.

My question is: suppose one now considers the product \begin{equation} \prod_{p \ \mbox{prime}}\zeta_{X\vert\mathbb{F}_{p^r}}(s), \end{equation} for a fixed positive integer $r$, is there also a nice identity involving (arithmetic) zeta functions?

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    $\begingroup$ I don't understand why you are taking the product over all primes. Do you mean that $X$ is a smooth projective scheme over $\operatorname{Spec} \mathbb Z$? (that seems very strong; I don't know many examples that have good reduction at all primes...) $\endgroup$ – R. van Dobben de Bruyn Nov 19 '15 at 16:26
  • $\begingroup$ @R. van Dobben de Bruyn: This is the standard definition of an arithmetic zeta function: en.wikipedia.org/wiki/Arithmetic_zeta_function. No good reduction hypotheses are required. The OP should really however be choosing a model for $X$ over $\mathbb{Z}$. $\endgroup$ – Daniel Loughran Nov 19 '15 at 21:27
  • $\begingroup$ I am fine with the arithmetic zeta function, but I was confused by the term nonsingular variety. For me, varieties are over a field, in which case there is at most one prime. But I guess the obvious modification is that $X$ is some scheme over $\operatorname{Spec} \mathbb Z$ whose generic fibre is smooth (i.e. nonsingular) (rather than the modification I first considered where $X \to \operatorname{Spec} \mathbb Z$ is smooth). $\endgroup$ – R. van Dobben de Bruyn Nov 20 '15 at 1:00
  • $\begingroup$ I am sorry for the confusion - it is enough to suppose that $X$ is a scheme of finite type over $\mathrm{Spec}(\mathbb{Z})$. $\endgroup$ – THC Nov 20 '15 at 9:06
  • $\begingroup$ When you replace $\mathbb{Z},\mathbb{Q}$ by $\mathbb{Z}[i],\mathbb{Q}(i)$ you are also replacing $\zeta_{X |\mathbb{F}_p}(s)$ by $\zeta_{X |\mathbb{F}_p^2}(s)$ for $p \equiv 3 \bmod 4$ and by $ \zeta_{X |\mathbb{F}_p}(s)^2$ for $p \equiv 1 \bmod 4$. $\endgroup$ – reuns Jan 3 '17 at 11:52

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