Simplifying the Proof of Dirichlet's Theorem

Abstract

Dirichlet showed that an arithmetic progression a, a + D, a + 2.D,. with > 1 and (a, D) = 1 contains infinitely many primes. Most of his argument is accessible to undergraduate mathematics majors, but a proof of the theorem is seldom presented to them because of the reputed difficulty of a key step-showing that certain infinite sums are non-zero. This note outlines a simple proof of the non-vanishing of these sums. The argument is very close to one given by Gelfond, [1], but is easier and works well in the classroom. The sums I'll treat may be described as follows. A character to the modulus D is a function X: Z -C satisfying: (1) If a b(D), then x(a) = x(b) (2) X(ab) = X(a)X(b) (3) x(a) = 0 if and only if (a, D) > 1. X is said to be real if it takes real values (which ca:n only be 1, -1, or 0), non-principal if it takes values other than 0 or 1. Suppose for example that is an odd prime. Then the Legendre symbol, taking each quadratic residue of to 1, each non-residue to -1 and each multiple of to 0 is a real non-principal character. For a non-principal X, EcX(n)/n converges. (This follows from summation by parts; see the argument given in the last paragraph of this note.) The usual approach to proving Dirichlet's theorem involves several standard analytic techniques (see [2], for example); the main non-formal step is showing that Exlx(n)/n = 0 whenever X is real and non-principal (the result is also needed for non-real X, but this is fairly easily handled). Dirichlet's original non-vanishing proof involved a detour through the theory of binary quadratic forms. Modern proofs generally use ideal theory in quadratic number fields or some complex variable theory. Elementary proofs are also known, but are more complicated than the one I'll now present. One begins by defining cn to be Ex(d) where d ranges over the positive divisors of n. Evidently Cpa = 1 + X(P) + x(p)2 + * +x(p)a > 0. It follows easily that Cn > 0 for all n. Furthermore, Cn = 1 whenever n is a power of a prime p dividing D. In particular ElTcn = ??' Next, following [1], one sets f(t) = E01X(n)tn/(1 tn). The series evidently converges in [0, 1). Expanding each tn/(1 tn) one finds that f(t) = Ex tn. The paragraph above shows that f(t) oo as t 1-. Suppose now that EwX(n)/n = 0. Then -f(t) = ESx(n)[{l/n(l t)} {tn/(l tn)}]; write this as Y2iX(n)bn. The critical observation is that b1 > b2 > b3 > * . . Note first that