Questions: Primes in Arithmetic Progressions (Dirichlet's Theorem)

Question 1

Suppose L(1, χ) = 0 for some non-principal Dirichlet character χ mod q. What would this imply about primes?

• A. The L-functions would diverge at s = 1, making the theorem unprovable
• B. The arithmetic progression corresponding to χ would contain only finitely many primes
• C. Primes would be equidistributed among fewer than φ(q) residue classes
• D. The Euler product for L(s, χ) would fail to converge

Answer: (B)

The non-vanishing of L(1, χ) is the technical heart of Dirichlet's proof. If any L(1, χ) were zero, the contribution from the corresponding residue class to Σ1/p would be finite — meaning only finitely many primes in that class. The proof proceeds by contradiction: assuming L(1, χ) = 0 leads to a contradiction with the divergence that arises when all characters are combined, forcing every L(1, χ) ≠ 0 and guaranteeing infinitely many primes in every valid residue class.

Question 2

Among all primes, how are they distributed across the residue classes mod 10 that are coprime to 10 (i.e., classes 1, 3, 7, 9)?

• A. Class 1 contains more primes, since 1 is the identity element mod 10
• B. Class 9 contains fewer primes, since 9 = 3² is 'more composite'
• C. Each class contains asymptotically 1/4 of all primes — equal density
• D. The distribution is irregular and depends on how far out you count

Answer: (C)

Dirichlet's theorem guarantees equidistribution: each of the φ(10) = 4 valid residue classes mod 10 contains exactly 1/4 of all primes in the sense of natural density. The apparent 'compositeness' of 9 or the 'primeness' of 1 is irrelevant — what matters is only whether gcd(a, q) = 1. This equidistribution is a deep regularity hidden beneath the apparent randomness of primes.

Question 3

Dirichlet's theorem applies to any arithmetic progression a, a+q, a+2q, ... as long as gcd(a, q) = 1.

• T. True
• F. False

Answer: True

Exactly right. The condition gcd(a, q) = 1 ensures the residue class a mod q is coprime to the modulus — otherwise the progression could contain at most one prime (namely a itself, if it happens to be prime). When this condition holds, Dirichlet guarantees infinitely many primes with asymptotic density 1/φ(q) among all primes.

Question 4

Dirichlet's theorem that arithmetic progressions contain infinitely many primes can be proved using primarily elementary combinatorial or algebraic arguments, without complex analysis.

• T. True
• F. False

Answer: False

This is a common misconception. The proof unavoidably requires analytic tools — specifically, complex analysis to establish the non-vanishing of L(1, χ) for non-principal characters. Elementary proofs of Dirichlet's theorem exist for special cases (e.g., primes ≡ 1 mod 4), but the general theorem resisted elementary proof for over a century after Dirichlet. This is what makes it a landmark result in analytic number theory: an elementary-sounding statement required genuinely analytic methods to establish.

Question 5

Why is the non-vanishing of L(1, χ) for non-principal characters the crucial step in Dirichlet's proof, and what would go wrong if any such L-function vanished at s = 1?

Model answer: Dirichlet's strategy isolates primes in a target residue class using character orthogonality: summing χ(a)^{-1} log L(s, χ) over all characters χ extracts the contribution from primes in class a. If L(1, χ) = 0 for some non-principal χ, its logarithm would be finite at s = 1, and the contribution from the target residue class to Σ1/p would converge — implying only finitely many primes in that class. But the sum over all residue classes must diverge (since Σ1/p over all primes diverges). The non-vanishing forces each class to contribute an infinite sum, guaranteeing infinitely many primes in each valid class.

The argument is a careful balancing act: the principal character contributes the Riemann zeta function (which diverges), and the non-principal characters must all have finite but nonzero contributions at s = 1 to avoid cancellations that would wrongly imply only finitely many primes in some classes. The entire proof hangs on this one non-vanishing condition, which is why it is the technical heart of the theorem.