#  $\mathfrak{m}$-Baer and $\mathfrak{m}$-Rickart lattices

Mauricio Medina-Bárcenas  Facultad de Ciencias, Universidad Nacional Autónoma
de México, Circuito Exterior, Ciudad Universitaria 04510, Ciudad de México,
México<EMAIL_ADDRESS>and  Hugo Rincón Mejía  Facultad de Ciencias,
Universidad Nacional Autónoma de México, Circuito Exterior, Ciudad
Universitaria 04510, Ciudad de México, México<EMAIL_ADDRESS>
###### Abstract.

In this paper we introduce the notions of Rickart and Baer lattices and their
duals. We show that part of the theory of Rickart and Baer modules can be
understood just using techniques from the theory of lattices. For, we use
linear morphisms introduced by T. Albu and M. Iosif. We focus on a submonoid
with zero $\mathfrak{m}$ of the monoid of all linear endomorphism of a lattice
$\mathcal{L}$ in order to give a more general approach and apply our results
in the theory of modules. We also show that $\mathfrak{m}$-Rickart and
$\mathfrak{m}$-Baer lattices can be characterized by the annihilators in
$\mathfrak{m}$ generated by idempotents as in the case of modules.

###### Key words and phrases:

Baer lattice, Rickart lattice, linear morphism, Baer module, Rickart module

###### 2010 Mathematics Subject Classification:

Primary 06C05, 06C15, 16D10; Secondary 08A35, 06B35

The first author was supported by the grant “Programa de Becas Posdoctorales
en la UNAM 2021” from the Universidad Nacional Autónoma de México (UNAM)

##  1\. Introduction

The study of rings and their modules is closely related with the study of
lattices. For any ring $R$, every $R$-module $M$ has associated a partial
ordered set with arbitrary suprema and infima, namely the lattice of
submodules of $M$. This lattice of submodules $\Lambda(M)$, is a complete
upper-continuous modular lattice for any $R$-module $M$. The greatest element
is $M$ and $0$ is the lowest. The supremum is given by the sum of submodules
and the infimum is given by the intersection. Much of the algebraic structure
of $M$ is codified by its lattice of submodules. Classical concepts in modules
are concepts which can be defined in any complete modular lattice. For
example, essential submodule, small submodule, finitely generated submodule,
finitely cogenerated submodule, Noetherian module, Artinian module,
distributive module and so on [4]. This is why sometimes the language and the
techniques from lattice theory gives a better context for algebraic
properties. Also, an advantage of working with lattices is the “point-free”
setting that it provides which allows us to apply the results to more general
situations. For example, it is known that the subobjects of an object in a
complete and cocomplete abelian category is a complete modular lattice [12,
Ch. IV]. The intention of this paper is to carry the notions of Baer and
Rickart modules to lattices. We want to show that these concepts and its
implications mostly depend just of the ordered structure of the lattice of
submodules and its relation with the homomorphisms. For, we will use linear
morphisms between lattices which were introduced by T. Albu and M. Iosif in
[1].

Given a ring $R$ and $X\subseteq R$, the left annihilator of $X$ is given by
the left ideal $\operatorname{Ann}^{\ell}(X)=\\{r\in R\mid rx=0\;\text{ for
all }\;x\in X\\}$. A ring $R$ is said to be _Baer_ (resp. _left Rickart_) if
$\operatorname{Ann}^{\ell}(X)$ (resp. $\operatorname{Ann}^{\ell}(x)$) is
generated by an idempotent as a left ideal for every $X\subseteq R$ (resp.
$x\in R$). We have that every idempotent $e^{2}=e$ in a ring $R$ induces a
decomposition on $R$ as left $R$-module $R=Re\oplus R(1-e)$ and every direct
summand of ${}_{R}R$ is generated by an idempotent. Moreover,
$\operatorname{Ann}^{\ell}(x)$ for an $x\in R$ can be seen as the kernel of
the $R$-endomorphism $\\_\cdot x:R\to R$ (multiply by $x$ from the right).
Hence,

######  Proposition 1.1.

The following conditions are equivalent for a ring $R$:

  1. _(a)_

$R$ is a Baer (resp. left Rickart) ring.

  2. _(b)_

$\operatorname{Ann}^{\ell}(X)$ (resp. $\operatorname{Ann}^{\ell}(x)$) is a
direct summand of ${}_{R}R$ for every $X\subseteq R$ (resp. $x\in R$).

  3. _(c)_

$\bigcap_{f\in X}\operatorname{Ker}f$ (resp. $\operatorname{Ker}f$) is a
direct summand of ${}_{R}R$ for every $X\subseteq\operatorname{End}_{R}(R)$
(resp. $f\in\operatorname{End}_{R}(R)$).

It is the last item which allows us to generalize the notion of Baer and
Rickart rings to modules, as it was made by G. Lee, S.T. Rizvi and C. Roman in
[11, 9]. An $R$-module $M$ with endomorphism ring $\operatorname{End}_{R}(M)$
is said to be _Baer_ (resp. _Rickart_) if $\bigcap_{f\in
X}\operatorname{Ker}f$ (resp. $\operatorname{Ker}f$) is a direct summand of
$M$ for every $X\subseteq\operatorname{End}_{R}(M)$ (resp.
$f\in\operatorname{End}_{R}(M)$). As it might be expected, the dual notions
can be defined similarly. The concepts of dual-Rickart and dual-Baer modules
were studied also by Lee, Rizvi and Roman [7] and by D.K. Tütüncü and R.
Tribak in [13] respectively. More recently, these notions have been taken to
the categorical context by S. Crivei and A. Kör in [5]. From these works, it
is clear that, independently of the context of generality, the lattice of
subojects is playing a principal rol. It is our intention to revel this rol
and show how the lattice techniques are enough to recover much of the theory
presented in [5, 9, 7] and [11].

In the theory of Baer and Rickart modules (or Baer and Rickart objects), it is
used all the endomorphims of a given module. In our case, given a lattice
$\mathcal{L}$ and the monoid $\operatorname{End}_{lin}(\mathcal{L})$ of all
linear endomorphism of $\mathcal{L}$, we will use submonoids of
$\operatorname{End}_{lin}(\mathcal{L})$ with certain characteristics and we
will develop the theory of Rickart and Baer lattices relative to these
submonoids. On one hand this gives a more general context for the theory and
on the other hand, with this approach we will be able to apply our results to
the case of modules and recover the known results. Since we are more
interested in module theory, our applications will be made for modules,
nevertheless it is not difficult to see that many of them also apply to the
context of abelian categories.

The paper is divided as follows: Section 1 is this introduction. In Section 2,
we present the linear morphisms and some properties of them which will be
needed along the paper. In particular we study the analogous of the
projections on direct summands of a module. We also show that an idempotent
linear morphism on a lattice $\mathcal{L}$ induces a decomposition of
$\mathcal{L}$ as in the case of modules (Proposition 2.10). The Section 3 is
the main part of this manuscript. Here we introduce the concepts of
$\mathfrak{m}$-Baer and $\mathfrak{m}$-Rickat lattices and their duals. The
prefix $\mathfrak{m}$ will stand for a submonoid with zero of the monoid
$\operatorname{End}_{lin}(\mathcal{L})$ of all linear endomorphisms of a
lattice $\mathcal{L}$. We show that for an $R$-module $M$, the endomorphism
ring $\operatorname{End}_{R}(M)$ can be seen as a submonoid of
$\operatorname{End}_{lin}(\Lambda(M))$ where $\Lambda(M)$ is the lattice of
submodules of $M$. With this identification we are able to recover many
results of Baer and Rickart modules from the results in lattices (see
Proposition 3.4). We determine when a initial interval of a
$\mathfrak{m}$-Baer (or $\mathfrak{m}$-Rickart or their duals) lattice
inherits the property (Proposition 3.16 and Proposition 3.17). It is proved
that the (resp. arbitrary) intersection of complements in a
$\mathfrak{m}$-Rickart lattice (resp. $\mathfrak{m}$-Baer lattice) is a
complement provided that the $\mathfrak{m}$ contains all the projections
(Proposition 3.22 and Proposition 3.24). Later, we study the behaviour of the
linear endomorphisms of these lattices. We give a lattice-counterpart of the
condition $D_{2}$ known for modules [10] and we use the condition $C_{2}$
defined in [2] to characterize $\mathfrak{m}$-Rickart lattices and
dual-$\mathfrak{m}$-Rickart lattices (Proposition 3.42 and Proposition 3.43).
Since $\mathfrak{m}$ is a monoid with zero, it possible to define
annihilators. Then, $\mathcal{L}$ is an $\mathfrak{m}$-Rickart lattice if and
only if the right annihilator of each $\varphi\in\mathfrak{m}$ is generated by
an idempotent and $\ker\varphi$ is $\mathfrak{m}$-$\mathcal{L}$-generated.
(Corollary 3.54 and its dual Corollary 3.55). In a similar way we characterize
$\mathfrak{m}$-Baer lattices and their dual (Proposition 3.60 and Proposition
3.62). At the end of this section we study when an $\mathfrak{m}$-Rickart
lattice is $\mathfrak{m}$-Baer. In Section 4, we translate to lattices the
concepts of $\mathcal{K}$-nonsingularity, $\mathcal{T}$-nonsingularity and
their duals [5, Definition 9.4]. With this, we are able to characterize
$\mathfrak{m}$-Baer and dual-$\mathfrak{m}$-Baer lattices. The main result of
this section states that $\mathcal{L}$ is $\mathfrak{m}$-Baer and
$\mathfrak{m}$-$\mathcal{K}$-cononsingular if and only if $\mathcal{L}$ is
$\mathfrak{m}$-$\mathcal{K}$-nonsingular and satisfies the condition $C_{1}$
(Corollary 4.22 and its dual Corollary 4.23). In the last section, we study
indecomposable Rickart lattices and their products. It is show that an
indecomposable lattice $\mathcal{L}$ is Rickart if and only if
$\mathcal{L}=2$, the complete lattice of two elements (Lemma 5.4). As
aplications we give some conditions to know when a Rickart $R$-module is
simple or semisimple (Corollary 5.6 and Corollary 5.7). Finally, we present
some results about the direct product of Rickart lattices.

##  2\. Preliminaries

Given a module $M$, the set $\Lambda(M)$ of submodules of $M$ is an upper-
continuous complete modular lattice where the supremum is given by the sum of
submodules and the infimum by the intersection of submodules [12, Ch. III]. An
element $a$ in a complete lattice
$(\mathcal{L},\leq,\bigvee,\bigwedge,\mathbf{0},\mathbf{1})$ is _complemented_
if there exists $b\in\mathcal{L}$ such that $a\wedge b=\mathbf{0}$ and $a\vee
b=\mathbf{1}$ where $\mathbf{0}$ and $\mathbf{1}$ denote the least and the
greatest elements of $\mathcal{L}$ respectively. The element $b$ is called _a
complement of $a$ in $\mathcal{L}$_. In the lattice $\Lambda(M)$,
$\mathbf{0}=0$ and $\mathbf{1}=M$. This implies that a submodule
$N\in\Lambda(M)$ is a complement if and only if $N$ is a direct summand of
$M$. Hence, an $R$-module $M$ with endomorphism ring
$\operatorname{End}_{R}(M)$ is _Baer_ (resp. _Rickart_) if and only if
$\bigcap_{f\in X}\operatorname{Ker}f$ (resp. $\operatorname{Ker}f$) has a
complement in $\Lambda(M)$ for every $X\subseteq\operatorname{End}_{R}(M)$
(resp. $f\in\operatorname{End}_{R}(M)$).

On the other hand, given a morphism $f:M\to N$ of $R$-modules, $f$ induces a
$\bigvee$-semilattice morphism $f_{\ast}:\Lambda(M)\to\Lambda(N)$ given by
$f_{\ast}(L)=f(L)$ for all $L\in\Lambda(M)$. Note that $f_{\ast}(0)=0$ and
$f_{\ast}(L+\operatorname{Ker}f)=f_{\ast}(L)$ for all $L\in\Lambda(M)$. Also,
$f_{\ast}$ induces an isomorphism of lattices between
$\Lambda(M/\operatorname{Ker}f)$ and $\Lambda(f(M))$. In [1], the authors
capture the properties of this morphism in the following definition.

######  Definition 2.1.

Let $\varphi:\mathcal{L}\to\mathcal{L}^{\prime}$ be a map between the bounded
lattices $\mathcal{L},\mathcal{L}^{\prime}$. $\varphi$ is called a _linear
morphism_ if there exist $\ker_{\varphi}\in\mathcal{L}$, called the _kernel_
of $\varphi$ and $a^{\prime}\in\mathcal{L}^{\prime}$ such that

  1. (1)

$\varphi(x)=\varphi(x\vee\ker_{\varphi})$ for all $x\in\mathcal{L}$.

  2. (2)

$\varphi$ induces an isomorphism of lattices
$\overline{\varphi}:[\ker_{\varphi},\mathbf{1}]\to[\mathbf{0},a^{\prime}]$
given by $\overline{\varphi}(x)=\varphi(x)$ for all
$x\in[\ker_{\varphi},\mathbf{1}]$.

######  Remark 2.2.

In the case that the lattices are complete, we will assume that the
isomorphism in item (2) of Definition 2.1 is an isomorphism of complete
lattices. Note that in this case a linear morphism commutes with arbitrary
joins [1, Proposition 1.3].

######  Example 2.3.

Let $M$ and $N$ be two $R$-modules. If $f:M\to N$ is a morphism of
$R$-modules, then $f_{\ast}:\Lambda(M)\to\Lambda(N)$ is a linear morphism of
lattices.

######  Proposition 2.4.

Let $\mathcal{L}$ be a bounded modular lattice. Let $x\in\mathcal{L}$ be an
element with complement $x^{\prime}$. Then, the map
$\pi_{x}:\mathcal{L}\to\mathcal{L}$ given by $\pi_{x}(a)=(a\vee
x^{\prime})\wedge x$ is a linear morphism.

###### Proof.

Let $a\in\mathcal{L}$. Then, $\pi_{x}(a\vee x^{\prime})=((a\vee
x^{\prime})\vee x^{\prime})\wedge x=(a\vee x^{\prime})\wedge x=\pi_{x}(a)$,
thus $x^{\prime}$ is a kernel for $\pi_{x}$. Note that
$\overline{\pi_{x}}:[x^{\prime},\mathbf{1}]\to[\mathbf{0},x]$ has the inverse
$(-)\vee x^{\prime}:[\mathbf{0},x]\to[x^{\prime},\mathbf{1}]$. Because if
$b\in[x^{\prime},\mathbf{1}]$ then $\overline{\pi_{x}}(b)=(x^{\prime}\vee
b)\wedge x=b\wedge x$, and so $(b\wedge x)\vee x^{\prime}=b\wedge(x\vee
x^{\prime})=b$, by modularity. If $c\in[\mathbf{0},x]$ then
$\overline{\pi_{x}}(c\vee x^{\prime})=((c\vee x^{\prime})\vee
x^{\prime})\wedge x=(c\vee x^{\prime})\wedge x=c\vee(x\wedge x^{\prime})=c$,
by modularity. Thus, $\pi_{x}$ is a linear morphism and
$\ker_{\pi_{x}}=x^{\prime}$. ∎

######  Corollary 2.5.

A bounded modular lattice $\mathcal{L}$ is boolean if and only if
$a\wedge\\_:\mathcal{L}\to\mathcal{L}$ is a linear morphism for all
$a\in\mathcal{L}.$

###### Proof.

$\Rightarrow$ Suppose $\mathcal{L}$ is boolean and let $a^{\prime}$ denote the
complement of $a$. For each $x\in\mathcal{L}$ we have that $x\wedge
a=\left(x\vee a^{\prime}\right)\wedge a$. Thus $a\wedge\\_=\pi_{a}$ which is a
linear morphism.

$\Leftarrow$ Let $a\in\mathcal{L}$ and $k=\ker_{a\wedge\\_}$. Then,
$a\wedge\left(x\vee k\right)=a\wedge x$ for all $x\in\mathcal{L}$, and
$a\wedge k=0$. Since $\overline{a\wedge\\_}:[k,\mathbf{1}]\to[\mathbf{0},a]$
is an isomorphism and $(k\vee a)\wedge a=a$, then $k\vee a=\mathbf{1}$. Thus
$k$ is a complement of $a$. Now, if $b\wedge a=0$ with $b\in\mathcal{L}$, then
$b\leq k$ because $k$ is the kernel of $a\wedge\\_$. It follows from [12, Ch.
III Proposition 4.4] that $\mathcal{L}$ is boolean. ∎

######  Definition 2.6.

Let $\mathcal{L}$ be a bounded modular lattice and $x\in\mathcal{L}$ be an
element with a complement. The linear morphism $\pi_{x}$ is called the
_projection on $x$_.

######  Remark 2.7.

Given an element $x$ with complement in a lattice $\mathcal{L}$ is possible
that $x$ has more than one complement. Therefore, $\pi_{x}$ depends of the
complement of $x$ which is taken.

######  Lemma 2.8.

Let $\mathcal{L}$ be a bounded modular lattice and $x,y\in\mathcal{L}$
elements with complements $x^{\prime}$ and $y^{\prime}$ respectively. Then
$\ker_{\pi_{y}\pi_{x}}=(x\wedge y^{\prime})\vee x^{\prime}$.

###### Proof.

It follows from the proof of [1, Lemma 2.1] that
$\ker_{\pi_{y}\pi_{x}}=\overline{\pi_{x}}^{-1}(\ker_{\pi_{y}}\wedge x)$. Since
$\pi_{x}$ induces the canonical isomorphism between $[x^{\prime},\mathbf{1}]$
and $[\mathbf{0},x]$, it follows that

$\ker_{\pi_{y}\pi_{x}}=\overline{\pi_{x}}^{-1}(\ker_{\pi_{y}}\wedge
x)=\overline{\pi_{x}}^{-1}(y^{\prime}\wedge x)=(y^{\prime}\wedge x)\vee
x^{\prime}.$

∎

######  Lemma 2.9.

Let $M$ be a module and $M=N\oplus N^{\prime}$ . Let $\pi:M\to N$ and
$\iota:N\to M$ be the canonical projection and the inclusion respectively.
Then $(\iota\pi)_{\ast}=\pi_{N}:\Lambda(M)\to\Lambda(M)$.

