# Some remarks on the ring of arithmetical functions

A. Goswami  [1] Department of Mathematics and Applied Mathematics, University
of Johannesburg, P.O. Box 524, Auckland Park 2006, South Africa. [2] National
Institute for Theoretical and Computational Sciences (NITheCS), South Africa.
<EMAIL_ADDRESS>
###### Abstract.

The aim of these notes is to study some of the structural aspects of the ring
of arithmetical functions. We prove that this ring is neither Noetherian nor
Artinian. Furthermore, we construct various types of prime ideals. We show
arithmetic ring has infinite Krull dimension and its associated prime ideal
set is nonempty.

###### Key words and phrases:

arithmetic ring, local ring, prime ideal, Krull dimension

###### 1991 Mathematics Subject Classification:

11A25; 13F25

##  1\. Introduction

The study of arithmetical functions is always a central theme of number
theory. One of the earliest work of Bell [2] on arithmetical functions
motivated rigorous study of these functions. A function
$f\colon\mathds{N}^{+}=\mathds{N}\backslash\\{0\\}\to\mathds{C}$ is called an
_arithmetical function_. According to this definition although any sequence of
complex numbers is indeed an arithmetical function, however, as Hardy & Wright
[8] mentioned that “functions $f(n)$ of the positive integer $n$ defined in a
manner which expresses some arithmetical property of $n$,” we expect these
functions are very much relevant to number theoretic study. Let us see some
examples of arithmetical functions. We give only the definitions. For further
study of them, we refer our reader to [1, 6, 11, 9, 12, 10]. Observe that any
positive integer $n>1$ can be written as $n=p_{1}^{a_{1}}\cdots
p_{k}^{a_{k}}$, where $p_{i}$’s are prime and $a_{j}$’s are positive integers.

(i) A _Möbius function_ $\mu$ is defined as follows:

$\mu(n)=\begin{cases}1,&n=1,\\\ (-1)^{k},&a_{1}=\cdots=a_{k}=1,\\\
0,&\text{otherwise}.\end{cases}$

(ii) The _Euler totient function_ $\phi(n)$ is defined to be the number of
positive integers not exceeding $n$ which are relatively prime to $n$, _i.e._
,

$\phi(n)=n\prod\limits_{p|n}\left(1-\dfrac{1}{p}\right).$

(iii) _Mangoldt’s function_ $\Lambda(n)$ is defined by

$\Lambda(n)=\begin{cases}\log p,&n=p^{m}\;\text{for some prime}\,p\,\text{and
some}\,m\geqslant 1,\\\ 0,&\text{otherwise.}\end{cases}$

(iv) _Liouville’s function_ $\lambda(n)$ is defined as

$\lambda(n)=\begin{cases}1,&n=1,\\\
(-1)^{a_{1}+\cdots+a_{k}},&\text{otherwise}.\end{cases}$

(v) _Ramanujan’s tau function_ is defined by

$\sum\limits_{n=1}^{\infty}\tau(n)x^{n}=x\prod\limits_{j=1}^{\infty}(1-x^{j})^{24},\,|x|<1.$

(vi) _Dedekind psi function_ $\psi(n)$ is defined by the formula:

$\psi(n)=n\prod\limits_{p|n}\left(1+\dfrac{1}{p}\right).$

(vii) The _prime divisor function_ $\Omega(n)$ is defined as the number of
prime factors of $n$ counted with multiplicities, i.e.,
$\Omega(n)=a_{1}+\cdots+a_{k},$ whereas, distinct prime divisor function
$\omega(n)$ is defined by $\omega(n)=k.$
(viii) A _$p$ -adic valuation_ $\nu_{p}$ is defined by
$\nu_{p}=\max\\{k\in\mathds{N}\mid p^{k}\,\text{divides}\,n\\},$ where $p$ is
a prime integer.
(ix) _Logarithmic function_ $\log$ is another example of an arithmetic
function.

We wish to do the structural study of the arithmetic ring in the next section.
First we review some known facts about this ring.

