René Schoof Roma, version June 2003 We prove the analogue of the Riemann hypothesis for ζ-functions associated to curves over finite fields. This result was claimed by André Weil in 1940. Weil published his proof in 1948. Our proof follows E. Bombieri’s 1973 Séminaire Bourbaki exposé of Stepanov’s proof.

Let F_{q} be a finite field of characteristic p. Let X be a smooth projective absolutely irreducible curve over F_{q} of genus g. Stepanov’s proof proceeds by constructing a rational function f on X that vanishes to high order m at the F_{q}-rational points of X, but whose number of poles is bounded by B. Since the number of poles of f is equal to its number of zeroes, we find that

m#X(F_{q}) ≤ B.

This implies an upper bound for #X(F_{q}). A modification of this idea leads to a lower bound and to an estimate for the absolute values of the reciprocal zeroes ϕ of the zeta functions Z_{X}(T) of X. The functional equation implies then that |ϕ| = \sqrt{q} which is the analogue of the Riemann hypothesis.

Let ∞ be a point in X(F_{q}). We study the spaces

H^{0}(m(∞)) = {f ∈ F_{q}(X) : ord_{∞}(f) ≥ −m}

for m ≥ 0. By the Riemann-Roch Theorem, the dimension of H^{0}(m(∞)) is equal to m − g + 1 whenever m > 2g − 2. We have that

… ⊂ H^{0}(m(∞)) ⊂ H^{0}((m + 1)(∞)) ⊂ …

Let π denote a uniformizer at ∞. The sequences

0 ⟶ H^{0}(m(∞)) ⟶ H^{0}((m + 1)(∞)) ⟶^{e} F_{q}

given by e(g) = (gπ^{m+1})(∞) are exact. Therefore the codimension of H^{0}(m(∞)) inside H^{0}((m + 1)(∞)) is at most 1. It follows that we can choose an F_{q}-basis e_{1}, …, e_{t} of H^{0}(m(∞)) for which

ord_{∞}(e_{1}) > ord_{∞}(e_{2}) > … > ord_{∞}(e_{t}).  (∗)

For a positive integer μ we let H^{0}(m(∞))^{p^{μ}} denote the F_{q}-vector space of functions of the form f^{p^{μ}} where f ∈ H^{0}(m(∞)). By H^{0}(a(∞))^{p^{μ}}H^{0}(b(∞))^{q} we denote the F_{q}-subspace of H^{0}((ap^{μ} + bq)(∞)) generated by products fg of functions f ∈ H^{0}(a(∞))^{p^{μ}} and g ∈ H^{0}(b(∞))^{q}.

Lemma 1. Let a, b > 2g − 2 and let p^{μ} be a power of p for which ap^{μ} < q. Let e_{1}, …, e_{a−g+1} and f_{1}, …, f_{b−g+1} denote bases as in (*) of the vector spaces H^{0}(a(∞)) and H^{0}(b(∞)) respectively. Then the products e_{i}^{p^{μ}}f_{j}^{q} for 1 ≤ i ≤ a−g +1 and 1 ≤ j ≤ b−g +1 form an F_{q}-basis for H^{0}(a(∞))^{p^{μ}}H^{0}(b(∞))^{q}.

Proof. Clearly the products e_{i}^{p^{μ}}f_{j}^{q} generate the vector space H^{0}(a(∞))^{p^{μ}}H^{0}(b(∞))^{q}. The point is to show that they are independent. To this end we observe that the functions e_{i}^{p^{μ}}f_{j}^{q} have poles of distinct orders at ∞. Indeed, if ord_{∞}(e_{i}^{p^{μ}}f_{j}^{q}) = ord_{∞}(e_{i′}^{p^{μ}}f_{j′}^{q}) we have that

p^{μ}(ord_{∞}(e_{i}) − ord_{∞}(e_{i′})) = q(ord_{∞}(f_{j′}) − ord_{∞}(f_{j}).

Since the absolute value of p^{μ}(ord_{∞}(e_{i})−ord_{∞}(e_{i′})) does not exceed p^{μ}a < q, we conclude that i = i′ and hence that j = j′.

The lemma now follows from the fact that any functions h_{i} ∈ H^{0}(m(∞)) with poles of distinct orders at ∞, are necessarily linearly independent. This proves the lemma.

Proposition 2. Let X be a curve of genus g over F_{q}. If q is a square and q > (2g + 2)^{2} then we have that

#X(F_{q}) < q + (3g + 1)\sqrt{q}.