###### Proof.

We just have to check the equality $\iota\pi(L)=(L+N^{\prime})\cap N$ for all
$L\leq M$ where $N^{\prime}$ is a complement of $N$. Let $n=\ell+n^{\prime}$
with $n\in N$, $n^{\prime}\in N^{\prime}$ and $\ell\in L$. Then,
$\ell=n-n^{\prime}$. This implies that $n=\iota\pi(\ell)$. Therefore,
$(L+N^{\prime})\cap N\subseteq\iota\pi(L)$. Now, let $\ell\in L$. Then, there
exist $n\in N$ and $n^{\prime}\in N^{\prime}$ such that $\ell=n+n^{\prime}$.
Hence, $n=\ell-n^{\prime}\in(L+N^{\prime})\cap N$. Since $n=\iota\pi(\ell)$,
$\iota\pi(L)\subseteq(L+N^{\prime})\cap N$. ∎

Given a lattice $\mathcal{L}$, the set of all linear morphisms
$\varphi:\mathcal{L}\to\mathcal{L}$ is a monoid with zero using the
composition of functions. This set will be denoted by
$\operatorname{End}_{lin}(\mathcal{L})$ and its elements will be called linear
endomorphisms of $\mathcal{L}$.

######  Proposition 2.10.

Let $\mathcal{L}$ be a bounded modular lattice and
$\varphi\in\operatorname{End}_{lin}(\mathcal{L})$. If $\varphi$ is idempotent,
then $\mathbf{0}=\ker_{\varphi}\wedge\varphi(\mathbf{1})$ and
$\mathbf{1}=\ker_{\varphi}\vee\varphi(\mathbf{1})$.

###### Proof.

Since $\varphi$ is a linear morphism, $\varphi$ induces an isomorphism
$\overline{\varphi}:[\ker_{\varphi},\mathbf{1}]\to[\mathbf{0},\varphi(\mathbf{1})]$.
Write $z=\ker_{\varphi}\vee\varphi(\mathbf{1})$. Then,

$\overline{\varphi}(z)=\varphi(z)=\varphi(\ker_{\varphi}\vee\varphi(\mathbf{1}))=\varphi\varphi(\mathbf{1})=\varphi(\mathbf{1})=\overline{\varphi}(\mathbf{1}).$

Hence, $z=\mathbf{1}$. Now, write $y=\ker_{\varphi}\wedge\varphi(\mathbf{1})$.
Then $y\in[0,\varphi(\mathbf{1})]$. Consider
$\overline{\varphi}^{-1}(y)\in[\ker_{\varphi},\mathbf{1}]$. It follows that

$\mathbf{0}=\varphi(y)=\varphi\overline{\varphi}\overline{\varphi}^{-1}(y)=\varphi\varphi\overline{\varphi}^{-1}(y)=\varphi\overline{\varphi}^{-1}(y)=y.$

∎

######  Proposition 2.11.

Let $\mathcal{L}$ be a bounded modular lattice and
$x,x^{\prime},y\in\mathcal{L}$ with $x^{\prime}$ a complement of $x$. Suppose
that there exists a linear morphism $\varphi:[\mathbf{0},x]\to[\mathbf{0},y]$.
Then $\varphi$ can be extended to a linear morphism
$\widehat{\varphi}:\mathcal{L}\to\mathcal{L}$.

###### Proof.

Define $\widehat{\varphi}:\mathcal{L}\to\mathcal{L}$ as
$\widehat{\varphi}(a)=\varphi(a\wedge x)$. Note that
$\widehat{\varphi}(\ker_{\varphi}\vee x^{\prime})=\varphi((\ker_{\varphi}\vee
x^{\prime})\wedge x)=\varphi(\ker_{\varphi})=\mathbf{0}$. The morphism
$\varphi$ induces an isomorphism
$\overline{\varphi}:[\ker_{\varphi},x]\to[\mathbf{0},\varphi(x)]$. On the
other hand, there is a canonical isomorphism
$[x^{\prime},\mathbf{1}]\to[\mathbf{0},x]$ given by $x\wedge\\_$. Therefore,
there are isomorphisms

$\textstyle{[\ker_{\varphi}\vee
x^{\prime},\mathbf{1}]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\scriptstyle{x\wedge\\_}$$\textstyle{[\ker_{\varphi},x]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\scriptstyle{\overline{\varphi}}$$\textstyle{[\mathbf{0},\varphi(x)].}$

Note that $\widehat{\varphi}(\mathbf{1})=\varphi(\mathbf{1}\wedge
x)=\varphi(x)$. Thus $\widehat{\varphi}$ is a linear morphism and
$\ker_{\widehat{\varphi}}=\ker_{\varphi}\vee x^{\prime}$. ∎

######  Definition 2.12.

Let $\mathcal{L}$ be a lattice and $x\in\mathcal{L}$. It is said that $x$ is
_fully invariant in $\mathcal{L}$_ if $\varphi(x)\leq x$ for all
$\varphi\in\operatorname{End}_{lin}(\mathcal{L})$.

######  Lemma 2.13.

Let $\mathcal{L}$ be a complete lattice and $\\{x_{i}\\}_{I}$ be a family of
elements of $\mathcal{L}$. If each $x_{i}$ is fully invariant in
$\mathcal{L}$, then so is $\bigvee_{i\in I}x_{i}$.

###### Proof.

Let $\varphi\in\operatorname{End}_{lin}(\mathcal{L})$. It follows from [1,
Proposition 1.3] and Remark 2.2 that

$\varphi\left(\bigvee_{i\in I}x_{i}\right)=\bigvee_{i\in
I}\varphi(x_{i})\leq\bigvee_{i\in I}x_{i}.$

∎

######  Proposition 2.14.

Let $\mathcal{L}$ be a complete modular lattice and $\\{a_{i}\\}_{I}$ be an
independent family such that $\mathbf{1}=\bigvee_{i\in I}a_{i}$.

  1. (1)

If $I$ is finite, then $x\leq\bigvee_{i\in I}\pi_{a_{i}}(x)$ for all
$x\in\mathcal{L}$.

  2. (2)

If $\mathcal{L}$ is upper-continuous, then $x\leq\bigvee_{i\in
I}\pi_{a_{i}}(x)$ for all $x\in\mathcal{L}$.

###### Proof.

(1) We will prove it first for $n=2$. Then $\mathbf{1}=a_{1}\vee a_{2}$ and
$\mathbf{0}=a_{1}\wedge a_{2}$. Let us compute the following:

$x\vee\pi_{a_{1}}(x)=x\vee((x\vee a_{2})\wedge a_{1})=(x\vee
a_{2})\wedge(x\vee a_{1}).$

Analogously,

$x\vee\pi_{a_{2}}(x)=(x\vee a_{1})\wedge(x\vee a_{2}).$

Note that $x\leq\pi_{a_{1}}(x)\vee\pi_{a_{2}}(x)$ if and only if
$\pi_{a_{1}}(x)\vee\pi_{a_{2}}(x)=\pi_{a_{1}}(x)\vee\pi_{a_{2}}(x)\vee
x=(x\vee a_{1})\wedge(x\vee a_{2})$. Suppose that
$x\nleq\pi_{a_{1}}(x)\vee\pi_{a_{2}}(x)$. Then
$\pi_{a_{1}}(x)\vee\pi_{a_{2}}(x)\neq(x\vee a_{1})\wedge(x\vee a_{2})$.
Therefore, we have the following pentagon

$\textstyle{x\vee
a_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(x\vee
a_{1})\wedge(x\vee
a_{2})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\pi_{a_{1}}(x)\vee\pi_{a_{2}}(x)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{a_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\pi_{a_{1}}(x).}$

Which is a contradiction. Thus, $x\leq\pi_{a_{1}}(x)\vee\pi_{a_{2}}(x)$. Now
suppose that the result is true for any complete modular lattice and any
independent family of length at most $n-1$. Set $b=\bigvee_{i=2}^{n}a_{i}$.
Then $\mathbf{0}=a_{1}\wedge b$ and $\mathbf{1}=a_{1}\vee b$. We have that
$\pi_{b}(x)=(x\vee a_{1})\wedge b$. Let $2\leq j\leq n$. Then,

$\pi_{b}(x)\vee a_{j}=((x\vee a_{1})\wedge b)\vee a_{j}=(x\vee a_{1}\vee
a_{j})\wedge b.$

This implies that $\pi_{b}(x)\vee a_{2}\vee\cdots a_{n-1}=(x\vee a_{1}\vee
a_{2}\vee\cdots a_{n-1})\wedge b$. Hence

$\begin{split}(\pi_{b}(x)\vee a_{2}\vee\cdots a_{n-1})\wedge a_{n}&=((x\vee
a_{1}\vee a_{2}\vee\cdots a_{n-1})\wedge b)\wedge a_{n}\\\ &=(x\vee a_{1}\vee
a_{2}\vee\cdots a_{n-1})\wedge a_{n}\\\ &=\pi_{a_{n}}(x)\end{split}$

Thus, for any $1\leq j\leq n$,

$\left(\pi_{b}(x)\vee\left(\bigvee_{i\neq j}^{2\leq i\leq
n}a_{i}\right)\right)\wedge a_{j}=\pi_{a_{j}}(x).$

Note that the left hand of the equality is the projection of $\pi_{b}(x)$ in
$a_{j}$ in the lattice $[\mathbf{0},b]$. By induction hypothesis

$\pi_{b}(x)\leq\bigvee_{j=2}^{n}\left(\left(\pi_{b}(x)\vee\left(\bigvee_{i\neq
j}^{2\leq i\leq n}a_{i}\right)\right)\wedge
a_{j}\right)=\bigvee_{i=2}^{n}\pi_{a_{i}}(x).$

On the other hand, also by induction hypothesis
$x\leq\pi_{a_{1}}(x)\vee\pi_{b}(x)$. Thus,

$x\leq\bigvee_{i=1}^{n}\pi_{a_{i}}(x).$

(2) For $\ell>0$, let $B=\\{b_{\ell}\\}$ be the family of elements of
$\mathcal{L}$ given by $b_{\ell}=a_{1}\vee\cdots\vee a_{\ell}$. Then $B$ is a
directed family and $\bigvee B=\bigvee_{i\in I}a_{i}=\mathbf{1}$. Since
$\mathcal{L}$ is upper-continuous,

$x=x\wedge\mathbf{1}=x\wedge\bigvee B=\bigvee\\{x\wedge b_{\ell}\mid
b_{\ell}\in B\\}.$

We have that $\mathbf{0}\leq x\wedge b_{\ell}\leq a_{1}\vee\cdots\vee
a_{\ell}$. By item (1),

$x\wedge b_{\ell}\leq\bigvee_{i=1}^{\ell}\pi_{a_{i}}(x\wedge
b_{\ell})\leq\bigvee_{i=1}^{\ell}\pi_{a_{i}}(x).$

Thus, $x\leq\bigvee_{i\in I}\pi_{a_{i}}(x)$. ∎

The next result appears in [3, Lemma 1.3] without a proof. We present a proof
of it as a corollary of last proposition.

######  Corollary 2.15.

Let $\mathcal{L}$ be a complete modular lattice and $x\in\mathcal{L}$. If $x$
is fully invariant in $\mathcal{L}$ then $x=\bigvee_{i=1}^{n}(x\wedge a_{i})$
for any independent family $\\{a_{1},...,a_{n}\\}$ in $\mathcal{L}$ such that
$\mathbf{1}=\bigvee_{i=1}^{n}a_{i}$. Moreover, $\pi_{a_{i}}(x)=x\wedge a_{i}$
for all $1\leq i\leq n$. In addition, if $\mathcal{L}$ is upper-continuous
then the result is true for any infinite family.

###### Proof.

Let $1\leq i\leq n$. By hypothesis, $\pi_{a_{i}}(x)\leq x$. Then

$\pi_{a_{i}}(x)\leq x\wedge a_{i}\leq\left(x\vee\bigvee_{i\neq
j}a_{j}\right)\wedge a_{i}=\pi_{a_{i}}(x).$

Thus, $\pi_{a_{i}}(x)=x\wedge a_{i}$ for all $1\leq i\leq n$. By Proposition
2.14,

$x\leq\bigvee_{i=1}^{n}\pi_{a_{i}}(x)=\bigvee_{i=1}^{n}x\wedge a_{i}.$

∎

##  3\. $\mathfrak{m}$-Rickart and $\mathfrak{m}$-Baer lattices

Given an $R$-module $M$ and an endomorphism $f:M\to M$, there is a linear
morphism $f_{\ast}:\Lambda(M)\to\Lambda(M)$ induced by $f$. Then, there is a
homomorphism of monoids with zero
$(-)_{\ast}:\operatorname{End}_{R}(M)\to\operatorname{End}_{lin}(\Lambda(M))$.
Let $\mathfrak{E}_{M}$ denote the image of $\operatorname{End}_{R}(M)$ under
$(-)_{\ast}$. Then $\mathfrak{E}_{M}$ is a submonoid with zero of
$\operatorname{End}_{lin}(\Lambda(M))$.

######  Definition 3.1.

Let $\mathcal{L}$ be a complete lattice and let $\mathfrak{m}$ be a submonoid
with zero of $\operatorname{End}_{lin}(\mathcal{L})$.

  * •

$\mathcal{L}$ is called _$\mathfrak{m}$ -Baer_ if $\bigwedge_{\varphi\in
X}\ker_{\varphi}$ has a complement in $\mathcal{L}$ for all
$X\subseteq\mathfrak{m}$.

  * •

$\mathcal{L}$ is called _$\mathfrak{m}$ -Rickart_ if $\ker_{\varphi}$ has a
complement in $\mathcal{L}$ for all $\varphi\in\mathfrak{m}$.

If the submonoid we are considering is
$\operatorname{End}_{lin}(\mathcal{L})$, we will omit the $\mathfrak{m}$.

######  Definition 3.2.

Let $\mathcal{L}$ be a complete lattice and let $\mathfrak{m}$ be a submonoid
with zero of $\operatorname{End}_{lin}(\mathcal{L})$.

  * •

$\mathcal{L}$ is called _dual- $\mathfrak{m}$-Baer_ if $\bigvee_{\varphi\in
X}\varphi(\mathbf{1})$ has a complement in $\mathcal{L}$ for all
$X\subseteq\mathfrak{m}$.

  * •

$\mathcal{L}$ is called _dual- $\mathfrak{m}$-Rickart_ if
$\varphi(\mathbf{1})$ has a complement in $\mathcal{L}$ for all
$\varphi\in\mathfrak{m}$.

If the submonoid we are considering is
$\operatorname{End}_{lin}(\mathcal{L})$, we will omit the $\mathfrak{m}$.

######  Remark 3.3.

It is clear that (dual-)$\mathfrak{m}$-Baer implies
(dual-)$\mathfrak{m}$-Rickart. Note that every complemented lattice is
$\mathfrak{m}$-Baer and dual-$\mathfrak{m}$-Baer for every submonoid
$\mathfrak{m}$. For example, the lattice of submodules of a semisimple
$R$-module.

The next result follows from the above definition and it will be the key to
apply our results in the context of modules.

######  Proposition 3.4.

Let $M$ be an $R$-module and $\Lambda(M)$ the lattice of submodules of $M$.
The following conditions are equivalent:

  1. _(a)_

$\Lambda(M)$ is (dual-)$\mathfrak{E}_{M}$-Rickart (resp.
(dual-)$\mathfrak{E}_{M}$-Baer).

  2. _(b)_

$M$ is a (dual-)Rickart (resp. (dual-)Baer) module.

###### Proof.

We will only prove the Rickart case, the others are similar.
(a)$\Rightarrow$(b) Let $f:M\to M$ be an endomorphism of $M$. Consider the
linear morphism $f_{\ast}:\Lambda(M)\to\Lambda(M)$. By hypothesis,
$\ker_{f_{\ast}}$ is a complement in $\Lambda(M)$. Note that
$\ker_{f_{\ast}}=\operatorname{Ker}f$, hence $\operatorname{Ker}f$ is a direct
summand of $M$.

(b)$\Rightarrow$(a) Let $\varphi\in\mathfrak{E}_{M}$. Then, there exists an
$R$-endomorphism $f:M\to M$ such that $f_{\ast}=\varphi$. Since $M$ is
Rickart, $\operatorname{Ker}f=\ker_{\varphi}$ is direct summand of $M$, that
is, $\ker_{\varphi}$ is complemented in $\Lambda(M)$. ∎

The following example shows that if we take the monoid
$\operatorname{End}_{lin}(\Lambda(M))$ then the implication
(b)$\Rightarrow$(a) of Proposition 3.4 might not be true.

######  Example 3.5.

Let $K$ be a field. Consider the ring $R=\left(\begin{smallmatrix}K&K\\\
0&K\end{smallmatrix}\right)$ and the left $R$-module
$M=\left(\begin{smallmatrix}0&K\\\ 0&K\end{smallmatrix}\right)$. Since
$\operatorname{End}_{R}(M)\cong K$, $M$ is a Rickart module. On the other
hand, the lattice of submodules of $M$ is
$\Lambda(M)=\\{0,N=\left(\begin{smallmatrix}0&K\\\
0&0\end{smallmatrix}\right),M\\}$. Consider the following linear morphism
$\varphi:\Lambda(M)\to\Lambda(M)$ given by $\varphi(0)=0$, $\varphi(N)=0$ and
$\varphi(M)=N$, represented in the figure below. Then $\ker_{\varphi}=N$ which
is not a complement in $\Lambda(M)$.

$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\bullet\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\bullet}$
Figure 1.

######  Remark 3.6.

Note that the lattice of submodules of the $\mathbb{Z}$-module
$\mathbb{Z}_{4}$ is isomorphic to the lattice $\Lambda(M)$ of the previous
example. Consider the homomorphism $f:\mathbb{Z}_{4}\to\mathbb{Z}_{4}$ given
by multiply by 2. Then $f_{\ast}=\varphi$, the linear morphism in Figure 1.

