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{\hfill{\small\rm C. Abdio\^{g}lu et al.}} % authors
{{\small\rm Submodules Are Projection Invariant and  Lifting Condition} \hfill} % abbreviated title

\showwiselinenumbers
\title{An ideal based  REGULAR DIGRAPH OF IDEALS OF COMMUTATIVE RINGS\\[2mm] Invariant and  the  Lifting Condition}

\bth
\author{C. Abdio\^{g}lu}

\no{\smaller Department of Mathematics,    Karamano\^{g}lu Mehmetbey
University,
Yunus Emre  Campus, Karaman,  Turkey.\\
Email: cabdioglu@kmu.edu.tr}

\dzh

\author{M. T. Ko\c san \ and \ S.  \c{S}ahinkaya}

\no{\smaller Department of Mathematics,   Gebze Institute of Techonology,
\c{C}ayirova  Campus, 41400 Gebze- Kocaeli,  Turkey.\\
Email: mtkosan@gyte.edu.tr; ssahinkaya@gyte.edu.tr}

\dzh

\no{Received 10 February 2008\\
Accepted 28 April 2010}
\flh
\no{Communicated by Nguyen Van Sanh}
\flh
\no{\smallerbf AMS Mathematics Subject Classification(2000):}{\smaller\ 16D99,
13C12, 13B20  }
\flh
\no{\smallerbf Abstract.}{\smaller\ The notion of projection invariant
subgroups was first introduced by Fuchs in \cite{Fuch}.
We will  define  the module-theoretic version of the projection
invariant subgroup.
In this note, we  give several characterizations
to these class of modules that generalize the recent results
in \cite{Ozcan}. We also
define and study the PI-lifting modules which is a generalization of
FI-lifting module.
It is shown that if each $M_i$ is a PI-lifting module for all
$1\leq i \leq n$, then
$M=\oplus_{i=1}^nM_i$ is a PI-lifting module. In particular,
we focus on  rings  satisfying
the following condition:
\begin{lieju}
\item[{\rm(*)}]Every submodule of $M$ is projection  invariant.
\end{lieju}
We prove that if $R$ has the $(*)$ property, then $R\oplus R$
does not satisfy  the $(*)$ property.}
\zyh
\no{\smallerbf Keywords:}{\smaller\ Fully invariant submodules; Projection
invariant submodules; Duo modules and rings;
Finite exchange property; Lifting modules.}
\zjq
\normalsize
\zhangjie{Introduction}
\no
Throughout this paper,  $R$ is an associative ring with identity
and all modules are unitary. For a right $R$-module $M$, we use $S={\rm End}_R (M_R)$ to denote
the endomorphism ring of $M$. Obviously, the module $M$ is an $(S,R)$-bimodule.

A submodule $N$ of $M$ is  said to be a {\em fully invariant} if $f(N)$ is contained in
$N$ for every $f\in S$. Clearly, $0$ and $M$ are fully invariant submodules of $M$. The
right $R$-module $M$ is said to be  {\em duo } provided every submodule of $M$ is
fully invariant.
It is clear that every simple right $R$-module is a duo module. Moreover, if the right $R$-module
$R$ is a duo module, then the ring $R$ is called {\em right duo}. Note that a ring $R$ is a right
duo ring if and only if every right ideal of $R$ is a two-sided ideal, equivalently $Ra$ is contained
in $aR$ for every element $a$ in $R$.

\example Let $\Bbb Z$ be the ring of integers, $n$ a positive integer and $p$ a prime integer.
Then, $\Bbb Z$ and $\Bbb Z/\Bbb Zp^n$ are duo $\Bbb Z$-modules, but the filed of rationals
is not a duo $\Bbb Z$-module.

\example
{\rm (see \cite[Example 6]{KH})}   Let $p$ be a
prime integer.Then, we have the following properties:
\begin{lieju}
\item[(1)]  The  $\Bbb Z$-module $\Bbb Z\oplus A$ is not duo for
any $\Bbb Z-$module $A$.
\item[(2)]  For any distinct prime
integers $p_i$ $(i=1,2,...,n)$, the $\Bbb Z-$module
$M=\oplus_{i=1}^n\Bbb Z/\Bbb Zp_{i}^{n_i}$ is a duo module for any
positive integers $n_i(i=1,2,...,n)$.
\item[(3)] The $\Bbb
Z$-module $\Bbb Q\oplus A$ is not a duo module for any $\Bbb
Z$-module $A$.
\end{lieju}
\dl

 An $R$-module $M$ is said to have the {\it summand sum property} if the sum of any two direct summands
of $M$ is a direct summand of $M$ ($SSP$).


$M$ is said to have the {\it summand intersection property} if
the intersection of any two direct summands of $M$ is a direct
summand of $M$ ($SIP$) (see \cite{GA},\cite{H}, \cite{Wi}).

\begin{thm}
{\rm (\cite [Theorem 5]{KH})}
Let $M$ be a duo module.  Then $M$ has the SIP and the SSP.
\end{thm}

In Section 2, we obtain some new properties of fully invariant
submodules and duo modules. In particular, it is shown that a
direct summand complement of a direct summand of $M$ is unique if
and only if it is a fully invariant submodule of $M$.

