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TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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The full product of Dirichlet L-functions has no zero on Re s = 1

Statement

Let

Fq(s):=χmodqL(s,χ).

Then Fq has no zero on the line Res=1. Moreover, Fq is meromorphic on a neighbourhood of the closed half-plane Res1, and any singularity at s=1 is at most a simple pole.

Facts & Assumptions

Given: A modulus q1 and the product Fq(s)=χmodqL(s,χ).

[L1]

For unit classes, the character sum χχ(u) is φ(q) when u=1 and 0 otherwise (Orthogonality relations for Dirichlet characters modulo q).

[L2]

Each L(s,χ) has its Euler product on Res>1 (Euler product for Dirichlet L-functions).

[L3]

The principal factor has one simple pole at 1 (The principal Dirichlet L-function factors through zeta).

[L4]

Every nonprincipal factor is holomorphic on Res>0 (Nonprincipal Dirichlet L-functions are holomorphic on Re s greater than 0).

[L5]

An Euler product whose logarithmic Dirichlet coefficients are nonnegative cannot vanish on Res=1 if it has at most a simple pole at 1 (Positive logarithmic Dirichlet series force boundary nonvanishing).

Proof

technique · direct
1.1

For Res>1, [L2] gives logFq(s)=χmodqp,m1χ(p)m/(mpms)=p,m1(mpms)1χmodqχ(pm). If pq, then every term is 0. If pq, then [L1] applied to the unit class of pm shows that the inner character sum is φ(q) when pm1(modq) and 0 otherwise. Hence the logarithmic coefficients of Fq are nonnegative.

L1L2givenalgebra
2.1

By [L3] and [L4], the product Fq is meromorphic on a neighbourhood of Res1, with at most a simple pole at 1 and no other singularities on the boundary line. Step 1.1 therefore places Fq under [L5], so Fq has no zero on Res=1. This proves both the nonvanishing claim and the stated meromorphic control at s=1.

step 1.1L3L4L5algebra

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