# Commutative rings whose proper ideals
are direct sum of cyclically presented modules††thanks: Key Words: Cyclic
modules, Cyclically presented modules, Semi-local rings, Principal ideal
rings, Prime ideals. ††thanks:  2010 Mathematics Subject Classification:
13C05, 13F10, 13H99.

R. Nikandisha, M.J. Nikmehrb and A. Yassineb

aDepartment of Mathematics, Jundi-Shapur University of Technology,
P.O. BOX 64615-334, Dezful, Iran
<EMAIL_ADDRESS>
bFaculty of Mathematics, K.N. Toosi University of Technology,
P.O. BOX 16315-1618, Tehran, Iran
<EMAIL_ADDRESS><EMAIL_ADDRESS>
Corresponding author

###### Abstract

A famous result due to I. M. Isaacs states that if a commutative ring $R$ has
the property that every prime ideal is principal, then every ideal of $R$ is
principal. This motivates ring theorists to study commutative rings for which
every ideal is a direct sum of cyclically presented modules. In this paper, we
study commutative rings whose ideals are direct sum of cyclically presented
modules.

##  1 Introduction

We assume throughout this paper that all rings are commutative with identity.
Let $R$ be a ring and $X$ be a nonempty subset of $R$. The set of prime ideals
of $R$, the set of maximal ideals of $R$, the set of nilpotent elements of
$R$, the set of zero-divisors of $R$, the set of integers, the annihilator of
$X$ and integers modulo $n$ are denoted by Spec$(R)$, Max$(R)$, Nil$(R)$,
$Z(R)$, $\mathbb{Z}$, Ann$(X)=\\{a\in R|aX=0\\}$ and $\mathbb{Z}_{n}$,
respectively. By a proper ideal $I$ of $R$, we mean an ideal with $I\neq R$. A
ring $R$ is called local (resp. semilocal) if $R$ has a unique maximal ideal
(resp. $R$ has a finite number of maximal ideals). For any undefined notation
or terminology in commutative ring theory, we refer the reader to [11].

Prime and maximal ideals play a central role in commutative ring theory and so
these notions have been studied to specify the structure of all ideals of the
rings. We note that two theorems from commutative algebra due to I. M. Isaacs
and I. S. Cohen state that, to check whether every ideal in a ring is cyclic
(resp. finitely generated), it suffices to test only the prime ideals (see [6,
p. 8, Exercise 10] and [4, Theorem 2]). It was shown by Kaplansky [7] that,
for a commutative Noetherian ring $R$, every maximal ideal of $R$ is principal
if and only if every ideal of $R$ is principal. The study of rings over which
modules are direct sums of cyclic modules has has been done by several
authors. It was shown by Köthe [8] and Cohen-Kaplansky [5] that “a commutative
ring $R$ has the property that every module is a direct sum of cyclic modules
if and only if $R$ is an Artinian principal ideal ring”. The study of
commutative rings $R$ such that the ideals of $R$ are direct sums of cyclic
modules was initiated by Behboodi et al. in [1, 2] and [3]. Later, Moradzadeh-
Dehkordi and Shojaee [10] investigated rings $R$ whose ideals are direct sums
of cyclically presented modules. Indeed, they proved the following theorem.

###### Theorem 1.1

([10, Theorem 2.16]) Let $(R,M)$ be a local ring. Then the following
statements are equivalent:

$(1)$ Every ideal of $R$ is a direct sum of cyclically presented $R$-modules.

$(2)$ Every ideal of $R$ is a direct summand of a direct sum of cyclically
presented $R$-modules.

$(3)$ Every prime ideal of $R$ is a direct sum of cyclically presented
R-modules.

$(4)$ Every prime ideal of $R$ is a direct summand of a direct sum of
cyclically presented R-modules.

$(5)$ Either $R$ is a principal ideal ring or $M=Rx\oplus Ry$, where
$x,y\notin$Nil$(R)$, $R/{\rm Ann}(x)$ and $R/{\rm Ann}(y)$ are principal ideal
rings.

So a natural question is posed: “What is the class of semi-local rings $R$ for
which every ideal of $R$ is a direct sum of cyclically presented $R$-modules”.
This paper is in this theme and it is devoted to study semi-local rings in
which every proper ideal is a direct sum of cyclically presented $R$-modules.
It is shown that if $R$ is a semi-local ring with maximal ideals
$\\{M_{1},\ldots,$ $M_{t}\\}$ and $x\in R$, then the cyclic $R$-module $Rx$ is
cyclically presented if and only if Ann$(x)$ is a principal ideal of $R$
(Proposition 2.3). We give a condition $(*)$ (Definition 2.1) and it is shown
that if $R$ is a semi-local ring that satisfies condition $(*)$ with maximal
ideals $\\{M_{1},\ldots,M_{t}\\}$ such that every prime ideal $P$ has the form
$\bigoplus_{i\in I}Rx_{i}$ where $Rx_{i}$ is a direct sum of cyclically
presented $R$-modules and $x_{i}\notin{\rm Nil}(R)$ for every $i\in I$ where
$I$ is an index set, then for every maximal ideal $M_{j}$ of $R$ either
$M_{j}$ is cyclic or $M_{j}=Rx_{1,j}\oplus Rx_{2,j}$ such that $Rx_{1,j}$,
$Rx_{2,j}$ are cyclically presented and $R/{\rm Ann}(x_{1,j})$, $R/{\rm
Ann}(x_{2,j})$ are principal ideal rings for every $j\in\\{1,\ldots,t\\}$
(Theorem 2.6). It is proved that to check whether every ideal of a semi-local
ring $R$ that satisfies condition $(*)$ is a direct sum of cyclically
presented modules, it suffices to test only the prime ideals or the structures
of the maximal ideals of $R$ (Theorem 2.11).

