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TheoremStatement: Literature-sourcedProof: AI-adaptedverified 2026-09-26 (gpt-6-sol)
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The principal Dirichlet L-function factors through zeta

Statement

Let χ0 be the principal Dirichlet character modulo q. Then on Re⁡s>1,

L(s,χ0)=ζ(s)∏p∣q(1−p−s).

Consequently, the meromorphic continuation of L(s,χ0) has a simple pole at s=1 with residue

∏p∣q(1−1p)=φ(q)q.

Facts & Assumptions

Given: The principal character χ0 modulo q.

[L1]

χ0(p)=0 for primes p∣q, and χ0(p)=1 for primes p∤q (The principal character modulo q).

[L3]

On Re⁡s>0, the meromorphic continuation of ζ has a single simple pole at 1 of residue 1 (For Re⁡s>0, zeta admits the fractional-part integral formula with a simple residue-one pole at 1).

Proof

technique · direct
1.1L1L2givenalgebra

By [L2], L(s,χ0)=∏p(1−χ0(p)p−s)−1. Using [L1], the Euler factors are 1 at primes dividing q and (1−p−s)−1 at all other primes, so L(s,χ0)=∏p∤q(1−p−s)−1=(∏p(1−p−s)−1)∏p∣q(1−p−s)=ζ(s)∏p∣q(1−p−s).

2.1L3step 1.1algebra∎

The right-hand side of step 1.1, using [L3], defines a meromorphic continuation of L(s,χ0) to Re⁡s>0, which contains a neighborhood of 1. Any further meromorphic continuation agrees with this germ. The finite factor ∏p∣q(1−p−s) is holomorphic at s=1 and has value ∏p∣q(1−p−1). Multiplying this with the residue-one pole from [L3] gives a simple pole of L(s,χ0) at 1 with residue ∏p∣q(1−p−1). Finally, φ(q)=q∏p∣q(1−p−1) by inclusion-exclusion on the finite set of primes dividing q (with empty product 1 when q=1), so the residue is φ(q)/q.

Depends on

Used by

Dependency tree · two levels

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Sources