# 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 $\mathcal{S}$ 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 $\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 |