Zeta function of a curve

1 Example

Consider the curve C with equation X^{3}Y +Y^{3}Z +Z^{3}X = 0 defined over k = F_{2}. Let N_{d} be the number of points with coordinates in F_{2^{d}}.

We have N_{1} = 3: there are three points defined over F_{2} = {0, 1}, namely (0, 0, 1), (0, 1, 0) and (1, 0, 0).

We have N_{2} = 5: there are five points defined over F_{4} = {0, 1, ω, ω^{2}}, namely three over F_{2} and the two points (1, ω, ω^{2}), (1, ω^{2}, ω).

We have N_{3} = 24: there are 24 points defined over F_{8} = {0, 1, ζ, ..., ζ^{6}} where ζ^{7} = 1, namely three over F_{2} and the 21 points (1, ζ^{i}, ζ^{−2i}α) where 0 ≤ i ≤ 6 and α^{3} + α + 1 = 0.

Continuing, we find

\begin{matrix}d & 1 & 2 & 3 & 4 & 5 & 6 & … & 9 \\ N_{d} & 3 & 5 & 24 & 17 & 33 & 38 & … & 528 \end{matrix}

and more generally: N_{d} = 2^{d} + 1 if 3 ∤ d, and N_{d} = 2^{d} + 1 − a_{d} if 3|d, where the a_{i} are found from a_{3} = −15, a_{6} = 27, a_{3k+6} + 5a_{3k+3} + 8a_{3k} = 0 (k ≥ 1).

Put Z(C, t) = exp(∑ \frac{1}{i}N_{i}t^{i}). Then in this example

Z(C, t) = \frac{1 + 5t^{3} + 8t^{6}}{(1 − t)(1 − 2t)},

a simple rational function that encodes the values of all N_{i}.

A simpler example is the projective line L. Over F_{q} there are N = q + 1 points. Now Z(L, t) = exp(∑ \frac{1}{i}(q^{i}+1)t^{i}). But ∑ \frac{1}{i}t^{i} = − log(1−t) (for |t| < 1), and ∑ \frac{1}{i}q^{i}t^{i} = − log(1 − qt) (for |qt| < 1), so Z(L, t) = 1/(1 − t)(1 − qt).

Comparing this with the previous we see that a zeta function Z(C, t) = (1 + 5t^{3} + 8t^{6})/(1 − t)(1 − 2t) corresponds to N_{i} = q^{i} + 1 when 3 ∤ i. The recurrence a_{3k+6} + 5a_{3k+3} + 8a_{3k} = 0 has solution a_{3k} = c_{1}α^{k} + c_{2}β^{k} if α, β are the two solutions of x^{2} +5x+8 = 0. From a_{0} = 6, a_{3} = −15, we see c_{1} = c_{2} = 3. Now −3 ∑ \frac{1}{3i}α^{i}t^{3i} = log(1−αt^{3}) so Z(C, t) = (1−αt^{3})(1−βt^{3})/(1−t)(1−2t) = (1 + 5t^{3} + 8t^{6})/(1 − t)(1 − 2t). In other words, the given expression for Z(C, t) is equivalent to the given values of N_{i}.

2 Zeta function

Let X be an absolutely irreducible algebraic curve defined over F_{q} with N_{i} points over F_{q^{i}}. The zeta function of X is defined as Z(X, t) := exp(∑ \frac{1}{i}N_{i}t^{i}).

Hasse (for g = 1) and Weil (for the general case) showed that this function is a rational function of the form P (t)/(1−t)(1−qt) where P (t) is a polynomial in t. The degree of P (t) is 2g, where g is the genus of X. The polynomial P (t) has the factorization P (t) = ∏_{i=1}^{2g}(1 − α_{i}t) where |α_{i}| = \sqrt{q} for all i.

Taking logarithms we find

N_{i} = 1 + q^{i} − ∑ α^{i}

(where α^{−1} runs through the 2g roots of P (t)). It follows that

|N_{1} − (q + 1)| ≤ 2g\sqrt{q}.

This Hasse-Weil bound was improved by Serre to

|N_{1} − (q + 1)| ≤ g[2\sqrt{q}].

(Proof: The α’s are algebraic integers and occur in complex conjugate pairs; if a runs through the g sums α + ᾱ, then for both choices of the sign the product ∏([2\sqrt{q}] + 1 ± a) is a positive integer, hence at least 1; by the arithmeticgeometric mean inequality we have ∑([2\sqrt{q}] + 1 ± a) ≥ g. □)

Ihara’s bound is better for g > (q − \sqrt{q})/2 :

N_{1} ≤ q + 1 − \frac{1}{2}g + \sqrt{2(q + \frac{1}{8})g^{2} + (q^{2} − q)g}.

(Proof: We have 1 + q − ∑ α = N_{1} ≤ N_{2} = 1 + q^{2} − ∑ α^{2}. The α’s occur in complex conjugate pairs with product q, and if a runs through the g sums α +ᾱ then 1 + q − ∑ a ≤ 1 + q^{2} + 2qg − ∑ a^{2}. Now use g ∑ a^{2} ≥ (∑ a)^{2}. □)

If g = (q − \sqrt{q})/2 then both Hasse-Weil and Ihara say N_{1} ≤ q\sqrt{q} + 1, and this upper bound is achieved (when q is a square) by the Hermitean curves X^{r+1} + Y^{r+1} + Z^{r+1} = 0 where q = r^{2} (and by no other curves).