##  2 Main results

Recently in [10], it has been given two criteria to check whether every ideal
of a commutative local ring $R$ is a direct sum of cyclically presented
modules. They described the ideal structure of commutative local rings for
which every ideal of $R$ is a direct sum of cyclically presented $R$-modules.
This turns out to be a key result to progress in the proof of commutative
semi-local rings. The other important goal of this paper is to show that
commutative semi-local rings whose ideals are direct sums of cyclically
presented modules are Noetherian. As a consequence, we describe the ideal
structure of commutative semi-local rings in which every ideal of $R$ is a
direct sum of cyclically presented $R$-modules. In light of [3, Proposition
3.4] and Nakayama’s Lemma, the following definition is given.

###### Definition 2.1

Let $R$ be a semi-local ring and Max$(R)=\\{M_{1},\ldots,$ $M_{t}\\}$. We say
that $R$ satisfies condition $(*)$ over $M_{i}$’s, if the following conditions
hold:

$(1)$ If $M_{i}=Rx_{i}\oplus L$ for some ideal $L$ of $R$ and some $x\in R$
such that $R/{\rm Ann}(x_{i})$ is a principal ideal ring, then every nonzero
element of $Rx_{i}$ is of the form $ax^{n}$ for some unit element $a\in R$ and
positive integer $n$.

$(2)$ If $M_{j}=Rx_{1,j}\oplus\cdots\oplus Rx_{n,j}$, then
Ann$(x_{i,j}^{\alpha})\subseteq M_{j}$ for every $i\in\\{1,\ldots,n\\}$,
$j\in\\{1,\ldots,t\\}$ and positive integer $\alpha$. Also, for every finitely
generated module $N$ over the ring $R$, if $I$ is an ideal of $R$ such that
$I\subseteq M_{j}$ for some $j\in\\{1,\ldots,t\\}$ and $N=IN$, then $N=0$.

###### Theorem 2.2

Let $R$ be a semi-local ring such that satisfies condition $(*)$,
Max$(R)=\\{M_{1},\ldots,$ $M_{t}\\}$ and every prime ideal is a direct sum of
cyclically presented $R$-modules. If $M_{i}=Rx_{i}$, then $R$ is a principal
ideal domain and Spec$(R)=\\{(0),M_{1},\ldots,$ $M_{t}\\}$. In particular,
every nontrivial ideal of $R$ is of the form $Rx_{i}^{n}$ for some positive
integer $n$ and $i\in\\{1,\ldots,t\\}$.

###### Proof.

Suppose that $M_{i}=Rx_{i}$ such that $x_{i}\notin$ Nil$(R)$ for every
$i\in\\{1,\ldots,t\\}$. It is clear that if $R$ has at least two distinct
maximal ideals $M_{1}=Rx_{1}$ and $M_{2}=Rx_{2}$, then $x_{1},x_{2}\notin$
Nil$(R)$, if not, $R$ is a local ring. Then Nil$(R)\neq M_{i}$ for each $i$.
So suppose that $P\in$ Spec$(R)\setminus\\{M_{1},\ldots,M_{t}\\}$. Then there
exists $i\in\\{1,\ldots,t\\}$ such that $P\subseteq M_{i}=Rx_{i}$, and so
$P=P\cap Rx_{i}$ and $x_{i}\notin P$, as $P\notin\\{M_{1},\ldots,M_{t}\\}$.
Since $P$ is prime, we conclude that $P=Px_{i}=PRx_{i}.$ Hence, $P=(0)$, and
therefore Spec$(R)=\\{(0),M_{1},\ldots,M_{t}\\}$. Thus, by [6, p. 8, Exercise
10], $R$ is a principal ideal ring and since $(0)$ is a prime ideal, $R$ is
also a principal ideal domain.

For the “in particular” statement suppose that $R$ satisfies condition $(*)$
and $I$ is a nontrivial ideal of $R$. Since $R$ is a principal ideal domain,
there exists an element $y\in R$ such that $I=Ry$. Assume that $I\subseteq
M_{i}$ for some $i\in\\{1,\ldots,t\\}$. Hence, $y=ax_{i}^{n}$ for some unit
element $a$ of $R$ and positive integer $n$, and thus $I=Rx^{n}$, for some
positive integer $n$.  $\Box$

In the following proposition, we show that if $R$ is a semi-local ring and
$a\in R$, then the cyclic $R$-module $Ra$ is cyclically presented if and only
if Ann$(a)$ is a principal ideal of $R$.

