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The principal Dirichlet L-function factors through zeta

Statement

Let $χ_{0}$ be the principal Dirichlet character modulo $q$ . Then on $Re s > 1$ ,

$L (s, χ_{0}) = ζ (s) \prod_{p ∣ q} (1 - p^{- s}) .$

Consequently, the meromorphic continuation of $L (s, χ_{0})$ has a simple pole at $s = 1$ with residue

$\prod_{p ∣ q} (1 - \frac{1}{p}) = \frac{φ (q)}{q} .$

Facts & Assumptions

Given: The principal character $χ_{0}$ modulo $q$ .

[L1]

$χ_{0} (p) = 0$ for primes $p ∣ q$ , and $χ_{0} (p) = 1$ for primes $p ∤ q$ (The principal character modulo q).

[L2]

Dirichlet $L$ -functions and $ζ$ have Euler products on $Re s > 1$ (Euler product for Dirichlet L-functions, The Riemann zeta function has its Euler product on the half-plane $Re s > 1$ ).

[L3]

The meromorphic continuation of $ζ$ has a single simple pole at $1$ of residue $1$ (The Riemann zeta function extends meromorphically to the complex plane with its only pole at $1$ ).

Proof

technique · direct

1.1

By [L2], $L (s, χ_{0}) = \prod_{p} (1 - χ_{0} (p) p^{- s})^{- 1}$ . Using [L1], the Euler factors are $1$ at primes dividing $q$ and $(1 - p^{- s})^{- 1}$ at all other primes, so $L (s, χ_{0}) = \prod_{p ∤ q} (1 - p^{- s})^{- 1} = (\prod_{p} (1 - p^{- s})^{- 1}) \prod_{p ∣ q} (1 - p^{- s}) = ζ (s) \prod_{p ∣ q} (1 - p^{- s})$ .

L1L2givenalgebra

2.1

The finite factor $\prod_{p ∣ q} (1 - p^{- s})$ is holomorphic at $s = 1$ and has value $\prod_{p ∣ q} (1 - p^{- 1})$ . Multiplying this with the residue-one pole from [L3] gives a simple pole of $L (s, χ_{0})$ at $1$ with residue $\prod_{p ∣ q} (1 - p^{- 1})$ . Finally, $φ (q) = q \prod_{p ∣ q} (1 - p^{- 1})$ by the standard totient product, so the residue is $φ (q) / q$ .

L3step 1.1algebra∎

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Sources

Kiran S. Kedlaya, Notes on Analytic Number Theory, Theorem 3.5 (standard reference, not scraped)
Andrew V. Sutherland, Number Theory I, Proposition 18.20 (standard reference, not scraped)