Notation: Let $\mathcal{L}$ be a complete modular lattice and
$a,x\in\mathcal{L}$. There are two canonical linear morphisms
$\iota_{x}:[\mathbf{0},x]\to\mathcal{L}$ the inclusion, and
$\rho_{a}:\mathcal{L}\to[a,\mathbf{1}]$ given by $\rho_{a}(y)=a\vee y$. Notice
that if $a$ has a complement $a^{\prime}$, then the interval
$[a^{\prime},\mathbf{1}]$ is canonically isomorphic to $[\mathbf{0},a]$ and so
$\pi_{a}=\iota_{a}(\\_\wedge a)\rho_{a^{\prime}}$.

Given a lattice $\mathcal{L}$ and a submonoid with zero,
$\mathfrak{m}\subseteq\operatorname{End}_{lin}(\mathcal{L})$, we will say that
$\mathfrak{m}$ _contains all the projections_ if $\pi_{a}\in\mathfrak{m}$ for
every complement $a\in\mathcal{L}$.

######  Remark 3.7.

Given an $R$-module $M$, the submonoid $\mathfrak{E}_{M}$ of
$\operatorname{End}_{lin}(\Lambda(M))$ contains all the projections by
Proposition 2.9.

######  Proposition 3.8.

Let $\mathcal{L}$ be a bounded modular lattice and
$\mathfrak{m}\subseteq\operatorname{End}_{lin}(\mathcal{L})$ be a submonoid
containing all the projections. The following conditions are equivalent:

  1. _(a)_

$\mathcal{L}$ is $\mathfrak{m}$-Rickart and for all $a\in\mathcal{L}$ there
exists (a unique) $b\in\mathcal{L}$ and an isomorphism
$\theta:[a,\mathbf{1}]\to[\mathbf{0},b]$ such that
$\iota_{b}\theta\rho_{a}\in\mathfrak{m}$.

  2. _(b)_

$\mathcal{L}$ is complemented (boolean).

###### Proof.

(a)$\Rightarrow$(b)Let us take $a,b\in\mathcal{L}$ as in the hypothesis. Then
the linear morphism,

$\textstyle{\mathcal{L}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho_{a}}$$\textstyle{[a,\mathbf{1}]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\theta}$$\scriptstyle{\cong}$$\textstyle{[\mathbf{0},b]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota_{b}}$$\textstyle{\mathcal{L}}$

is in $\mathfrak{m}$ and has kernel $a$. Since $\mathcal{L}$ is
$\mathfrak{m}$-Rickart, $a$ has a complement in $\mathcal{L}$. Now suppose
that $b$ is unique an let $a^{\prime}$ be a complement of $a$. Then, there is
a canonical isomorphism $[a,\mathbf{1}]\cong[\mathbf{0},a^{\prime}]$ given by
$\\_\wedge a^{\prime}$. Therefore, $\iota_{a^{\prime}}(\\_\wedge
a^{\prime})\rho_{a}=\pi_{a^{\prime}}\in\mathfrak{m}$. By hypothesis,
$b=a^{\prime}$. Since $L$ is modular and every element has a unique
complement, $\mathcal{L}$ is boolean [6, Ch. II, Sec. 1, Corollary 3].

(b)$\Rightarrow$(a) Since $\mathcal{L}$ is complemented, $\mathcal{L}$ is
$\mathfrak{m}$-Rickart. If $a\in\mathcal{L}$ and $a^{\prime}$ is a complement
of $a,$ then $a^{\prime}\wedge\\_:[a,\mathbf{1}]\to[\mathbf{0},a^{\prime}]$ is
an isomorphism, and
$\iota_{a^{\prime}}(a^{\prime}\wedge\\_)\rho_{a}=\pi_{a^{\prime}}\in\mathfrak{m}$.
If $\mathcal{L}$ is boolean, then $a^{\prime}$ is unique. ∎

######  Corollary 3.9.

The following conditions are equivalent for a module $M$:

  1. _(a)_

$M$ is Rickart and for all $N\leq M$ there exists a (unique) $L\leq M$ such
that $M/N\cong L$.

  2. _(b)_

$M=\bigoplus_{I}S_{i}$ with $S_{i}$ simple ($S_{i}\ncong S_{j}$ for all $i\neq
j\in I$).

######  Proposition 3.10.

Let $\mathcal{L}$ be a complete modular lattice and $x\in\mathcal{L}$ be a
fully invariant element. If $\mathcal{L}$ is Rickart and every endomorphism
$\varphi\in\operatorname{End}_{lin}([\mathbf{0},x])$ can be extended to an
endomorphism $\widehat{\varphi}\in\operatorname{End}_{lin}(\mathcal{L})$, then
$[\mathbf{0},x]$ is a Rickart lattice.

###### Proof.

Let $\varphi:[\mathbf{0},x]\to[\mathbf{0},x]$ be a linear morphism. By
hypothesis, $\varphi$ can be extended to a linear morphism
$\widehat{\varphi}:\mathcal{L}\to\mathcal{L}$. Then $\ker_{\widehat{\varphi}}$
is a complement in $\mathcal{L}$. Since the restriction of $\widehat{\varphi}$
to $[\mathbf{0},x]$ is $\varphi$,
$\ker_{\varphi}=\ker_{\widehat{\varphi}}\wedge x$. Let $y$ be a complement of
$\ker_{\widehat{\varphi}}$ in $\mathcal{L}$. Then,
$\mathbf{1}=\ker_{\widehat{\varphi}}\vee y$ and
$\mathbf{0}=\ker_{\widehat{\varphi}}\wedge y$. It follows from Corollary 2.15
that

$x=x\wedge(\ker_{\widehat{\varphi}}\vee
y)=(x\wedge\ker_{\widehat{\varphi}})\vee(x\wedge y)=\ker_{\varphi}\vee(x\wedge
y).$

It is clear that $\ker_{\varphi}\wedge(x\wedge y)=\mathbf{0}$. Therefore,
$\ker_{\varphi}$ is a complement in $[\mathbf{0},x]$. Thus $[\mathbf{0},x]$ is
a Rickart lattice. ∎

######  Lemma 3.11.

Let $\mathcal{L}$ be a complete modular lattice. The following conditions are
equivalent for $a\in\mathcal{L}$:

  1. _(a)_

$a$ is a complement in $\mathcal{L}$.

  2. _(b)_

the inclusion $\iota_{a}:[\mathbf{0},a]\to\mathcal{L}$ splits.

  3. _(c)_

the projection $\rho_{a}:\mathcal{L}\to[a,\mathbf{1}]$ splits.

###### Proof.

(a)$\Rightarrow$(b) Suppose that $a$ is a complement in $\mathcal{L},$ with
complement $a^{\prime}$. Consider the projection $\pi_{a}$ over $a$ and let us
see that it is a splitting for the inclusion of $[\mathbf{0},a]$ in
$\mathcal{L}.$ By modularity, we have that

$\begin{array}[]{ccccc}[\mathbf{0},a]&\rightarrowtail&\mathcal{L}&\overset{\pi_{a}}{\rightarrow}&[\mathbf{0},a]\\\
x&\mapsto&x&\text{$\mapsto$}&\left(x\vee a^{\prime}\right)\wedge
a=x.\end{array}$

(a)$\Rightarrow$(c) Now, for $\rho_{a}$ consider
$\varphi:[a,\mathbf{1}]\to\mathcal{L}$ defined as $\varphi(x)=x\wedge
a^{\prime}$. Then $\varphi(\mathbf{1})=a^{\prime}$. It is clear that $\varphi$
induces an isomorphism
$\overline{\varphi}:[a,\mathbf{1}]\to[\mathbf{0},\varphi(\mathbf{1})]$. Thus
$\varphi$ is a linear morphism with $\ker_{\varphi}=a$. It follows that
$\varphi\circ\rho_{a}=Id_{[a,\mathbf{1}]}$.

(b)$\Rightarrow$(a) Assume that
$\varphi:\text{$\mathcal{L\rightarrow\mathit{[\mathbf{0},a]}}$}$ is a
splitting for $\iota_{a},$ i.e. $\varphi\circ\iota_{a}=Id_{[\mathbf{0},a]}.$
We will see that $k=\ker_{\varphi}$ is a complement for $a.$ For,
$\mathbf{0}=\varphi(k)=\varphi\left(\left(k\wedge a\right)\vee
k\right)=\varphi\left(k\wedge a\right)=k\wedge a.$ On the other hand, as
$\bar{\varphi}:[k,\mathbf{1}]\rightarrow[\mathbf{0},a]$ is an isomorphism and
$\bar{\varphi}{\left(\mathbf{1}\right)=a=\varphi\left(a\right)=\varphi}\left(a\vee
k\right),$ then $a\vee k=\mathbf{1}.$

(c)$\Rightarrow$(a) Assume that $\varphi:[a,\mathbf{1}]\to\mathcal{L}$ is a
splitting for $\rho_{a}$ i.e. $\rho_{a}\circ\varphi=Id_{[a,\mathbf{1}]}.$ We
will see that $b=\varphi(\mathbf{1})$ is a complement for $a.$ For,
$\mathbf{1}=\rho_{a}(\varphi(\mathbf{1}))=b\vee a$. On the other hand, as
$\bar{\varphi}:[a,\mathbf{1}]\rightarrow[\mathbf{0},b]$ is an isomorphism and
$\overline{\varphi}^{-1}(a\wedge
b)=\overline{\varphi}^{-1}(a)\wedge\overline{\varphi}^{-1}(b)=a\wedge(a\vee
b)=a=\overline{\varphi}^{-1}\left(\mathbf{0}\right)$. Thus,
$\mathbf{0}=a\wedge b$. ∎

######  Corollary 3.12.

Let $\mathcal{L}$ be a complete modular lattice and $a,b\in\mathcal{L}$. If
$a$ is a complement in $[\mathbf{0},b]$ and $b$ is a complement in
$\mathcal{L}$ then $a$ is a complement in $\mathcal{L}.$

######  Proposition 3.13.

Let $\mathcal{L}$ be a complete modular lattice and $a\in\mathcal{L}$ be an
element with complement $a^{\prime}$. Let $\mathfrak{m}$ be a submonoid of
$\operatorname{End}_{lin}(\mathcal{L})$ and let $\mathfrak{n}$ be a submonoid
of $\operatorname{End}_{lin}([\mathbf{0},a])$ such that
$\iota\psi\pi_{a}\in\mathfrak{m}$ for every element $\psi\in\mathfrak{n}$. If
$\mathcal{L}$ is $\mathfrak{m}$-Rickart, then $[\mathbf{0},a]$ is a
$\mathfrak{n}$-Rickart lattice.

###### Proof.

Let $\psi:[\mathbf{0},a]\longrightarrow[\mathbf{0},a]$ be a linear morphism in
$\mathfrak{n}$. Let us take the projection
$\pi_{a}:\mathcal{L\text{$\rightarrow$}L}$ and let
$\pi_{a}^{\upharpoonright}:\mathcal{L\mathit{\longrightarrow[\mathbf{0},a]}}$
denote the corestriction to $[\mathbf{0},a]$ of $\pi_{a}.$ By hypothesis
$\begin{array}[]{ccccccc}\mathcal{L}&\overset{\pi_{a}^{\upharpoonright}}{\rightarrow}&[\mathbf{0},a]&\overset{\psi}{\rightarrow}&[\mathbf{0},a]&\overset{\iota}{\rightarrow}&\mathcal{L}\end{array}$
is a linear morphism in $\mathfrak{m}$ with kernel $k,$ say, and let us denote
$b=\ker_{\psi}$. We have that $k=\left(\bar{\pi}\right)^{-1}\left(b\right).$
As $\mathcal{L}$ is $\mathfrak{m}$-Rickart, $k$ is a complement in
$\mathcal{L}.$ Since $\left(a^{\prime}\vee b\right)\wedge a=b$ with
$a^{\prime}$ a complement of $a$, $k=a^{\prime}\text{$\vee$}b.$ Let
$k^{\prime}$ denote a complement of $k$ in $\mathcal{L}.$ Then,

$\psi\left(a\right)=\left(\psi\pi_{a}\right)\left(1\right)=\left(\psi\pi_{a}\right)\left(k\vee
k^{\prime}\right)=\left(\psi\pi_{a}\right)\left(a^{\prime}\vee b\vee
k^{\prime}\right)=\psi\left(b\vee k^{\prime}\right).$

As $\overline{\psi}:[b,a]\to[\mathbf{0},\psi\left(a\right)]$ is an
isomorphism, then $b\vee k^{\prime}=a.$ Thus, $\left(a\wedge
k^{\prime}\right)\vee b=a\wedge\left(k^{\prime}\vee b\right)=a.$ Besides,
$\left(k^{\prime}\wedge a\right)\wedge b\leq k^{\prime}\wedge b\leq
k^{\prime}\wedge\left(b\vee a^{\prime}\right)=k^{\prime}\wedge k=0.$ Hence,
$k^{\prime}\wedge a$ is a complement of $b$ in $[\mathbf{0},a].$ Thus
$[\mathbf{0},a]$ is $\mathfrak{n}$-Rickart. ∎

We have the dual version of last proposition.

######  Proposition 3.14.

Let $\mathcal{L}$ be a complete modular lattice and $a\in\mathcal{L}$ be an
element with complement $a^{\prime}$. Let $\mathfrak{m}$ be a submonoid of
$\operatorname{End}_{lin}(\mathcal{L})$ and let $\mathfrak{n}$ be a submonoid
of $\operatorname{End}_{lin}([\mathbf{0},a])$ such that
$\iota\psi\pi_{a}\in\mathfrak{m}$ every element $\psi\in\mathfrak{n}$. If
$\mathcal{L}$ is dual-$\mathfrak{m}$-Rickart, then $[\mathbf{0},a]$ is a
dual-$\mathfrak{n}$-Rickart lattice.

######  Corollary 3.15.

Let $M$ be an $R$-module and $N$ be a direct summand of $M$. If $M$ is a
(dual-)Rickart module, so is $N$.

######  Proposition 3.16.

Let $\mathcal{L}$ be a complete modular lattice and $x\in\mathcal{L}$ be an
element with complement $x^{\prime}$. Let $\mathfrak{m}$ be a submonoid of
$\operatorname{End}_{lin}(\mathcal{L})$ and let $\mathfrak{n}$ be a submonoid
of $\operatorname{End}_{lin}([\mathbf{0},x])$ such that the extension given by
Proposition 2.11 of every element $\psi\in\mathfrak{n}$ is in $\mathfrak{m}$.
If $\mathcal{L}$ is $\mathfrak{m}$-Baer, then $[\mathbf{0},x]$ is a
$\mathfrak{n}$-Baer lattice.

###### Proof.

Consider a family of linear morphisms
$\\{\varphi_{i}:[\mathbf{0},x]\to[\mathbf{0},x]\mid\varphi_{i}\in\mathfrak{n}\\}_{I}$
and let $k=\bigwedge_{I}\ker_{\varphi_{i}}$. By hypothesis, the extension
$\widehat{\varphi}_{i}:\mathcal{L}\to\mathcal{L}$ with
$\ker_{\widehat{\varphi}_{i}}=\ker_{\varphi_{i}}\vee x^{\prime}$ is in
$\mathfrak{m}$ for all $i\in I$. Since $\\_\wedge
x:[x^{\prime},\mathbf{1}]\to[\mathbf{0},x]$ is an isomorphism with inverse
$\\_\vee x^{\prime}$,

$k=(k\vee x^{\prime})\wedge x\leq\bigwedge_{I}(\ker_{\varphi_{i}}\vee
x^{\prime})\wedge x=k.$

Therefore, $k\vee x^{\prime}=\bigwedge_{I}(\ker_{\varphi_{i}}\vee
x^{\prime})$. By hypothesis, $k\vee
x^{\prime}=\bigwedge_{I}(\ker_{\varphi_{i}}\vee x^{\prime})$ has a complement
in $\mathcal{L}$. Thus, $k$ has a complement in $[\mathbf{0},x]$. ∎

######  Proposition 3.17.

Let $\mathcal{L}$ be a complete modular lattice and $x\in\mathcal{L}$ be an
element with complement $x^{\prime}$. Let $\mathfrak{m}$ be a submonoid of
$\operatorname{End}_{lin}(\mathcal{L})$ and let $\mathfrak{n}$ be a submonoid
of $\operatorname{End}_{lin}([\mathbf{0},x])$ such that the extension given by
Proposition 2.11 of every element $\psi\in\mathfrak{n}$ is in $\mathfrak{m}$.
If $\mathcal{L}$ is dual-$\mathfrak{m}$-Baer, then $[\mathbf{0},x]$ is a
dual-$\mathfrak{n}$-Baer lattice.

######  Corollary 3.18.

Let $M$ be a (dual-)Baer module. Then every direct summand $N$ of $M$ is also
a (dual-)Baer module.

We have to notice the short proof we present in Proposition 3.16 which gives
as corollary [11, Theorem 2.17].

######  Definition 3.19.

A complete lattice $\mathcal{L}$ satisfies _the_ (resp. _strong_) _complement
intersection property_ CIP (resp. SCIP) if the infimum of any finite family
(resp. any family) of complements in $\mathcal{L}$ is a complement.

######  Definition 3.20.

A complete lattice $\mathcal{L}$ satisfies _the_ (resp. _strong_) _complement
supremum property_ CSP (resp. SCSP) if the supremum of any finite (resp. any
family) of complements in $\mathcal{L}$ is a complement.

######  Remark 3.21.

For a module $M$, the lattice $\Lambda(M)$ satisfies (resp. SCIP) CIP is
equivalent to say that $M$ satisfies the (resp. strong) summand intersection
property (SIP) (resp. SSIP). The dual concept is called the (resp. strong)
summand sum property (resp. SSSP) SSP.

######  Proposition 3.22.

Let $\mathcal{L}$ be a complete lattice and let $\mathfrak{m}$ be a submonoid
of $\operatorname{End}_{lin}(\mathcal{L})$ containing all the projections.
Then $\mathcal{L}$ is $\mathfrak{m}$-Rickart (resp.
dual-$\mathfrak{m}$-Rickart) and satisfies SCIP (resp. SCSP) if and only if
$\mathcal{L}$ is $\mathfrak{m}$-Baer (resp. dual-$\mathfrak{m}$-Baer).

