I'm looking for examples of the Weil conjecturesspecifically rationality of the zeta functionthat can be appreciated with minimal background in algebraic geometry. Are there varieties for which one can easily calculate the numbers of points over finite fields and witness the rationality directly? Of course, there are entirely straightforward examples coming from projective space and Grassmannians (or anything with a paving by affines).

5$\begingroup$ I'm not sure if this is too elementary, but the case of elliptic curves can be worked out pretty concretely. Silverman does this in his first book on elliptic curves, for example. $\endgroup$– RamseyJan 2, 2013 at 21:40

9$\begingroup$ Artin computed a bunch of examples by hand in his thesis. In Weil's paper, he does "Fermat" varieties (sum of nth powers). The latter is reproduced in detail in Ireland and Rosen. None of these examples need any algebraic geometry, but they might require a bit of number theory. $\endgroup$– Felipe VolochJan 2, 2013 at 21:45

3$\begingroup$ Even the Grassmannian example is not entirely trivial, since you get almost for free the generating function (generating polynomial, in this case) for the number of affine cells of each dimension. $\endgroup$– Noam D. ElkiesJan 2, 2013 at 22:14

4$\begingroup$ See the exercises at the end of chapter 11 in Ireland and Rosen's "A Classical Introduction to Modern Number Theory" $\endgroup$– StoppleJan 2, 2013 at 23:34

1$\begingroup$ There is an elementary proof of the Riemann hypothesis for curves (of any genus) over a finite field by Stepanov. See the Bourbaki talk by Bombieri (June 1973). $\endgroup$– Damian RösslerJan 3, 2013 at 18:38
6 Answers
Weil himself verifies the conjectures "by hand" for diagonal hypersurfaces, that is, hypersurfaces defined by an equation of the form
$$a_0x_0^{n_0}+a_1x_1^{n_1}+\cdots+a_kx_k^{n_k}=b.$$
The argument is pretty elementaryit essentially uses only character theory. It seems to me likely that this argument heavily influenced Dwork's original proof of rationality.
The paper is quite readable; I learned of it from Akshay Venkatesh.
One elementary example comes from the theory of elliptic curves, as in Silverman's book. Namely, given an elliptic curve over a finite field $\mathbb{F}_q$, the number of $\mathbb{F}_q$rational points is the degree of $1F$ for $F$ the Frobenius  that is, it's the kernel of the isogeny $1F$. Now you can compute the degree as $(1F)(1  F)^t$ (where $F^t$ is the dual isogeny) and this is $1  (F + F^t) + q$ since $q$ is the degree of the Frobenius. So the key quantity to compute is $F + F^t$, which is an integer  it's secretly the trace of $F$ on $l$adic cohomology. If you replace $F$ by $F^n$, this gives you the rationality of the zeta function of an elliptic curve. You can also get the Riemann hypothesis by purely elementary means by using the fact that the degree is a positive definite quadratic form and a CauchySchwarz type inequality. My guess is that something like this should work with higherdimensional abelian varieties (the $l$adic cohomology is an exterior algebra, as with the topological cohomology of a torus) as well.

1$\begingroup$ Indeed this does work in general for Abelian varieties; one may also deduce the Weil conjectures for curves this way by applying similar arguments to their Jacobians. That said, my feeling is that imitating Tate's thesis is the most "elementary" way to see rationality for curvessince developing the theory of Jacobians is pretty involved. $\endgroup$ Jan 2, 2013 at 21:46

$\begingroup$ @Daniel: Ah, so was this Weil's argument? I idly wondered about this once but never looked into it too carefully. $\endgroup$ Jan 2, 2013 at 22:04

2$\begingroup$ The argument in Mumford's Abelian Varieties book (which I presume was essentially the same as Weil's) does use the trace of Frobenius on the $\ell$adic Tate module (which is essentially what you are doing); of course this is secretly dual to the $\ell$adic $H^1$. That said, if all you care about is rationality and the functional equation for curves, you only need the idele class group. $\endgroup$ Jan 2, 2013 at 22:09