  Let $M$ be a module and $N$ be a submodule of $M$.
We call $N$ {\it a projection invariant submodule} of $M$ if every
projection $\pi$ of $M$ onto a direct summand maps $N$ into
itself, i.e. $N$ is invariant under any projection of $M$.
Clearly, each fully invariant submodule of a module is a
projection invariant submodule.

 In the first part of Section 3, we obtain some basic properties  of
 projection invariant submodules and also we study the interrelation between these
submodules and the finite exchange property .

In the last part of Section 3, we consider  the condition $(*)$.

We will prove that the condition $(*)$ holds for every direct summand of $M= \oplus_{i\in
I} M_i$ where each  $M_i$ $(i\in I)$ is an  indecomposable submodule of $M$ if and only if
\begin{lieju}
\item[{\rm(1)}]$Hom(M_i,M_j)=0$ for all distinct $i,j\in I$, and
\item[{\rm(2)}] For every direct summand $N$ of $M$, there exist a (finite)
subset $I'$ of $I$ such that $N=\oplus_{i\in I'} (N\cap M_i)$.
\end{lieju}
Let $M$ be an $R$-module and $N$ be a submodule of $N$. $N$  is
called {\it small}, written $N \ll M$, if $M\neq N+L$ for every
proper submodule $L$ of $M$. Properties of small submodules are
given in \cite[ Lemma 4.2]{MM} and \cite[ Proposition 19.3]{Wi}.
Let $M$ be a module. $M$ is said to be a  {\it lifting module}, if
for every submodule $N$ of $M$, $M$ has a decomposition
$M=M_1\oplus M_2$ with $M_1\leq N$ and $M_2\cap N$ small in $M_2$,
i.e. if for every submodule $A$ of $M$ there exists a direct
summand $B$ of $M$ such that $B\leq A$ and $A/B$ is small in
$M/B$.

According to Ko\c{s}an \cite{Ta}, the module $M$ is called {\it
FI-lifting } if for every fully invariant submodule $A$ of $M$,
there exists a direct summand $B$ of $M$ such that $B \subseteq A$
and $A/B$ small in $M/B$ as a generalization of lifting module. By
\cite{Ta}, if $X$ is a  fully invariant submodule of a FI-lifting
module $M$ then $M/X$ is FI-lifting. In this section, similar to
FI-lifting modules, we define PI-lifting modules. $M$ is called a {\
PI-lifting module } if for every projection invariant submodule $A$
of $M$, there exists a direct summand $B$ of $M$ such that $B
\subseteq A$ and $A/B$ small in $M/B$. This definition is not
meaningless, that is not every  PI-lifting module is a lifting
module. Let $M_{\Bbb Z}={\Bbb Z}/2{\Bbb Z}\oplus {\Bbb Z}/8{\Bbb
Z}$. Then $M_{\Bbb Z}$ is a FI-lifting module  by \cite[Corollary
3.5]{Ta}. Therefore, $M$ is a PI-lifting module. We note that
$M_{\Bbb Z}$ is not a lifting module by \cite[ Example 1 ]{Der}.
On the other hand,
\begin{lieju}
\item[{\rm(1)}]  $M$ is a PI-lifting module if and only if for every
projection invariant submodule $A$ of $M$ there exist a
decomposition $M=M_1\oplus M_2$ such that $M_1\leq A$ and $M_2\cap
A$ is small in $M_2$.
\item[{\rm(2)}] By definitions, every lifting modules are FI-lifting and PI-lifting. One may suspect that if $M$ is an FI-lifting module then it is also a PI-lifting module. But the following example eliminates this possibility: Let $R$ be a simple domain that is not a division ring (e.g. the first Weyl
Algebra over a field of characteristic 0). Then the only fully invariant right
ideals of $R$ are the trivial ones, so $R_R$ is FI-lifting. Since the only idempotents
of $R$ are 0 and 1 any right ideal of $R$ is projection invariant; but $J(R) = 0$,
so that $R_R$ is not PI-lifting.
\end{lieju}
In  Section 4, we obtain some basic properties of projection
invariant lifting modules. In particular, it is shown that if each
$M_i$ is a PI-lifting module, then $M=\oplus_{i=1}^nM_i$ is a
PI-lifting module.
\zjq
\zhangjie{Fully Invariant Submodules}
\no
The next results are well known facts proved for
groups in   Lemma 9.5, Theorem~9.6 and Corollary 9.7 of
\cite{Fuch}, respectively.

\begin{lem}
Let $M = M_1\oplus M_2$ be a decomposition of $M$ with
associated projections $\pi_i:M\rightarrow M_i$ (for $i=1, 2$). If
we also have $M = M_1\oplus M_3$ with projections
$\pi_1':M\rightarrow M_1$ and $\pi_3:M\rightarrow M_3$, then, for
some endomorphism $\phi$ of $M$, we have
\begin{linenomath*} \begin{align}
\pi'_1=\pi_1+\pi_1\phi\pi_2 \quad\quad  \mbox{and} \quad\quad
 \pi_3= \pi_2-\pi_1\phi\pi_2
\tag{2.1}
\end{align}\end{linenomath*}
Conversely, if $\pi_1'$ and $\pi_3$ are
endomorphisms of $M$ satisfying $(2.1)$ for some $\phi\in End(M)$,
then $M = M_1\oplus \pi_3(M)$.
\end{lem}
\begin{proof}
Let $\phi=\pi_2-\pi_3$. Then $M_1\leq Ker(\phi)$.
 Since $\phi\pi_1=0$ and $1_M=\pi_1+\pi_2$,
$\phi=\phi\pi_1+\phi\pi_2=\phi\pi_2$. Let $m=m_1+m_2=m'_1+m_3\in
M$ where $m_1,m'_1\in M_1$, $m_2\in M_2$, and $m_3\in M_3$. Then
$\phi(m)=(\pi_2-\pi_3)m=m_2-m_3=m'_1-m_1\in M_1$. Hence
$\pi_1\phi(m)=\phi(m)$ for all $m\in M$. Thus $\phi=\pi_1\phi$.
Since $\phi=\pi_1\phi=\phi\pi_2$,  we have $\phi=\pi_1\phi\pi_2$
and $\pi_3=\pi_2-\phi=\pi_2-\pi_1\phi\pi_2$. Since
$1_M=\pi'_1+\pi_3$, then $\pi'_1=1_M-\pi_3=\pi_1+\pi_2-\pi_3=
\pi_1+\pi_1\phi\pi_2$.

