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Primitive Dirichlet L Functions and Functional Equations — Examples

1 · Prerequisites

2 · Summary

These computations distinguish a displayed modulus from a conductor and keep the additive-character convention visible in every Gauss-sum phase.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Primitive ancestors of small characters

Example

The odd character χ4 modulo 4 is primitive. Its induction to modulus 12 is zero at multiples of 2 or 3, including 3, although χ4(3)=−1. The induction of χ4 to modulus 8 has conductor 4; the primitive quadratic character of conductor 8 has values 1,−1,−1,1 on 1,3,5,7.

Facts & Assumptions

Given: The unique-ancestor theorem (Unique primitive ancestor of a Dirichlet character).

Verification

technique · direct
1.1given

The displayed values are multiplicative on the listed unit groups; their zero extensions are therefore characters.

2.1step 1.1given∎

Direct reduction of their unit values identifies the stated ancestor, and uniqueness in the given theorem fixes the conductors.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Finite Euler factor from modulus 4 to 12

Example

If χ modulo 12 is induced from χ4, then for Re⁡s>1, L(s,χ)=L(s,χ4)(1+3−s), because χ4(3)=−1.

Facts & Assumptions

Given: The finite-factor formula (Finite Euler factors under character induction).

Verification

technique · direct
1.1given

The primes dividing 12 but not 4 consist only of 3.

2.1step 1.1givenalgebra∎

Substitute p=3 and χ4(3)=−1 in the given formula.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedverified 2026-09-26 (gpt-6-sol)Open item page →

Gauss sum for the character modulo 4

Example

For χ4(1)=1, χ4(3)=−1, one has τ(χ4)=2i and ε(χ4)=1.

Facts & Assumptions

Given: The displayed Gauss normalization (Gauss sum of a Dirichlet character) and odd parity convention (Parity of a Dirichlet character). The root number is ε(χ)=(−i)aτ(χ)/q for primitive χ of parity a (Functional equation for primitive Dirichlet L-functions).

Verification

technique · direct
1.1givenalgebra

The two nonzero terms are e(1/4)−e(3/4)=i−(−i)=2i.

2.1step 1.1givenalgebra∎

Its squared modulus is 4, agreeing with the primitive norm theorem; χ4 is primitive of parity a=1, so the cited root-number formula gives ε(χ4)=(−i)τ(χ4)/2=1.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Even and odd theta kernels

Example

For the even character modulo 1, the kernel is ∑ne−πn2t. For odd χ4, it is ∑nnχ4(n)e−πn2t/4; the extra n produces the −i in its transformation.

Facts & Assumptions

Given: Twisted Poisson summation (Twisted Poisson summation) and the functional equation (Functional equation for primitive Dirichlet L-functions).

Verification

technique · direct
1.1given

The parity-a Mellin kernel is ∑nχ(n)nae−πn2t/q, so the two displayed kernels are its a=0,1 cases.

2.1step 1.1given∎

Applying the given transformation yields phase (−i)a, agreeing with the stated functional equations.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Trivial zeros of the beta function

Example

The beta function L(s,χ4) has forced zeros at −1,−3,−5,…. No assertion at s=0 follows from this odd-parity rule.

Facts & Assumptions

Given: χ4 is odd (Gauss sum for the character modulo 4) and the trivial-zero corollary (Parity-forced trivial zeros).

Verification

technique · direct
1.1given

Odd parity means a=1.

2.1step 1.1given∎

The given corollary gives L(−1−2m,χ4)=0 for m≥0, which is exactly the displayed list.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

A modulus need not be a conductor

Statement refuted

“Every character modulo q has conductor q.”

Facts & Assumptions

Given: Induction (Induced Dirichlet characters) and unique primitive ancestors (Unique primitive ancestor of a Dirichlet character).

Counterexample

technique · direct
1.1given

Induce χ4 to modulus 12; on units it agrees with reduction modulo 4 and it vanishes at every nonunit modulo 12.

2.1step 1.1given∎

Its primitive ancestor is χ4, so uniqueness in the given theorem makes its conductor 4, not 12.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Gauss-sum phase requires a convention

Statement refuted

“The displayed Gauss-sum phase is independent of the additive-character convention.”

Facts & Assumptions

Given: Under e(x)=exp⁡(2πix), τ(χ4)=2i (Gauss sum for the character modulo 4).

Counterexample

technique · direct
1.1givenalgebra

Replace e(x) by e(−x). The two terms become e(−1/4)−e(−3/4)=−i−i=−2i.

2.1step 1.1algebra∎

This is the conjugate phase, not 2i, although its squared modulus remains 4.

Sources