##  2\. Structures of $\mathcal{A}$

###  2.1. Local domain

Let $\mathcal{A}$ be the set of arithmetical functions. We define addition and
multiplication (Dirichlet convolution) on $\mathcal{A}$ respectively by
$(f+g)(n)=f(n)+g(n)$ and $(f*g)(n)=\sum_{ij=n}f(i)g(j).$ It is well-known (see
Cashwell & Everett [5]) that $(\mathcal{A},+,*)$ is a commutative ring with
identity $e$ defined by

$e(n)=\begin{cases}1,&n=1,\\\ 0,&\text{otherwise.}\end{cases}$

The ring $\mathcal{A}=(\mathcal{A},+,*)$ is called the _arithmetic ring_ or
the _ring of arithmetical functions_. It is also easy to prove (see Apostol
[1]) that an element $f\in\mathcal{A}$ is invertible if and only if $f(1)\neq
0.$ Furthermore, the set of invertible elements of $\mathcal{A}$ has the
following structure.

######  Lemma 2.1.

Let $\mathcal{G}=\\{f\in\mathcal{A}\mid f(1)\neq 0\\}$, the set of units of
the arithmetic ring $\mathcal{A}$. Then $(\mathcal{G},*)$ is an abelian group.

###### Proof.

First we observe that if $f,$ $g\in\mathcal{G},$ then $(f*g)(1)=f(1)g(1)\neq
0$, i.e., the set $\mathcal{G}$ is closed under the operation $*$. The rest
follows from the fact that ($\mathcal{A},*$) is a commutative monoid, whereas
existence of inverses of an $f$ follows from the hypothesis: $f(1)\neq 0.$ ∎

######  Example 2.2.

_From the invertibility criteria of an element in $\mathcal{A}$, it follows
that the arithmetical functions $\Omega,$ $\omega,$ $\Lambda,$ $\nu_{p},$
$\log$ are non-invertible; whereas the functions $\mu,$ $\phi,$ $\lambda,$
$\tau,$ $\psi$ are invertible._

Recall that an arithmetical function $f$ is called _additive_ (_completely
additive_) if $f(mn)=f(m)+f(n)$ for coprime positive integers (all positive
integers) $m$ and $n$. We observe that if $f$ is additive or completely
additive, then

$f(1)=f(1\cdot 1)=f(1)+f(1)=2f(1)$

implying that $f(1)=0.$ Therefore, we have

######  Proposition 2.3.

If $f$ is additive or completely additive, then $f$ is non-invertible.

The converse of the above statement may not be true. For example, the
functions $\Lambda$ and $\log$ are non-invertible but they are not additive
(and hence not completely additive). Moreover, it is well-known that

######  Proposition 2.4.

The set $\mathtt{Add}(\mathcal{A})$ of additive arithmetical functions forms a
group under pointwise addition.

######  Theorem 2.5.

The arithmetic ring $\mathcal{A}$ is a local ring.

###### Proof.

Let $\mathfrak{m}=\mathcal{A}\backslash\mathcal{G}=\\{f\in\mathcal{A}\mid
f(1)=0\\}.$ Since $\mathcal{G}$ is the set of units of $\mathcal{A}$ (see
Lemma 2.1), by Bourbaki [3, Chapter II, §3, no.1, Proposition 1], it is
sufficient to show that $\mathfrak{m}$ is an ideal of $\mathcal{A}.$ Suppose
$f,f^{\prime}\in\mathfrak{m}$ and $g\in G.$ Then
$(f+f^{\prime})(1)=f(1)+f^{\prime}(1)=0$ and $(f*g)(1)=f(1)g(1)=0\cdot
g(1)=0.$ Therefore, $f+f^{\prime},$ $f*g\in\mathfrak{m}$. ∎

Note that the unique maximal ideal $\mathfrak{m}$ of the local ring
$(\mathcal{A},\mathfrak{m})$ is the Jacobson radical of $\mathcal{A}$. The
arithmetic ring is not only a local ring but also an integral domain. In fact,
Cashwell & Everett [5] proved that the formal power series over $\mathds{C}$
in a countable set of indeterminates is isomorphic to the arithmetic ring, and
hence from this isomorphism we obtain the following

######  Theorem 2.6.

The arithmetic ring $\mathcal{A}$ is a unique factorization domain.