Conversely, assume that $\pi'_1$ and $\pi_3$ are of the form $(2.1)$.
We add the equalities $(2.1)$ side by side to get
$\pi'_1+\pi_3=\pi_1+\pi_2=1_M$. Also it is easy to check that
$\pi'_1$ and $\pi_3$ are orthogonal idempotents in $S$. Then
$M=\pi'_1(M)\oplus \pi_3(M)$. By $(2.1)$, $\pi'_1(M)\leq\pi_1(M)$, and
since $\pi_1(M)= M_1$ and $\pi_3(M)\cap M_1=0$,  we have
$M=M_1\oplus \pi_3(M)$.
\end{proof}
\fudl
\begin{thm}
If $M_1$ is a direct summand of the module $M$, then
the intersection of all direct summand complements of $M_1$ in $M$
is the maximal fully invariant submodule of $M$ that has the zero
intersection with $M_1$.
\end{thm}
\begin{proof}
Let $K$ denote the intersection of all direct summand
complements of $M_1$ in $M$. Let $M=M_1\oplus M_2$ and both $\pi_1$
and $\pi_2$ be projections of $M$ along  $M_1$ and $M_2$ respectively,
and let $\phi\in S=End(M)$. By Lemma 2.1, $M_3=(\pi_2-\pi_1\phi\pi_2)(M)$ is again
a direct summand complement of $M_1$ in $M$. Let $x\in K$. Since
$K\leq M_2\cap M_3$, $(\pi_2-\pi_1\phi\pi_2)(x)=x$ and $\pi_2(x)=x$. Hence $0=(\pi_1\phi\pi_2)(x)
=(\pi_1\phi)(x)$. Thus, $\phi(x)\in M_2,$ for all direct summand complement of $M_2$ in $M$.
It follows that $\phi(x)\in K$. Now clearly $M_1\cap K=0$. If $L$ is any fully invariant submodule
of $M$ with $L\cap M_1=0$ and $M=M_1\oplus M_2$,  then, if $x\in L$ with $x = m_1 + m_2$
where $m_1\in M_1$ and $m_2\in M_2$, we have $m_1 = \pi_1(x) \in M_1\cap L =0$, and so
$L=(L\cap M_1)\oplus (L\cap M_2)=L\cap M_2$. Hence $L\leq M_2$ for all direct summand complements
of $M_2$ in $M$.Thus, $L\leq K$. This completes the proof.
\end{proof}
\fudl
\begin{cor}
Let $M$ be a module. A direct summand complement of a direct
summand of $M$ is unique if and only if it is a fully invariant
submodule of $M$.
\end{cor}

Following Warfield \cite{Wa}, we say that a ring $R$ is  {\it
exchange } in case the regular right $R$-module $R_R $ satisfies the
(finite) exchange property, that is, for every $R$-module $M$ and
decompositions 
\begin{linenomath*} $$M=X\oplus Y=\oplus_{i\in I} M_i
$$ \end{linenomath*}
with $X\cong R_R$
(and $I$ finite), there exist submodules $N_i\subseteq M_i$ such
that
\begin{linenomath*} $$
M=X\oplus (\oplus_{i\in I} N_i).
$$\end{linenomath*}

Some kinds of generalized
exchange rings were studied by Chen Huayin in \cite{chen}. We remark here that   GM-condition
on a ring $R$ was also stated by H.Chen  and M.Chen  which generalizes the known  unit 1-stable
range condition \cite{chen-chen}. By using this GM-condition, they investigated the exchange rings with Artinian
primitive factors satisfying the GM-condition.

It is well know that  all
continuous modules have the full exchange property (see \cite{MM}). The following theorem is a slight version of this result on  quasi-injective modules.
For some new results on injective module and quasi  injective modules, the readers are referred to \cite{Sanh 2},  \cite {Hai} and \cite {Sanh}.


