Dirichlet's theorem promises infinitely many primes in every progression a, a+q, a+2q, \dots with \gcd(a,q)=1. The whole machinery of characters and Dirichlet $L$-functions is set up to isolate one residue class — but the argument only closes if a single, stubborn analytic fact holds:

That's it — one number, for each non-principal character, must miss zero. It looks almost trivial next to the grand statement about primes, yet it is the only hard step in the whole proof, and the one case that resisted Dirichlet longest still resists everyone today. This page is about why that tiny inequality carries the entire theorem, and why one half of it is easy and the other half genuinely deep.

Why exactly this fact is what's needed

The engine of Dirichlet's proof is the logarithm of an L-function. Taking logs of the Euler product L(s,\chi)=\prod_p (1-\chi(p)p^{-s})^{-1} and keeping only the leading prime term gives, as s \to 1^+,

Now average against the characters to pick out the class p \equiv a \pmod q. Orthogonality of characters turns the weighted sum of these logs into

Split the character sum into the principal character and the rest. The principal one, \chi_0, behaves like \zeta: it has a pole at s=1, so \log L(s,\chi_0) \to +\infty — this is the term that forces divergence and hence infinitely many primes. Every other term is \log L(s,\chi) for \chi \ne \chi_0, and here is the crux:

• If L(1,\chi) \ne 0 (and finite, which it is), then \log L(s,\chi) stays bounded as s\to 1^+. The sum \sum_p \chi(p)/p converges to a finite number — it contributes noise, not signal.
• If some L(1,\chi)=0, then \log L(s,\chi)\to -\infty, and this negative infinity could cancel the +\infty from \chi_0, leaving the prime sum finite — Dirichlet's conclusion collapses.

So the entire theorem reduces to: the non-principal terms must stay bounded, i.e. L(1,\chi)\ne 0. Grant that, and only the pole survives:

The product over all characters — the master identity

Everything hinges on one product. Multiply the L-functions of all \varphi(q) characters mod q together. The result is the Dedekind zeta function of the cyclotomic field \mathbb{Q}(\zeta_q), up to a handful of Euler factors:

Two facts about this product are the whole game:

• It has a simple pole at s=1, inherited from the single factor L(s,\chi_0) (which is \zeta(s) with the finitely many primes dividing q removed). One pole, order exactly one.
• Its Dirichlet coefficients are non-negative, and in fact \zeta_q(s) \ge 1 for real s>1. Written as a Dirichlet series it is \sum_n a_n n^{-s} with every a_n \ge 0 and a_1=1 — it counts something (ideals), so it cannot be small.

A product that is \ge 1 and has a genuine pole at s=1 cannot vanish there. If even one factor L(s,\chi) hit zero at s=1, it would kill the pole or drag the product to 0 — contradiction. That single observation is the skeleton of every non-vanishing proof; the two cases below just supply the missing bookkeeping.

The easy half: complex characters

Suppose \chi is complex — meaning \chi \ne \overline{\chi}, so it takes non-real values. Then \chi and its conjugate \overline{\chi} are two different non-principal characters, both appearing in the product \zeta_q(s). Because the coefficients of L(s,\chi) are the conjugates of those of L(s,\overline{\chi}), we have L(1,\overline{\chi}) = \overline{L(1,\chi)}. Hence:

One assumed zero instantly gives two zeros in the product — a double zero, of order \ge 2, at s=1.

Worked example — running the contradiction. Near s=1 track the order of vanishing (a pole counts as a negative-order zero) of each factor of \zeta_q(s)=\prod_\chi L(s,\chi):

Adding the orders, the product \zeta_q(s) vanishes to order at least -1+1+1 = 1 at s=1. So \zeta_q(1)=0: instead of a pole, the product has a zero. But we just saw \zeta_q(s)\ge 1 for real s>1, and it has an honest pole (it certainly does not tend to 0). Contradiction. Therefore L(1,\chi)\ne 0 for every complex \chi. Clean, and the conjugate partner did all the work.

The blue curve is the true product: a simple pole, blowing up to +\infty as s\to 1^+. The orange curve is what a double zero would force instead — the product bending down to 0, annihilating the pole. Both cannot describe the same function, and only the blue one is consistent with \zeta_q(s)\ge 1.

The hard half: real (quadratic) characters

Now let \chi be real: \chi=\overline{\chi}, taking only the values +1,-1,0 (these are exactly the quadratic characters, the Kronecker/Legendre symbols). The pairing trick dies: the "conjugate partner" of \chi is \chi itself, so an assumed zero of L(1,\chi) gives only a single zero, of order +1. Redo the bookkeeping: -1 + 1 = 0. The zero exactly cancels the pole, leaving a finite non-zero limit — no contradiction. The easy argument fails completely, and this is the case Dirichlet had to fight for. Two classic ways to win it:

Both routes reach the same wall: a real zero at s=1 would over-cancel a pole that provably survives. The arithmetic route is the more remarkable — it says the analytic quantity L(1,\chi) is secretly counting something (ideal classes), and you can't count fewer than 1 of a thing that exists.

The link to class numbers and Siegel zeros

The identity L(1,\chi)=(\text{positive constant})\cdot h is a two-way street. Read left to right it proves non-vanishing. Read right to left it turns an analytic lower bound on L(1,\chi) into a lower bound on the class number — the origin of the class number problem (how small can h be for imaginary quadratic fields?). Non-vanishing is only the qualitative version; the quantitative question "how far from zero?" is where the modern difficulty lives.

And it lives entirely in the real characters. A real L(s,\chi) could, as far as anyone can prove, have a real zero \beta extremely close to 1 — a so-called Siegel (exceptional) zero. We know each L(1,\chi)\ne 0, but Siegel's theorem gives only an ineffective lower bound L(1,\chi)\gg_\varepsilon q^{-\varepsilon}: it forbids a zero too close to 1 without telling you where the boundary is. Complex characters can never host such a zero (their partners would double it, exactly the easy contradiction). The exceptional zero, if it exists, must sit on a single real quadratic character — the same character whose non-vanishing was the hard case.