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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 $χ_{4}$ modulo $4$ is primitive. Its induction to modulus $12$ is zero at multiples of $2$ or $3$ , including $3$ , although $χ_{4} (3) = - 1$ . The induction of $χ_{4}$ 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

1.1

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

given

2.1

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

step 1.1given∎

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 $χ_{4}$ , then for $Re s > 1$ , $L (s, χ) = L (s, χ_{4}) (1 + 3^{- s}),$ because $χ_{4} (3) = - 1$ .

Facts & Assumptions

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

Verification

technique · direct

1.1

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

given

2.1

Substitute $p = 3$ and $χ_{4} (3) = - 1$ in the given formula.

step 1.1givenalgebra∎

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

Gauss sum for the character modulo 4

Example

For $χ_{4} (1) = 1$ , $χ_{4} (3) = - 1$ , one has $τ (χ_{4}) = 2 i$ and $ε (χ_{4}) = 1$ .

Facts & Assumptions

Given: The displayed Gauss normalization (Gauss sum of a Dirichlet character) and odd parity convention (Parity of a Dirichlet character).

Verification

technique · direct

1.1

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

givenalgebra

2.1

Its squared modulus is $4$ , agreeing with the primitive norm theorem; with $a = 1$ , $(- i) τ (χ_{4}) / 2 = 1$ .

step 1.1algebra∎

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 $\sum_{n} e^{- π n^{2} t}$ . For odd $χ_{4}$ , it is $\sum_{n} n χ_{4} (n) e^{- π n^{2} t / 4}$ ; 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

1.1

The parity- $a$ Mellin kernel is $\sum_{n} χ (n) n^{a} e^{- π n^{2} t / q}$ , so the two displayed kernels are its $a = 0, 1$ cases.

given

2.1

Applying the given transformation yields phase $(- i)^{a}$ , agreeing with the stated functional equations.

step 1.1given∎

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 $L (s, χ_{4})$ has forced zeros at $- 1, - 3, - 5, \dots$ . No assertion at $s = 0$ follows from this odd-parity rule.

Facts & Assumptions

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

Verification

technique · direct

1.1

Odd parity means $a = 1$ .

given

2.1

The given corollary gives $L (- 1 - 2 m, χ_{4}) = 0$ for $m \geq 0$ , which is exactly the displayed list.

step 1.1given∎

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

1.1

Induce $χ_{4}$ to modulus $12$ ; on units it agrees with reduction modulo $4$ and it vanishes at every nonunit modulo $12$ .

given

2.1

Its primitive ancestor is $χ_{4}$ , so uniqueness in the given theorem makes its conductor $4$ , not $12$ .

step 1.1given∎

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 $e (x) = \exp (2 π i x)$ , $τ (χ_{4}) = 2 i$ (Gauss sum for the character modulo 4).

Counterexample

technique · direct

1.1

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

givenalgebra

2.1

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

step 1.1algebra∎

Primitive Dirichlet L Functions and Functional Equations — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Primitive ancestors of small characters

Example

Facts & Assumptions

Verification

Finite Euler factor from modulus 4 to 12

Example

Facts & Assumptions

Verification

Gauss sum for the character modulo 4

Example

Facts & Assumptions

Verification

Even and odd theta kernels

Example

Facts & Assumptions

Verification

Trivial zeros of the beta function

Example

Facts & Assumptions

Verification

A modulus need not be a conductor

Statement refuted

Facts & Assumptions

Counterexample

Gauss-sum phase requires a convention

Statement refuted

Facts & Assumptions

Counterexample

Sources