Weil cohomology in practice

The hypothesis implies that

converges in the open disc \(|T|\lt q^{-d/2}\) (say, by the root test). In particular, the power series

converges uniformly on \(|T|\lt q^{-d/2}\) (i.e., there are no poles), so that \(|\alpha_i|\geq q^{-1/2}\text{.}\) The functional equation then tells us that \(\alpha_{2g-i} = \alpha_i/q\text{,}\) and hence we obtain \(|\alpha_i| = q^{-1/2}\) for all \(i\text{.}\)

There is nothing to verify if \(\#X(\FF_q)\) is empty, so assume there is an \(\FF_q\)-rational point on \(X\) and call it \(\infty\text{.}\) The goal is to write down a rational function on \(X\) with a controlled pole at \(\infty\) and with zeroes at \(X(\FF_q)\setminus \{\infty\}\text{;}\) this would then imply \(\#X(\FF_q)\leq 1+P\text{,}\) where \(P\) denotes the pole order of the function at \(\infty\text{.}\) To this end, let

Let us consider a function

with \(\nu_i\in H_l^{p^\mu}\) and \(s_i\in H^m\text{.}\) Suppose that \(f\) is not identically zero and that \(\delta(f) = \sum \nu_i s_i = 0\text{.}\) It follows that \(f\) vanishes on \(X(\FF_q)\setminus \{\infty\}\text{.}\) If we assume moreover that \(p^\mu\lt q\text{,}\) then \(f\) is a perfect \(p^\mu\)-th power and hence vanishes to order \(p^\mu\) at each of its zeroes; in particular, we obtain

Now we examine when such an \(f\) exists. By polar expansion around infinity, one may show that the map

is in fact a well-defined linear morphism; moreover, if we assume additionally that \(lp^u\lt q\text{,}\) a straightforward calculation gives an isomorphism \(H_l^{p^\mu}\cdot H_m^q\cong H_l^{p^\mu}\otimes H_m^q\text{,}\) and hence

By Riemann–Roch,

Hence \(\delta\) will have a nontrivial kernel whenever

To optimize this, choose \(\mu = s/2\) and \(m=\sqrt{q}+2g\text{.}\) All of the requisite conditions will be satisfied if we can choose an integer \(l\) for which

This is possible so long as \(q>(g+1)^4\text{;}\) with these choices of \(l,m,\mu\) the bound reduces to

as desired.

Let \(\infty\) be a point counted by \(N(X/S,\sigma)\) (if there are no such points there is nothing to prove). Consider the endomorphism \(\phi\colonequals\sigma^{-1}\circ \text{Frob}\) on \(X\text{;}\) it suffices to bound the fixed points of \(\overline{\phi}\) on \(X_{\overline{\FF}_q}\text{.}\)

Maintaining the notation of Theorem 5.1.5, any nonconstant function in \(\overline{\phi}^*(H_m)\) has a pole solely at \(\infty\text{,}\) since \(\overline{\phi}^*(H_m)\subset H_{qm}\text{.}\) Consider \(f = \sum v_i\overline{\phi}^*(s_i)\) in \(H_l^{p^\mu}\overline{\phi}^*(H_m)\text{,}\) and set \(\delta(f) = \sum v_is_i\text{.}\) As in the previous proof, if there exists a nonzero function \(f\) for which \(\delta(f) = 0\text{,}\) it follows that \(f\) vanishes at all points counted by \(N(X/S,\sigma)\text{.}\) One then proceeds as before to show that \(\delta\) must have a nontrivial kernel once \(q\) is suitably chosen with respect to \(g\text{.}\)

Both \(\sum_{\sigma \in \Gal(X/S)} N(X/S, \sigma)\) and \(\# \Gal(X/S) \# S(\FF_q)\) can be written as a sum over points \(P_0 \in S(\FF_q)\text{.}\) If \(P_0\) is not a branch point of \(\pi\text{,}\) then \(P_0\) makes identical contributions to both quantities. Thus the discrepancy comes only from fibers containing branch points, the number of which is controlled by the Riemann–Hurwitz formula.

If \(X\) itself can be written as a Galois cover of \(\PP^1\) via some map \(\pi\text{,}\) then Lemma 5.1.8 implies that an upper bound on \(N(X/S, \sigma)\) for each nontrivial automorphism \(\sigma\) implies a lower bound on \(N(X/S, \mathrm{id}_X) = \#X(\FF_q)\text{.}\) So in this case, we just apply Lemma 5.1.7 and we are done.

In general, \(X\) cannot always be written as a Galois cover of \(\PP^1\) (e.g., if it has trivial automorphism group and positive genus). However, we can always choose a finite separable morphism \(X\to \PP^1\) (perhaps after extending the base field, although this isn't really needed) and then take its Galois closure to obtain a Galois cover \(Z\to X\) for which \(Z\to X\to \PP^1\) is also Galois. By applying Lemma 5.1.8 to both \(Z \to X\) and \(Z \to \PP^1\text{,}\) we may again reduce the desired lower bound to some instances of Lemma 5.1.7.

See [63], Theorem~V.1.9.

For the first statement, we may take an orthogonal decomposition of the space of divisors to write \(D = aH+bE\text{,}\) where \(E\cdot H=0\text{.}\) Then \(D^2H^2 = ((aH)^2+(bE)^2)H^2\text{.}\) By Theorem 5.2.2, \((bE)^2\leq 0\text{,}\) so \(D^2H^2\leq (aH)^2H^2\text{.}\) But now \((aH)^2H^2 = (H\cdot(aH+bE))^2 = (D\cdot H)^2\) as desired. For the second statement, apply the first statement to the ample divisor \(\sigma(1,1) = \sigma(0,1)+\sigma(1,0)\text{.}\)