\begin{thm}
Every quasi-injective duo module has the finite exchange
property.
\end{thm}
\begin{proof}
Let $M$ be a quasi-injective duo module with $S=End_R(M)$. Note that
every module is a submodule of a quasi-injective module. Let $N$
be a right  $R$-module and $g: N\rightarrow M$ be a monomorphism.
Then we may assume $g(N)$ is a fully invariant submodule of $M$.
Since $M$ is a quasi-injective module, for $\alpha, \beta \in
S'=End_R(N)$ with $\alpha +\beta =1_N$, there exist a $f\in S$
such that $g\alpha =fg$. It is easy to see that
$g\beta=(1_M-\alpha)g$ and so $\alpha +(1_M-\alpha)=1_M$. Now,
since $S/J(S)$ is regular and self-injective,  the ring $S$ is an
exchange ring by \cite[Theorem 29.2]{T}. By \cite[Theorem
29.1]{T}, we have $e_1\in \alpha S$ with $e_1^2=e_1$ and $e_2\in
(1_M -\alpha) S$ with $e_2^2=e_2$ such that $e_1+e_2=1_M$. Let
$e_1=\alpha s_1$ and $e_2=\alpha s_2$ for some $s_1,s_2\in S$.
Since $g(N)$ is a fully invariant submodule of $M$, there are
unique $h_1,h_2,t_1,t_2\in S'$ such that $gh_1=e_1g$, $gh_2=e_2g$
, $gt_1=s_1g$ and $gt_2=s_2g$. Then $h_1,h_2$ are idempotents and
$h_1+h_2=1_N$. Since $g: N\rightarrow M$ is  monomorphism, we have
$h_1=\alpha t_1$ and $h_2=\beta t_2$. Now, by \cite[Theorem 29.1]{T},
the ring $S'$ is an
exchange ring. This implies that $N$ has the finite exchange property.
\end{proof}


\zhangjie{Projection Invariant Submodules}
\no
We list below some of the basic properties of  projection
invariant submodules that will be needed in this paper.

\begin{prop}
\label{oz}
Let $M$ be a module and $N$
be a submodule of $M$. Then;

$(1)$ $N$ is a projection invariant submodule of $M$  if and only
if $\pi(N)=N\cap\pi(M)$ for every projection $\pi$ of $M$.

$(2)$ $N$ is a projection invariant submodule of $M$ if and only
if $N$ is an intersection of projection invariant submodules of
$M$.

$(3)$ Any sum and intersection of projection  invariant submodules
of $M$ is again a projection invariant submodule of $M$.

$(4)$ A projection invariant direct summand of $M$ is a fully
invariant submodule of $M$.

$(5)$ Let $M=M_1\oplus M_2$ be a decomposition and $N$ be any
projection invariant submodule of $M$. Then $N=(N\cap M_1)\oplus
(N\cap M_2)$.

$(6)$ If $M=\oplus_{i\in I}M_i$ and $N$ is a projection invariant
submodule of $M$, then $N=\oplus_{i\in I}\pi_i(N)=\oplus_{i\in
I}(M_i\cap N)$, where $\pi_i$ is the $i$-th  projection
homomorphism of $M$ along $M_i$.
\end{prop}
\begin{proof}
$(1)$ Assume that $N$ is a projection invariant submodule of $M$.
Let $\pi$ be a projection of $M$. Then  $\pi(N)\leq
N\cap\pi(M)\leq \pi(N)$. Since  $N\cap\pi(M)\leq \pi(N)$ always
holds, then
$\pi(N) = N\cap\pi(M)$. The converse is clear.

$(2)$ Assume that $N$ is a projection invariant
submodule of $M$. Note that $M$ is a  projection invariant
submodule of $M$. Since $N=N\cap M$, then $N$ is the intersection
of projection invariant submodules $N$ and $M$. Conversely, let
$N=\cap_{i\in I} N_i$ where $N_i$ $(i\in I)$ are projection
invariant submodules of $M$ and let $\pi$ be a projection of $M$.
Then $\pi(N)=\pi(\cap_{i\in I} N_i)\leq \cap_{i\in I} \pi(N_i)\leq
\cap_{i\in I} N_i=N$.
Hence $\pi(N)\leq N$.

(3) This is similar to \cite[ Lemma 3.2]{Ta}.

(4) Let $M_1$ be a projection invariant direct summand
of $M$, $f\in S=End(M)$ and $M=M_1\oplus M_2$. Let $\pi_1$ and
$\pi_2$ be projections of $M$ onto $M_1$ and $M_2$, respectively.
Let $\phi$ be any element in $S$. By Lemma 2.1, we obtain that
$\pi_3=\pi_1-\pi_2\phi\pi_1$ is a projection of $M$. By
hypothesis, $\pi_3(M_1)\leq M_1$. Let $x\in M_1$. Then
$\pi_3(x)=x-(\pi_2\phi)(x)\in M_1$. Hence $(\pi_2\phi)(x)=0$. Thus
$\phi(x)\in M_1$.

$(5)$  Let $\pi_1$ and $\pi_2$ be projections of $M$
along with $M_1$ and $M_2$ respectively. Then, for any
$m=m_1+m_2\in M$ where $m_1\in M_1$ and $m_2\in M_2$, we have
$\pi_1(m)=m_1$ and $\pi_2(m)=m_2$. Let $n=n_1+n_2\in N$ where
$n_1\in M_1$ and $n_2\in M_2$. By hypothesis, we obtain that
$\pi_1(n)=n_1\in N$ and $\pi_2(n)=n_2\in N$, and so
$\pi_1(n)=n_1\in N\cap M_1$ and $\pi_2(n)=n_2\in N\cap M_2$. Then
$n=n_1+n_2\in N\cap M_1+ N\cap
M_2$. Hence $N\leq N\cap M_1+ N\cap M_2$. The rest is clear.

