Arithmetic functions form a commutative ring under pointwise addition and Dirichlet convolution

Because $\mathbb{C}$ is a field by $\mathbb{C}=\mathbb{R}[x]\mathrm{/}(x^2+1)$ is a field, every element is uniquely $a+bi$, and every nonzero element has inverse $(a-bi)\mathrm{/}(a^2+b^2)$, the pointwise formulas for $f+g$, $0$, and $-f$ define arithmetic functions and satisfy the abelian-group laws at each positive integer. The convolution $f\ast g$ is also an arithmetic function by Dirichlet convolution of arithmetic functions.

For convolution commutativity, $(f\ast g)(n)={\sum }_{d\mid n}f(d)g(n\mathrm{/}d)={\sum }_{d\mid n}f(n\mathrm{/}d)g(d)=(g\ast f)(n)$, where the middle equality reindexes the finite divisor sum by the involution $d\mapsto n\mathrm{/}d$ using Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule.

For associativity, expand both sides and rewrite them as the same finite sum over ordered factorizations $abc=n$: $(f\ast (g\ast h))(n)={\sum }_{abc=n}f(a)g(b)h(c)=((f\ast g)\ast h)(n)$. The passage from nested divisor sums to the triple sum uses the finite Fubini and reindexing rules of Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule.

Distributivity follows by expanding a finite divisor sum termwise in $\mathbb{C}$. For the identity, only the divisor $d=n$ contributes in $(f\ast \epsilon )(n)$ and only $d=1$ contributes in $(\epsilon \ast f)(n)$, because The Dirichlet-convolution identity and the constant-one function makes $\epsilon (m)=0$ for $m>1$. Thus $(f\ast \epsilon )(n)=f(n)=(\epsilon \ast f)(n)$.

Steps 1.1, 1.2, 1.3, and 1.4 give all ring axioms, and step 1.2 also shows the multiplication is commutative.