######  Remark 2.7.

_Since the arithmetic ring $\mathcal{A}$ is an integral domain, it follows
that $\mathcal{A}$ does not have any non-trivial idempotent elements. Here is
an alternative proof of this fact without using the domain property of
$\mathcal{A}$. Suppose $f$ is a non-trivial idempotent element of the
arithmetic ring $\mathcal{A}$. Then for any $n\in\mathds{N}^{+}$, we must
have_

$f^{2}(n)=2f(1)f(n)+\sum_{\begin{subarray}{c}ij=n\\\ i\neq 1,\,j\neq
n\end{subarray}}f(i)f(j)=f(n),$

_and that implies $f(1)\neq 0$. By Lemma 2.1, this means $f$ is invertible, a
contradiction._

Next, we wish to prove that the arithmetic ring $\mathcal{A}$ is not Artinian,
and for that we first need the following definition. A _norm_ on $\mathcal{A}$
is a map $\mathfrak{N}\colon\mathcal{A}\to\mathds{Z}$ defined by

$\mathfrak{N}(f)=\begin{cases}0,&f=0;\\\ n,&n\;\text{is the least positive
integer with}\;f(n)\neq 0.\end{cases}$

######  Theorem 2.8.

For each positive integer $n,$ the subset

$\mathcal{I}_{n}=\\{f\in\mathcal{A}\mid\mathfrak{N}(f)\geqslant
n\\}\cup\\{0\\}$

is an ideal of $\mathcal{A}$, and
$\mathcal{I}_{1}\supseteq\mathcal{I}_{2}\supseteq\mathcal{I}_{3}\supseteq\cdots$
is an infinite descending chain of ideals. Therefore,
$(\mathcal{A},\mathfrak{m})$ is not Artinian.

###### Proof.

Let $f,$ $g\in\mathcal{I}_{n}.$ Then $\mathfrak{N}(f)\geqslant n$ and
$\mathfrak{N}(g)\geqslant n.$ Since $(f+g)(n)=f(n)+g(n),$ at once we have
$\mathfrak{N}(f+g)\geqslant n,$ and hence $f+g\in\mathcal{I}_{n}.$ Now, let
$h\in\mathcal{A}.$ To show $h*f\in\mathcal{I}_{n},$ we consider the following
two possibilities for $h.$

(i) $h=0$: Obviously $h*f=0\in\mathcal{I}_{n}.$

(ii) $h\neq 0$: Since $f(k)=0$ for all $k\leqslant n-1,$ we also have

$(h*f)(k)=\sum_{ij=k}h(i)f(j)=0$

for all $k\leqslant n-1.$ Since $\mathcal{A}$ is an integral domain, this
implies there exists $k\geqslant n$ such that $\mathfrak{N}(h*f)=k.$

The second part of the proposition follows from the fact that
$\mathfrak{N}(f)\geqslant n$ implies $\mathfrak{N}(f)\geqslant n-1$. ∎

A norm $\mathfrak{N}$ is not a ring homomorphism, however, $\mathfrak{N}$
preserves the (multiplicative) monoid structure of the arithmetic ring
$\mathcal{A}$, that is,

######  Proposition 2.9.

The norm $\mathfrak{N}$ is a (multiplicative) monoid homomorphism.