Proof. We may assume that X(F_{q}) ≠ ∅ and choose a point ∞ ∈ X(F_{q}). We consider the space H^{0}(a(∞))^{p^{μ}}H^{0}(b(∞))^{q} of Lemma 1 with p^{μ} = \sqrt{q} and a = \sqrt{q} − 1. Then a > 2g − 2. We choose b > 2g − 2 later. These choices imply that the dimension of the space H^{0}(a(∞))^{p^{μ}}H^{0}(b(∞))^{q} is equal to

dim H^{0}(a(∞)) × dim H^{0}(b(∞)) = (a − g + 1)(b − g + 1).

We define an F_{q}-linear homomorphism

H^{0}(a(∞))^{p^{μ}}H^{0}(b(∞))^{q} ⟶ H^{0}((ap^{μ} + b)(∞))

by mapping the basis vectors e_{i}^{p^{μ}}f_{j}^{q} to e_{i}^{p^{μ}}f_{j}. This homomorphism is well defined by Lemma 1. The dimension of H^{0}((ap^{μ} + b)(∞)) is equal to ap^{μ} + b − g + 1. Therefore, if

(a − g + 1)(b − g + 1) > ap^{μ} + b − g + 1,

there is a non-zero function f = ∑_{ij} c_{ij}e_{i}^{p^{μ}}f_{j}^{q} in the kernel of this homomorphism. This function f has the property that

f(P) = ∑_{ij} c_{ij}e_{i}(P)^{p^{μ}}f_{j}(P)^{q} = ∑_{ij} c_{ij}e_{i}(P)^{p^{μ}}f_{j}(P) = 0

whenever P ∈ X(F_{q}). Since p^{μ} < q, this shows that f has a zero of multiplicity at least p^{μ} in every P ∈ X(F_{q}) − {∞}. On the other hand the order of its pole at ∞ is at most ap^{μ} + bq. It follows that

p^{μ}(#X(F_{q}) − 1) ≤ #{zeroes of f} = #{poles of f} ≤ ap^{μ} + bq.

Substituting p^{μ} = \sqrt{q} and a = \sqrt{q} − 1 this means that

#X(F_{q}) ≤ (b + 1)\sqrt{q}.

Now we choose b. This number should be at least 2g −2 and satisfy (a−g +1)(b−g +1) > ap^{μ} − g + 1. This means that any choice of b must satisfy

b > g − 1 + \frac{(q − 1)\sqrt{q}}{\sqrt{q} − 1 − g} .

Note that the right hand side of this inequality is larger than 2g − 2. We choose b as small as possible. This b satisfies b ≤ g + \frac{(q−1)\sqrt{q}}{\sqrt{q}−1−g} so that

#X(F_{q}) ≤ (1 + g + \frac{(q − 1)\sqrt{q}}{\sqrt{q} − 1 − g})\sqrt{q} ≤ \sqrt{q}(g + 1) + q + \frac{g}{\sqrt{q} − 1 − g}

which is at most q + (3g + 1)\sqrt{q} since \sqrt{q} > 2g + 2. This proves the Proposition.

Theorem 3. Let X be a curve of genus g over a finite field k. Then there is an extension field k ⊂ F_{q} for which q is a square larger than (2g + 2)^{2} and

#X(F_{q^{k}}) = q^{k} + O(q^{k/2})

for all k > 0. Here the O-symbol depends only on the curve X.

Proof. Let f ∈ F_{q}(X) be a function that is not a p-th power. Then the field extension F_{q}(f) ⊂ F_{q}(X) is separable and corresponds to a non-constant separable morphism π_{X} : X ⟶ P^{1}. Let K be a finite extension of F_{q}(X) that is also a Galois extension of F_{q}(f). Then K is the function field of an absolutely irreducible projective curve Y over F_{q^{k}} for some k ≥ 1. It may happen that the field of constants F_{q^{k}} of K is strictly larger than F_{q}. Extending the field of constants of F_{q}(f) and F_{q}(X) if necessary, we may assume that all three curves P^{1}, X and Y have the same field of constants, which we denote by F_{q} again.

Corresponding to the field inclusions F_{q}(f) ⊂ F_{q}(X) ⊂ F_{q}(Y) = K we have the F_{q}-morphisms

Y ⟶^{π} X ⟶^{π_{X}} P^{1}.

Let π_{Y} = π_{X} · π be the composite morphism Y ⟶ P^{1}. Let G = Gal(K/F_{q}(f)) = Gal(Y/P^{1}) and let H denote its subgroup Gal(K/F_{q}(X)) = Gal(Y/X).