###### Proof.

Let $X\subseteq\mathfrak{m}$. Since $\mathcal{L}$ is $\mathfrak{m}$-Rickart,
$\ker_{\varphi}$ is a complement in $\mathcal{L}$ for every $\varphi\in X$. By
hypothesis, $\bigwedge_{\varphi\in X}\ker_{\varphi}$ is also a complement.
Thus, $\mathcal{L}$ is $\mathfrak{m}$-Baer. Convesely, let $\\{c_{i}\\}_{I}$
be a family of elements of $\mathcal{L}$ with complement and let
$c_{i}^{\prime}$ be a complement of $c_{i}$ for all $i\in I$. By hypothesis,
$\pi_{c_{i}^{\prime}}\in\mathfrak{m}$. Therefore
$\bigwedge_{I}c_{i}=\bigwedge_{I}\ker_{\pi_{c_{i}^{\prime}}}$ is complemented
in $\mathcal{L}$. This proves that $\mathcal{L}$ satisfies SCIP. It is clear
that $\mathcal{L}$ is $\mathfrak{m}$-Rickart. ∎

######  Lemma 3.23.

Let $\mathcal{L}$ be a complete modular lattice. Let $a,b,c\in\mathcal{L}$
such that $a\wedge b=\mathbf{0}$, $(a\vee b)\wedge c=\mathbf{0}$. Then
$a\wedge(b\vee c)=\mathbf{0}$.

######  Proposition 3.24.

Let $\mathcal{L}$ be a complete modular lattice and let $\mathfrak{m}$ be a
submonoid of $\operatorname{End}_{lin}(\mathcal{L})$ containing all the
projections. If $\mathcal{L}$ is $\mathfrak{m}$-Rickart (resp.
dual-$\mathfrak{m}$-Rickart) then $\mathcal{L}$ satisfies CIP (resp. SSP).

###### Proof.

Let $x$ and $y$ be two complements in $\mathcal{L}$. Let $x^{\prime}$ and
$y^{\prime}$ the complements of $x$ and $y$ respectively. By hypothesis and
Lemma 2.8 $(x\wedge y)\vee x^{\prime}$ has a complement in $\mathcal{L}$. That
is, there exists $z\in\mathcal{L}$ such that $((x\wedge y)\vee x^{\prime})\vee
z=\mathbf{1}$ and $((x\wedge y)\vee x^{\prime})\wedge z=\mathbf{0}$. By Lemma
3.23 $x^{\prime}\vee z$ is a complement of $x\wedge y$. ∎

######  Corollary 3.25.

The following conditions are equivalent for an Artinian (resp. Noetherian)
complete modular lattice $\mathcal{L}$ and a submonoid
$\mathfrak{m}\subseteq\operatorname{End}_{lin}(\mathcal{L})$ containing all
the projections.

  1. _(a)_

$\mathcal{L}$ is $\mathfrak{m}$-Rickart (resp. dual-$\mathfrak{m}$-Rickart).

  2. _(b)_

$\mathcal{L}$ is $\mathfrak{m}$-Baer (resp. dual-$\mathfrak{m}$-Baer).

###### Proof.

Let $X\subseteq\mathfrak{m}$ and let $k=\bigwedge_{\varphi\in
X}\ker_{\varphi}$. Since $\mathcal{L}$ is Artinian, so is $[k,\mathbf{1}]$. It
follows from [4, Proposition 1.3] that there exists a finite subset
$F\subseteq X$ such that $k=\bigwedge_{\varphi\in F}\ker_{\varphi}$. Hence $k$
is a complement by Proposition 3.24. The converse is trivial. ∎

In [9, Corollary 2.23], it is proved that if the module $M\oplus M$ has SIP,
then $M$ is a Rickart module. The lattice counterpart of this result is not
true in general as the following example shows.

######  Example 3.26.

Consider the following lattice $\mathcal{L}$

$\textstyle{\mathbf{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{a\vee
c\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{c\vee
b\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{a\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{c\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{b\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{k\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{f\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathbf{0}}$

Then $\mathcal{L}=[\mathbf{0},a\vee b]$ with $a\wedge b=\mathbf{0}$ and the
lattices $[\mathbf{0},a]$ and $[\mathbf{0},b]$ are isomorphic. The elements
with complement of $\mathcal{L}$ are $\\{\mathbf{0},\mathbf{1},a,b\\}$,
therefore $\mathcal{L}$ has the CIP. Consider the linear morphism
$\varphi:[\mathbf{0},a]\to[\mathbf{0},b]$ given by
$\varphi(\mathbf{0})=\mathbf{0}$, $\varphi(k)=\mathbf{0}$ and $\varphi(a)=f$
(illustrated in Figure 1). Then, $\ker_{\varphi}=k$ but $k$ is not a
complement in $[\mathbf{0},a]$.

######  Definition 3.27.

Let $\mathcal{L}$ be a modular lattice. Let $C(\mathcal{L})$ denote the set of
complements in $\mathcal{L}$, that is,

$C(\mathcal{L})=\\{a\in\mathcal{L}\mid a\;\text{has a complement}\\}.$

######  Corollary 3.28.

Let $\mathcal{L}$ be a complete modular lattice. If $\mathcal{L}$ is Baer or
dual-Baer then $C(\mathcal{L})$ is a complete complemented lattice.

######  Proposition 3.29.

Let $\mathcal{L}$ be a complete modular lattice and suppose that
$C(\mathcal{L})$ is a sublattice of $\mathcal{L}$. The following conditions
are equivalent:

  1. _(a)_

$\pi_{x}(y)=\pi_{y}(x)=x\wedge y$ for all $x,y\in C(\mathcal{L})$.

  2. _(b)_

$C(\mathcal{L})$ is a Boolean algebra.

###### Proof.

(a)$\Rightarrow$(b) Since $C(\mathcal{L})$ is a complemented modular lattice,
we just have to prove that $C(\mathcal{L})$ is distributive. Let $x,y,z\in
C(\mathcal{L})$. Then,

$x\wedge(y\vee z)=\pi_{x}(y\vee z)=\pi(y)\vee\pi_{x}(z)=(x\wedge
y)\vee(x\wedge z).$

(b)$\Rightarrow$(a) Let $x,y\in C(\mathcal{L})$. Then

$\pi_{x}(y)=(y\vee x^{\prime})\wedge x=(y\wedge x)\vee(x\wedge
x^{\prime})=y\wedge x$

where $x^{\prime}$ is a complement of $x$. Analogously, $\pi_{y}(x)=x\wedge
y$. ∎

Note that if $M$ is an $R$-module, the set $C(\Lambda(M))$ consists of all
direct summands of $M$.

######  Corollary 3.30.

Let $M$ be an $R$-module and suppose that $C(\Lambda(M))$ is a sublattice of
$\Lambda(M)$. The following conditions are equivalent:

  1. _(a)_

$\pi_{N}(L)=\pi_{L}(N)=N\cap L$ for all direct summands $N$ and $L$ of $M$.

  2. _(b)_

$C(\Lambda(M))$ is a Boolean algebra.

######  Corollary 3.31.

The following conditions are equivalent for a semisimple $R$-module $M$:

  1. _(a)_

$\pi_{N}(L)=\pi_{L}(N)=N\cap L$ for all submodules $N$ and $L$ of $M$.

  2. _(b)_

$M=\bigoplus_{I}S_{i}$ with $S_{i}\ncong S_{j}$ for every $i\neq j$.

######  Proposition 3.32.

Let $\mathcal{L}$ be a complete modular lattice. The following conditions are
equivalent for $\varphi\in\operatorname{End}_{lin}(\mathcal{L})$:

  1. _(a)_

$\ker_{\varphi}$ is a complement in $\mathcal{L}$.

  2. _(b)_

There exists a complement $x\in\mathcal{L}$ such that $\varphi=\varphi\pi_{x}$
and $\ker_{\varphi}\wedge x=\mathbf{0}$.

###### Proof.

(a)$\Rightarrow$(b) By hypothesis there exists $x\in\mathcal{L}$ such that $x$
is a complement of $\ker_{\varphi}$. Let $\ker_{\varphi}\leq y\leq\mathbf{1}$.
By the modular property, $y=y\wedge\mathbf{1}=y\wedge(\ker_{\varphi}\vee
x)=\ker_{\varphi}\vee(y\wedge x)$. Then $\varphi(y)=\varphi(y\wedge x)$. Let
$a\in\mathcal{L}$. Then,

$\varphi\pi_{x}(a)=\varphi((a\vee\ker_{\varphi})\wedge
x)=\varphi(a\vee\ker_{\varphi})=\varphi(a).$

(b)$\Rightarrow$(a) Let $x\in\mathcal{L}$ be a complement such that
$\varphi=\varphi\pi_{x}$ and $x\wedge\ker_{\varphi}=\mathbf{0}$. Let
$x^{\prime}$ be a complement of $x$. Then,
$\varphi(x^{\prime})=\varphi\pi_{x}(x^{\prime})=\varphi(\mathbf{0})=\mathbf{0}$.
This implies that $x^{\prime}\leq\ker_{\varphi}$. By the modularity,
$\ker_{\varphi}=x^{\prime}$. ∎

######  Corollary 3.33.

Let $M$ be an $R$-module. The following conditions are equivalent for
$f\in\operatorname{End}_{R}(M)$:

  1. _(a)_

$\operatorname{Ker}f$ is a direct summand of $M$.

  2. _(b)_

There exists a direct summand $N$ of $M$ such that
$N\cap\operatorname{Ker}f=0$ and $f(L)=f\pi(L)$ for all $L\leq M$ where
$\pi:M\to N$ is the canonical projection.

######  Corollary 3.34.

The following conditions are equivalent for a complete modular lattice
$\mathcal{L}$ and a submonoid
$\mathfrak{m}\subseteq\operatorname{End}_{lin}(\mathcal{L})$ containing all
the projections:

  1. _(a)_

$\mathcal{L}$ is $\mathfrak{m}$-Rickart.

  2. _(b)_

For every $\varphi\in\mathfrak{m}$ there exists a complement $x\in\mathcal{L}$
such that $\varphi=\varphi\pi_{x}$ and $x\wedge\ker_{\varphi}=\mathbf{0}$.

######  Corollary 3.35.

The following conditions are equivalent for an $R$-module $M$:

  1. _(a)_

$M$ is Rickart.

  2. _(b)_

For every $f\in\operatorname{End}_{R}(M)$ there exists a direct summand $N$ of
$M$ such that $N\cap\operatorname{Ker}f=0$ and $f(L)=f\pi(L)$ for all $L\leq
M$ where $\pi:M\to N$ is the canonical projection.

######  Lemma 3.36.

Let $\mathcal{L}$ be a complete modular lattice, $a,x\in\mathcal{L}$ and
$\theta:[a,\mathbf{1}]\to[\mathbf{0},x]$ be an isomorphism of lattices. Then
$\iota_{x}\theta\rho_{a}:\mathcal{L}\to\mathcal{L}$ is a linear morphism.

###### Proof.

Define $\varphi:\mathcal{L}\to\mathcal{L}$ as
$\varphi(y)=\iota_{x}\theta\rho_{a}(y)$. Note that $\varphi(y\vee
a)=\theta(y\vee a\vee a)=\theta(y\vee a)=\varphi(y)$. Since
$\theta:[a,\mathbf{1}]\to[\mathbf{0},x]$ is an isomorphism, it follows that
$\ker_{\varphi}=a$. ∎

######  Definition 3.37.

Let $\mathcal{L}$ be a complete modular lattice, $a,x\in\mathcal{L}$ and
$\mathfrak{m}\subseteq\operatorname{End}_{lin}(\mathcal{L})$ be a submonoid.
It is said that $\mathcal{L}$ satisfies _$\mathfrak{m}$ -$D_{2}$ condition_ if
whenever there is an isomorphism
$\theta:[a,\mathbf{1}]\overset{\cong}{\rightarrow}[\mathbf{0},x]$ with $x$ a
complement in $\mathcal{L}$ and $\iota_{x}\theta\rho_{a}$ in $\mathfrak{m}$,
implies that $a$ is a complement.

Recall that an $R$-module $M$ satisfies the $(D_{2})$ condition if whenever
$N\cong M/L$ with $L,N\leq M$ and $N$ a direct summand of $M$ then $L$ is a
direct summand of $M$ [10, pp. 57]. For example, every projective $R$-module
satisfies ($D_{2}$).

######  Lemma 3.38.

The following conditions are equivalent for an $R$-module $M$:

  1. _(a)_

$M$ satisfies $(D_{2})$ condition.

  2. _(b)_

$\Lambda(M)$ satisfies $\mathfrak{E}_{M}$-$D_{2}$ condition.

###### Proof.

(a)$\Rightarrow$(b) Let $N,L\in\Lambda(M)$ with $N$ a direct summand of $M$
and $\theta:[L,M]\to[0,N]$ an isomorphism of lattices such that
$\iota_{N}\theta\rho_{L}\in\mathfrak{E}_{M}$. Then, there exists an
endomorphism $f\in\operatorname{End}_{R}(M)$ such that
$\iota_{N}\theta\rho_{L}=f_{\ast}$. Let $g:M/L\to N$ defined as $g(m+L)=f(m)$.
Since $f_{\ast}(L)=0$, $g$ is well defined. Suppose $g(m+L)=0$. This implies
that $f_{\ast}(Rm)=0$. Since $\theta$ is an isomorphism, $Rm\subseteq L$. Thus
$g$ is inyective. Let $n\in N$. Consider $Rn$. Then, there exists $K$ with
$L\leq K\leq M$ such that $\theta(K)=Rn$. Since $\rho_{L}$ is induced by the
canonical projection $M\to M/L$, $f_{\ast}(K)=Rn$. This implies that there
exists $k\in K$ such that $f(k)=n$. Hence $g$ is an isomorphism. By
hypothesis, $L$ is a direct summand of $M$. Thus $\Lambda(M)$ satisfies
$\mathfrak{E}_{M}$-$D_{2}$.

(b)$\Rightarrow$(a) Let $N,L$ be submodules of $M$ with $N$ a direct summand
and suppose that there exists an isomorphism $f:M/L\to N$. Then $f$ induces an
isomorphism of lattices $\theta:[L,M]\to[0,N]$. Note that
$\iota_{N}\theta\rho_{L}(K)=i_{N}fp_{L}(K)$ where $i_{N}:N\to M$ and
$p_{L}:M\to M/L$ are the canonical inclusion and the canonical projection
respectively. Therefore $\iota_{N}\theta\rho_{L}=(i_{N}fp_{L})_{\ast}$, that
is, $\iota_{N}\theta\rho_{L}\in\mathfrak{E}_{M}$. By hypothesis, $L$ is a
complement in $\Lambda(M)$, that is, $L$ is a direct summand. ∎

######  Definition 3.39.

Let $\mathcal{L}$ be a complete modular lattice,
$a,x,x^{\prime}\in\mathcal{L}$ with $x^{\prime}$ a complement of $x$. It is
said that $\mathcal{L}$ satisfies _$\mathfrak{m}$ -$C_{2}$ condition_ if
whenever there is an isomorphism
$\theta:[\mathbf{0},x]\overset{\cong}{\rightarrow}[\mathbf{0},a]$ and
$\iota_{a}\theta(x\wedge\\_)\rho_{x^{\prime}}$ is in $\mathfrak{m}$, implies
that $a$ is a complement.

######  Remark 3.40.

In the case when $\mathfrak{m}=\operatorname{End}_{lin}(\mathcal{L})$ the
above definition agrees with the condition $(C_{2})$ given in [2, Definition
1.1].

Recall that an $R$-module $M$ satisfies the $(C_{2})$ condition if whenever
$N\cong L$ with $L,N\leq M$ and $N$ a direct summand of $M$ then $L$ is a
direct summand of $M$ [10, pp. 18]. For example, every injective $R$-module
satisfies ($C_{2}$).

######  Lemma 3.41.

The following conditions are equivalent for an $R$-module $M$.

  1. _(a)_

$M$ satisfies $(C_{2})$ condition.

  2. _(b)_

$\Lambda(M)$ satisfies $\mathfrak{E}_{M}$-$C_{2}$ condition.

###### Proof.

(a)$\Rightarrow$(b) Let $N,L\in\Lambda(M)$ with $N$ a direct summand of $M$
with complement $N^{\prime}$, and $\theta:[0,N]\to[0,L]$ an isomorphism of
lattices such that
$\iota_{L}\theta(N\cap\\_)\rho_{N^{\prime}}\in\mathfrak{E}_{M}$. Then, there
exists an endomorphism $f\in\operatorname{End}_{R}(M)$ such that
$\iota_{L}\theta(N\cap\\_)\rho_{N^{\prime}}=f_{\ast}$. Let $g:N\to L$ defined
as $g(n)=f(n)$. Suppose $g(n)=0$. This implies that
$f_{\ast}(Rn)=0=\theta(N\cap(Rn+N^{\prime}))=\theta(Rn+(N\cap
N^{\prime})=\theta(Rn)$. Since $\theta$ is an isomorphism, $Rn=0$. Thus $g$ is
inyective. Let $l\in L$. Consider $Rl$. Then, there exists $K$ with $0\leq
K\leq N$ such that $\theta(K)=Rl$. Since $\rho_{N^{\prime}}$ is induced by the
canonical projection $M\to M/N^{\prime}$,
$f_{\ast}(K)=\theta(N\cap(K+N^{\prime}))=\theta(K+(N\cap
N^{\prime}))=\theta(K)=Rl$. This implies that there exists $k\in K$ such that
$f(n)=l$. Hence $g$ is an isomorphism. By hypothesis, $L$ is a direct summand
of $M$. Thus $\Lambda(M)$ satisfies $\mathfrak{E}_{M}$-$C_{2}$.

