Theorem 1   The set $\mathcal{S}$ of arithmetic functions forms a commutative ring with unity under the operations of element-by-element addition and Dirichlet convolution, i.e. under

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

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

That $\mathcal{S}$ is an abelian group under $+$ is obvious; the only interesting point 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 $\varepsilon$ 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 $\varepsilon$ is the multiplicative identity, note that$$(\varepsilon*f)(n)=\sum_{d|n} \varepsilon(d)f\left(\frac{n}{d}\right)=\varepsilon(1)f(n)=f(n$$ and thus $\varepsilon*f=f$ .

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 representation 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

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 $f(1)\neq 0$ .