###### Proof.

We observe that $\mathfrak{N}(e)=1.$ We wish to prove

$\mathfrak{N}(f*g)=\mathfrak{N}(f)\mathfrak{N}(g)$ (2.1)

for all $f,$ $g\in\mathcal{A}.$ If $f$ or $g$ is zero, then the condition
(2.1) is satisfied trivially. Therefore, assume that $\mathfrak{N}(f)=i$ and
$\mathfrak{N}(g)=j$, where $i$ and $j$ are positive integers. If $n=ij$ then

$\displaystyle(f*g)(n)$ $\displaystyle=(f*g)(ij)$
$\displaystyle=\sum_{st=n}f(s)g(t)=f(1)g(st)+\cdots+f(i)g(j)+\cdots+f(st)g(1)$
$\displaystyle=f(i)g(j),$

where the last equality follows from the facts: (i) $f(k)=0$ for all $k<i$ and
$k$ is a divisor of $n$, (ii) $g(l)=0$ for all $l<j$ and $l$ is a divisor of
$n.$ From this it follows that $\mathfrak{N}(f*g)=ij,$ and hence we have the
claim (2.1). ∎

###  2.2. Prime ideals of $\mathcal{A}$

Since $\mathcal{A}$ is an integral domain, the zero ideal is a prime ideal of
$\mathcal{A}$. Also, the unique maximal ideal $\mathfrak{m}$ of $\mathcal{A}$
is a prime ideal. For $g,h\in\mathcal{A}$ and $g*h\in\mathfrak{m}$ implies
that $f(1)g(1)=0,$ whence $f(1)=0$ or $g(1)=0,$ i.e., either
$f\in\mathfrak{m}$ or $g\in\mathfrak{m}$. Moreover, we use the following
result from Sivaramakrishnan [11, Chapter II, Theorem 14] to obtain prime
ideals of $\mathcal{A}$.

######  Theorem 2.10.

An arithmetic function $f$ for which $\mathfrak{N}(f)$ is a prime integer is a
prime element in $\mathcal{A}$,

######  Proposition 2.11.

If $\mathfrak{N}(f)$ is a prime integer, then $\langle f\rangle$ is a prime
ideal of $\mathcal{A}.$

###### Proof.

Note that $f$ is a prime element of $\mathcal{A}$ if and only if $\langle
f\rangle$ is a prime ideal of $\mathcal{A}$ (see [4, p. 98]), and hence we
have the desired claim. ∎

In fact, we can have more. For each (positive) prime integer, we can identify
a prime ideal of $\mathcal{A}$ which we obtain in

######  Theorem 2.12.

The Krull dimension of $\mathcal{A}$ is infinite.

###### Proof.

Let $p$ be a positive prime integer. Define

$\mathfrak{a}_{p}=\\{f\in\mathcal{A}\mid f(p)\neq 0\\}.$

Observe that by definition of a prime ideal, if $f\in\mathfrak{a}_{p},$ then
$f(1)=0.$ For every positive prime number $p$, we claim that
$\mathfrak{a}_{p}$ is a prime ideal of $\mathcal{A}$. To show this, let
$g,h\in\mathcal{A}$ such that $g,$ $h\notin\mathfrak{a}_{p}.$ Since $g(p)=0$
and $h(p)=0,$ we must have

$(g*h)(p)=g(p)h(1)+g(1)h(p)=0,$

whence $g(p)=0.$ In other words, $g*h\notin\mathfrak{a}_{p}.$

Now we generalize the above construction of $\mathfrak{a}_{p}$ to obtain an
infinite chain of prime ideals. Suppose $p_{1},\ldots,p_{n}$ be positive prime
integers and define

$\mathfrak{a}_{p_{1},\ldots,p_{n}}=\\{f\in\mathcal{A}\mid f(p_{1})\neq
0\,\text{or}\,f(p_{2})\neq 0\;\text{or}\,\ldots\,\text{or}\,f(p_{n})\neq
0\\}.$

Following the above argument, it follows that
$\mathfrak{a}_{p_{1},\ldots,p_{n}}$ is also a prime ideal of $\mathcal{A}$.
Furthermore,

$\cdots\supsetneq\mathfrak{a}_{p_{1},\ldots,p_{n}}\supsetneq\mathfrak{a}_{p_{1},\ldots,p_{n-1}}\supsetneq\cdots\supsetneq\mathfrak{a}_{p_{1},p_{2}}\supsetneq\mathfrak{a}_{p_{1}}$

is an infinite (ascending) chain of prime ideals of $\mathcal{A}.$ ∎

######  Corollary 2.13.

The ring of arithmetical functions is not Noetherian.