(b)$\Rightarrow$(a) Let $N,L$ be submodules of $M$ with $N$ a direct summand
and suppose that there exists an isomorphism $f:N\to L$. Then $f$ induces an
isomorphism of lattices $\theta:[0,N]\to[0,L]$. Note that
$\iota_{L}\theta(N\cap\\_)\rho_{N^{\prime}}(K)=i_{L}fp_{N}(K)$ where
$i_{L}:L\to M$ and $p_{N}:M\to N$ are the canonical inclusion and the
canonical projection respectively. Therefore
$\iota_{L}\theta(N\cap\\_)\rho_{N^{\prime}}=(i_{L}fp_{N})_{\ast}$, that is,
$\iota_{L}\theta(N\cap\\_)\rho_{N^{\prime}}\in\mathfrak{E}_{M}$. By
hypothesis, $L$ is a complement in $\Lambda(M)$, that is, $L$ is a direct
summand. ∎

######  Proposition 3.42.

The following conditions are equivalent for a complete modular lattice
$\mathcal{L}$ and a submonoid
$\mathfrak{m}\subseteq\operatorname{End}_{lin}(\mathcal{L})$ containing all
the projections:

  1. _(a)_

$\mathcal{L}$ is an $\mathfrak{m}$-Rickart Lattice.

  2. _(b)_

$\mathcal{L}$ satisfies $\mathfrak{m}$-$D_{2}$ condition and for every
$\varphi\in\mathfrak{m}$ there exists an isomorphism
$\theta:[\mathbf{0},\varphi(\mathbf{1})]\to[\mathbf{0},x]$ with $x$ a
complement in $\mathcal{L}$ such that $\iota_{x}\theta\varphi\in\mathfrak{m}$.

###### Proof.

(a)$\Rightarrow$(b) Let $a\in\mathcal{L}$ such that there exists an
isomorphism $\theta:[a,\mathbf{1}]\to[\mathbf{0},x]$ with $x$ a complement in
$\mathcal{L}$ and $\iota_{x}\theta\rho_{a}$ in $\mathfrak{m}$. By Lemma 3.36
and since $\mathcal{L}$ is $\mathfrak{m}$-Rickart, $a$ is a complement. On the
other hand, let $\varphi\in\mathfrak{m}$. By hypothesis $\ker_{\varphi}$ is a
complement in $\mathcal{L}$. Let $k^{\prime}$ be a complement of
$\ker_{\varphi}$. Then
$[\mathbf{0},k^{\prime}]\cong[\ker_{\varphi},\mathbf{1}]$ and
$\overline{\varphi}:[\ker_{\varphi},\mathbf{1}]\to[\mathbf{0},\varphi(\mathbf{1})]$
is an isomorphism. Define
$\theta:[\mathbf{0},\varphi(\mathbf{1})]\to[\mathbf{0},k^{\prime}]$ as
$\theta=(\\_\wedge k^{\prime})\overline{\varphi}^{-1}$. Then,
$\iota_{k^{\prime}}\theta\varphi(a)=(a\vee\ker_{\varphi})\wedge
k^{\prime}=\pi_{k^{\prime}}(a)$. Since $\mathfrak{m}$ contains all the
projections, $\iota_{k^{\prime}}\theta\varphi\in\mathfrak{m}$.

(b)$\Rightarrow$(a) Let $\varphi:\mathcal{L}\to\mathcal{L}$ be a linear
morphism in $\mathfrak{m}$ with kernel $\ker_{\varphi}$. Then, there is an
isomorphism
$\overline{\varphi}:[\ker_{\varphi},\mathbf{1}]\to[\mathbf{0},\varphi(\mathbf{1})]$.
By hypothesis, $[\mathbf{0},\varphi(\mathbf{1})]\cong_{\theta}[\mathbf{0},x]$
with $x$ a complement in $\mathcal{L}$ such that
$\iota_{x}\theta\varphi\in\mathfrak{m}$. Consider the isomorphism
$\theta\overline{\varphi}$. We claim that
$\iota_{x}\theta\overline{\varphi}\rho_{\ker_{\varphi}}\in\mathfrak{m}$. For
every $a\in\mathcal{L}$, we have that
$\iota_{x}\theta\overline{\varphi}\rho_{\ker_{\varphi}}(a)=\iota_{x}\theta\varphi(a)$.
Thus $\iota_{x}\theta\overline{\varphi}\rho_{\ker_{\varphi}}\in\mathfrak{m}$.
By $\mathfrak{m}$-$D_{2}$ condition, $\ker_{\varphi}$ is complemented in
$\mathcal{L}$. Thus, $\mathcal{L}$ is $\mathfrak{m}$-Rickart. ∎

######  Proposition 3.43.

The following conditions are equivalent for a complete modular lattice
$\mathcal{L}$ and a submonoid
$\mathfrak{m}\subseteq\operatorname{End}_{lin}(\mathcal{L})$ containing all
the projections:

  1. _(a)_

$\mathcal{L}$ is a dual-$\mathfrak{m}$-Rickart Lattice.

  2. _(b)_

$\mathcal{L}$ satisfies $\mathfrak{m}$-$C_{2}$ condition and for every
$\varphi\in\mathfrak{m}$ there exists an isomorphism
$\theta:[\mathbf{0},x]\to[\mathbf{0},\varphi(\mathbf{1})]$ with $x$ a
complement in $\mathcal{L}$ such that
$\iota_{\varphi(\mathbf{1})}\theta(x\wedge\\_)\rho_{x^{\prime}}\in\mathfrak{m}$
where $x^{\prime}$ is a complement of $x$.

###### Proof.

(a)$\Rightarrow$(b) Let $a\in\mathcal{L}$ such that there exists an
isomorphism $\theta:[\mathbf{0},x]\to[\mathbf{0},a]$ with $x$ a complement in
$\mathcal{L}$ and $\iota_{a}\theta(x\wedge\\_)\rho_{x^{\prime}}$ in
$\mathfrak{m}$ where $x^{\prime}$ is a complement of $x$. By Lemma 3.36 and
since $\mathcal{L}$ is dual-$\mathfrak{m}$-Rickart, $a$ is a complement. On
the other hand, let $\varphi\in\mathfrak{m}$. By hypothesis
$\varphi(\mathbf{1})$ is a complement in $\mathcal{L}$. Let $b$ be a
complement of $\varphi(\mathbf{1})$. Take
$\theta=Id_{[\mathbf{0},\varphi(\mathbf{1})]}$. Thus
$\iota_{\varphi(\mathbf{1})}\theta(\varphi(\mathbf{1})\wedge\\_)\rho_{b}=\pi_{\varphi(\mathbf{1})}\in\mathfrak{m}$.

(b)$\Rightarrow$(a) Let $\varphi:\mathcal{L}\to\mathcal{L}$ be a linear
morphism in $\mathfrak{m}$. By hypothesis,
$[\mathbf{0},x]\cong_{\theta}[\mathbf{0},\varphi(\mathbf{1})]$ with $x$ a
complement in $\mathcal{L}$ such that
$\iota_{\varphi(\mathbf{1})}\theta(x\wedge\\_)\rho_{x^{\prime}}\in\mathfrak{m}$
where $x^{\prime}$ is a complement of $x$. It follows from the
$\mathfrak{m}$-$C_{2}$ condition, that $\varphi(\mathbf{1})$ is complemented
in $\mathcal{L}$. Thus, $\mathcal{L}$ is dual-$\mathfrak{m}$-Rickart. ∎

######  Corollary 3.44 ([9, Proposition 2.11]).

The following conditions are equivalent for an $R$-module $M$.

  1. _(a)_

$M$ is a Rickart module.

  2. _(b)_

$M$ satisfies $(D_{2})$ and for every $f\in\operatorname{End}_{R}(M)$ there is
an isomorphism $g:f(M)\to N$ with $N$ a direct summand of $M$.

######  Corollary 3.45 ([7, Proposition 2.21]).

The following conditions are equivalent for an $R$-module $M$.

  1. _(a)_

$M$ is a dual-Rickart module.

  2. _(b)_

$M$ satisfies $(C_{2})$ and for every $f\in\operatorname{End}_{R}(M)$ there is
an isomorphism $g:N\to f(M)$ with $N$ a direct summand of $M$.

######  Definition 3.46.

Let $\mathcal{L}$ be a modular lattice and $\mathfrak{m}$ be a submonoid of
$\operatorname{End}_{lin}(\mathcal{L})$. The lattice $\mathcal{L}$ is
$k$-local $\mathfrak{m}$-retractable if for each $\varphi\in\mathfrak{m}$ and
for each $b\leq\ker_{\varphi}$, there exists
$\psi_{b}:\mathcal{L}\rightarrow[\mathbf{0},\ker_{\varphi}]$ such that
$b\leq\psi_{b}(\mathbf{1})$ and
$\iota_{\ker_{\varphi}}\psi_{b}\in\mathfrak{m}$.

######  Proposition 3.47.

Let $\mathcal{L}$ be a complete modular lattice and
$\mathfrak{m}\subseteq\operatorname{End}_{lin}(\mathcal{L})$ be a submonoid
containig all the projections. If $\mathcal{L}$ is $\mathfrak{m}$-Rickart,
then $\mathcal{L}$ is $k$-local $\mathfrak{m}$-retractable.

###### Proof.

Let $\varphi:\mathcal{L}\rightarrow\mathcal{L}$ be a linear morphism. Since
$\mathcal{L}$ is $\mathfrak{m}$-Rickart, we can consider the projection
$\pi_{\ker_{\varphi}}:\mathcal{L}\to[\mathbf{0},\ker_{\varphi}]$ co-
restricted. Then for $b\leq\ker_{\varphi}$, we have that
$b\leq\ker_{\varphi}=\pi_{\ker_{\varphi}}(\mathbf{1})$. ∎

######  Definition 3.48.

Let $\mathcal{L}$ be a complete modular lattice, $\mathfrak{m}$ be a submonoid
of $\operatorname{End}_{lin}(\mathcal{L})$ and $x\in\mathcal{L}$. It is said
that $x$ is $\mathfrak{m}$-$\mathcal{L}$-generated if there exists a family
$\\{\varphi_{i}:\mathcal{L}\to[\mathbf{0},x]\\}_{i\in I}$ of linear morphisms
such that $\iota_{x}\varphi_{i}\in\mathfrak{m}$ for all $i\in I$ and
$x=\bigvee_{i\in I}\varphi_{i}(\mathbf{1})$.

######  Remark 3.49.

It is clear that if $\mathcal{L}$ is $k$-local $\mathfrak{m}$-retractable,
then $\ker_{\varphi}$ is $\mathfrak{m}$-$\mathcal{L}$-generated for all
$\varphi\in\mathfrak{m}$.

######  Definition 3.50.

Let $\mathcal{L}$ be a complete modular lattice, $\mathfrak{m}$ be a submonoid
of $\operatorname{End}_{lin}(\mathcal{L})$ and $x\in\mathcal{L}$. It is said
that $x$ is _$\mathfrak{m}$ -$\mathcal{L}$-cogenerated_ if there exists a
family $\\{\varphi_{i}:[x,\mathbf{1}]\to\mathcal{L}\\}_{i\in I}$ of linear
morphisms such that $\varphi_{i}\rho_{x}\in\mathfrak{m}$ for all $i\in I$ and
$x=\bigwedge_{i\in I}\ker_{\varphi_{i}}$.

######  Proposition 3.51.

Let $\mathcal{L}$ be a complete modular lattice and $\mathfrak{m}$ be a
submonoid of $\operatorname{End}_{lin}(\mathcal{L})$ containing all the
projections. The following conditions are equivalent for
$\varphi\in\mathfrak{m}$:

  1. _(a)_

$\ker_{\varphi}$ has a complement in $\mathcal{L}$.

  2. _(b)_

$\ker_{\varphi}$ is $\mathfrak{m}$-$\mathcal{L}$-generated and
$\operatorname{Ann}^{r}(\varphi)=\varepsilon\mathfrak{m}$ for some idempotent
$\varepsilon\in\mathfrak{m}$.

###### Proof.

(a)$\Rightarrow$(b) Since $\ker_{\varphi}$ is a complement in $\mathcal{L}$,
$\ker_{\varphi}=\pi_{\ker_{\varphi}}(\mathbf{1})$. Thus, $\ker_{\varphi}$ is
$\mathfrak{m}$-$\mathcal{L}$-generated. By hypothesis, there exists
$x\in\mathcal{L}$ such that $\mathbf{1}=\ker_{\varphi}\vee x$ and
$\mathbf{0}=\ker_{\varphi}\wedge x$. We have that
$\pi_{\ker_{\varphi}}(\mathbf{1})=\ker_{\varphi}$, so
$\varphi\pi_{\ker_{\varphi}}(\mathbf{1})=\mathbf{0}$. This implies that
$\pi_{\ker_{\varphi}}\in\operatorname{Ann}^{r}(\varphi)$. Now, let
$\psi\in\operatorname{Ann}^{r}(\varphi)$, that is, $\varphi\psi=0$. Hence
$\varphi\psi(\mathbf{1})=\mathbf{0}$. Therefore,
$\psi(\mathbf{1})\leq\ker_{\varphi}=\pi_{\ker_{\varphi}}(\mathbf{1})$. Let
$a\in\mathcal{L}$. Then $\psi(a)=(\psi(a)\vee
x)\wedge\ker_{\varphi}=\pi_{\ker_{\varphi}}\psi(a)$. Thus
$\pi_{\ker_{\varphi}}\psi=\psi$. This implies that
$\operatorname{Ann}^{r}(\varphi)=\pi_{\ker_{\varphi}}\mathfrak{m}$.

(b)$\Rightarrow$(a) By hypothesis there exists $\varepsilon\in\mathfrak{m}$
idempotent such that
$\operatorname{Ann}^{r}(\varphi)=\varepsilon\mathfrak{m}$. Then
$\varphi\varepsilon(\mathbf{1})=\mathbf{0}$. Thus
$\varepsilon(\mathbf{1})\leq\ker_{\varphi}$. On the other hand, there exists
linear morphisms $\psi_{i}:\mathcal{L}\to[\mathbf{0},\ker_{\varphi}]$ ($i\in
I$) such that $\bigvee_{i\in I}\psi_{i}(\mathbf{1})=\ker_{\varphi}$. This
implies that $\varphi\psi_{i}=0$ for all $i\in I$ and so
$\psi_{i}\in\operatorname{Ann}^{r}(\varphi)$ for all $i\in I$. Therefore
$\varepsilon\psi_{i}=\psi_{i}$ for all $i\in I$. Then

$\ker_{\varphi}=\bigvee_{i\in I}\psi_{i}(\mathbf{1})=\bigvee_{i\in
I}\varepsilon\psi_{i}(\mathbf{1})\leq\varepsilon(\mathbf{1}).$

Thus, $\varepsilon(\mathbf{1})=\ker_{\varphi}$. Since $\varepsilon$ is
idempotent, $\ker_{\varphi}$ is a complement in $\mathcal{L}$. ∎

######  Proposition 3.52.

Let $\mathcal{L}$ be a complete modular lattice and $\mathfrak{m}$ be a
submonoid of $\operatorname{End}_{lin}(\mathcal{L})$ containing all the
projections. The following conditions are equivalent for
$\varphi\in\mathfrak{m}$:

  1. _(a)_

$\varphi(\mathbf{1})$ has a complement in $\mathcal{L}$.

  2. _(b)_

$\varphi(\mathbf{1})$ is $\mathfrak{m}$-$\mathcal{L}$-cogenerated and
$\operatorname{Ann}^{\ell}(\varphi)=\mathfrak{m}\varepsilon$ for some
idempotent $\varepsilon\in\mathfrak{m}$.

###### Proof.

(a)$\Rightarrow$(b) Since $\varphi(\mathbf{1})$ is complemented in
$\mathcal{L}$, there exists $x\in\mathcal{L}$ such that
$\mathbf{1}=\varphi(\mathbf{1})\vee x$ and
$\mathbf{0}=\varphi(\mathbf{1})\wedge x$. Then
$\varphi(\mathbf{1})=\ker_{\iota_{x}(x\wedge\\_)}$. Thus,
$\varphi(\mathbf{1})$ is $\mathfrak{m}$-$\mathcal{L}$-cogenerated. We have
that $\pi_{x}\varphi(\mathbf{1})=\mathbf{0}$. This implies that
$\pi_{x}\in\operatorname{Ann}^{\ell}(\varphi)$. Now, let
$\psi\in\operatorname{Ann}^{\ell}(\varphi)$, that is, $\psi\varphi=0$. Hence
$\psi\varphi(\mathbf{1})=\mathbf{0}$. Therefore,
$\varphi(\mathbf{1})\leq\ker_{\psi}$. Let $a\in\mathcal{L}$. Then
$\psi\pi_{x}(a)=\psi((a\vee\varphi(\mathbf{1}))\wedge
x)\leq\psi((a\vee\varphi(\mathbf{1})))\wedge\psi(x)=\psi(a)\wedge\psi(x)=\psi(a)\wedge\psi(\mathbf{1})=\psi(a)$.
On the other hand, $a\leq\pi_{\varphi(\mathbf{1})}(a)\vee\pi_{x}(a)$. Hence
$\psi(a)\leq\psi(\pi_{\varphi(\mathbf{1})}(a)\vee\pi_{x}(a))=\psi(\pi_{\varphi(\mathbf{1})}(a))\vee\psi(\pi_{x}(a))=\psi\pi_{x}(a)$.
Thus $\psi\pi_{x}=\psi$. This implies that
$\operatorname{Ann}^{\ell}(\varphi)=\mathfrak{m}\pi_{x}$.

