3
$\begingroup$

The Hasse-Weil zeta function is defined only for arithmetic schemes. By an arithmetic scheme I will mean a scheme $X$ together with a morphism of finite type $X\rightarrow S$, where $S$ is an affine Dedekind scheme (a $0$ or $1$ dimensional nonsingular affine scheme). Actually, in the case where $S$ is $0$-dimensional, I believe it is only defined for $S$ being $Spec$ of a finite field.

The way the Hasse-Weil zeta function is defined is like so: first you define it for varieties over finite fields, and then if $S$ is one dimensional, you define the zeta function as the product of the zeta function of every fiber.

It seems rather arbitrary for it to be defined only in these cases. Is there a definition of a zeta function of a morphism of finite type (or maybe flat?) in general, even when $S$ is $\geq 2$ dimensional? I would be surprised if there isn't, but I've never heard of such an entity.

Improve this question
$\endgroup$
6
  • $\begingroup$ Have you seen the discussion at golem.ph.utexas.edu/category/2010/07/… ? $\endgroup$
    – Qiaochu Yuan
    Commented Aug 9, 2011 at 17:34
  • $\begingroup$ Do you mean "even when S is not $\ge2$ dimensional", or do you mean "even when S is $\ge2$ dimensional"? $\endgroup$
    – Joe Silverman
    Commented Aug 9, 2011 at 18:28
  • 3
    $\begingroup$ 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. $\endgroup$
    – Joe Silverman
    Commented Aug 9, 2011 at 18:31
  • $\begingroup$ 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. $\endgroup$
    – Moosbrugger
    Commented Aug 9, 2011 at 18:55
  • 4
    $\begingroup$ 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 $X$. $\endgroup$
    – Moosbrugger
    Commented Aug 9, 2011 at 23:38

0

Reset to default

You must log in to answer this question.