2$\begingroup$ Akhil, the endnotes of Milne's article on Jacobian varieties in the CornellSilverman book "Arithmetic Geometry" give the fascinating history. The functional equation and rationality for curves were known via RR before Weil, whose adelic contribution was an adelic proof of RR (preTate!). The real contribution was RH, which he proved via intersection theory on $X \times X$ (and wrote his book to make intersection theory rigorous in positive characteristic). His 2nd proof was via Jacobians, for which he created yet more fundamental ideas (birational group laws, abstract varieties, etc.). $\endgroup$ Jan 3, 2013 at 0:25
It's possible to give a semester course in number theory, free from overt algebraic geometry, that handles the zeta functions of curves over finite fields, proving the functional equation and Weil's RH for them. (And also does MordellWeil for elliptic curves over number fields). I taught such a course to second year grad students some 20 or 30 years ago.
One gets around the geometry of curves in 19th century Dedekind fashion, working with their function fieldsfinite extensions of k(t) and valuations, just as one does with number fields. RiemannRoch is done as in Chevalley's book using "repartitions". Rationality and the functional equation follow directly from RiemannRoch. The RH is done by Bombieri's elegant techniquefirst one uses RiemannRoch to get a good upper bound for the number of rational points, then one combines this upper bound with the functional equation to get a good lower bound. (Ayanta thinks the proof is miraculous but uninformative; this may be true of Stepanov's original version, but I find Bombieri's argument to be natural).
Of course there are defects to this approach. Algebraic geometry is far more enlightening. But it can be learned later, in whatever form the student finds appropriate. And there are also advantages. To do Weilstyle algebraic geometry one would have to worry about "fields of definition". And Grothendieck's version (which continues to intimidate me), would only appeal to the rare student at this level. That such beautiful mathematics can be presented in such an accessible fashion seems to me a boon.
The grandfather of all examples is by Gauss:
http://en.wikipedia.org/wiki/Weil_conjectures#Background_and_history
Of course Gauss didn't mention finite fields other than the prime field. I think it is in the nature of a remark that the method carries over, but I haven't written it down.
Any variety which admits a cell decomposition gives an example. Cell decomposition of $X$ is a stratification $X=\bigcup\limits_n X^n$ such that $X^n\setminus X^{n1}=\coprod\limits_{i=1}^{k_n}\mathbb{A}^n$ is a disjoint union of affine spaces.
Then $\#X(\mathbb{F}_q)=k_0+k_1q+\dots+k_dq^d$ so zetafunction of $X$ is a product of zetafunctions of affine spaces hence rational.
Examples of such varieties are given by flag varieties of reductive groups(Bruhat decomposition).
More generally, rationality of zetafunctions of moduli spaces of some objects can be proven elementary by counting these objects over finite field. For example, for quiver varieties it is done by V. Kac in "Root systems, representation of quivers and invariant theory".
Edit: Another, less elementary example is given by rational surfaces. For them Chow groups satisfiy the axioms of Weil cohomology theory, as explained in Manin's book "Cubic forms", so the rationality follows.
Perhaps the following two examples would be of interest to you; my apologies if they are too simple.
Notation: (in accordance with Koblitz's padic Numbers, padic Analysis, and ZetaFunctions)
Given $f \in \mathbb{F}_{q}[X_1, \ldots, X_n]$ let us define a sequence $N_s = (H_{f} (\mathbb{F}_{q^s})$, where $H_f(K) := \{(x_1, \ldots, x_n) \in \mathbb{A}^{n}_{K}  f(x_1, \ldots, x_n) = 0\}$.
The zeta function is then defined for a hyperplane $H_f$ and field $\mathbb{F}_q$ by:
$$Z(T) := \exp\big(\sum_{s=1}^{\infty} N_s T^s /s\big)$$
Before giving a few examples of the rationality of $Z(T)$, we recall the Maclaurin series
$$\log(1  T) = \sum_{s=1}^{\infty}T^s / s$$
Example 1. $f(x_1, \ldots, x_n) \equiv 0$. Then $N_s ={\mathbb{A}}_{\mathbb{F}_{q^s}}^{n} = q^{ns}$, so that we find $Z(T)$ becomes
$$\exp\big(\sum_{s=1}^{\infty} N_s T^s /s\big) = \exp\big(\sum_{s=1}^{\infty} (q^n T)^s /s\big) = \exp(\log(1q^n T)) = 1/(1  q^n T)$$
Example 2. Let $f = x_1 x_4  x_2 x_3  1$. We now consider two cases:
Case 1. $x_3 = 0$. Then $x_1 x_4  x_2 x_3 = 1$ becomes $x_1 x_4 = 1$. Since $x_2$ is out of the equation, it can be any element of $\mathbb{F}_{q^s}$. Thus, there are $q^s$ choices for $x_2$. Meanwhile, $x_1$ can be any nonzero element of $\mathbb{F}_{q^s}$, and in each case this will determine $x_4$. Hence there are $q^s(q^s  1) = q^{2s}  q^s$ points in $H_f$ when $x_3 = 0$.
Case 2. $x_3 \neq 0$. Then $x_1$ and $x_4$ can be any elements of $F_{q^s}$, and $x_3$ can be any nonzero element of $F_{q^s}$. But this completely determines $x_2$, so that there are $q^s q^s (q^s  1) = q^{3s}  q^{2s}$ points in $H_f$ when $x_3 \neq 0$.
Therefore, $N_s = q^{3s}  q^{2s} + q^{2s}  q^{s} = q^{3s}  q^{s}$, whence the zetafunction $Z(T)$ is
$$\frac{\exp(\sum_{s=1}^{\infty}q^{3s}T^s /s)}{\exp(\sum_{s=1}^{\infty}q^s T^s /s)} = \frac{1  qT}{1  q^3 T}$$
Note: The case for an affine variety $H_{f_1, \ldots, f_m}$ follows from the affine hypersurface $H_f$ case by a simple application of the Inclusion/Exclusion Principle. Bearing this in mind, it shouldn't be too hard to construct examples similar to those above over a variety for which the rationality can be witnessed directly. I imagine this would be somewhat tedious, though.

2$\begingroup$ Both of these examples fall into the case of "paving by affines" (or affines minus several points) that Jonathan mentions in his question. Your note is a nice pointthis reduction is essential to Dwork's proof of rationality. I see this as an indication that the hypersurface case is quite hard, though. $\endgroup$ Jan 4, 2013 at 6:07

3$\begingroup$ Dear Benjamin, One way to think about your example 2 is that you write your 3fold as $(\mathbb A^1 \setminus \{0\})\times \mathbb A^1 \coprod \mathbb A^2 \times (\mathbb A^1 \setminus \{0\}) = (\mathbb A^2 \setminus \mathbb A^1) \coprod (\mathbb A^3 \setminus \mathbb A^2) = \mathbb A^3 \setminus \mathbb A^1,$ and so the zetafunction is the ratio of the zeta function of $\mathbb A^3$ by that of $\mathbb A^1$. (In other words, zeta functions behave like a multiplicative version of Euler chars.: they are multiplicative w.r.t. disjoint unions.) Regards, Matthew $\endgroup$– EmertonJan 4, 2013 at 6:45