arithmetic functions form a ringArithmeticFunctionsFormARing2006-12-24 21:11:302007-04-19 07:04:06Theoremarithmetic function% this is the default PlanetMath preamble. as your knowledge
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\begin{thm} 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
\begin{align*}(f+g)(n)&=f(n)+g(n)\\
(f*g)(n)&=\sum_{d|n} f(d)g\left(\frac{n}{d}\right)
\end{align*}
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 $f(1)\neq 0$.
\end{thm}
\textbf{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 \PMlinkescapetext{point} 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 \PMlinkescapetext{representation} 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{multline*}(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)=\\
\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{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 $f$ with $f(1)\neq 0$.