$(6)$ This is similar to \cite[ Lemma 3.2]{Ta}.
\end{proof}


Let $M$ and $N$ be two submodules with $S=End_R(M_R)$ and
$S'=End_R(N_R)$. For a right $R$-homomorphism $g: N\rightarrow M$,
we consider the set $I=\{f\in S: gf=0\}$. It is easy to see that
$I$ is a right ideal of $S$.

\begin{prop}
Let $M$ be a quasi-projective module. With the above notation, if $I$ is a projection
invariant direct summand, then $S/I\cong S'$.
\end{prop}
\begin{proof}
Let $M$ be a quasi-projective module  and
$I=\{f\in S: gf=0\}$ for any right $R$-homomorphism $g:
N\rightarrow M$. By Proposition 3.1,  we may assume that $I$ is
fully invariant. Then we have the $R$-module homomorphism $h:
M/Rad (M)\rightarrow S $ such that $hg=gf$. Now $\alpha :
S\rightarrow S'$, defined by $\alpha (f)=h_f$ where $h_f$ depends
on  any $f\in S$, is a homomorphism. Since $M$ is a
quasi-projective module, for any $\beta \in S'$, there exist a
$f'\in S$ such that $\beta g=gf'$. It is easy to see that $\alpha
$ is a surjective homomorphism and $Ker (\alpha )=I$.
\end{proof}
\fudl
\begin{thm} Let $M$ be a quasi-projective module and $N$ be a module. With the above notations,
let $I$ be a projection invariant direct summand. If $M$ has the
finite exchange property, then
\begin{lieju}
\item[{\rm(1)}]$N$ has the finite exchange property.
\item[{\rm(2)}] $I$ is an exchange ring.
\item[{\rm(3)}] For any $f\in S$, if $Im (f-f^2)\subseteq I$ then there
exists an idempotent $e\in S$ such that $Im (f-e)\subseteq I$.
\end{lieju}
\end{thm}
\begin{proof}
$(1)$ Assume that $M$ has the finite exchange property. By
\cite[Theorem~28.7]{T}, the ring $S$ is an exchange ring. Then, by
Proposition 3.2, we have $S/I\cong S'$. By \cite[Theorem 2.2]{A},
the ring $S'$ is an exchange ring. Then $N$ has the finite
exchange property.

$(2)$ Clear.

 $(3)$ For any $f\in S$, there
exist an idempotent $e\in S$ such that $f-e=(f-f^2)f'$ for some
$f'\in S$ by \cite[Theorem 29.1]{T}. This implies that
$g(f-e)=g(f-f^2)f'=0$, i.e., $Im (f-e)\subseteq I$.
\end{proof}


We consider the condition  $(*)$ for an $R$-module $M$.

Clearly, duo modules satisfy the $(*)$-condition. If $M$
is a right $R$-module, then $M$ satisfies the $(*)$-condition
because $M$ is a duo module.


\begin{prop}
\label{dik}
\begin{lieju}
\item[{\rm(1)}]If a module satisfies   the
$(*)$-condition, then any direct summand of it  also satisfies
the $(*)$-condition.
\item[{\rm(2)}]Let $M$ be an $R$-module. Assume that the $(*)$-condition
holds for every summands of $M$, i.e. all direct summands of $M$
are projection invariant. Then $M$ has the SIP and SSP properties.
\end{lieju}
\end{prop}
\begin{proof}
$(1)$ Assume $M$ satisfies  the
$(*)$-condition and $M=M^\prime\oplus M^{\prime\prime}$ with
$M^\prime, M^{\prime\prime}$ submodules of $M$.
 Let $\pi_{M^\prime}: M\rightarrow M^\prime$ be the canonical projection and
 let $N$ be any submodule of $M^\prime$. Suppose that $\pi$ is a
 projection of $M^\prime$, i.e. $\pi: M^\prime= M^\prime\oplus (0) \rightarrow
 M^\prime$.  Then $p=\pi \pi_{M^\prime}$ is a projection of $M$ and $\pi(N)=p(N)$ which is contained
 in $N$ because $M$ satisfies  the
$(*)$-condition. It follows that $M^\prime$ satisfies  the
$(*)$-condition.

$(2).$  Let $M_1$ and $M_2$ be direct summands of $M$. Note that,
$M_1$ and $M_2$ are fully invariant submodules of $M$ by
Prop.~3.1(4). For some submodule $M_2 '$ of $M$, let
$M=M_2\oplus M_2 '$. By assumption and Prop.~3.1(6),
we have $M_1=(M_1 \cap M_2)\oplus (M_1\cap M_2 ')$. Clearly,
$M_1\cap M_2$ is a direct summand of $M$, i.e. $M$ has the SIP
property. Since $M_1+M_2=M_2 \oplus (M_1\cap M_2 ')$ and $M_1\cap
M_2 '$ is a direct summand of $M_2 '$, then $M_1\cap M_2 '$ is a
direct summand of $M$, i.e., $M$ has the SSP property.
\end{proof}
\fudl
\remark
Note that Proposition 3.5(2) also follows
from Proposition 3.1.(4) and Theorem 1.3.

