The Pólya-Vinogradov Inequality

  • Davenport H
N/ACitations
Citations of this article
3Readers
Mendeley users who have this article in their library.
Get full text

Abstract

Let χ : Z → C be a primitive Dirichlet character modulo m. χ being a Dirichlet character modulo m means that χ(kn) = χ(k)χ(n) for all k, n, that χ(n + m) = χ(n) for all n, and that if gcd(n, m) > 1 then χ(n) = 0. χ being primitive means that the conductor of χ is m. The conductor of χ is the smallest defining modulus of χ. If m is a divisor of m, m is said to be a defining modulus of χ if gcd(n, m) = 1 and n ≡ 1 (mod m) together imply that χ(n) = 1. If n ≡ 1 (mod m) then χ(n) = 1 (sends multiplicative identity to multiplicative identity), so m is a defining modulus, so the conductor of a Dirichlet character modulo m is less than or equal to m. We shall prove the Polya-Vinogradov inequality for primitive Dirchlet characters. The same inequality holds (using an O term rather than a particular constant) for non-primitive Dirichlet characters. The proof of that involves the fact [1, p. 152, Proposition 8] that a divisor m of m is a defining modulus for a Dirichlet character χ modulo m if and only if there exists a Dirichlet character χ modulo m such that χ(n) = χ 0 (n) · χ (n) n ∈ Z, where χ 0 is the principal Dirichlet character modulo m. (The principal Dirichlet character modulo m is that character such that χ(n) = 0 if gcd(n, m) > 1 and χ(n) = 1 otherwise.) If χ is a Dirichlet character modulo m, define the Gauss sum G(·, χ) : Z → C corresponding to this character by G(n, χ) = m−1 k=0 χ(k)e 2πikn/m , n ∈ Z. The Polya-Vinogradov inequality states that if χ is a primitive Dirichlet character modulo m, then n≤N χ(n) < √ m log m.

Cite

CITATION STYLE

APA

Davenport, H. (1980). The Pólya-Vinogradov Inequality (pp. 135–137). https://doi.org/10.1007/978-1-4757-5927-3_23

Register to see more suggestions

Mendeley helps you to discover research relevant for your work.

Already have an account?

Save time finding and organizing research with Mendeley

Sign up for free