Motivation for zeta function of an algebraic variety

Question

If $p$ is a prime then the zeta function for an algebraic curve $V$ over $\mathbb{F}_p$ is defined to be $$\zeta_{V,p}(s) := \exp\left(\sum_{m\geq 1} \frac{N_m}{m}(p^{-s})^m\right). $$ where $N_m$ is the number of points over $\mathbb{F}_{p^m}$.

I was wondering what is the motivation for this definition. The sum in the exponent is vaguely logarithmic. So maybe that explains the exponential?

What sort of information is the zeta function meant to encode and how does it do it? Also, how does this end up being a rational function?

Answer 1

The definition using exponential of such an ad hoc looking series is admittedly not too illuminating. You mention that the series looks vaguely logarithmic, and that's true because of denominator $m$. But then we can ask, why include $m$ in the denominator?

A "better" definition of a zeta function of a curve (more generally a variety) over $\mathbb F_p$ involves an Euler product. The product will be over all points $P$ of $V$ which are defined over the algebraic closure $\overline{\mathbb F_p}$ (this isn't exactly true, see below). Any such point has a minimal field of definition, namely the field $\mathbb F_{p^n}$ generated by the coordinates of this point. We shall define the norm of this point $P$ as $|P|=p^n$. Then we can define $$\zeta_{V,p}(s)=\prod_P(1-|P|^{-s})^{-1}.$$ (again, this is not quite right) Why would this definition be equivalent to yours? It's easiest to see by taking the logarithms. Then for a point $P$, the logarithm of the corresponding factor of the product will contribute $$-\log(1-|P|^{-s})=\sum_{k=1}^\infty\frac{1}{k}|P|^{-ks}=\sum_{k=1}^\infty\frac{1}{k}p^{-nks}=\sum_{k=1}^\infty\frac{n}{nk}(p^{nk})^{-s}.$$ In the last step I have multiplied the numerator and the denominator by $n$, because the point $P$ contributes precisely to numbers $N_{nk}$, since $P$ is defined over all the fields $\mathbb F_{p^{nk}}$.

But we see a problem - this way, we have counted each point $n$ times because of $n$ in the numerator. The resolution is rather tricky - instead of taking a product over points, we take a product over Galois orbits of the points - if we have a point $P$ minimally defined over $\mathbb F_{p^n}$, then there are exactly $n$ points (conjugates of $P$) which we can reach from $P$ by considering the automorphisms of $\mathbb F_{p^n}$. If we were to write $Q$ for this set of conjugates, and we define $|Q|=p^n$, then repeating the calculation above we see that we always count $Q$ $n$ times - which is just right, since it consists of $n$ points! Thus we arrive at the following (this time correct) definition of the zeta function: $$\zeta_{V,p}(s)=\prod_Q(1-|Q|^{-s})^{-1},$$ the product this time over Galois orbits.

Apart from being (in my opinion) much better motivated, it has other advantages. For instance, from the product formula it is clear that the series has integer coefficients. Further, it highlights the similarity with the Riemann zeta function, which has a very similar Euler product. Both of those are generalized to the case of certain arithmetic schemes, but that might be a story for a different time.

As for your last question, regarding rationality, this is a rather nontrivial result, even, as far as I know, for curves. If you are interested in details, then I recommend taking a look at Koblitz's book "$p$-adic numbers, $p$-adic analysis and zeta functions". There he proves, using moderately elementary $p$-adic analysis, rationality of zeta functions of arbitrary varieties.

As KConrad says in the comment, the proof of rationality actually is much simpler for curves than for general varieties. I imagine it is by far more illuminating than Dwork's proof as presented in Koblitz.

Answer 2

Exercise 4.8 of Enumerative Combinatorics, vol. 1, second ed., and Exercise 5.2(b) in volume 2 give an explanation of sorts for general varieties over finite fields. According to Exercise 4.8, a generating function $\exp \sum_{n\geq 1} a_n\frac{x^n}{n}$ is rational if and only if we can write $$ a_n=\sum_{i=1}^r\alpha_i^n-\sum_{j=1}^s \beta_j^n, $$ for nonzero complex numbers $\alpha_i$, $\beta_j$ (independent of $n$). This is stronger than saying that $\sum_{n \geq 1}a_nx^n$ is rational. Moreover, if the variety $V$ is defined over $\mathbb{F}_q$ and $N_n$ is the number of points over $\mathbb{F}_{q^n}$, then the solution to Exercise 5.2(b) is a simple argument showing that $\exp \sum_{n\geq 1}N_n\frac{x^n}{n}$ has integer coefficients. It corresponds to partitioning the rational points into their Galois orbits.

Answer 3

Here's the simple motivation going back to the beginning with Riemann of the cottage industry of zeta functions.

The Euler product for the Riemann zeta function is

$$\zeta(s) = \prod_p \frac{1}{1-p^{-s}}$$

with the product over all primes. Then

$$\ln(\zeta(s)-1) = \sum_p -\ln(1-p^{s-1}) = \sum_p \sum_{n > 0} \frac{p^{n(s-1)}}{n}= \sum_{n > 0} \frac{1}{n}\sum_p p^{n(s-1)}$$

for $Re(s) < 0$.

From this expression it's easy to derive Riemann's staircase for the primes as I sketched in "Dirac’s Delta Function and Riemann’s Jump Function J(x) for the Primes". The pdf "An overview of Deligne's proof of the Riemann hypothesis for varieties over finite fields" by Nicholas Katz begins also with this formulation just as Wojowu does.

The OP's exponential formula is a variant of the generating function of the compositional cycle index partition polynomials of the symmetric groups $S_n$, a.k.a. the refined Stirling polynomials of the first kind (sometimes called the one-part Schur polynomials), of OEIS A036039, a core construct in symmetric function theory along with the Faber polynomials of A263916 (see, e.g., this MO-Q), used, among other things, to relate determinants of square matrices (elementary or complete symmetric polynomials) to their traces (power sum symmetric polynomials), so not so ad hoc. The OEIS entry and the MO-Q "Cycling through the Zeta Garden: Zeta functions for graphs, cycle index polynomials, and determinants" has quite a lot of occurrences of this type of generating function with numerous refs in the comments.

Answer 4

In this article of Kapranov, you will find the motiv-ation of the zeta function of an algebraic curve.