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Aug 9, 2011 at 23:38 comment added Moosbrugger If $X$ and $S$ are finite type over $\operatorname{Spec}(\mathbb{Z})$, then what you've said is indeed well-defined and is just the zeta function of $X$.
Aug 9, 2011 at 22:08 comment added James D. Taylor I meant is. Fixed. Moosbrugger, What I mean is for example: Let $S$ be a surface. Then I define the zeta function of $X\rightarrow S$ to be the product of the zeta function of every fiber. Does this have good properties? Is it used? Also, if the residue fields of the points of $S$ are not finite, then the way I just defined the zeta function of that morphism isn't well defined (because we don't have a notion of say the zeta function of $Y\rightarrow Spec(\mathbb{C})$). Is it going to be meromorphic? I don't know. Hopefully. Have people studied this?
Aug 9, 2011 at 22:04 history edited James D. Taylor CC BY-SA 3.0
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Aug 9, 2011 at 18:55 comment added Moosbrugger What is the shape of an answer you are hoping for? E.g., should it be a holomorphic function? For "relative" algebraic geometry, you typically want something of local nature on the base, and given the form of usual zeta functions (which you seem to want as the result when $S=\operatorname{Spec}(\mathbb{Z})$), it's hard to see what form it would take.
Aug 9, 2011 at 18:31 comment added Joe Silverman Probably not what you want, but if one views the zeta function over a finite field as encoding the number of fixed points of iterates of the Frobenius map, then one is led do the same thing for the fixed points of the iterates of an arbitrary map. These types of zeta functions have been studied in dynamical systems.
Aug 9, 2011 at 18:28 comment added Joe Silverman Do you mean "even when S is not $\ge2$ dimensional", or do you mean "even when S is $\ge2$ dimensional"?
Aug 9, 2011 at 17:34 comment added Qiaochu Yuan Have you seen the discussion at golem.ph.utexas.edu/category/2010/07/… ?
Aug 9, 2011 at 17:12 history edited James D. Taylor CC BY-SA 3.0
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Aug 9, 2011 at 17:07 history asked James D. Taylor CC BY-SA 3.0