###### Proposition 2.3

Let $R$ be a semi-local ring, Max$(R)=\\{M_{1},\ldots,$ $M_{t}\\}$ and $x\in
R$. Then the cyclic $R$-module $Rx$ is cyclically presented if and only if
Ann$(x)$ is a principal ideal of $R$.

###### Proof.

Let $R$ be a semi-local ring and Max$(R)=\\{M_{1},\ldots,$ $M_{t}\\}$ and
$x\in R$ such that the cyclic $R$-module $Rx$ is cyclically presented. Then
$Rx\cong R/I$ for some principal ideal $I$ of $R$. Now, we consider the
following diagram:

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\rm
Ann}(x)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$$\scriptstyle{f}$$\textstyle{R\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{||}$$\scriptstyle{g}$$\textstyle{R/{\rm
Ann}(x)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\wr}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{I\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f^{\prime}}$$\textstyle{R\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g^{\prime}}$$\textstyle{R/I\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

Therefore, by [9, Lemma 5.1], we conclude that $R\oplus{\rm Ann}(x)\cong
R\oplus I$. But $R$ is semi-local; hence finitely generated projective modules
cancel from direct sums. Thus Ann$(x)\cong I$ which is a principal ideal of
$R$. Conversely, since $Rx\cong R/{\rm Ann}(x)$, the proof is clear.  $\Box$

In the following result the spectrum of $R$ is described in case $R$ is a
semi-local ring satisfies condition $(*)$ for which every maximal ideal is not
cyclic.

###### Theorem 2.4

Let $R$ be a semi-local ring such that satisfies condition $(*)$,
Max$(R)=\\{M_{1},\ldots,M_{t}\\}$, every prime ideal $P$ has the form
$\bigoplus_{i\in I}Rx_{i}$ where $Rx_{i}$ is a direct sum of cyclically
presented $R$-modules and $x_{i}\notin{\rm Nil}(R)$ for every $i\in I$ where
$I$ is an index set and $M_{i}$’s are not cyclic. Then the following
statements hold:

$(1)$ Spec$(R)=\\{M_{1},\ldots,M_{t},P_{k,j}\\}_{k\in
I,j\in\\{1,\ldots,t\\}}$, where $P_{k,j}=\bigoplus_{i\in
I\setminus\\{k\\}}Rx_{i,j}$ such that $x_{i,j}\in R$.

$(2)$ $R/$Ann$(x_{k,j})$ is a principal ideal ring for every $k\in I$ and
$j\in\\{1,\ldots,t\\}$.

###### Proof.

$(1)$ Suppose that $R$ is a semi-local ring such that satisfies condition
$(*)$, Max$(R)=\\{M_{1},\ldots,M_{t}\\}$ and every prime ideal $P$ has the
form $\bigoplus_{i\in I}Rx_{i}$, where $Rx_{i}$ is a direct sum of cyclically
presented $R$-modules for every $i\in I$ and $M_{i}$’s are not cyclic. Then
for every $j\in\\{1,\ldots,t\\}$ we have $M_{j}=\bigoplus_{i\in I}Rx_{i,j}$,
where $I$ is an index set, $x_{i,j}\notin{\rm Nil}(R)$ and $Rx_{i,j}$ is
cyclically presented for each $i\in I$ and $j\in\\{1,\ldots,t\\}$. It is clear
that Nil$(R)\neq M_{i}$ for each $i$, and so suppose that $P$ is a prime ideal
of $R$ such that $P\notin\\{M_{1},\ldots,M_{t}\\}$. Then there exists
$j\in\\{1,\ldots,t\\}$ such that $P\subset M_{j}$. Since $x_{i,j}x_{k,j}=0\in
P$, for every $i\neq k$ we deduce that either $x_{i,j}\in P$ or $x_{k,j}\in
P$. Therefore there exists $k\in I$ such that $x_{k,j}\notin P$ and
$x_{i,j}\in P$ for every $i\neq k$. This means that $\bigoplus_{i\in
I\setminus\\{k\\}}Rx_{i,j}\subseteq P$. Since $P$ is prime, we conclude that

$P=P\cap M_{j}=(P\cap Rx_{k,j})\oplus(\bigoplus_{i\in
I\setminus\\{k\\}}Rx_{i,j})=Px_{k,j}\oplus(\bigoplus_{i\in
I\setminus\\{k\\}}Rx_{i,j}).$

