Let \(\mathcal A\) be the set of functions \(f:\NN\to\CC\), with pointwise addition and Dirichlet convolution \[ (f*g)(n)=\sum_{d\mid n}f(d)g(n/d). \] Show that \(\mathcal A\) is a commutative ring. What further structure can be described?
Solution
Pointwise addition makes \(\mathcal A\) an abelian group. We verify the multiplicative axioms for Dirichlet convolution.
Commutativity follows by replacing each divisor \(d\mid n\) by its complementary divisor \(n/d\): \[ (f*g)(n) =\sum_{d\mid n}f(d)g(n/d) =\sum_{d\mid n}g(d)f(n/d) =(g*f)(n). \]
For associativity, \[ ((f*g)*h)(n) =\sum_{ab\mid n}f(a)g(b)h(n/ab) =\sum_{abc=n}f(a)g(b)h(c), \] and the same triple sum equals \[ (f*(g*h))(n). \]
The multiplicative identity is \[ \delta_1(n)= \begin{cases} 1,&n=1,\\ 0,&n>1. \end{cases} \] Indeed, \[ (f*\delta_1)(n)=f(n). \] Distributivity over pointwise addition follows term-by-term from the defining divisor sum. Hence \(\mathcal A\) is a commutative ring.
An arithmetic function \(f\) is a unit exactly when \[ f(1)\ne0. \] If \(f(1)\ne0\), its inverse is determined recursively by \[ g(1)=f(1)^{-1}, \] and for \(n>1\), \[ g(n) =-\frac1{f(1)}\sum_{\substack{d\mid n\\ d>1}}f(d)g(n/d). \]
This ring is the classical ring of arithmetic functions under Dirichlet convolution. Formally, Dirichlet series turn convolution into multiplication: \[ \left(\sum_{n\ge1}\frac{f(n)}{n^s}\right) \left(\sum_{n\ge1}\frac{g(n)}{n^s}\right) = \sum_{n\ge1}\frac{(f*g)(n)}{n^s}, \] whenever analytic convergence is available, and otherwise as a formal identity of coefficients.