# Bounding the first betti number

For some reason I am annoyed with the use of the structure of Jacobians and abelian varieties in the proof of finiteness of the first l-adic betti nr of a curve. Here is a silly argument to get around this.

Let k be an algebraically closed field. Let X be a smooth projective curve over k. We want to prove the number of finite etale G = Z/lZ covers is finite. In fact the argument will work for any finite group G.

If not then we get p_n : Y_n —> X, n = 1, 2, 3, 4, … which are finite etale covers with Galois group G and which are pairwise nonisomorphic. By Riemann-Hurwitz each of the curves Y_n has the same genus, call it g. Choose an integer d bigger than 2g and prime to |G|. Fix closed points x_i in X where i = 1, …, d. Set D = \sum x_i as a divisor on X. For each n pick points y_{i, n} in Y_n mapping to x_i. Then we can find a rational function f_n on Y_n which has poles exactly at the set of points y_{1, n}, …, y_{d, n}. Then we get a morphism

(f_n, p_n) : Y_n —–> P^1 x X

which is birational onto its image. By our careful choice of f_n the divisor class of the image is the class of a fixed line bundle, namely L = OP^1(|G|) ⊗ OX(D). Thus in the linear system |L| we get an infinitude of curves whose normalization is finite etale Galois over X with group G. By standard (but nontrivial) arguments, we get an actual family of such curves which contains an infinity of our Y_n. However, there are no moduli of finite \’etale Galois covers (another standard fact). Hence infinitely many Y_n are isomorphic (as Galois covers) which is a contradiction.

Of course this is quite useless!