Therefore, $Px_{k,j}=Px^{2}_{k,j}=Px_{k,j}Rx_{k,j}$. But $Px_{k,j}$ is a
direct sum of cyclic modules and $R$ satisfies condition $(*)$, hence
$Px_{k,j}=0$. Thus $M_{j}=P\oplus Rx_{k,j}$, and so
Spec$(R)\subseteq\\{M_{1},\ldots,M_{t},P_{k,j}\\}_{k\in
I,j\in\\{1,\ldots,t\\}}$, where $P_{k,j}=\bigoplus_{i\in
I\setminus\\{k\\}}Rx_{i,j}$. Conversely, suppose that $ab\in P_{k,j}$, for
some $a,b\in R$ such that $a,b\notin P_{k,j}$. Since $M_{j}$ is maximal,
assume that $a\in M_{j}$, and so $a=r_{k}x_{k,j}+p_{k,j}$ for some $r_{k}\in
R$ and $p_{k,j}\in P_{k,j}$. But $R$ satisfies condition $(*)$, which means
that $a=cx_{k,j}^{n}+p_{k,j}$ for some unit element $c\in R$ and positive
integer $n$. Since $ab\in P_{k,j}$, we deduce that $cbx_{k,j}^{n}+bp_{k,j}\in
P_{k,j}$, and so $cbx_{k,j}^{n}\in P_{k,j}$. Therefore $cbx_{k,j}^{n}\in
P_{k,j}\cap Rx_{k,j}=0$. Since $R$ satisfies condition $(*)$, $b\in{\rm
Ann}(x_{k,j}^{n})\subseteq M_{j}$. Hence $b=dx_{k,j}^{n}+p_{2}$ for some unit
element $d\in R$, positive integer $m$ and $p_{2}\in P_{k,j}$. Thus
$ab=cdx_{k,j}^{m+n}\in P_{k,j}\cap Rx_{k,j}=0$. This means that
$x_{k,j}\in{\rm Nil}(R)$, a contradiction. Hence either $a\in P_{k,j}$ or
$b\in P_{k,j}$ and thus Spec$(R)=\\{M_{1},\ldots,M_{t},P_{k,j}\\}_{k\in
I,j\in\\{1,\ldots,t\\}}$, where $P_{k,j}=\bigoplus_{i\in
I\setminus\\{k\\}}Rx_{i,j}$.

$(2)$ By Part $(1)$, Spec$(R)=\\{M_{1},\ldots,M_{t},P_{k,j}\\}_{k\in
I,j\in\\{1,\ldots,t\\}}$, where $P_{k,j}=\bigoplus_{i\in
I\setminus\\{k\\}}Rx_{i,j}$, and so [6, p. 8, Exercise 10] implies that
$R/P_{k,j}$ is a principal ideal ring for each $k\in I$ and
$j\in\\{1,\ldots,t\\}$, as every prime ideal of $R/P_{k,j}$ is cyclic. Also,
$P_{k,j}\subset$ Ann$(x_{k,j})$ for each $k\in I$ and $j\in\\{1,\ldots,t\\}$.
Therefore, $R/$Ann$(x_{k,j})$ is a homomorphic image of $R/P_{k,j}$ and hence
$R/$Ann$(x_{k,j})$ is a principal ideal ring for each $k\in I$ and
$j\in\\{1,\ldots,t\\}$.  $\Box$

The following lemma is needed for the proof of Theorem 2.6.

###### Lemma 2.5

(See [2, Proposition 2.7]). Let $R$ be a semi-local ring and let $M\in$
Max$(R)$. If $M=\bigoplus_{i\in I}Rx_{i}=\bigoplus_{j\in J}Ry_{j}$ where $I,J$
are index sets and $Rx_{i},Ry_{j}$ are nonzero cyclic $R$-modules, then
$|I|=|J|$.

###### Theorem 2.6

Let $R$ be a semi-local ring such that satisfies condition $(*)$,
Max$(R)=\\{M_{1},\ldots,M_{t}\\}$, every prime ideal $P$ has the form
$\bigoplus_{i\in I}Rx_{i}$ where $Rx_{i}$ is a direct sum of cyclically
presented $R$-modules and $x_{i}\notin{\rm Nil}(R)$ for every $i\in I$ where
$I$ is an index set. Then for every maximal ideal $M_{j}$ of $R$ either
$M_{j}$ is cyclic or $M_{j}=Rx_{1,j}\oplus Rx_{2,j}$ such that $Rx_{1,j}$ and
$Rx_{2,j}$ are cyclically presented. Moreover, $R/{\rm Ann}(x_{1,j})$ and
$R/{\rm Ann}(x_{2,j})$ are principal ideal rings for every
$j\in\\{1,\ldots,t\\}$.

