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?


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 involves an Euler product. The product will be over all points $P$ of $V$ which are defined 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 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 $P$ mininally 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.

| cite | improve this answer | |
  • 4
    $\begingroup$ It's not "tricky" to use Galois orbits: $V$ is a variety over $\mathbf F_p$ and using Galois orbits is how we make $P$ into a point over $\mathbf F_p$: the point is the orbit of $P$ by ${\rm Gal}(\overline{{\mathbf F}_p}/\mathbf F_p)$. This need not be a rational point over $\mathbf F_p$ (a point of degree 1, i.e., a classical point over $\mathbf F_p$), but it should be regarded as a point over $\mathbf F_p$ nonetheless. If you want to view $V$ as a variety over a larger field $\mathbf F_{p^k}$ then its zeta-function uses orbits for the absolute Galois group of $\mathbf F_{p^k}$. $\endgroup$ – KConrad Mar 11 '19 at 15:56
  • 8
    $\begingroup$ The rationality for the zeta-function of curves over finite fields is a lot simpler than for higher-dimensional varieties, and it was known to F. K. Schmidt in the early 1930s using Riemann-Roch, long before there were any Weil conjectures. See mathoverflow.net/questions/14627/…. $\endgroup$ – KConrad Mar 11 '19 at 16:01
  • $\begingroup$ @KConrad Thank you for both of those comments. I actually didn't know that there was a simpler proof for curves, so thank you for giving me something to read up on :) $\endgroup$ – Wojowu Mar 11 '19 at 17:17
  • $\begingroup$ @KConrad, when treating orbits as "points" over $\mathbb{F}_p$, can one formulate the corresponding zeta function in terms of irreducible subschemes of $V$? $\endgroup$ – Rdrr Mar 11 '19 at 18:05
  • 1
    $\begingroup$ @Rdrr Viewing $V$ as a scheme over $\mathbb F_p$, the Galois orbits correspond bijectively to closed points of $V$. More specifically, Galois orbits of points minimally definable over $\mathbb F_{p^n}$ correspond to closed points with residue field isomorphic to $\mathbb F_{p^n}$. This is the way you generalize the zeta function to other schemes - you take the product over closed points with finite residue fields, and as the norm of a point you take the size of the residue field. $\endgroup$ – Wojowu Mar 11 '19 at 18:37

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.

| cite | improve this answer | |
  • 1
    $\begingroup$ This is nice because it explains both why we care that it is rational, and why we care about the absolute values of roots and poles (i.e. the "Riemann hypothesis") $\endgroup$ – Denis Nardin Mar 11 '19 at 20:46
  • $\begingroup$ @Dennis Nardin Could you explain why the above explains why we care about the absolute values of the roots and poles and how this is a "Riemann hypothesis" $\endgroup$ – Rdrr Mar 11 '19 at 22:36
  • $\begingroup$ @Rdrr The roots and poles of the rational function are the $\alpha_i$ and $\beta_j$ in the formula above, so we care about their absolute value because it describes the growth of $a_n$. It is called the "Riemann hypothesis" because it is vaguely analoguous to the Riemann hypothesis for the zeta function of a ring of integers (in both cases it's saying something about the real part of the roots of the zeta function: recall that the logarithm converts absolute values into real parts) $\endgroup$ – Denis Nardin Mar 12 '19 at 6:06
  • $\begingroup$ @Rdrr make the change of variables $x = q^{-s}$, which turns the zeta-function into a Dirichlet series in $s$. Then $|x| = q^{-{\rm Re}(s)}$, so knowing the absolute value of $x$ is the same as knowing the real part of $s$. For zeta-functions of curves over $\mathbf F_q$ this converts the statement that all zeros of that zeta-function as a function of $x$ (initially only defined for $|x| < 1/q$, but extended to all $x$ in $\mathbf C$ as a rational function of $x$) have absolute value $1/\sqrt{q}$ into the statement that all zeros of that zeta-function as a function of $s$ have real part $1/2$. $\endgroup$ – KConrad Mar 12 '19 at 7:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.