Fix k ≥ 1. Let A denote the set of unramified points P ∈ Y (\overline{F}_{q}) whose image π_{Y} (P) is in P^{1}(F_{q^{k}}). Since P^{1} has q^{k} + 1 rational points over F_{q^{k}}, it is immediate that

#A = (q^{k} + 1)#G + O(1)

where O(1) indicates a number that is at most the number of ramification points of the covering Y of P^{1} over \overline{F}_{q}. In particular, it is bounded independently of k.

Let φ denote the Frobenius morphism. It raises the coordinates of a point to the power q. The F_{q}-morphisms π, π_{X} and π_{Y} introduced above all commute with φ. It follows that f or every unramified point P ∈ A the conjugate point φ(P) maps to the same point in P^{1} as P does. Therefore there is a unique σ ∈ G such that φ(P) = σ(P). For every σ ∈ G put

A_{σ} = {P ∈ A: φ(P) = σ(P)}.

The set A is a disjoint union of the A_{σ}’s.

By taking q so large that q > (2g + 2)^{2} we may assume that there is a point ∞ ∈ Y (F_{q}) and that the conditions of Proposition 2 are satisfied. Let σ ∈ G. If σ = id_{Y}, Proposion 2 implies that #A_{σ} ≤ q^{k} + (3g + 2)\sqrt{q^{k}}. We modify the proof of Prop. 2 and show that the same inequality holds for every σ. We let a = \sqrt{q} − 1 and p^{μ} = \sqrt{q} and consider the vector space H^{0}(a(∞))^{p^{μ}}H^{0}(b(∞))^{q} of functions in F_{q^{k}}(Y). We map it to H^{0}(p^{μ}a(∞) + b(σ^{−1}(∞)) by mapping, in the notation of the proof of Prop. 2, the basis vector e_{i}^{p^{μ}}f_{j}^{q} to e_{i}^{p^{μ}} · (f_{j} ∘ σ). Any function f = ∑_{ij} c_{ij}e_{i}^{p^{μ}}f_{j}^{q} in the kernel of this map has the property that

f(P) = ∑_{ij} c_{ij}e_{i}(P)^{p^{μ}}f_{j}(P)^{q} = ∑_{ij} c_{ij}e_{i}(P)^{p^{μ}}f_{j}(φ(P)) = ∑_{ij} c_{ij}e_{i}(P)^{p^{μ}}f_{j}(σ(P)) = 0.

for every point P ∈ A_{σ}. The same arguments as in Prop. 2 now show that

#A_{σ} ≤ q^{k} + (2g + 3)\sqrt{q^{k}}

for every σ ∈ G. Since

∑_{σ∈G} #A_{σ} = #A = (q + 1)#G + O(1),

there is a constant C > 0 so that #A_{σ} ≥ q^{k} + 1 − Cq^{k/2}. Since ∪_{σ∈H}A_{σ} is precisely the set of unramified points P ∈ Y (\overline{F}_{q}) for which π(P) ∈ X(F_{q}), we see that

#H · #X(F_{q}) = ∑_{σ∈H} #A_{σ} = #H(q^{k} + 1) + O(q^{k/2})

as k → ∞. This proves the Theorem.

Corrolary 3. (A. Weil, 1948) Let X be a curve of genus g over a finite field F_{q}. Then the reciprocal roots ϕ ∈ C of the function Z_{X}(T) satisfy

|ϕ| = \sqrt{q}.

Proof. Since the reciprocal zeroes of the zeta function of X over F_{q^{m}} are the m-th powers of the zeroes of Z_{X}(T), it suffices to give the proof for a power of q. We will call this power q again and choose it so large that the condition of Proposition 2 is satisfied: q is a square exceeding (2g + 2)^{2} and the curves X, Y and the morphisms π_{Y}, π_{X} and π of Proposition 2 are all defined over F_{q}. We deduce that for large enough k we have that

#X(F_{q^{k}}) = q^{k} + 1 + O(q^{k/2}).

and therefore, with the usual notation, that

∑_{ϕ} ϕ^{k} = O(q^{k/2}).

This implies that the function f(z) = ∑_{ϕ} \frac{1}{1−ϕz} has a radius of convergence at least as large as q^{−1/2}. Therefore |ϕ| ≤ \sqrt{q} for each ϕ. The theorem now follows from the functional equation satisfied by Z_{X}(T): when ϕ is a reciprocal root, so is q/ϕ and it follows that |ϕ| = \sqrt{q} for all ϕ as required.