###### Proof.

Let $R$ be a semi-local ring such that satisfies condition $(*)$,
Max$(R)=\\{M_{1},\ldots,M_{t}\\}$ every prime ideal $P$ has the form
$\bigoplus_{i\in I}Rx_{i}$ where $Rx_{i}$ is a direct sum of cyclically
presented $R$-modules and $x_{i}\notin{\rm Nil}(R)$ for every $i\in I$ where
$I$ is an index set. Suppose that $M_{j}$ is not cyclic for some
$j\in\\{1,\ldots,t\\}$. Then by Theorem 2.4 Part $(2)$, $R/$Ann$(x_{i,j})$ is
a principal ideal ring and $Rx_{1,j}$ and $Rx_{2,j}$ are cyclically presented
for each $j\in\\{1,\ldots,t\\}$. Since $M_{j}$ is not cyclic, one may assume
that $x=x_{1,j}$ and $y=x_{2,j}$ are nonzero. Hence, $M_{j}=Rx\oplus Ry\oplus
L$, where $L=\bigoplus_{i\in I\setminus\\{1,2\\}}Rx_{i,j}$. We claim that
Ann$(x)=Ry\oplus L$. It is clear that $Ry\oplus L\subseteq$ Ann$(x)$. Suppose
that $a\in$ Ann$(x)$. Since $R$ satisfies condition $(*)$, $a=rx+z\in$
Ann$(x)$, where $r\in R$ and $z\in Ry\oplus L$. If $rx=0$, then
Ann$(x)=Ry\oplus L$. So, suppose that $rx\neq 0$, then $rx=bx^{n}$ for some
unit element $b$ and positive integer $n$, as $R$ satisfies condition $(*)$
and $R/$Ann$(x)$ is a principal ideal ring. This means that
$0=ax=(rx+z)x=rx^{2}=bx^{n+1}$. Hence $x^{n+1}=0$, and thus $x\in$ Nil$(R)$
which is a contradiction. This follows that Ann$(x)=Ry\oplus L$ and so the
claim is proved. Since $Rx$ is cyclically presented, Proposition 2.3 follows
that Ann$(x)$ is cyclic, and so Ann$(x)=R\alpha$ for some $\alpha\in R$.
Therefore $M_{j}=R\alpha\oplus Rx$. Thus Lemma 2.5 implies that $|I|=2$. This
completes the proof.  $\Box$

###### Corollary 2.7

Let $R$ be a semi-local ring such that satisfies condition $(*)$,
Max$(R)=\\{M_{1},\ldots,M_{t}\\}$ and every prime ideal $P$ has the form
$\bigoplus_{i\in I}Rx_{i}$ where $Rx_{i}$ is a direct sum of cyclically
presented $R$-modules and $x_{i}\notin{\rm Nil}(R)$ for every $i\in I$ where
$I$ is an index set. Then for every maximal ideal $M_{j}$ of $R$ either
$M_{j}$ is cyclic or $M_{j}=Rx_{1,j}\oplus Rx_{2,j}$ such that $Rx_{1,j}$ and
$Rx_{2,j}$ are cyclically presented. Moreover,

$(1)$ If $M_{j}=Rx_{j}$ such that $x_{j}\notin Nil(R)$ for every
$j\in\\{1,\ldots,t\\}$, then Spec$(R)=\\{(0),M_{1},\ldots,M_{t}\\}$.

$(2)$ If $M_{j}=Rx_{1,j}\oplus Rx_{2,j}$ such that $x_{1,j},x_{2,j}\notin{\rm
Nil}(R)$ for every $j\in\\{1,\ldots,t\\}$, then
Spec$(R)=\\{Rx_{1,j},Rx_{2,j},M_{1},\ldots,M_{t}\\}_{j\in\\{1,\ldots,t\\}}$.

###### Proof.

$(1)$ follows from Theorem 2.2.

$(2)$ follows from Theorem 2.4.

$\Box$

Part $(2)$ of Corollary 2.7 implies that $R$ is a local ring. Because, if $R$
has at least two maximal ideals of the form $M_{1}=P_{1,1}\oplus P_{2,1}$ and
$M_{2}=P_{1,2}\oplus P_{2,2}$ where $P_{i,j}=Rx_{i,j}$ for $1\geq i,j\geq 2$,
then $P_{1,1}P_{2,1}=0\subseteq P_{1,2},P_{2,2}$. So, without loss of
generality, $P_{1,1}\subseteq P_{1,2}$ and $P_{2,1}\subseteq P_{2,2}$, since
$P_{1,2}\cap P_{2,2}=0$. This follows that $M_{1}\subseteq M_{2}$, a
contradiction.

###### Corollary 2.8

Let $R$ be a semi-local ring such that satisfies condition $(*)$,
Max$(R)=\\{M_{1},\ldots,M_{t}\\}$, every prime ideal $P$ has the form
$\bigoplus_{i\in I}Rx_{i}$ where $Rx_{i}$ is a direct sum of cyclically
presented $R$-modules and $x_{i}\notin{\rm Nil}(R)$ for every $i\in I$ where
$I$ is an index set. Then $R$ is Noetherian.

###### Proof.

By Corollary 2.7, every prime ideal of $R$ is finitely generated and so Cohen
Theorem (see [4, Theorem 2]) implies that $R$ is Noetherian.  $\Box$

Next a class of semi-local rings for which every ideal is a direct sum of
cyclically presented $R$-modules is given.

###### Theorem 2.9

Let $R$ be a semi-local ring such that satisfies condition $(*)$,
Max$(R)=\\{M_{1},\ldots,M_{t}\\}$, $M_{j}=Rx_{1,j}\oplus Rx_{2,j}$ where
$x_{1,j},x_{2,j}\notin{\rm Nil}(R)$ and let $R/{\rm Ann}(x_{1,j})$ and $R/{\rm
Ann}(x_{2,j})$ are principal ideal rings for every $j\in\\{1,\ldots,t\\}$.
Then every ideal of R is a direct sum of cyclically presented R-modules.

