arithmetic functions form a ring

Theorem 1

The set $\mathrm{S}$ of arithmetic functions forms a commutative ring with unity under the operations of element-by-element addition and Dirichlet convolution, i.e. under

$(f+g)(n)$ $=f(n)+g(n)$

$(f*g)(n)$ $={\displaystyle \sum _{d|n}}f(d)g\left({\displaystyle \frac{n}{d}}\right)$

The $\mathrm{0}$ of the ring is the function $z$ such that $z\mathit{}\mathrm{(}n\mathrm{)}\mathrm{=}\mathrm{0}$ for all positive integers $n$, the $\mathrm{1}$ of the ring is the convolution identity function $\epsilon$, and the units of the ring are those arithmetic functions $f$ such that $f\mathit{}\mathrm{(}\mathrm{1}\mathrm{)}\mathrm{\ne }\mathrm{0}$.

Proof. This is essentially a triviality and a little bit of computation.

That $𝒮$ is an abelian group under $+$ is obvious; the only interesting is noting that indeed $z$ is the identity of the group (the $0$ of the ring).

Many of the ring identities are also obvious. We will prove that $\epsilon$ is the multiplicative identity, that $*$ is commutative and associative, that $*$ distributes over $+$, and that the units of the ring are as stated.

To see that $\epsilon$ is the multiplicative identity, note that

$$(\epsilon *f)(n)=\sum _{d|n}\epsilon (d)f\left(\frac{n}{d}\right)=\epsilon (1)f(n)=f(n)$$

and thus $\epsilon *f=f$.

To see that $*$ is commutative, note that $f*g$ can also be written as

$$(f*g)(n)=\sum _{ab=n}f(a)g(b)$$

Commutativity is obvious from this of the operation.

Associativity follows similarly. Note that

$$((f*g)*h)(n)=\sum _{ra=n}(f*g)(r)h(a)=\sum _{ra=n}h(a)\sum _{bc=r}f(b)g(c)=\sum _{abc=n}f(b)g(c)h(a)$$

If one expands $(f*(g*h))(n)$ similarly, the resulting sum is identical, so the two are equal.

Distributivity follows since

$$\begin{array}{c}(f*(g+h))(n)=\sum _{d|n}f(d)\left(g+h\right)\left(\frac{n}{d}\right)=\sum _{d|n}f(d)\left(g\left(\frac{n}{d}\right)+h\left(\frac{n}{d}\right)\right)=\hfill \\ \hfill \sum _{d|n}f(d)g\left(\frac{n}{d}\right)+\sum _{d|n}f(d)h\left(\frac{n}{d}\right)=((f*g)+(f*h))(n)\end{array}$$

The units of the ring are simply the invertible functions; the entry on convolution inverses for arithmetic functions shows that the invertible functions are those functions $f$ with $f(1)\ne 0$.

Title	arithmetic functions form a ring
Canonical name	ArithmeticFunctionsFormARing
Date of creation	2013-03-22 16:30:28
Last modified on	2013-03-22 16:30:28
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	6
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 11A25
Related topic	ConvolutionInversesForArithmeticFunctions