In  \cite[ Proposition 1.3]{Ozcan}, it is proved that any direct
summand of a duo module is also a duo module.

\begin{prop}
Any direct summand of a duo module is
also a duo module.
\end{prop}
\begin{proof} The proof is clear from Props.~3.4 and 3.1.
\end{proof}
\fudl
\begin{prop}
\label{hom}
Let $M$ be an $R$-module.
\begin{lieju}
\item[{\rm(1)}]Assume that $M$ has a decomposition  $M=M_1\oplus M_2$ for some
 submodules $M_1$,$M_2$ of $M$. If $M_1$ is a projection invariant
submodule of $M$, then $Hom(M_1, M_2)=0$.
\item[{\rm(2)}]Assume that the $(*)$-condition holds for every direct
summand of $M$. If $M$ has a decomposition $M=M_1\oplus M_1$ for
some  submodules $M_1$,$M_2$ of $M$, then $Hom(M_1,M_2)=0$.
\end{lieju}
\end{prop}
\begin{proof}
$(1).$  By Prop.~3.1, we can suppose  that
$M_1$ is a fully invariant submodule of $M$. Let $f:
M_1\rightarrow M_2$ be any homomorphism. Let $p_1:M\rightarrow
M_1$ denote the canonical projection and let $i_2:M_2\rightarrow
M$ denote inclusion. Then $f^*=i_2fp_1$ is an endomorphism of $M$.
By hypothesis, $f^*(M_1)\subseteq M_1$, so that $f(M_1)\subseteq
M_1\cap M_2=0$. It follows that $f=0$.

$(2)$. By $(1)$ and Prop.~3.1(4).
\end{proof}
\fudl
\begin{thm}
\label{44}
Let a module $M=\oplus_{i\in I} M_i$
 be a direct sum of submodules $M_i$ $(i\in I)$. Then, the  $(*)$-condition holds for every direct
summand of $M$ if and only if
\begin{lieju}
\item[{\rm(1)}]The $(*)$-condition holds for every direct summand of
$M_i$ for all $i\in I$,
\item[{\rm(2)}]$Hom(M_i,M_j)=0$ for all distinct $i,j\in I$,
\item[{\rm(3)}]$N=\oplus_{i\in I} (N\cap M_i)$ for every  direct summand $N$
of $M$.\end{lieju}
\end{thm}
\begin{proof}
\textit{Sufficiency.} It is clear from by Props.~3.1 and
3.7.

\textit{(Necessity).} Suppose that  $M$ satisfies (1), (2) and (3). Let
$K$ be any direct summand of $M$ and let $f$ be any endomorphism of $M$. For each
$j$ in $I$, let $p_j:M\rightarrow M_j$ denote the canonical projection and
let $i_j: M_j\rightarrow M$ denote the inclusion.Then,by (1), we have  $p_jf{i_j}(K\cap M_j)\subseteq
K\cap M_j$ for all $j\in I.$ Because every projection-invariant direct summand of $M_j$ is a
fully invariant submodule by Prop.~3.1(4). Moreover,  we have $p_kfi_j (K\cap M_j)=0$ for
all distinct $j,k\in I$ by (2).
Now ,(3) gives $f(K)=\sum_{j\in I}f(K\cap M_j)\subseteq \sum _{j\in I} p_jfi_j (K\cap M_j) \subseteq
\sum_{j\in I}(K\cap M_j)\subseteq K$. Thus, $K$ is a fully invariant  submodule of $M$ and so a projection
invariant submodule of $M$.
 \end{proof}
\fudl
\begin{cor}
\label{45}
Let a module $M=\oplus_{i\in I} M_i$
 be a direct sum of indecomposable submodules $M_i$ $(i\in I)$. Then the $(*)$-condition
holds for every direct summand of $M$ if and only if
\begin{lieju}
\item[{\rm(1)}]$Hom(M_i,M_j)=0$ for all distinct $i,j\in I$.
\item[{\rm(2)}]For every direct summand $N$ of $M$, there exist a (finite)
subset $I'$ of $I$ such that $N=\oplus_{i\in I'} (N\cap M_i)$.
\end{lieju}
\end{cor}
\begin{proof}
\textit{(Sufficiency).} Clear from by Thm.~3.8 and
Prop.~3.4.

\textit{(Necessity).} By Thm.~3.8.
 \end{proof}


Let $R$ be a ring and let $M$ be a right $R$-module. For any non-empty subset $S$ of $M$,
the {\em annihilator} of $S$ (in $R$) will be denoted by $ann(S)$, i.e.$ann(S)=\{r\in R: sr=0$
for all $s$ in $S \}$. In case $S=\{ m\}$, then we write $ann(m)$ for $ann(\{ m\})$. We now
prove another basic fact about direct sum decompositions.