###### Proof.

Let $I$ be a nonzero ideal of $R$. Then $I\subseteq M_{j}=Rx_{1,j}\oplus
Rx_{2,j}$ for some $j\in\\{1,\ldots,t\\}$. If $I\subseteq Rx_{1,j}$, then $I$
is cyclic, as $Rx_{1,j}\cong R/{\rm Ann}(x_{1,j})$ is a principal ideal ring,
and hence $I=Rz$ for some element $z$ of $R$. Since $R$ satisfies condition
$(*)$, $z=ax_{1,j}^{n}$ for some unit element $a$ and positive integer $n$.
Thus $I=Rx_{1,j}^{n}$. It is shown that Ann$(x_{1,j}^{n})=Rx_{2,j}$. It is
easy to see that $Rx_{2,j}\subseteq$ Ann$(x_{1,j}^{n})$. So suppose that
$\alpha\in$ Ann$(x_{1,j}^{n})$. Then $\alpha=rx_{1,j}+sx_{2,j}$ where $r,s\in
R$, since $R$ satisfies condition $(*)$. If $rx_{1,j}\neq 0$, again by
condition $(*)$, $rx_{1,j}=bx_{1,j}^{m}$ for some unit element $b$ and
positive integer $m$. This means that
$bx_{1,j}^{m+n}=(rx_{1,j}+sx_{2,j})x_{1,j}^{n}=0$. Thus, $x_{1,j}^{n+m}=0$, a
contradiction since $x_{1,j}\notin{\rm Nil}(R)$, and hence
$I=Rx_{1,j}^{n}\cong R/{\rm Ann}(x_{1,j}^{n})=R/Rx_{2,j}$ which is cyclically
presented. If $I\subseteq Rx_{2,j}$, a similar argument shows that $I$ is
cyclically presented. Therefore, assume that $I\nsubseteq Rx_{1,j}$ and
$I\nsubseteq Rx_{2,j}$. Now, suppose that $0\neq w=rx_{1,j}+sx_{2,j}\in I$
where $r,s\in R$. By condition $(*)$, $w=ax_{1,j}^{n}+bx_{2,j}^{m}$ for some
unit elements $a,b\in R$ and positive integers $n$ and $m$, and so
$ax_{1,j}^{n+1}=wx_{1,j}$ and $bx_{2,j}^{m+1}=wx_{2,j}$. Let $e$ and $t$ be
the smallest positive integers such that $x_{1,j}^{e}\in I$ and
$x_{2,j}^{t}\in I$. By a similar argument as above one may see that
Ann$(x_{1,j}^{e})=Rx_{2,j}$ and Ann$(x_{2,j}^{t})=Rx_{1,j}$, which means that
$Rx_{1,j}^{e}$ and $Rx_{2,j}^{t}$ are cyclically presented $R$-modules. If
$I=Rx_{1,j}^{e}\oplus Rx_{2,j}^{t}$, then $I$ is a direct sum of cyclically
presented $R$-modules. Hence, without loss of generality assume that $I\neq
Rx_{1,j}^{e}\oplus Rx_{2,j}^{t}$. It is easy to see that $Rx_{1,j}^{e}\oplus
Rx_{2,j}^{t}\subset I$, and thus there exists an element
$u=a_{1}x_{1,j}^{n_{1}}+b_{1}x_{2,j}^{m_{1}}\in I\setminus Rx_{1,j}^{e}\oplus
Rx_{2,j}^{t}$ for some unit elements $a_{1},b_{1}\in R$ and positive integers
$n_{1}$ and $m_{1}$. We show that $n_{1}=e-1$ and $m_{1}=t-1$. First, suppose
that $n_{1}\geq e$ and $m_{1}\geq t$. Then $u\in Rx_{1,j}^{e}\oplus
Rx_{2,j}^{t}$, a contradiction. If $n_{1}\geq e$ and $m_{1}<t$, then
$a_{1}x_{1,j}^{n_{1}}\in Rx_{1,j}^{e}\subset I$, and so
$b_{1}x_{2,j}^{m_{1}}\in I$. This means that $x_{2,j}^{m_{1}}\in I$, a
contradiction. If $m_{1}\geq t$ and $n_{1}<e$, then $b_{1}x_{2,j}^{m_{1}}\in
Rx_{2,j}^{t}\in I$, and so $a_{1}x_{1,j}^{n_{1}}\in I$. Therefore
$x_{1,j}^{n_{1}}\in I$, a contradiction. Thus, $n_{1}<e$ and $m_{1}<t$. If
$n_{1}<e-1$, then one may see that $ux_{1,j}=a_{1}x_{1,j}^{n_{1}+1}\in I$.
Therefore, $x_{1,j}^{n_{1}+1}\in I$ which is a contradiction. Hence
$n_{1}=e-1$. A similar argument shows that $m_{1}=t-1$, and thus
$u=a_{1}x_{1,j}^{e-1}+b_{1}x_{2,j}^{t-1}$. In the following, we show that
$I=Ru$. One can easily see that $Rx_{1,j}^{e}\oplus Rx_{2,j}^{t}\subseteq
Ru\subseteq I$. Conversely, suppose that $v\in I\setminus(Rx_{1,j}^{e}\oplus
Rx_{2,j}^{t})$. As above, we have $v=a_{2}x_{1,j}^{e-1}+b_{2}x_{2,j}^{t-1}$
for some unit elements $a_{2},b_{2}\in R$. Thus,
$v=a_{2}a_{1}^{-1}u+(b_{2}-a_{2}a_{1}^{-1})x_{2,j}^{t-1}\in Ru\oplus
Rx_{2,j}^{t-1}$. Hence $I\oplus Rx_{2,j}^{t-1}\subseteq Ru\oplus
Rx_{2,j}^{t-1}$, and thus $I\oplus Rx_{2,j}^{t-1}=Ru\oplus Rx_{2,j}^{t-1}$.
But $R$ is semi-local; hence finitely generated projective modules cancel from
direct sums. Therefore $I=Ru$. It is enough to prove that $Ru$ is cyclically
presented. Clearly, $Rx_{1,j}\cap{\rm Ann}(Rx_{1,j})=0$ and $Rx_{2,j}\cap{\rm
Ann}(Rx_{2,j})=0$. We claim that Ann$(u)=0$. Suppose to the contrary, $0\neq
w\in{\rm Ann}(u)$. Then $w=a_{3}x_{1,j}^{n_{3}}+b_{3}x_{2,j}^{m_{3}}$ for some
unit elements $a_{3},b_{3}\in R$ and positive integers $n_{3}$ and $m_{3}$
since Ann$(u)\subseteq{\rm Ann}(Ru)={\rm Ann}(Rx_{1,j})\cap{\rm
Ann}(Rx_{2,j})\subseteq M_{j}$. This means that
$wa_{1}x_{1,j}^{n_{1}}=a_{1}a_{3}x_{1,j}^{n_{1}+n_{3}}\in Rx_{1,j}\cap{\rm
Ann}(u)\subseteq Rx_{1,j}\cap{\rm Ann}(Rx_{1,j})\cap{\rm Ann}(Rx_{2,j})=0$. By
a similar argument as above one may see that
$wb_{1}x_{2,j}^{m_{1}}=b_{1}b_{3}x_{2,j}^{m_{1}+m_{3}}\in Rx_{2,j}\cap{\rm
Ann}(u)\subseteq Rx_{2,j}\cap{\rm Ann}(Rx_{1,j})\cap{\rm Ann}(Rx_{2,j})=0$.
Thus $x_{1,j},x_{2,j}\in$ Nil$(R)$, a contradiction. Therefore, Ann$(u)=0$,
and hence $I=Ru\cong R/(0)$ is cyclically presented. $\Box$