(b)$\Rightarrow$(a) By hypothesis there exists $\varepsilon\in\mathfrak{m}$
idempotent such that
$\operatorname{Ann}^{\ell}(\varphi)=\mathfrak{m}\varepsilon$. Then
$\varepsilon\varphi(\mathbf{1})=\mathbf{0}$. Thus
$\varphi(\mathbf{1})\leq\ker_{\varepsilon}$. On the other hand, there exists
linear morphisms $\psi_{i}:[\varphi(\mathbf{1}),\mathbf{1}]\to\mathcal{L}$
($i\in I$) such that $\bigwedge_{i\in I}\ker_{\psi_{i}}=\varphi(\mathbf{1})$
and $\psi_{i}\rho_{\varphi(\mathbf{1})}\in\mathfrak{m}$. This implies that
$\psi_{i}\rho_{\varphi(\mathbf{1})}\varphi=0$ for all $i\in I$ and so
$\psi_{i}\rho_{\varphi(\mathbf{1})}\in\operatorname{Ann}^{\ell}(\varphi)$ for
all $i\in I$. Therefore
$\psi_{i}\rho_{\varphi(\mathbf{1})}\varepsilon=\psi_{i}\rho_{\varphi(\mathbf{1})}$
for all $i\in I$. Then, for each $i\in I$,

$\psi_{i}(\ker_{\varepsilon})=\psi_{i}(\ker_{\varepsilon}\vee\varphi(\mathbf{1}))=\psi_{i}\rho_{\varphi(\mathbf{1})}(\ker_{\varepsilon})=\psi_{i}\rho_{\varphi(\mathbf{1})}\varepsilon(\ker_{\varepsilon})=\mathbf{0}.$

This implies that $\ker_{\varepsilon}\leq\bigwedge_{i\in
I}\ker_{\psi_{i}}=\varphi(\mathbf{1})$. Thus,
$\varphi(\mathbf{1})=\ker_{\varepsilon}$. Since $\varepsilon$ is idempotent,
$\varphi(\mathbf{1})$ is complemented in $\mathcal{L}$. ∎

######  Definition 3.53.

A monoid $E$ with zero element is called _right Rickart_ if for every
$\varphi\in E$ there exists an idempotent element $\varepsilon\in E$ such that
$\operatorname{Ann}^{r}(\varphi)=\\{\psi\in E\mid\varphi\psi=0\\}=\varepsilon
E$. A _left Rickart_ monoid is defined similarly.

It is clear that if a ring $R$ is right Rickart if and only if $R$ as monoid
with the multiplication is right Rickart in the sense of the above definition.
In [9] it is proved that, if $M$ is a Rickart module, then
$\operatorname{End}_{R}(M)$ is a right Rickart ring. The next results explore
the lattice-counterpart of this fact.

######  Corollary 3.54.

The following conditions are equivalent for a complete modular lattice
$\mathcal{L}$ and a submonoid
$\mathfrak{m}\subseteq\operatorname{End}_{lin}(\mathcal{L})$ containing all
the projections:

  1. _(a)_

$\mathcal{L}$ is $\mathfrak{m}$-Rickart.

  2. _(b)_

The monoid $\mathfrak{m}$ is right Rickart and $\mathcal{L}$ is $k$-local
$\mathfrak{m}$-retractable.

  3. _(c)_

The monoid $\mathfrak{m}$ is right Rickart and $\ker_{\varphi}$ is
$\mathfrak{m}$-$\mathcal{L}$-generated for all $\varphi\in\mathfrak{m}$.

######  Corollary 3.55.

The following conditions are equivalent for a complete modular lattice
$\mathcal{L}$ and a submonoid
$\mathfrak{m}\subseteq\operatorname{End}_{lin}(\mathcal{L})$ containing all
the projections:

  1. _(a)_

$\mathcal{L}$ is dual-$\mathfrak{m}$-Rickart.

  2. _(b)_

The monoid $\mathfrak{m}$ is left Rickart and $\varphi(\mathbf{1})$ is
$\mathfrak{m}$-$\mathcal{L}$-cogenerated for all $\varphi\in\mathfrak{m}$.

######  Corollary 3.56 (Theorem 3.9, [9]).

The following conditions are equivalent for an $R$-module $M$:

  1. _(a)_

$M$ is Rickart.

  2. _(b)_

The ring $\operatorname{End}_{R}(M)$ is right Rickart and $M$ is $k$-local-
retractable.

  3. _(c)_

The ring $\operatorname{End}_{R}(M)$ is right Rickart and
$\operatorname{Ker}\varphi$ is $M$-generated for all
$\varphi\in\operatorname{End}_{R}(M)$.

######  Corollary 3.57.

The following conditions are equivalent for an $R$-module $M$:

  1. _(a)_

$M$ is dual-Rickart.

  2. _(b)_

The ring $\operatorname{End}_{R}(M)$ is left Rickart and
$\operatorname{Coker}\varphi$ is $M$-cogenerated for all
$\varphi\in\operatorname{End}_{R}(M)$.

######  Definition 3.58.

A monoid $E$ with zero element is called _right Baer_ if for every $X\subseteq
E$ there exists an idempotent element $\varepsilon\in E$ such that
$\operatorname{Ann}^{r}(X)=\\{\psi\in E\mid\varphi\psi=0\;\forall\varphi\in
X\\}=\varepsilon E$. A _left Baer_ monoid is defined similarly.

It is clear that if a ring $R$ is Baer if and only if $R$ as monoid with the
multiplication is Baer in the sense of the above definition, moreover the Baer
condition on a ring is left-right symmetric. Below we prove that for the case
of the monoid $\operatorname{End}_{lin}(\mathcal{L})$ with $\mathcal{L}$ a
complete modular lattice, the right and left Baer conditions are equivalent
(Compare).

######  Lemma 3.59.

The following conditions are equivalent for a complete modular lattice
$\mathcal{L}$ and a submonoid
$\mathfrak{m}\subseteq\operatorname{End}_{lin}(\mathcal{L})$ containing all
the projections:

  1. _(a)_

For every $X\subseteq\mathfrak{m}$,
$\operatorname{Ann}^{\ell}(X)=\\{\psi\in\mathfrak{m}\mid\psi\varphi=0\;\forall\varphi\in
X\\}=\mathfrak{m}\zeta$ for some idempotent $\zeta\in\mathfrak{m}$.

  2. _(b)_

For every $Y\subseteq\mathfrak{m}$,
$\operatorname{Ann}^{r}(Y)=\\{\psi\in\mathfrak{m}\mid\varphi\psi=0\;\forall\varphi\in
X\\}=\varepsilon\mathfrak{m}$ for some idempotent
$\varepsilon\in\mathfrak{m}$.

###### Proof.

Since the conditions are symmetric, we only prove (a)$\Rightarrow$(b). Let
$Y\subseteq\mathfrak{m}$. Then
$\operatorname{Ann}^{r}(\operatorname{Ann}^{\ell}(\operatorname{Ann}^{r}(Y)))=\operatorname{Ann}^{r}(Y)$.
By hypothesis, there exists an idempotent $\zeta\in\mathfrak{m}$ such that
$\operatorname{Ann}^{\ell}(\operatorname{Ann}^{r}(Y))=\mathfrak{m}\zeta$. By
Proposition 2.10, $\ker_{\zeta}$ is a complement in $\mathcal{L}$. Consider
$\pi_{\ker_{\zeta}}\in\mathfrak{m}$. We claim that
$\operatorname{Ann}^{r}(Y)=\operatorname{Ann}^{r}(\mathfrak{m}\zeta)=\pi_{\ker_{\zeta}}\mathfrak{m}$.
Since $\zeta\pi_{\ker_{\zeta}}=0$, then
$\pi_{\ker_{\zeta}}\mathfrak{m}\subseteq\operatorname{Ann}^{r}(Y)$. Now, let
$\psi\in\operatorname{Ann}^{r}(Y)=\operatorname{Ann}^{r}(\mathfrak{m}\zeta)$.
Then $\zeta\psi=0$. This implies that $\psi(\mathbf{1})\leq\ker_{\zeta}$.
Hence $\pi_{\ker_{\zeta}}(\psi(a))=\psi(a)$ for all $a\in\mathcal{L}$. Thus,
$\psi\in\pi_{\ker_{\zeta}}\mathfrak{m}$. This proves the claim. ∎

In [11, Theorem 4.1] it is proved that, if $M$ is a Baer module, then
$\operatorname{End}_{R}(M)$ is a Baer ring. The next results explore the
lattice-counterpart of this fact.

######  Proposition 3.60.

Let $\mathcal{L}$ be a complete modular lattice and $\mathfrak{m}$ be a
submonoid of $\operatorname{End}_{lin}(\mathcal{L})$ containing all the
projections. The following conditions are equivalent:

  1. _(a)_

$\mathcal{L}$ is $\mathfrak{m}$-Baer.

  2. _(b)_

For every $a\in\mathcal{L}$ there exists an idempotent
$\varepsilon\in\mathfrak{m}$ such that
$\\{\varphi\in\mathfrak{m}\mid\varphi(a)=\mathbf{0}\\}=\mathfrak{m}\varepsilon$.

  3. _(c)_

The monoid $\mathfrak{m}$ is Baer and $\bigwedge_{\varphi\in X}\ker_{\varphi}$
is $\mathfrak{m}$-$\mathcal{L}$-generated for all $X\subseteq\mathfrak{m}$.

###### Proof.

(a)$\Rightarrow$(b) Let $a\in\mathcal{L}$. Consider
$X_{a}=\\{\varphi\in\mathfrak{m}\mid\varphi(a)=\mathbf{0}\\}$ and
$k=\bigwedge_{\varphi\in X_{a}}\ker_{\varphi}$. Note that
$X_{a}=\\{\varphi\in\mathfrak{m}\mid\varphi(k)=\mathbf{0}\\}$. By hypothesis,
$k$ has a complement $k^{\prime}$ in $\mathcal{L}$. Then $\pi_{k^{\prime}}\in
X_{a}$. If $\psi\in\mathfrak{m}$, then $\psi\pi_{k^{\prime}}(k)=\mathbf{0}$.
Thus, $\mathfrak{m}\pi_{k^{\prime}}\subseteq X_{a}$. Now, let $\varphi\in
X_{a}$ and $x\in\mathcal{L}$. We have that $\varphi(k)=\mathbf{0}$, that is,
$k\leq\ker_{\varphi}$. Also, $\varphi(\mathbf{1})=\varphi(k\vee
k^{\prime})=\varphi(k^{\prime})$. Then

$\varphi\pi_{k^{\prime}}(x)=\varphi((x\vee k)\wedge
k^{\prime})\leq\varphi(x\vee
k)\wedge\varphi(k^{\prime})=\varphi(x)\wedge\varphi(k^{\prime})=\varphi(x).$

On the other hand, it follows from Proposition 2.14 that
$x\leq\pi_{k}(x)\vee\pi_{k^{\prime}}(x)$. Therefore,
$\varphi(x)\leq\varphi(\pi_{k}(x)\vee\pi_{k^{\prime}}(x))=\varphi(\pi_{k}(x))\vee\varphi(\pi_{k^{\prime}}(x))=\varphi\pi_{k^{\prime}}(x)$.
Thus, $\varphi=\varphi\pi_{k^{\prime}}$ and so $X_{a}=\mathfrak{m}\varepsilon$
with $\varepsilon=\pi_{k^{\prime}}$.

(b)$\Rightarrow$(c) Let $X\subseteq\mathfrak{m}$ and $k=\bigwedge_{\varphi\in
X}\ker_{\varphi}$. By hypothesis there exists an idempotent
$\varepsilon\in\mathfrak{m}$ such that
$X_{k}=\\{\psi\in\mathfrak{m}\mid\psi(k)=\mathbf{0}\\}=\mathfrak{m}\varepsilon$.
Suppose $\varepsilon(a)=\mathbf{0}$ for some $a\in\mathcal{L}$. Since
$X\subseteq X_{k}$, $\varphi(a)=\varphi\varepsilon(a)=\mathbf{0}$. Hence
$a\leq k$. Thus, $k=\ker_{\varepsilon}$. This proves that $k$ is
$\mathfrak{m}$-$\mathcal{L}$-generated. Consider $\pi_{k}$. Since
$\varepsilon\pi_{k}=0$,
$\pi_{k}\mathfrak{m}\subseteq\operatorname{Ann}^{r}(X)$. Now, let
$\xi\in\operatorname{Ann}^{r}(X)$, that is, $\varphi\xi=0$ for all $\varphi\in
X$. This implies that $\xi(\mathbf{1})\leq k$ and so
$\varepsilon\xi(\mathbf{1})\leq\varepsilon(k)=\mathbf{0}$. Therefore,
$\xi(\mathbf{1})\leq k$. It follows that $\xi=\xi\pi_{k}$. Thus,
$\operatorname{Ann}^{r}(X)=\pi_{k}\mathfrak{m}$.

(c)$\Rightarrow$(a) A small modification of the proof (b)$\Rightarrow$(a) of
Proposition 3.51 works here.

Let $X\subseteq\mathfrak{m}$. By hypothesis there exists
$\varepsilon\in\mathfrak{m}$ idempotent such that
$\operatorname{Ann}^{r}(X)=\varepsilon\mathfrak{m}$. Then
$\varphi\varepsilon(\mathbf{1})=\mathbf{0}$ for all $\varphi\in X$. Thus
$\varepsilon(\mathbf{1})\leq\ker_{\varphi}$ for all $\varphi\in X$. On the
other hand, there exists linear morphisms
$\psi_{i}:\mathcal{L}\to[\mathbf{0},\bigwedge_{\varphi\in X}\ker_{\varphi}]$
($i\in I$) such that $\bigvee_{i\in
I}\psi_{i}(\mathbf{1})=\bigwedge_{\varphi\in X}\ker_{\varphi}$. This implies
that $\varphi\psi_{i}=0$ for all $i\in I$ and for all $\varphi\in X$, and so
$\psi_{i}\in\operatorname{Ann}^{r}(X)$ for all $i\in I$. Therefore
$\varepsilon\psi_{i}=\psi_{i}$ for all $i\in I$. Then

$\bigwedge_{\varphi\in X}\ker_{\varphi}=\bigvee_{i\in
I}\psi_{i}(\mathbf{1})=\bigvee_{i\in
I}\varepsilon\psi_{i}(\mathbf{1})\leq\varepsilon(\mathbf{1}).$

Thus, $\varepsilon(\mathbf{1})=\bigwedge_{\varphi\in X}\ker_{\varphi}$. Since
$\varepsilon$ is idempotent, $\bigwedge_{\varphi\in X}\ker_{\varphi}$ is a
complement in $\mathcal{L}$. ∎

######  Corollary 3.61.

The following conditions are equivalent for an $R$-module $M$ with
$S=\operatorname{End}_{R}(M)$:

  1. _(a)_

$M$ is Baer.

  2. _(b)_

For every $N\leq M$ there exists an idempotent $e\in S$ such that $\\{f\in
S\mid f(N)=0\\}=Se$

  3. _(c)_

$S$ is a Baer ring and $\bigwedge_{f\in X}\operatorname{Ker}f$ is
$M$-generated for all $X\subseteq S$.

######  Proposition 3.62.

Let $\mathcal{L}$ be a complete modular lattice and $\mathfrak{m}$ be a
submonoid of $\operatorname{End}_{lin}(\mathcal{L})$ containing all the
projections. The following conditions are equivalent:

  1. _(a)_

$\mathcal{L}$ is dual-$\mathfrak{m}$-Baer.

  2. _(b)_

For every $a\in\mathcal{L}$ there exists an idempotent
$\varepsilon\in\mathfrak{m}$ such that
$\\{\varphi\in\mathfrak{m}\mid\varphi(\mathbf{1})\leq
a\\}=\varepsilon\mathfrak{m}$.

  3. _(c)_

The monoid $\mathfrak{m}$ is Baer and $\bigvee_{\varphi\in
X}\varphi(\mathbf{1})$ is $\mathfrak{m}$-$\mathcal{L}$-cogenerated for all
$X\subseteq\mathfrak{m}$.

###### Proof.

(a)$\Rightarrow$(b) Let $a\in\mathcal{L}$. Consider
$D_{a}=\\{\varphi\in\mathfrak{m}\mid\varphi(\mathbf{1})\leq a\\}$ and
$d=\bigvee_{\varphi\in D_{a}}\varphi(\mathbf{1})$. Note that
$D_{a}=\\{\varphi\in\mathfrak{m}\mid\varphi(\mathbf{1})\leq d\\}$. By
hypothesis, $d$ has a complement $d^{\prime}$ in $\mathcal{L}$. Then
$\pi_{d}\in D_{a}$. If $\psi\in\mathfrak{m}$, then
$\pi_{d}\psi(\mathbf{1})\leq d$. Thus, $\pi_{d}\mathfrak{m}\subseteq D_{a}$.
Now, let $\varphi\in D_{a}$ and $x\in\mathcal{L}$. We have that
$\varphi(x)\leq d$. Then $\pi_{d}\varphi(x)=(\varphi(x)\vee d^{\prime})\wedge
d=\varphi(x)$ by modularity. Thus, $D_{a}=\varepsilon\mathfrak{m}$ with
$\varepsilon=\pi_{d}$.

(b)$\Rightarrow$(a)(c) Let $X\subseteq\mathfrak{m}$ and $d=\bigvee_{\varphi\in
X}\varphi(\mathbf{1})$. Consider
$D_{d}=\\{\varphi\in\mathfrak{m}\mid\varphi(\mathbf{1})\leq d\\}$. By
hypothesis, $D_{d}=\varepsilon\mathfrak{m}$ for some idempotent $\varepsilon$.
It follows that $\varepsilon(\mathbf{1})\leq d$. On the other hand, if
$\varphi\in D_{d}$, then $\varphi=\varepsilon\varphi$. Therefore
$\varphi(\mathbf{1})=\varepsilon\varphi(\mathbf{1})\leq\varepsilon(\mathbf{1})$.
This implies $d=\bigvee_{\varphi\in
X}\varphi(\mathbf{1})\leq\varepsilon(\mathbf{1})$. Thus
$\varepsilon(\mathbf{1})=d$ and by Proposition 2.10, $d$ is complemented in
$\mathcal{L}$. This proves (a). Now, consider the linear morphism
$[d,\mathbf{1}]\to\mathcal{L}$ given by the composition of the isomorphism
$\ker_{\varepsilon}\wedge\\_:[d,\mathbf{1}]\to[\mathbf{0},\ker_{\varepsilon}]$
and the canonical inclusion $\iota_{\ker_{\varepsilon}}$. Then
$\iota_{\ker_{\varepsilon}}(\ker_{\varepsilon}\wedge\\_)\rho_{d}=\pi_{\ker_{\varepsilon}}\in\mathfrak{m}$.
This implies that $d$ is $\mathfrak{m}$-$\mathcal{L}$-cogenerated. Since
$\varphi=\varepsilon\varphi$, $\pi_{\ker_{\varepsilon}}\varphi=0$. Then
$\mathfrak{m}\pi_{\ker_{\varepsilon}}\subseteq\operatorname{Ann}^{\ell}(X)$.
Let $\psi\in\operatorname{Ann}^{\ell}(X)$, that is, $\psi\varphi=0$ for all
$\varphi\in X$. It follows that $\psi\varphi(\mathbf{1})=\mathbf{0}$ for all
$\varphi\in X$ and so $\psi(d)=\mathbf{0}$. Hence $\psi\varepsilon=0$. By
Proposition 2.14.(1), $\psi(a)\leq\psi\pi_{\ker_{\varepsilon}}(a)$ for all
$a\in\mathcal{L}$. On the other hand,

$\psi\pi_{\ker_{\varepsilon}}(a)=\psi((a\vee
d)\wedge\ker_{\varepsilon})\leq\psi(a\vee
d)\wedge\psi(\ker_{\varepsilon})\leq\psi(a)\vee\psi(d)=\psi(a).$

Thus, $\psi=\psi\pi_{\ker_{\varepsilon}}$. This implies that
$\operatorname{Ann}^{\ell}(X)=\mathfrak{m}\pi_{\ker_{\varepsilon}}$.

