arithmetic functions form a ring

Theorem 1

The set 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) =d|nf(d)g(nd)

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 ε, and the units of the ring are those arithmetic functions f such that f(1)0.

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

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

Many of the ring identities are also obvious. We will prove that ε is the multiplicative identityPlanetmathPlanetmath, that * is commutativePlanetmathPlanetmath and associative, that * distributes over +, and that the units of the ring are as stated.

To see that ε is the multiplicative identity, note that


and thus ε*f=f.

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


Commutativity is obvious from this of the operation.

Associativity follows similarly. Note that


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)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