Let X be a projective nonsingular curve over Fp. The genus of X is g. Define the set of Fq-points X(Fq) to be Hom(Spec(Fq),X) and denote the size of the set by N, here q=pm.
We will prove the theorem in several steps. The first thing we need to do is relate N with the intersection multiplicity.
Definition 1. The Galois group Gal(Fqˉ/Fq) action on X as follows: For each σ∈Gal(Fqˉ/Fq), f:XFqˉ→XFqˉ is the fibre product σ×idX.
The geometric Frobenius morphism φ is the automorphism associated to the Frobenius on Fq.
Lemma 1. Let X is reduced k-scheme and Y is separated k-scheme. f:X→Y is a morphism. Let Γf(X) be the closed subscheme with reduced subscheme structure. Then Γf(X)≅X.
Proof
Proof. Note that Γf is obtained by pulling back the diagonal:
Δ is closed immersion so Γf is a closed immersion, hence Γf is an isomorphism to the schematic image Γf(X). Since X is reduced, we may factor Γf through Γf(X), which induces an morphism from Γf(X) to Γf(X). From the definition of reduced subscheme, Γf(X)≅Γf(X). ◻
Theorem 2. Denote the image of Γf and Δ by Γf and Δ. Then the intersection multiplicity Γf⋅Δ=N.
Proof
Proof. The fixed point of X under Galois group action Gal(Fqˉ/Fq) is exactly the Fq-points X(Fq). So x is Fq-point if and only if it’s in the intersection of Γf⋅Δ. It suffices to show each point has multiplicity 1. From the lemma, Γf and Δ are all isomorphic to X, so they are nonsingular. For every point (φ(x),x)∈Γf∩Δ, the tangent space is the product of corresponding tangent space X. Note (dφ)x=0, Γf and Δ intersect transversally, so each point has single multiplicity. ◻
To get the estimation of Γf⋅Δ, we need the Hodge index theorem. Recall some facts in the intersection theory: Fix an very ample line sheaf OX×X(1), and denote the corresponding hyperplane section by H.
Definition 2. Let L be a line sheaf on X. The Hilbert polynomial χ(L(n)) is a quadratic function 2α2(L)n2+α1(L)n+α0(L). The degree of L is defined to be α1(L)−α1(OX).
If D is a divisor of X, by Hirzebruch-Riemann-Roch theorem, D⋅H=deg(OX(D)), we define the number be the degree of the divisor D. The degree of D is a additive function on the divisor group Div(X) since the first chern class c1 is additive on tensor products.
Definition 3. Let f:X→Y be a proper morphism of varieties. Define the degree of f be the number [K(X),K(f(Y))]. If dimf(Y)<dimX, define the degree be 0.
Using the projection formula f∗(x⋅f∗y)=f∗x⋅y, we can show f∗C⋅f∗D=deg(f)C⋅D. For any line bundle L, deg(f∗L)=deg(f)deg(L). Note the degree is always a finite number.
Lemma 2. Let L1, L2 be two line sheaves on a variety X. Then Hom(L1,L2)=0 provided degL1>degL2.
Proof
Proof. Since rk(L1)=α2(OX)α2(L1) and rk(L2)=α2(OX)α2(L2), we have α2(L1)=α2(L2). The Hilbert polynomial P(L1)(n)<P(L2)(n). Suppose there is a morphism φ:L1→L2. Let K, F be the kernel and the image of φ, respectively.
We first show the support of F is X. Assume there is a point x∈X that is not in the support of F, since SuppF is closed, theres’s an open neighborhood U that has empty intersection with SuppF. We can choose an affine open set W=SpecA that has nonempty intersection with X−U and U. Pick an nonzero global section s supported on X−U and f is an element in ideal which define W∩Supps. Then s is zero in the localization of D(f)⊂W, so sfn=0 for some n. However, s is also in L1(W), which is torsion-free, contradict to the above argument.