(c)$\Rightarrow$(a) It follows from a small modification on the proof
(b)$\Rightarrow$(a) of Proposition 3.52.

Let $X\subseteq\mathfrak{m}$ and $d=\bigvee_{\varphi\in
X}\varphi(\mathbf{1})$. By hypothesis there exists
$\varepsilon\in\mathfrak{m}$ idempotent such that
$\operatorname{Ann}^{\ell}(X)=\mathfrak{m}\varepsilon$. Then
$\varepsilon\varphi(\mathbf{1})=\mathbf{0}$ for all $\varphi\in X$. Thus
$\varphi(\mathbf{1})\leq\ker_{\varepsilon}$ for all $\varphi\in X$. On the
other hand, there exists linear morphisms
$\psi_{i}:[d,\mathbf{1}]\to\mathcal{L}$ ($i\in I$) such that $\bigwedge_{i\in
I}\ker_{\psi_{i}}=d$ and $\psi_{i}\rho_{d}\in\mathfrak{m}$. This implies that
$\psi_{i}\rho_{d}\varphi=0$ for all $i\in I$ and for all $\varphi\in X$, and
so $\psi_{i}\rho_{d}\in\operatorname{Ann}^{\ell}(\varphi)$ for all $i\in I$.
Therefore $\psi_{i}\rho_{d}\varepsilon=\psi_{i}\rho_{d}$ for all $i\in I$.
Then, for each $i\in I$,

$\psi_{i}(\ker_{\varepsilon})=\psi_{i}(\ker_{\varepsilon}\vee
d)=\psi_{i}\rho_{d}(\ker_{\varepsilon})=\psi_{i}\rho_{d}\varepsilon(\ker_{\varepsilon})=\mathbf{0}.$

This implies that $\ker_{\varepsilon}\leq\bigwedge_{i\in I}\ker_{\psi_{i}}=d$.
Thus, $d=\ker_{\varepsilon}$. Since $\varepsilon$ is idempotent, $d$ is
complemented in $\mathcal{L}$. ∎

######  Corollary 3.63.

The following conditions are equivalent for an $R$-module $M$ with
$S=\operatorname{End}_{R}(M)$:

  1. _(a)_

$M$ is dual-Baer.

  2. _(b)_

For every $N\leq M$ there exists an idempotent $e\in S$ such that $\\{f\in
S\mid f(M)\leq N\\}=eS$

  3. _(c)_

$S$ is a Baer ring and the factor module $M/\bigvee_{f\in X}f(M)$ is
$M$-cogenerated for all $X\subseteq S$.

######  Corollary 3.64.

The following conditions are equivalent for an $R$-module $M$ with
$S=\operatorname{End}_{R}(M)$:

  1. _(a)_

$M$ is Baer and dual-Baer.

  2. _(b)_

$S$ is a Baer ring, $\bigwedge_{f\in X}\operatorname{Ker}f$ is $M$-generated
and $M/\bigvee_{f\in X}f(M)$ is $M$-cogenerated for all $X\subseteq S$.

In this case the set of direct summnads of $M$ is a complete complemented
sublattice of $\Lambda(M)$.

The following propositions give conditions in order to the concepts of Rickart
and Baer (resp. dual-Rickart and dual-Baer) coincide. As corollaries we get
[7, Theorem 4.2] and [9, Theorem 4.5]

######  Proposition 3.65.

Let $\mathcal{L}$ be a complete modular lattice such that $C(\mathcal{L})$
satisfies ACC and $\mathfrak{m}$ be a submonoid of
$\operatorname{End}_{lin}(\mathcal{L})$ containing all the projections. The
following conditions are equivalent:

  1. _(a)_

$\mathcal{L}$ is $\mathfrak{m}$-Rickart.

  2. _(b)_

$\mathcal{L}$ is $\mathfrak{m}$-Baer.

###### Proof.

Suppose $\mathcal{L}$ is $\mathfrak{m}$-Rickart. For $a\in\mathcal{L}$ let
$X_{a}$ denote the set
$\\{\varphi\in\mathfrak{m}\mid\varphi(a)=\mathbf{0}\\}$. If $\varphi\in
X_{a}$, then $a\leq\ker_{\varphi}$. Let $k^{\prime}$ be a complement of
$\ker_{\varphi}$ in $\mathcal{L}$. Then $\pi_{k^{\prime}}(a)=\mathbf{0}$, that
is, $\pi_{k^{\prime}}\in X_{a}$ with $k^{\prime}\in C(\mathcal{L})$.
Therefore, if $0\neq X_{a}$ then there always exists $\mathbf{0}\neq x\in
C(\mathcal{L})$ such that $\pi_{x}\in X_{a}$. Let $\Gamma=\\{x\in
C(\mathcal{L})\mid\pi_{x}\in X_{a}\\}$. Then $\Gamma\neq\emptyset$ and by
hypothesis $\Gamma$ has maximal elements. Let $k\in\Gamma$ be a maximal and
$k^{\prime}$ be a complement of $k$. We claim that $X_{a}\cap X_{k}=0$.
Suppose $0\neq X_{a}\cap X_{k}=X_{a\vee k}$. By the comment above, there
exists $\mathbf{0}\neq x\in C(\mathcal{L})$ such that $\pi_{x}\in X_{a\vee
k}$. Let $x^{\prime}$ be a complement of $x$ in $\mathcal{L}$. Hence $k\leq
x^{\prime}$. Note that $a\leq k^{\prime}\wedge x^{\prime}$. Since
$\mathcal{L}$ is Rickart, $x^{\prime}\wedge k^{\prime}\in C(\mathcal{L})$ and
following the proof of Proposition 3.24, a complement for $k^{\prime}\wedge
x^{\prime}$ is $k\vee x$. It follows that $k<k\vee x$ and $k\vee x\in\Gamma$
which is a contradiction. Thus, $X_{a}\cap X_{k}=0$. Now, since $\pi_{k}\in
X_{a}$, $a\leq\pi_{k}(a)\vee\pi_{k^{\prime}}(a)=\pi_{k^{\prime}}(a)\leq
k^{\prime}$. This implies that $\pi_{k^{\prime}}(a)=a$. Let $\varphi\in
X_{a}$. Then $\varphi\pi_{k^{\prime}}(a)=\varphi(a)=\mathbf{0}$. Hence
$\varphi\pi_{k^{\prime}}\in X_{a}\cap X_{k}=0$. Thus,
$k^{\prime}\leq\ker_{\varphi}$ for all $\varphi\in X_{a}$. Then
$k^{\prime}=\bigwedge_{\varphi\in X_{a}}\ker_{\varphi}$ and so $\mathcal{L}$
is Baer by Proposition 3.60 ∎

######  Proposition 3.66.

Let $\mathcal{L}$ be a complete modular lattice such that $C(\mathcal{L})$
satisfies ACC and $\mathfrak{m}$ be a submonoid of
$\operatorname{End}_{lin}(\mathcal{L})$ containing all the projections. The
following conditions are equivalent:

  1. _(a)_

$\mathcal{L}$ is dual-$\mathfrak{m}$-Rickart.

  2. _(b)_

$\mathcal{L}$ is dual-$\mathfrak{m}$-Baer.

###### Proof.

Suppose $\mathcal{L}$ is dual-$\mathfrak{m}$-Rickart. Let
$X\subseteq\mathfrak{m}$ and $a=\bigvee_{\varphi\in X}\varphi(\mathbf{1})$.
Consider $\Gamma=\\{x\in C(\mathcal{L})\mid x\leq a\\}$. Then
$\Gamma\neq\emptyset$ and by hypothesis $\Gamma$ has maximal elements. Let
$\ell\in\Gamma$ be a maximal. Then $\ell\leq a$. Suppose that $\ell<a$. This
implies that there exists $\psi\in\mathfrak{m}$ such that
$\psi(\mathbf{1})\leq a$ and $\psi(\mathbf{1})\nleq\ell$. By Proposition 3.24,
$\ell\vee\psi(\mathbf{1})\in C(\mathcal{L})$ and
$\ell<\ell\vee\psi(\mathbf{1})\leq a$. This is a contradiction. Then
$\ell=a=\bigvee_{\varphi\in X}\varphi(\mathbf{1})$ and so $\mathcal{L}$ is
dual-$\mathfrak{m}$-Baer. ∎

##  4\. (Dual-)Baer and $\mathcal{K}$-(co)nonsingularity

In the module-theoretic context there are the following definitions which have
appeared in the literature with different names (for example see [11] and
[13]). For the convenience of the reader, we present the definitions with the
names from [5, Definition 9.4] where the definitions are given in the context
of abelian categories.

######  Definition 4.1.

Let $M$ be an $R$-module.

  * •

$M$ is $\mathcal{K}$-nonsingular if whenever $\operatorname{Ker}\varphi$ is
essential in $M$ with $\varphi\in\operatorname{End}_{R}(M)$ it follows that
$\varphi=0$.

  * •

$M$ is $\mathcal{K}$-cononsingular if whenever $\varphi(N)\neq 0$ for all
$\varphi\in\operatorname{End}_{R}(M)$ it follows that $N$ is essential in $M$.

  * •

$M$ is $\mathcal{T}$-nonsingular if for every nonzero endomorphism $\varphi$
of $M$, $\varphi(M)$ is not small in $M$.

  * •

$M$ is $\mathcal{T}$-cononsingular if for every non-small submodule $N$ of
$M$, there exists a nonzero endomorphism $\varphi$ of $M$ such that
$\varphi^{-1}(N)=M$.

Baer and dual-Baer modules have been characterized using these notions [11,
Theorem 2.12] and [13, Theorem 2.14]. Also, the counterpart in abelian
categories was presented in [5, Theorem 9.5]. As it can be seen, with the
notion of linear morphism, it is possible to carry the above definitions to
the lattice context.

######  Definition 4.2.

Let $\mathcal{L}$ be a complete modular lattice and let $\mathfrak{m}$ be a
submonoid with zero of $\operatorname{End}_{lin}(\mathcal{L})$.

  1. (1)

$\mathcal{L}$ is called _$\mathfrak{m}$ -$\mathcal{K}$-nonsingular_ if
whenever $\ker_{\varphi}$ is essential in $\mathcal{L}$ with
$\varphi\in\mathfrak{m}$ it follows that $\varphi=0$.

  2. (2)

$\mathcal{L}$ is called _$\mathfrak{m}$ -$\mathcal{T}$-nonsingular_ if
whenever $\varphi(\mathbf{1})$ is superfluous in $\mathcal{L}$ with
$\varphi\in\mathfrak{m}$ it follows that $\varphi=0$.

If the submonoid we are considering is
$\operatorname{End}_{lin}(\mathcal{L})$, we will omit the $\mathfrak{m}$.

######  Definition 4.3.

Let $\mathcal{L}$ be a complete modular lattice and let $\mathfrak{m}$ be a
submonoid with zero of $\operatorname{End}_{lin}(\mathcal{L})$.

  1. (1)

$\mathcal{L}$ is called _$\mathfrak{m}$ -$\mathcal{K}$-cononsingular_ if
whenever $\varphi(a)\neq\mathbf{0}$ for all $0\neq\varphi\in\mathfrak{m}$ it
follows that $a$ is essential in $\mathcal{L}$.

  2. (2)

$\mathcal{L}$ is called _$\mathfrak{m}$ -$\mathcal{T}$-cononsingular_ if
$\varphi(\mathbf{1})\nleq a$ for all $0\neq\varphi\in\mathfrak{m}$, then $a$
is superfluous in $\mathcal{L}$.

If the submonoid we are considering is
$\operatorname{End}_{lin}(\mathcal{L})$, we will omit the $\mathfrak{m}$.

The next result follows from the above definitions.

######  Proposition 4.4.

Let $M$ be an $R$-module. Then,

  1. (1)

$M$ is $\mathcal{K}$-nonsingular if and only if $\Lambda(M)$ is
$\mathfrak{E}_{M}$-$\mathcal{K}$-nonsingular.

  2. (2)

$M$ is $\mathcal{T}$-nonsingular if and only if $\Lambda(M)$ is
$\mathfrak{E}_{M}$-$\mathcal{T}$-nonsingular.

  3. (3)

$M$ is $\mathcal{K}$-cononsingular if and only if $\Lambda(M)$ is
$\mathfrak{E}_{M}$-$\mathcal{K}$-cononsingular.

  4. (4)

$M$ is $\mathcal{T}$-cononsingular if and only if $\Lambda(M)$ is
$\mathfrak{E}_{M}$-$\mathcal{T}$-cononsingular.

######  Definition 4.5.

Let $\mathcal{L}$ be a lattice.

  1. (1)

$\mathcal{L}$ satisfies the condition _( $C_{1}$)_ if for every
$x\in\mathcal{L}$ there exists $c\in C(\mathcal{L})$ such that $x$ is
essential in the interval $[\mathbf{0},c]$ [2, Definition 1.1].

  2. (2)

$\mathcal{L}$ satisfies the condition _( $D_{1}$)_ if for every
$x\in\mathcal{L}$ there exists $c\in C(\mathcal{L})$ with complement
$c^{\prime}$ such that $c\leq x$ and $x\wedge c^{\prime}$ is superfluous in
$\mathcal{L}$.

######  Lemma 4.6.

Let $\mathcal{L}$ be a complete modular lattice and let $\mathfrak{m}$ be a
submonoid of $\operatorname{End}_{lin}(\mathcal{L})$ containing all the
projections. If $\mathcal{L}$ satisfies ($C_{1}$) then $\mathcal{L}$ is
$\mathfrak{m}$-$\mathcal{K}$-cononsingular.

###### Proof.

Let $a\in\mathcal{L}$ such that $\varphi(a)\neq\mathbf{0}$ for all
$\varphi\in\mathfrak{m}$. By hypothesis, there exists $c\in C(\mathcal{L})$
such that $a$ is essential in $[\mathbf{0},c]$. Let $c^{\prime}$ be a
complement of $c$ in $\mathcal{L}$ and consider the projection
$\pi_{c^{\prime}}$. Then $\pi_{c^{\prime}}(a)=\mathbf{0}$. This implies that
$\pi_{c^{\prime}}=0$ and so $c^{\prime}=\mathbf{0}$. Therefore $c=\mathbf{1}$
and hence $a$ is essential in $\mathcal{L}$. ∎

######  Corollary 4.7 ([11, Lemma 2.13]).

Let $M$ be an $R$-module. If $M$ satisfies ($C_{1}$), then $M$ is
$\mathcal{K}$-cononsingular.

######  Lemma 4.8.

Let $\mathcal{L}$ be a complete modular lattice and let $\mathfrak{m}$ be a
submonoid of $\operatorname{End}_{lin}(\mathcal{L})$ containing all the
projections. If $\mathcal{L}$ satisfies ($D_{1}$) then $\mathcal{L}$ is
$\mathfrak{m}$-$\mathcal{T}$-cononsingular.

###### Proof.

Let $a\in\mathcal{L}$. By hypothesis, there exists $c\in C(\mathcal{L})$ with
complement $c^{\prime}$ such that $c\leq a$ and $a\wedge c^{\prime}$ is
superfluous in $\mathcal{L}$. Consider the projection $\pi_{c}$. Then
$\pi_{c}(1)=\pi_{c}(c)=c\leq a$. This implies that $\pi_{c}=0$ and so
$c=\mathbf{0}$. Therefore $c^{\prime}=\mathbf{1}$ and hence $a$ is superfluous
in $\mathcal{L}$. ∎

######  Corollary 4.9 ([13, Lemma 2.12]).

Let $M$ be an $R$-module. If $M$ satisfies ($D_{1}$), then $M$ is
$\mathcal{T}$-cononsingular.

######  Lemma 4.10.

Let $\mathcal{L}$ be a complete modular lattice and let $\mathfrak{m}$ be a
submonoid of $\operatorname{End}_{lin}(\mathcal{L})$ containing all the
projections. If $\mathcal{L}$ is $\mathfrak{m}$-$\mathcal{K}$-nonsingular and
satisfies ($C_{1}$) then $\mathcal{L}$ is $\mathfrak{m}$-Baer.

###### Proof.

Let $\\{\varphi_{i}\\}_{I}\subseteq\mathfrak{m}$ and let
$k=\bigwedge_{I}\ker_{\varphi_{i}}$. By hypothesis there exists an element
$c\in\mathcal{L}$ with complement $c^{\prime}$ such that $k$ is essential in
$[\mathbf{0},c]$. This implies that $\ker_{\varphi_{i}}\wedge c$ is also
essential in $[\mathbf{0},c]$ for all $i\in I$. Therefore
$\ker_{\varphi_{i}\pi_{c}}=(\ker_{\varphi_{i}}\wedge c)\vee c^{\prime}$ is
essential in $\mathcal{L}$ for all $i\in I$. Since $\mathcal{L}$ is
$\mathfrak{m}$-$\mathcal{K}$-nonsingular, $\varphi_{i}\pi_{c}=0$ for all $i\in
I$. Hence $c\leq\ker_{\varphi_{i}}$ for all $i\in I$, that is, $c\leq k$.
Thus, $k=c$. This proves that $\mathcal{L}$ is $\mathfrak{m}$-Baer. ∎

######  Corollary 4.11 ([11, Lemma 2.14]).

Let $M$ be an $R$-module. If $M$ is $\mathcal{K}$-nonsingular and satisfies
($C_{1}$), then $M$ is Baer.