######  Remark 2.14.

_It is easy to see that if $f$ is invertible, then $\langle f\rangle$ is not a
prime ideal. Therefore, the set of principal prime ideals is a (proper) subset
of the set of non-invertible elements of $\mathcal{A}$. Notice that the unique
maximal ideal $\mathfrak{m}$ of $\mathcal{A}$ is not finitely-generated.
Otherwise, all ideals of $\mathcal{A}$ would also be finitely-generated, a
contradiction to the fact that $\mathcal{A}$ is not Notherian (see Corollary
2.13)._

######  Proposition 2.15.

If $f$ and $g$ are non-invertible elements of $\mathcal{A}$, then
$\mathfrak{N}(f*g)$ cannot be a prime integer.

###### Proof.

Suppose $f,$ $g\in\mathcal{A}\setminus\mathcal{G}.$ Since
$(f*g)(p)=f(1)g(p)+f(p)g(1)=0$ for all prime integers $p$, the claim follows
immediately. ∎

We now talk about a special type of prime ideals of $\mathcal{A}.$ Suppose $M$
is an $\mathcal{A}$-module and $m\in M.$ The _annihilator_ of $m$ is the ideal
$\mathcal{A}$ defined by

$\mathtt{Ann}_{\mathcal{A}}(m)=\\{a\in\mathcal{A}\mid am=0\\}.$

Following Dutton [7], an ideal $\mathfrak{a}$ of a ring $R$ is said to be
_associated with_ an $R$-module $M$ if $\mathfrak{a}=\mathtt{Ann}_{R}(M).$ The
set of prime ideals associated with $M$, as is customary, is denoted by
$\mathtt{Ass}_{R}(M)$. It is known (see Bourbaki [3, Chapter IV, §1, Corollary
1]) that if $R$ is Noetherian then $M\neq\\{0\\}$ is equivalent to
$\mathtt{Ass}_{R}(M)\neq\emptyset$. Although $\mathcal{A}$ is not Noetherian,
nevertheless, we have the following

######  Proposition 2.16.

Whenever $M$ is nonzero, $\mathtt{Ass}_{\mathcal{A}}(M)$ is nonempty.

###### Proof.

Suppose $M$ is an $\mathcal{A}$-module and

$\mathfrak{p}=\max\limits_{m\neq 0}\\{\mathtt{Ann}_{\mathcal{A}}(m)\mid m\in
M\\}.$

We claim that $\mathfrak{p}$ is prime. Let $x\in\mathfrak{p}$ and
$a\in\mathcal{A}.$ Then

$x(am)=(xa)m=(ax)m=a(xm)=a0=0,$

implying $x\in\mathtt{Ann}_{\mathcal{A}}(am)$. Therefore,
$\mathfrak{p}\subseteq\mathtt{Ann}_{\mathcal{A}}(am).$ Let $a,$
$a^{\prime}\in\mathcal{A}$ such that $aa^{\prime}\in\mathfrak{p}$. If
possible, let $a\notin\mathfrak{p}$ and $a^{\prime}\notin\mathfrak{p}.$ Now,
$aa^{\prime}\in\mathfrak{p}$ implies
$a\in\mathtt{Ann}_{\mathcal{A}}(a^{\prime}m)$ and
$a^{\prime}\in\mathtt{Ann}_{\mathcal{A}}(am)$ i.e.,
$\mathfrak{p}\subsetneq\mathtt{Ann}_{\mathcal{A}}(am)$ and
$\mathfrak{p}\subsetneq\mathtt{Ann}_{\mathcal{A}}(a^{\prime}m)$, contradicts
the maximality of $\mathfrak{p}.$ ∎

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