Consequently, in view of the proof of Theorem 2.9, we state the following
corollary.

###### Corollary 2.10

Let $R$ be a semi-local ring such that satisfies condition $(*)$,
Max$(R)=\\{M_{1},\ldots,M_{t}\\}$, $M_{j}=Rx_{1,j}\oplus Rx_{2,j}$ where
$x_{1,j},x_{2,j}\notin{\rm Nil}(R)$ and let $R/{\rm Ann}(x_{1,j})$ and $R/{\rm
Ann}(x_{2,j})$ are principal ideal rings for every $j\in\\{1,\ldots,t\\}$.
Then every ideal of $R$ is one of the following forms $R$, $Rx_{1,j}^{n}$ for
some positive integer $n$ and $j\in\\{1,\ldots,t\\}$, $Rx_{2,j}^{m}$ for some
positive integer $m$ and $j\in\\{1,\ldots,t\\}$, and $Rx_{1,j}\oplus Rx_{2,j}$
for some $j\in\\{1,\ldots,t\\}$.

We are now ready for the main result of this paper. In fact, we give two
gauges to check whether every ideal of a semi-local ring $R$ is a direct sum
of cyclically presented modules, it suffices to test only the prime ideals or
structures of the maximal ideals of $R$.

###### Theorem 2.11

Let $R$ be a semi-local ring such that satisfies condition $(*)$,
Max$(R)=\\{M_{1},\ldots,M_{t}\\}$ and if $M_{j}=\bigoplus_{i\in I}Rx_{i,j}$,
then $x_{i,j}\notin{\rm Nil}(R)$ for every $i\in I$. Then the following
statements are equivalent:

$(1)$ Every ideal of R is a direct sum of cyclically presented $R$-modules.

$(2)$ Every ideal of $R$ is a direct summand of a direct sum of cyclically
presented $R$-modules.

$(3)$ Every prime ideal of $R$ is a direct sum of cyclically presented
R-modules.

$(4)$ Every prime ideal of $R$ is a direct summand of a direct sum of
cyclically presented R-modules.