The Hilbert polynomial of F is also a quadratic function. Since X is integral, the rank of F is integer, and it is 1. Thus rk(K)=0, which means α2(K)=0, and dimSuppK≤1. Using a similar argument, we can prove that K=0. Therefore, P(L1)(n)=P(F)(n)>P(L2)(n). This contradicts the fact that F is a subsheaf of L2. ◻
Lemma 3. Let {Di} be a set of divisor on X×X with bounded degree. Then {h0(OX×X(Di))} is bounded.
Proof
Proof. Assume degDi<m. We first reduce to the case degDi is a negative fixed number. For any divisor D in the family and the corresponding line sheaf OX(D)=L, consider the exact sequence
0→L(−m)⋅sL→L∣C→0
where s is a regular section of L and C is the support of s, by Bertini theorem, such s exists. Then h0(L)≤h0(L(−m))+h0(L∣C). Note that degC(L∣C)=degC(D∩C)=degX×X(D). The genus gC is independent on the choice of C, and is given by the adjunction formula gC=1+21(K⋅C+C⋅C).
If degC(L∣C)>2g−2, then h0(L∣C)=degC(L∣C)+1−g≤m+1−g.
If degC(L∣C)≤2g−2, then h0(L∣C)≤h0(L∣C(2g−1−m))=g.
Then we have h0(L∣C) bounded and we reduce to the case deg(Di) is negative constant by twisting a small number. Since a global section in h0(OX×X(D)) will induce a morphism from OX×X to OX×X(D), by above lemma, OX×X(Di) has no global sections. This finishes the proof. ◻
Theorem 3 (Hodge Index Theorem). Let D be a divisor on X×X. If deg(D)=0, then D⋅D<0.
Proof
Proof. By above lemma, {h0(nD)} and {h0(K−nD)} are all bounded. Using Riemann-Roch theorem on surface,
h0(nD)−h1(nD)+h0(K−nD)=21nD(nD−K)+1+pa
For large n, lhs is negative, so n2D⋅D<0. ◻
We next compute the self intersection number of Γf. Let deg(f)=d. Using the adjunction formula, Γf⋅(Γf+K)=2g−2. Since ωX×X=p1∗ωX⊗p2∗ωX, KX×X=p1∗KX+p2∗KX. Γf⋅p1∗KX is the same as the degΓf(p1∗ωX⊗OΓf)=degΓf(Γf∗p1∗ωX), since p1∘Γf=f, it is also the same as degΓf(f∗ωX)=deg(f)deg(ωX)=d(2g−2). Similarly, Γf⋅p2∗KX=2g−2. So Γf⋅Γf=−d(2g−2).
Let f=id, the self intersection number Δ⋅Δ=2−2g.
Let d1(D)=D⋅(P×X) and d2(D)=D⋅(X×P′) for some P, P′ closed points in X. Applying Bezout theorem, d1 and d2 are independent of the choice of P and P′.
Theorem 4. $$D\cdot D\leq 2d_{1}(D)d_{2}(D)$$
Proof
Proof. Let V be the vector space generated by basis P×X, X×P′ and D. The intersection product defines a quadratic form
M=⎝⎛01d1(D)10d2(D)d1(D)d2(D)D⋅D⎠⎞
det(M)=2d1(D)d2(D)−D⋅D. If det(M)<0, since M is indefinite, we may choose A1,A2,A3 such that Ai⋅Aj=0, and AiTMAi=ai such that a1,a2>0, a3<0. Then appropriate linear combination can give a degree 0 divisor with positive self intersection. Contradict to Hodge index theorem. ◻
By the above theorem, if we define D∗E=d1(E)d2(D)+d1(D)d2(E)−D⋅E, then D∗D≥0. Using the Cauchy inequality for bilinear forms, we have ∣Γf∗Δ∣≤(Δ∗Δ)(Γf∗Γf), that is
∣Γf⋅Δ−1−deg(f)∣≤2gdeg(f)
It remains to show the degree of geometric Frobenius is q. The degree is a generic condition so we may focus on an affine open set. The case K(X)=Fq(t) is clear, the morphism is given by σ:Fq(t)→Fq(tq). For the general case, K(X) is finite algebraic over Fq(t). Since the action on K(X) restrict on k(t) is the action on k(t), [K(X):k(t)]=[σK(X):σk(t)], and the results follow from the multiplicity of extension degree.