Simple Things We Don’t Know

This is a quite faithful rendering of a Colloquio De Giorgi I had the honor to give at Scuola Normale Superiore on March 21, 2012. The idea was to explain some open problems in arithmetic algebraic geometry which are simple to state but which remain shrouded in mystery. 1. An interactive game: dimension zero Suppose I give you an integer N ≥ 2, and tell you that I am thinking of a monic integer polynomial f(X) ∈ Z[X] whose discriminant ∆(f) divides some power of N . I tell you further, for every prime number p not dividing N , the number np(f) := #{x ∈ Fp|f(x) = 0 in Fp} of its solutions in the prime field Fp := Z/pZ. You must then tell me the degree of the polynomial f . In this “infinite” version, where I tell you the np(f) for every good prime, your task is simple; the degree of f is simply the largest of the np(f). Indeed, np(f) = deg(f) precisely when p is a prime which splits completely in the number field Kf := Q(the roots of f). By Chebotarev, this set of primes is infinite, and has density 1/#Gal(Kf/Q). If you do not have infinite patience, you may hope that you can specify a constantXN , depending only onN , such that it will be enough for me to tell you np(f) only for the good primes which are ≤ XN . Alas, this cannot be done. Whatever constant XN you choose, I will pick an integer a ≥ 2 such that N > XN , and take for my f the cyclotomic polynomial ΦNa(X), whose roots are the primitive N ’th roots of unity. With this choice of f , we have np(f) = 0 for all good primes p ≤ N. Indeed, with this choice of f , np(f) vanishes for a good prime p unless p ≡ 1 mod N, in which case np(f) = deg(f)(= φ(N) = Na−1φ(N), φ being Euler’s φ function.). But the condition that p be congruent to 1 mod N certainly forces p > N. In particular, we have np(f) = 0 for all good primes p ≤ N. 1We will call such a prime a good prime (for this problem). 1