PlanetMath: arithmetic functions form a ring

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arithmetic functions form a ring

(Theorem)

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

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

The 0 of the ring is the function such that for all positive integers , the of the ring is the convolution identity function $\varepsilon$ , and the units of the ring are those arithmetic functions 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 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

$\displaystyle (\varepsilon*f)(n)=\sum_{d\vert n} \varepsilon(d)f\left(\frac{n}{d}\right)=\varepsilon(1)f(n)=f(n)$

and thus $\varepsilon*f=f$ .

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

$\displaystyle (f*g)(n)=\sum_{ab=n}f(a)g(b)$

Commutativity is obvious from this representation of the operation.

Associativity follows similarly. Note that

$\displaystyle ((f*g)*h)(n)= \sum_{ra=n}(f*g)(r)h(a)=\sum_{ra=n}h(a)\sum_{bc=r}f(b)g(c)=\sum_{abc=n}f(b)g(c)h(a)$

If one expands

similarly, the resulting sum is identical, so the two are equal.

Distributivity follows since

$\begin{multline*}(f*(g+h))(n)=\sum_{d\vert n}f(d)\left(g+h\right)\left(\frac{n}{... ...t)+\sum_{d\vert n}f(d)h\left(\frac{n}{d}\right)=((f*g)+(f*h))(n) \end{multline*}$

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 with $f(1)\neq 0$ .

"arithmetic functions form a ring" is owned by rm50.

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See Also: convolution inverses for arithmetic functions

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Cross-references: convolution inverses for arithmetic functions, invertible functions, distributivity, sum, expands, distributes over, associative, commutative, group, identity, obvious, abelian group, proof, units, convolution identity function, integers, positive, function, ring, Dirichlet convolution, addition, operations, unity, commutative ring, arithmetic functions
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This is version 3 of arithmetic functions form a ring, born on 2006-12-24, modified 2007-04-19.
Object id is 8683, canonical name is ArithmeticFunctionsFormARing.
Accessed 1069 times total.

Classification:

AMS MSC:

11A25 (Number theory :: Elementary number theory :: Arithmetic functions; related numbers; inversion formulas)

Pending Errata and Addenda

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