\begin{lem}
\label{Ozcan 2.4}
{\rm (\cite[ Lemma 2.4]{Ozcan})} Let a module $M=\oplus _{i\in I} M_i$ be
 a direct sum of submodules $M_i$ $(i\in I)$.
 Then the following statements are equivalent.
\begin{lieju}
\item[{\rm(1)}]$R=ann(m_i)+ann(m_j)$ for all $m_i\in M_i$, $m_j\in M_j$, for
all $i\neq j$ in $I$.
\item[{\rm(2)}]$N=\oplus_{i\in I} (N\cap M_i)$ for every (cyclic)
submodule $N$ of $M$.
\end{lieju}
Moreover, in this case $Hom(M_i,M_j)=0$ for  all distinct $i,j $
in $I$.
\end{lem}
\fudl
\begin{thm}
\label{46}
Let a module $M=\oplus_{i\in I} M_i$
 be a direct sum of  submodules $M_i$ $(i\in I)$.
Then $M$ satisfies the $(*)$-condition  if and only if
\begin{lieju}
\item[{\rm(1)}]$M_i$  satisfies the $(*)$-condition for all $i\in I$, and
\item[{\rm(2)}]$N=\oplus_{i\in I} (N\cap M_i)$ for every submodule $N$ of
$M$.
\end{lieju}
\end{thm}
\begin{proof}
Using Lemma~3.10, the proof is similar to that of  Cor.~3.9.
 \end{proof}
\fudl
\begin{cor}
Let a module $M=\oplus_{i\in I}M_i$ be a direct sum of submodules
$M_i$ $(i\in I)$. Then $M$ satisfies the $(*)$-condition  if and
only if $M_i\oplus M_j$ satisfies the $(*)$-condition for all
distinct $i,j$ in $I$.
\end{cor}
\begin{proof}
\textit{(Sufficiency).} By Prop.~3.4.

\textit{(Necessity).} Let $M_i\oplus M_j$  satisfy the $(*)$-condition
for all distinct $i\neq j$ in $I$. Then $M_i$ satisfies the
$(*)$-condition for all $i\in I$, by Prop.~3.4.
Moreover, for all $i\neq j$ in $I$, $R=ann(m_i)+ann(m_j)$ for all
$m_i\in M_i$, $m_j\in M_j$ by Prop.~3.1 and
Lemma~3.10.  Hence $M$ satisfies the $(*)$-condition by
 Lemma~3.10 and Theorem~3.11.
\end{proof}


If the right $R$-module $R$ has the $(*)$ property, we say $R$ has
the $(*)$ property on the right side. Clearly commutative rings
and  division rings satisfy the   $(*)$ property on the right side.

\remark
\begin{lieju}
\item[{\rm(1)}]If $R$ has the $(*)$ property, then
$R\oplus R$ does not satisfy  the $(*)$ property.
\item[{\rm(2)}]Any $2\times 2$ matrix ring over division rings does not
satisfy the $(*)$ property.
\end{lieju}
\dl
\begin{proof}
Let $A$ and $B$ be right $R$-modules and $f:A\rightarrow B$ be an
epimorphism. Then $A$ is not projection invariant in $M=A\oplus
B$, because if $p:M\rightarrow A$ denotes the canonical
projection, then $fp:M\rightarrow B$ is a projection to the direct
summand B, but $fp(A)=B$ is not contained in A. Hence the module M
does not have property $(*)$. In particular, for any non-zero
module M, the module $M \oplus M$ does not have property $(*)$,
independent of M having property $(*)$ or not. This shows that
neither $R\oplus R$ nor the ring of $2\times 2$ matrices over any
ring $R$ ( $R$ can be even a field) satisfies $(*)$.
\end{proof}
\fudl
\yhgs{Question}
Let $R$ be a ring and $R'$ be a proper
subring of $R$. Does  $R'_R$ satisfy the $(*)$-property or not ?
\zjq
\zhangjie{The Lifting Condition}
\no
Following \cite{Ta}, the module $M$ is called {\ FI-lifting } if
for every fully invariant submodule $A$ of $M$, there exists a
direct summand $B$ of $M$ such that $B \subseteq A$ and $A/B$
small in $M/B$.

\begin{definition}
A right $R$-module  $M$ is called  {\ PI-lifting } if for every
projection invariant submodule $A$ of $M$, there exists a direct
summand $B$ of $M$ such that $B \subseteq A$ and $A/B$ small in
$M/B$.
\end{definition}
\fudl
\begin{lem}
The following statements are equivalent for a right $R$-module $M$.
\begin{lieju}\item[{\rm(1)}]$M$ is a  PI-lifting module.
\item[{\rm(2)}]For every projection invariant submodule $A$ of $M$ there is
a decomposition $A=N\oplus S$ with $N$ a direct summand of $M$ and
$S$ small in $M$.
\item[{\rm(3)}]For every projection invariant submodule $X$ of $M$, there
exists an idempotent homomorphism $e:M\rightarrow X$ such that
$(1-e)(X)\leq (1-e)(M)$.
\end{lieju}
\end{lem}
\begin{proof}
$(1)\Rightarrow (2)$. Let $A$ be a projection invariant submodule
of $M$. Since $M$ is a PI-lifting module, there exist a
decomposition $M=M_1\oplus M_2$ such that $M_1\leq A$ and
$M_2\cap A$ small in $M_2$. Therefore $A=M_1\oplus (A\cap M_2)$, as required.

$(2)\Rightarrow (1)$. Assume that every projection invariant
submodule has the stated decomposition. Let $A$ be a projection
invariant submodule of $M$. By hypothesis, there exist a direct
summand $N$ of $M$ and a small submodule $S$ of $M$ such that
$A=N\oplus S$. Now $M=N\oplus N'$ for some submodule $N'$ of $M$.
Consider the natural epimorphism $\pi : M\longrightarrow M/N$.
Then $\pi (S)=(S+N)/N=A/N$ small in $M/N$. Therefore $M$ is a
PI-lifting module.
$(1)\Leftrightarrow (3)$. Clear.
\end{proof}
\fudl
\begin{thm}
Let $M=\oplus_{i=1}^nM_i$. If
each $M_i$ is a PI-lifting module, then $M$ is a  PI-lifting
module
\end{thm}
\begin{proof}
Let $N$ be a projection invariant submodule of $M$.
It is easy to see that for every $1\leq i\leq n$, $N\cap M_i$ is
projection invariant in $M_i$ by Lemma 3.1. Since $M_i$ is a
PI-lifting module for every $i$, there exist a direct summand
$K_i$ of $M_i$ such that $K_i\leq N\cap M_i$ and $(N\cap M_i)/K_i$
is small in $M_i/K_i$ for every $i$. Clearly, $K=\oplus_{i=1}^n
K_i$ is a direct summand of $M$ and $K\subseteq \oplus_{i=1}^n
(N\cap M_i)$. We know that $\oplus_{i=1}^n (N\cap M_i)= M$ by
Lemma 3.1. Now consider the homomorphism $\beta :\oplus_{i=1}^n

