# arithmetic functions form a ring

###### 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

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

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 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 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

 $\displaystyle(f*(g+h))(n)=\sum_{d|n}f(d)\left(g+h\right)\left(\frac{n}{d}% \right)=\sum_{d|n}f(d)\left(g\left(\frac{n}{d}\right)+h\left(\frac{n}{d}\right% )\right)=\\ \displaystyle\sum_{d|n}f(d)g\left(\frac{n}{d}\right)+\sum_{d|n}f(d)h\left(% \frac{n}{d}\right)=((f*g)+(f*h))(n)$

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 $f(1)\neq 0$.

Title arithmetic functions form a ring ArithmeticFunctionsFormARing 2013-03-22 16:30:28 2013-03-22 16:30:28 rm50 (10146) rm50 (10146) 6 rm50 (10146) Theorem msc 11A25 ConvolutionInversesForArithmeticFunctions