$(5)$ Either $R$ is a principal ideal ring or $M_{j}=Rx_{1,j}\oplus Rx_{2,j}$,
where $R/{\rm Ann}(x_{1,j})$ and $R/{\rm Ann}(x_{2,j})$ are principal ideal
rings for every $j\in\\{1,\ldots,t\\}$.

###### Proof.

$(5)\Rightarrow(1)$ It follows from Theorem 2.9.

$(3)\Rightarrow(5)$ It follows from Theorem 2.6, Corollary 2.7 and [6, p. 8,
Exercise 10].

$(1)\Leftrightarrow(2)$ and $(3)\Leftrightarrow(4)$ follow from [12,
Proposition 3].  $\Box$

We close this paper with the following example.

###### Example 2.12

Let $F$ be a field and $R:=R_{1}\oplus R_{2}$ such that
$R_{1}=F[[X,Y]]/{\rm<}XY{\rm>}$ and $R_{2}$ is a field where $F[[X,K]]$ is a
formal power series ring. It is easy to see that $R_{1}$ and $R_{2}$ are local
rings with maximal ideals $M_{1}=R_{1}x\oplus R_{1}y$ and $M_{2}=\\{0\\}$
respectively, where $x=X+{\rm<}XY{\rm>}$, $y=Y+{\rm<}XY{\rm>}$ and $x,y\notin$
Nil$(R_{1})$. Also, Max$(R)=\\{M_{1}\oplus R_{2},R_{1}\oplus M_{2}\\}$. First
we show that $R$ satisfies condition $(*)$. Suppose that $a\in R_{1}x$ such
that $a\neq 0$. Since $R_{1}$ is a local ring and $R_{1}/{\rm Ann}(x)$ is a
principal ideal ring by [10, Example 2.17], it follows from [3, Proposition
3.4] that $a=bx^{n}$ for some unit $b$ and positive integer $n$. Hence the
axiom $(1)$ in condition $(*)$ is satisfied, and so if $M_{i}=Rx_{i}\oplus L$
for some ideal $L$ of $R$ and some nonunit $x_{i}\in R$ such that $R/{\rm
Ann}(x_{i})$ is a principal ideal ring, then every nonzero element of $Rx_{i}$
is of the form $ax^{n}$ for some unit $a$ and positive integer $n$. Now, since
$x,y$ are nonunit in $R_{1}$ and $x,y\notin$ Nil$(R_{1})$, we conclude that
Ann$(x_{i}^{\alpha_{i}})\subset M_{1}$ for every $x_{i}\in\\{x,y\\}$ and
positive integer $\alpha_{i}$. This means that
Ann$(x_{i}^{\alpha_{i}})\subseteq M_{1}\oplus R_{2}$ for every
$x_{i}\in\\{x,y\\}$ and positive integer $\alpha_{i}$. It remains only for us
show that for every finitely generated module $N$ over the commutative ring
$R$, if $I$ is an ideal of $R$ such that $I\subseteq M_{j}$ for some
$M_{j}\in$ Max$(R)=\\{M_{1}\oplus R_{2},R_{1}\oplus M_{2}\\}$ and $N=IN$, then
$N=0$. Suppose contrary that $N\neq 0$ and look for a contradiction. Assume
that $\\{a_{1},\ldots,a_{n}\\}$ is a minimal generating set for the module $N$
over the commutative ring $R$ and $I\subseteq M_{1}\oplus R_{2}$ such that
$N=IN$. Since $a_{1}\in IN$, we have $a_{1}=\sum_{i=1}^{n}a_{i}b_{i}$ for some
$b_{1},\ldots,b_{n}\in I$, and so
$(1-b_{1})a_{1}=a_{2}b_{2}+\cdots+a_{n}b_{n}$. Suppose that $b_{1}=m+r_{2}\in
M_{1}\oplus R_{2}$ for some $m\in M_{1}$ and $r_{2}\in R_{2}$. If $r_{2}\in
M_{2}$, then $b_{1}\in M_{1}\oplus M_{2}\subseteq R_{1}\oplus\\{0\\}$, and so
$1-b_{1}$ is a unit of $R$ by [11, Lemma 3.17]. Thus $N$ is generated by
$\\{a_{2},\ldots,a_{n}\\}$, a contradiction. If $r_{2}\notin M_{2}$, then
$r_{2}$ is a unit of $R_{2}$. This means that
$(1-m_{1})a_{1}-r_{2}a_{1}=\sum_{i=2}^{n}a_{i}b_{i}$. Hence
$(r_{2}-r_{2}^{2})a_{1}=\sum_{i=2}^{n}a_{i}b_{i}r_{2}$. Since $R_{2}$ is a
field we conclude that $(r_{2}-r_{2}^{2})$ is unit, and hence $N$ is generated
by $\\{a_{2},\ldots,a_{n}\\}$, a contradiction. Thus $N=0$. In the other hand,
it follows from Theorem 2.11 that every ideal of $R$ is a direct sum of
cyclically presented modules.

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