Dirichlet's theorem tells us that whenever \gcd(a,q)=1, the progression a, a+q, a+2q, \dots contains infinitely many primes. That is a yes/no answer to a counting question. But it leaves the sharper, more human question wide open: how many? If you split the primes up to a million by their remainder mod 10, do the four allowed endings — 1, 3, 7, 9 — get roughly a quarter each, or does one ending hog the primes and starve the others?

The answer is one of the most satisfying facts in analytic number theory: the primes are equidistributed. Each of the \varphi(q) admissible classes gets, asymptotically, the same share — 1/\varphi(q) of all the primes. This is the Prime Number Theorem for arithmetic progressions (PNT for APs), the quantitative refinement of Dirichlet, and this page is about what it says, why it is true, and the deep subtlety hiding in its error term.

From "infinitely many" to "how many"

Write \pi(x;q,a) for the number of primes p \le x with p \equiv a \pmod q. The ordinary Prime Number Theorem says \pi(x) \sim x/\log x. PNT for APs says those primes are shared out evenly among the \varphi(q) classes coprime to q.

Here \psi(x;q,a) = \sum_{\substack{n \le x \\ n \equiv a \,(q)}} \Lambda(n) is the summatory von Mangoldt function restricted to the class, where \Lambda(n) = \log p if n = p^k and 0 otherwise. The \psi form is the one the analysis actually proves; the \pi form follows by partial summation. The clean constant 1/\varphi(q) is the whole story: it does not depend on a. Nature does not play favourites among the coprime classes.

Worked example: primes mod 4

Take q = 4. The residues coprime to 4 are 1 and 3, so \varphi(4) = 2. PNT for APs predicts that the primes split evenly between the two shapes:

In words: about half the odd primes are of the form 4k+1 (5, 13, 17, 29, \dots) and about half are of the form 4k+3 (3, 7, 11, 19, \dots). The residue 2 \bmod 4 is not coprime to 4, so it hosts no odd primes at all — it simply does not enter the count. Two admissible classes, so each collects a \tfrac12 share; nothing is left over.

A more refined yardstick than x/\log x is the logarithmic integral \operatorname{Li}(x) = \int_2^x \frac{dt}{\log t}. The sharp expectation is that both counts hug \tfrac12 \operatorname{Li}(x). Below, the two real prime counts (computed by sieving every integer up to x) are plotted against that common target.

The two prime counts and their shared prediction are almost indistinguishable at this scale — visual proof of equidistribution. Look very closely, though, and the 3 \bmod 4 curve tends to sit a hair above the 1 \bmod 4 curve. That persistent tilt is not noise; it has a name.

The engine: characters turn a class into a sum

How do you prove a statement about one residue class using Dirichlet L-functions, which are global objects summed over all integers? The bridge is character orthogonality. The Dirichlet characters mod q form a group of exactly \varphi(q) functions, and they act as a set of filters. Their key property, for \gcd(a,q)=1, is

This indicator lets us detect the class a. Multiply by \Lambda(n) and sum over n \le x:

We have traded one hard sum over a residue class for \varphi(q) sums \psi(x,\chi), one per character — and each of those is governed by a single L-function. That is the entire strategic move.

Each character sum comes from its L-function

The twisted sum \psi(x,\chi) is extracted analytically exactly as in the ordinary PNT. The logarithmic derivative of L(s,\chi) is a Dirichlet series whose coefficients are precisely \Lambda(n)\chi(n):

A Perron contour integral then recovers the partial sum:

Push the contour left and pick up residues. Everything now hinges on which character you have:

• The principal character \chi_0. Here L(s,\chi_0) is essentially the Riemann zeta function (times a few Euler factors), so it inherits \zeta's simple pole at s=1. That pole contributes a residue of x, giving \psi(x,\chi_0) \sim x. This is the main term.
• Every non-principal character \chi \ne \chi_0. Here L(s,\chi) has no pole — it is entire. So there is no x term at all; \psi(x,\chi) = o(x) is pure error.

Feed this back into the character expansion. Only the \chi_0 term survives to leading order, and \overline{\chi_0}(a) = 1 for any admissible a:

There it is: the pole of the single principal character supplies the main term x/\varphi(q), identical for every class; the non-principal characters supply only error. That is why the share is exactly 1/\varphi(q) and the same for all admissible a.

Why the error is small: L(1,χ)≠0 and zero-free regions

Two facts make the non-principal terms genuinely negligible, not just hopefully so.

First, L(1,\chi) \ne 0. This is the same non-vanishing that powered Dirichlet's original theorem. If some L(1,\chi) were zero, the contour argument would spring a spurious pole and the error term could rival the main term — the class might even be starved of primes. Non-vanishing at s=1 is what keeps -L'/L regular on the edge \Re(s)=1.

Second, a zero-free region. Just as PNT needs \zeta(s) \ne 0 on the line \Re(s)=1 and a little to its left, PNT for APs needs L(s,\chi) \ne 0 in a similar region. A classical zero-free region of the shape

for the zeros \rho = \sigma + it pushes the contour left of \Re(s)=1 and turns the qualitative o(x) into a quantitative saving, an error like x\,\exp(-c\sqrt{\log x}). Zeros of L-functions control the error exactly as zeros of \zeta control the error in ordinary PNT — and the Generalized Riemann Hypothesis, that all these zeros lie on \Re(s)=\tfrac12, would give the best possible error O(\sqrt{x}\,\log^2 x).

What "equidistributed" really means

Equidistribution is a statement about proportions in the limit, and it is worth being precise about what it does and does not claim.

• It claims: the fraction of primes up to x falling in each admissible class tends to 1/\varphi(q). Long-run, the classes are indistinguishable by density.
• It does not claim: the classes are equal at any finite x, nor that they take turns, nor that the small differences between them are random. The Chebyshev bias is a real, structured, persistent inequality living entirely inside the error term — the leading term is blind to it.

So "the primes don't care about last digits" is true asymptotically and false in the fine print. The leading term is perfectly even; the character sums \psi(x,\chi) for \chi\ne\chi_0 encode every deviation from that perfect evenness, and their size is dictated by where the zeros of L(s,\chi) sit.