TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Euler product for Dirichlet L-functions

Statement

For every Dirichlet character $χ$ and every $s$ with $Re s > 1$ ,

$L (s, χ) = \prod_{p} \frac{1}{1 - χ (p) p^{- s}},$

and this product is nonzero on $Re s > 1$ .

Facts & Assumptions

Given: A Dirichlet character $χ$ and a complex number $s$ with $Re s > 1$ .

[L1]

The Dirichlet $L$ -function is $\sum_{n \geq 1} χ (n) n^{- s}$ (Dirichlet L-functions).

[L2]

A completely multiplicative arithmetic function has the geometric Euler product $\sum_{n \geq 1} f (n) n^{- s} = \prod_{p} (1 - f (p) p^{- s})^{- 1}$ at points of absolute convergence (Completely multiplicative Dirichlet series have geometric Euler factors).

Proof

technique · direct

1.1

A Dirichlet character is completely multiplicative on $Z$ , because it is multiplicative on unit classes and both sides vanish when a nonunit factor is present. Hence [L2] applied to $f = χ$ and [L1] give the Euler product formula on $Re s > 1$ .

L1L2givenalgebra

2.1

Write $σ = Re s > 1$ . Then $∣ χ (p) p^{- s} ∣ \leq p^{- σ}$ , so $\sum_{p} ∣ χ (p) p^{- s} ∣$ converges. Therefore $\sum_{p, m \geq 1} χ (p)^{m} / (m p^{m s})$ converges absolutely, and $- \log (1 - χ (p) p^{- s})$ is the absolutely convergent local series. Summing over $p$ shows that the Euler product of step 1.1 is the exponential of a convergent complex series, so it cannot vanish.

step 1.1algebra∎

Depends on

Used by

Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Kiran S. Kedlaya, Notes on Analytic Number Theory, Definition 3.4 (standard reference, not scraped)
Andrew V. Sutherland, Number Theory I, Definition 18.19 (standard reference, not scraped)