Dan, There was an extensive discussion about this very topic on mathoverflow about a year ago: http://mathoverflow.net/questions/15220/is-there-an-elementary-proof-of-the-... All the slick modern versions of the infinitude of primes in an arithmetic progression all proceed in the following way: Let N be the modulus of the progression, and chi: Z --> C^* be a non-trivial character (i.e. chi(a*b) = chi(a)chi(b), and chi(a) = 0 if and only if gcd(a,N) > 1). Then if we set pi(x,chi) = sum_{p <= x} chi(p), we need to prove that pi(x,chi) = o(pi(x)). We do this by showing that L(1+it,chi) != 0, where t is real, and L(s,chi) = sum_{n=1}^{infinty} chi(n)/n^s, which only converges absolutely when Re(s) > 1. The technical trick to go from that statement to the statement pi(x,chi) = o(x) is via a Tauberian theorem (a theorem that deduces the behavior of its function from the behavior of an average). Victor On Sat, Jan 28, 2012 at 12:46 AM, Dan Asimov <dasimov@earthlink.net> wrote:
Dirichlet's theorem on primes in arithmetic progressions says that for any sequence s_N := aN + b, with a,b in Z+ with GCD(a,b) = 1, there exists an infinite number of primes s_N.
I'm reading detailed versions of his proof, circa 1837.
I'm deeply impressed by the complexity of this proof, making use of Fourier analysis on finite abelian groups as well as the theory of (analytic) functions. The various modern versions I'm reading seem to agree almost exactly on the reasoning involved, so I presume that there isn't much to modernize.
Question: Is it really true that there isn't an "elementary" proof that avoids analytic functions? It appears that at least such do exist for a number of specific arithmetic progressions, like 6N+1 or 6N-1.
Somewhat related question: Given a finite abelian group G, its group G^ of "characters" is the group of all homomorphisms Hom(G,R/Z) from G to the circle group R/Z (the group operation is pointwise multiplication in R/Z). It's easy to see that |G^| = |G|, but to see that these two groups are isomorphic seems to be proved using the classification theorem for finite abelian groups.
Is there a proof that G^ is isomorphic to G that doesn't rely on this classification theorem?
--Dan
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