(N_i/K_i)\rightarrow (\oplus_{i=1}^n M_i)/K$ with $(m_1+K_1, ...,
m_n+K_n)\rightarrow (\Sigma_{i=1}^n m_i)+K_i$, where $m_i\in M_i$
for $1\leq i\leq n$. Then $\beta (\oplus_{i=1}^n ((N\cap
M_i)/K_i))=(\oplus_{i=1}^n (N\cap M_i))/K$.  Since any finite sum
of small submodules again a small submodule,
$\oplus_{i=1}^n((N\cap M_i)/K_i)$ is small in
$\oplus_{i=1}^n(M_i/K_i)$. Then by \cite[ Lemma 4.2]{MM},
$(\oplus_{i=1}^n(N\cap M_i))/K$ is small in $M/K$.
\end{proof}

We do not know if any direct sum of PI-lifting
modules is a PI-lifting module.

\begin{cor}
If $M$ is a finite direct sum
of lifting (or hollow) modules, then $M$ is a PI-lifting
module.
\end{cor}
\fudl
\example
Let $R$ be a PID and $M$ be any finitely generated $R$-module.
We consider the  torsion submodule $Tor (M)$ of $M$.  Since $Tor
(M )$ is a finite direct sum of  hollow $R$-modules, then $Tor
(M)$ is a PI-lifting module by Corollary 4.4.
\dl



Let $M$ be a lifting module. By \cite[ Corollary 2.2]{Ta}, for every
fully invariant submodule $Y$ of $M$, $M/Y$ is a lifting module. Let
$X$ be a fully invariant submodule of $M$. If $M$ is an FI-lifting
module then $M/X$ is an FI-lifting module (see
\cite[Proposition 3.3]{Ta}).

\begin{prop}
Let $M$ be a module and
$X$ be a projection invariant submodule of $M$. Assume that $X'/X$
is a projection invariant submodule of $M/X$ where $X\leq X'\leq
M$. Then $X'$ is a projection invariant submodule of $M$. If $M$
is a PI-lifting module then $M/X$ is a PI-lifting module.
\end{prop}
\begin{proof}
Let $Y$ be a submodule of $M$ with
$X\subseteq Y$ and let $Y/X$ be a projection  invariant submodule
of $M/X$. By assumption, $Y$ is a projection invariant submodule
of $M$. Since $M$ is a PI-lifting module, there exist a direct
summand $D$ of $M$ such that $D\leq Y$ and $Y/D$ is small in
$M/D$. Assume $M=D\oplus D'$ for some submodule $D'$ of $M$. Let
$\pi$ be the projection with the kernel $D$ and $i: D'\rightarrow
M$ the inclusion map. Now, $\alpha=i\pi : M\rightarrow M$ be a
homomorphism of $M$. Since $X$ and $Y$ are projection invariant
submodules of $M$, then $\alpha (X)\subseteq X$ and $\alpha
(Y)\subseteq Y$. It is easy to see that $Y=\alpha^{-1} (Y)$. Now,
$\alpha^{-1}(X)\subseteq Y=\alpha^{-1}(Y)$. Let $K$ be a submodule
of $M$ with $\alpha^{-1}(X)\subseteq K$ and
$M/\alpha^{-1}(X)=(Y/\alpha^{-1}(X))+(K/\alpha^{-1}(X))$. Then
$M=Y+K$ and since $Y/D$ is small in $M/D$, $M=K$. Therefore
$Y/\alpha^{-1}(X)$ is small in $M/\alpha^{-1}(X)$, namely
$(Y/X)/(\alpha^{-1}(X)/X)<< (M/X)/(\alpha^{-1}(X)/X)$. Now, we
want to show that $\alpha^{-1}(X)/X$ is a direct summand of $M/X$.
Since $M=D\oplus D'$, then $M=\alpha^{-1}(X)+D'$. Therefore
$M/X=(\alpha^{-1}(X)/X)+ (D'+X)/X$. Since $\alpha^{-1}(X)\cap
(D'+X)=X+(\alpha^{-1}(X)\cap D')=X$, then $\alpha^{-1}(X)/X$ is a
direct summand of $M/X$. Hence $M/X$ is a PI-lifting module.
\end{proof}
\fudl
\begin{thm}
Let $M=M_1\oplus M_2$ be a module with the (*)-condition.
Then $M$ is a  PI-lifting module if and only if each $M_i$
  is a PI-lifting module for $i=1,2$.
\end{thm}
\begin{proof}
By Theorem 4.3 and Proposition 4.6.
\end{proof}
\fudl
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