Definition 2.2.1.
An arithmetic function \(f\) is multiplicative when it satisfies:- \(f\) is not identically zero, and
- if \((m,n) = 1\) then \(f(mn) = f(m)f(n)\text{.}\)
Section 2.2 Multiplicative functions
Remark 2.2.2.
Crisman allowed the identically zero function to be considered multiplicative. That forces a lot of statements to say “if \(f\) is multiplicative and not identically zero, then...”.
We saw that \(\phi\) is multiplicative. However it is not completely multiplicative: \(\phi(p^2) \neq \phi(p)^2\text{.}\)
The Möbius function is multiplicative, but not completely multiplicative, since \(\mu(p) = -1\) but \(\mu(p^2) = 0\text{.}\) The Liouville function \(\lambda(p_1^{a_1} \dotsm p_k^{a_k}) = (-1)^{a_1+\dotsb+a_k}\) is completely multiplicative.
Observe that for a multiplicative (or completely multiplicative) function \(f\) we must have \(f(1) = 1\text{.}\) Indeed, \(f(1) = f(1 \cdot 1) = f(1)^2\text{,}\) and the only solutions are \(f(1) = 1\) or \(0\text{.}\) However if \(f(1)=0\) (and still assuming \(f\) is multiplicative or completely multiplicative) then \(f(n)=0\) for all \(n\text{,}\) because \(f(n) = f(1 \cdot n) = f(1)f(n) = 0\text{,}\) using that \((1,n) = 1\text{.}\) That is, if \(f\) has the multiplicative property and \(f(1)=0\text{,}\) then \(f\) is identically zero. Since we require “multiplicative” functions to not be identically zero, then they must have \(f(1) \neq 0\text{.}\) So it must be \(f(1)=1\text{.}\)
A completely multiplicative function \(f\) is entirely determined by its values at primes. That is, the values \(f(p)\) for \(p\) prime are enough to determine \(f(n)\) for all \(n \geq 1\text{.}\) Indeed, for \(n = p_1^{a_1} \dotsm p_k^{a_k}\) we must have \(f(n) = f(p_1)^{a_1} \dotsm f(p_k)^{a_k}\text{.}\)
A multiplicative function \(f\) is entirely determined by its values at powers of primes, because \(f(p_1^{a_1} \dotsc p_k^{a_k}) = \prod f(p_i^{a_i})\text{.}\)
The functions \(1\text{,}\) \(N\text{,}\) and \(e\) are completely multiplicative. The functions \(\sigma_s\) are multiplicative, but not completely multiplicative. (For example \(\sigma_s(2)=1+2^s\text{,}\) but \(\sigma_s(4)=1+2^s+4^s \neq \sigma_s(2)^s\text{.}\)) The proof that \(\sigma_s\) are multiplicative will be given in the next section, as part of a more general statement about Dirichlet products of multiplicative functions.