######  Lemma 4.12.

Let $\mathcal{L}$ be a complete modular lattice and let $\mathfrak{m}$ be a
submonoid of $\operatorname{End}_{lin}(\mathcal{L})$ containing all the
projections. If $\mathcal{L}$ is $\mathfrak{m}$-$\mathcal{T}$-nonsingular and
satisfies ($D_{1}$) then $\mathcal{L}$ is dual-$\mathfrak{m}$-Baer.

###### Proof.

Let $\\{\varphi_{i}\\}_{I}\subseteq\mathfrak{m}$ and let
$a=\bigvee_{I}\varphi_{i}(\mathbf{1})$. By hypothesis there exists an element
$c\in\mathcal{L}$ with complement $c^{\prime}$ such that $c\leq a$ and
$a\wedge c^{\prime}$ is superfluous in $\mathcal{L}$. This implies that
$\pi_{c^{\prime}}\varphi_{i}(\mathbf{1})=(\varphi_{i}(\mathbf{1})\vee c)\wedge
c^{\prime}$ is also superfluous in $\mathcal{L}$ for all $i\in I$. Since
$\mathcal{L}$ is $\mathfrak{m}$-$\mathcal{T}$-nonsingular,
$\pi_{c^{\prime}}\varphi_{i}=0$ for all $i\in I$. Hence
$\varphi_{i}(\mathbf{1})\leq c$ for all $i\in I$, that is, $a\leq c$. Thus,
$a=c$. This proves that $\mathcal{L}$ is dual-Baer. ∎

######  Corollary 4.13 ([13, Lemma 2.11]).

Let $M$ be an $R$-module. If $M$ is $\mathcal{T}$-nonsingular and satisfies
($D_{1}$), then $M$ is dual-Baer.

######  Lemma 4.14.

Let $\mathcal{L}$ be a complete modular lattice and let $\mathfrak{m}$ be a
submonoid of $\operatorname{End}_{lin}(\mathcal{L})$. If $\mathcal{L}$ is
$\mathfrak{m}$-Rickart then $\mathcal{L}$ is
$\mathfrak{m}$-$\mathcal{K}$-nonsingular.

###### Proof.

Let $\varphi\in\mathfrak{m}$ such that $\ker_{\varphi}$ is essential in
$\mathcal{L}$. Since $\mathcal{L}$ is $\mathfrak{m}$-Rickart, $\ker_{\varphi}$
has a complement. This implies that $\ker_{\varphi}=\mathbf{1}$ and hence
$\varphi=0$. ∎

######  Corollary 4.15 ([11, Lemma 2.15]).

Let $M$ be an $R$-module. If $M$ is Baer, then $M$ is
$\mathcal{K}$-nonsingular.

######  Lemma 4.16.

Let $\mathcal{L}$ be a complete modular lattice and let $\mathfrak{m}$ be a
submonoid of $\operatorname{End}_{lin}(\mathcal{L})$. If $\mathcal{L}$ is
dual-$\mathfrak{m}$-Rickart then $\mathcal{L}$ is
$\mathfrak{m}$-$\mathcal{T}$-nonsingular.

###### Proof.

Let $\varphi\in\mathfrak{m}$ such that $\varphi(\mathbf{1})$ is superfluous in
$\mathcal{L}$. Since $\mathcal{L}$ is dual-Rickart, $\varphi(\mathbf{1})$ has
a complement. This implies that $\varphi(\mathbf{1})=\mathbf{0}$ and hence
$\varphi=0$. ∎

######  Corollary 4.17 ([13, Corollary 2.3]).

Let $M$ be an $R$-module. If $M$ is dual-Baer, then $M$ is
$\mathcal{T}$-nonsingular.

######  Lemma 4.18.

Let $\mathcal{L}$ be a complete modular lattice and let $\mathfrak{m}$ be a
submonoid of $\operatorname{End}_{lin}(\mathcal{L})$ containing all the
projections. If $\mathcal{L}$ is $\mathfrak{m}$-Baer and
$\mathfrak{m}$-$\mathcal{K}$-cononsingular then $\mathcal{L}$ satisfies
($C_{1}$).

###### Proof.

Let $a\in\mathcal{L}$. By Proposition 3.60, there exists an idempotent
$\varepsilon\in\mathfrak{m}$ such that $\varepsilon(a)=\mathbf{0}$. Then
$a\leq\ker_{\varepsilon}$ and $\ker_{\varepsilon}\in C(\mathcal{L})$. We claim
that $a$ is essential in $[\mathbf{0},\ker_{\varepsilon}]$. Let $k^{\prime}$
be a complement of $\ker_{\varepsilon}$. Let $\varphi\in\mathfrak{m}$ such
that $\varphi(a\vee k^{\prime})=\mathbf{0}$. Then
$\varphi(k^{\prime})=\mathbf{0}$ and $\varphi(a)=\mathbf{0}$. Since
$\varphi(a)=\mathbf{0}$, $\varphi=\varphi\varepsilon$. Therefore,

$\varphi(\mathbf{1})=\varphi(\ker_{\varepsilon}\vee
k^{\prime})=\varphi(\ker_{\varepsilon})\vee\varphi(k^{\prime})=\varphi\varepsilon(\ker_{\varepsilon})\vee\varphi(k^{\prime})=\mathbf{0}.$

Thus, $\varphi=0$. By hypothesis, $a\vee k^{\prime}$ is essential in
$\mathcal{L}$. Now, let $\mathbf{0}\leq b\leq\ker_{\varepsilon}$ such that
$a\wedge b=\mathbf{0}$. Then $(a\vee k^{\prime})\wedge b\leq(a\vee
k^{\prime})\wedge\ker_{\varepsilon}=a$. Hence $(a\vee k^{\prime})\wedge b\leq
a\wedge b=\mathbf{0}$. This implies that $b=\mathbf{0}$ proving the claim. ∎

######  Corollary 4.19 ([11, Lemma 2.16]).

Let $M$ be an $R$-module. If $M$ is Baer and $\mathcal{K}$-cononsingular, then
$M$ satisfies ($C_{1}$).

######  Lemma 4.20.

Let $\mathcal{L}$ be a complete modular lattice and let $\mathfrak{m}$ be a
submonoid of $\operatorname{End}_{lin}(\mathcal{L})$ containing all the
projections. If $\mathcal{L}$ is dual-$\mathfrak{m}$-Baer and
$\mathfrak{m}$-$\mathcal{T}$-cononsingular then $\mathcal{L}$ satisfies
($D_{1}$).

###### Proof.

Let $a\in\mathcal{L}$. By Proposition 3.62, there exists an idempotent
$\varepsilon\in\mathfrak{m}$ such that $\varepsilon(\mathbf{1})\leq a$ and
$\varepsilon(\mathbf{1})\in C(\mathcal{L})$. We have to show that
$a\wedge\ker_{\varepsilon}$ is superfluous in $\mathcal{L}$. Let
$\varphi\in\mathfrak{m}$ such that $\varphi(\mathbf{1})\leq
a\wedge\ker_{\varepsilon}$. Then $\varphi=\varepsilon\varphi$. Therefore
$\varphi(\mathbf{1})=\varepsilon\varphi(\mathbf{1})\leq\varepsilon(a\wedge\ker_{\varepsilon})=\mathbf{0}$.
Thus $\varphi=0$. Since $\mathcal{L}$ is $\mathcal{T}$-cononsingular,
$a\wedge\ker_{\varepsilon}$ is superfluous in $\mathcal{L}$. ∎

######  Corollary 4.21 ([13, Proposition 2.13]).

Let $M$ be an $R$-module. If $M$ is dual-Baer and $\mathcal{T}$-cononsingular,
then $M$ satisfies ($D_{1}$).

######  Corollary 4.22.

The following conditions are equivalent for a complete modular lattice
$\mathcal{L}$ and a submonoid $\mathfrak{m}$ of
$\operatorname{End}_{lin}(\mathcal{L})$ containing all the projections:

  1. _(a)_

$\mathcal{L}$ is $\mathfrak{m}$-$\mathcal{K}$-nonsingular and satisfies
($C_{1}$).

  2. _(b)_

$\mathcal{L}$ is $\mathfrak{m}$-Baer and
$\mathfrak{m}$-$\mathcal{K}$-cononsingular.

######  Corollary 4.23.

The following conditions are equivalent for a complete modular lattice
$\mathcal{L}$ and a submonoid $\mathfrak{m}$ of
$\operatorname{End}_{lin}(\mathcal{L})$ containing all the projections:

  1. _(a)_

$\mathcal{L}$ is $\mathfrak{m}$-$\mathcal{T}$-nonsingular and satisfies
($D_{1}$).

  2. _(b)_

$\mathcal{L}$ is dual-$\mathfrak{m}$-Baer and
$\mathfrak{m}$-$\mathcal{T}$-cononsingular.

##  5\. Decomposition and direct products of Rickart lattices

######  Proposition 5.1.

Let $\mathcal{L}$ be a complete modular lattice and $\\{a_{1},...,a_{n}\\}$ be
a independent family of fully invariant elements of $\mathcal{L}$ such that
$\mathbf{1}=\bigvee_{i=1}^{n}a_{i}$. Then, $\mathcal{L}$ is Rickart if and
only if $[\mathbf{0},a_{i}]$ are Rickart lattices for all $1\leq i\leq n$.

###### Proof.

$\Rightarrow$ It follows from Proposition 3.13.

$\Leftarrow$ Suppose $[\mathbf{0},a_{i}]$ is a Rickart lattice for every
$1\leq i\leq n$. Let $\varphi:\mathcal{L}\to\mathcal{L}$ be a linear morphism.
For any $1\leq i\leq n$, $\varphi(a_{i})\leq a_{i}$. This implies that
$\varphi|:[\mathbf{0},a_{i}]\to[\mathbf{0},a_{i}]$ is a linear morphism.
Therefore, $\ker_{\varphi}\wedge a_{i}$ is a complement in
$[\mathbf{0},a_{i}]$. On the other hand,

$\varphi(\pi_{a_{i}}(\ker_{\varphi}))=\varphi\left(\left(\ker_{\varphi}\vee\bigvee_{i\neq
j}a_{j}\right)\wedge
a_{i}\right)\leq\varphi\left(\ker_{\varphi}\vee\bigvee_{i\neq
j}a_{j}\right)=\varphi\left(\bigvee_{i\neq j}a_{j}\right).$

Hence,

$\varphi(\pi_{a_{i}}(\ker_{\varphi}))\leq\varphi\left(\bigvee_{i\neq
j}a_{j}\right)\wedge\varphi(a_{i})\leq\left(\bigvee_{i\neq
j}a_{j}\right)\wedge a_{i}=\mathbf{0}.$

Thus, $\pi_{a_{i}}(\ker_{\varphi})\leq\ker_{\varphi}\wedge a_{i}$ for all
$1\leq i\leq n$. This implies that
$\ker_{\varphi}=\bigvee_{i=1}^{n}\ker_{\varphi}\wedge a_{i}$. Then,
$\ker_{\varphi}$ is a complement in
$[\mathbf{0},\bigvee_{i=1}^{n}a_{i}]=\mathcal{L}$. ∎

######  Lemma 5.2 ([4, Proposition 6.8]).

Let $\mathcal{L}$ be a complete Artinian or Noetherian lattice. Then there
exists an independent family $\\{a_{i}\\}_{i=1}^{n}$ of elements of
$\mathcal{L}$ such that $\mathbf{1}=\bigvee_{i=1}^{n}a_{i}$ and
$[\mathbf{0},a_{i}]$ is indecomposable for all $1\leq i\leq n$.

######  Definition 5.3.

Let $\mathcal{L}$ be a complete lattice. The _Socle_ and the _Radical_ of
$\mathcal{L}$ are the elements defined as follows:

$\operatorname{Soc}(\mathcal{L})=\bigvee\\{x\in\mathcal{L}\mid x\text{ is an
atom}\\},$ $\operatorname{Rad}(\mathcal{L})=\bigwedge\\{x\in\mathcal{L}\mid
x\text{ is a coatom}\\}.$

######  Lemma 5.4.

Let $\mathcal{L}$ be a complete modular lattice suppose
$Soc(\mathcal{L})\neq\mathbf{0}$ and
$\operatorname{Rad}(\mathcal{L})\neq\mathbf{1}$. Then $\mathcal{L}$ is
indecomposable Rickart if and only if $\mathcal{L}\cong 2$ where $2$ is the
complete lattice of two elements.

###### Proof.

By hypothesis, there exist $x,y\in\mathcal{L}$ such that $x$ is a coatom and
$y$ is an atom. Then it can be defined the linear morphism
$\varphi:\mathcal{L}\to\mathcal{L}$ as

$\varphi(a)=\begin{cases}\mathbf{0}&\text{if }a\leq x\\\ y&\text{if
}a=\mathbf{1}\end{cases}$

Hence, $\ker_{\varphi}=x$. By hypothesis $x$ has to be a complement but
$\mathcal{L}$ is indecomposable. Thus $x=\mathbf{0}$ and $y=\mathbf{1}$, that
is, $\mathcal{L}\cong 2$. The converse is obvious. ∎

######  Proposition 5.5.

Let $\mathcal{L}$ be a complete modular lattice. Then $\mathcal{L}$ is Rickart
of finite length if and only if there exists an independent family
$\\{a_{i}\\}_{i=1}^{n}$ of elements of $\mathcal{L}$ such that
$\mathbf{1}=\bigvee_{i=1}^{n}a_{i}$ and $[\mathbf{0},a_{i}]\cong 2$ for all
$1\leq i\leq n$.

###### Proof.

$\Rightarrow$ By Lemma 5.2, there exists an independent family
$\\{a_{i}\\}_{i=1}^{n}$ of elements of $\mathcal{L}$ such that
$\mathbf{1}=\bigvee_{i=1}^{n}a_{i}$ and each interval $[\mathbf{0},a_{i}]$ is
indecomposable. By Proposition 3.13 and Lemma 5.4, $[\mathbf{0},a_{i}]\cong 2$
for all $1\leq i\leq n$.

$\Leftarrow$ It is clear that $\mathcal{L}$ has finite length. We claim that
$\mathcal{L}$ is complemented. We prove the claim by induction in the length
of the independent family $\\{a_{i}\\}_{i=1}^{n}$. If $n=1$, then
$\mathcal{L}=[\mathbf{0},a_{1}]\cong 2$ which is complemented. Now, suppose
that the interval $[\mathbf{0},a_{1}\vee\cdots\vee a_{n-1}]$ is complemented
and $\bigvee_{i=1}^{n}a_{i}=\mathbf{1}$. Let $x\in\mathcal{L}$. Set
$y=a_{1}\vee\cdots\vee a_{n-1}$ and consider $x\vee y$. We have that $y\leq
x\vee y\leq\mathbf{1}$ and by hypothesis
$[y,\mathbf{1}]\cong[\mathbf{0},a_{n}]\cong 2$. This implies that $x\leq y$ or
$x\vee y=\mathbf{1}$. If $x\leq y$ then $x$ has a complement by induction
hypothesis. Suppose $x\vee y=\mathbf{1}$. Since $x\wedge y\leq y$ then there
exists $\mathbf{0}\leq b\leq y$ such that $b\wedge(x\wedge y)=\mathbf{0}$ and
$b\vee(x\wedge y)=y$. Therefore $b\wedge x=b\wedge y\wedge x=\mathbf{0}$ and
$\mathbf{1}=x\vee y=x\vee(b\vee(x\wedge y))=x\vee((b\vee x)\wedge b)=x\vee b$.
Thus, $x$ is a complement proving the claim. ∎

The last proofs of Lemma 5.4 and Proposition 5.5 can be applied to modules in
particular cases as follows.

######  Corollary 5.6.

Suppose that

  1. (1)

$R$ is a left local ring and $M$ is an $R$-module with
$\operatorname{Soc}(M)\neq 0$ and $\operatorname{Rad}(M)\neq M$, or

  2. (2)

$M$ is a Kasch $R$-module and $\operatorname{Rad}(M)\neq M$, or

  3. (3)

$M$ contains an $M$-generated simple module.

Then, $M$ is indecomposable Rickart if and only if $M$ is simple.

######  Corollary 5.7.

Suppose that

  1. (1)

$R$ is a local ring and $M$ is an $R$-module of finite length, or

  2. (2)

$M$ is a Kasch $R$-module of finite length.

Then, $M$ is Rickart if and only if $M$ is semisimple.

###### Proof.

Since $M$ has finite length, $M=M_{1}\oplus\cdots\oplus M_{\ell}$ for some
$\ell>0$ with each $M_{i}$ an indecomposable Rickart module of finite length.
If $R$ is local, the result is clear. Suppose $M$ is a Kasch module and let
$N$ be a maximal submodule of $M_{i}$. Then, there exists an embedding
$M_{i}/N\to M$. It follows from [8, Theorem 2.6] (see also Proposition 5.12
below) that $M_{i}$ is $M$-Rickart, hence $N$ is direct summand of $M_{i}$.
Thus, $M_{i}$ is simple. ∎

######  Corollary 5.8.

Let $\mathcal{L}$ be a finite modular lattice and $\\{a_{1},...,a_{n}\\}$ be a
independent family of elements of $\mathcal{L}$ such that
$\mathbf{1}=\bigvee_{i=1}^{n}a_{i}$. The following conditions are equivalent:

  1. _(a)_

$\mathcal{L}$ is Rickart.

  2. _(b)_

$[\mathbf{0},a_{i}]$ are Rickart lattices for all $1\leq i\leq n$.

###### Proof.

(a)$\Rightarrow$(b) is clear. For the converse, since each interval
$[\mathbf{0},a_{i}]$ is a Rickart lattice of finite length, there exists an
independent family $\\{b_{ij}\\}_{j=1}^{\ell_{i}}$ in the interval
$[\mathbf{0},a_{i}]$ such that $\bigvee_{j=1}^{\ell_{i}}b_{ij}=a_{i}$ and
$[\mathbf{0},b_{ij}]\cong 2$ for all $1\leq j\leq\ell_{i}$ by Proposition 5.5.
Then, the family $\\{b_{ij}\mid 1\leq i\leq n,\;1\leq j\leq\ell_{i}\\}$ is an
independent family of elements of $\mathcal{L}$ such that
$\bigvee\\{b_{ij}\mid 1\leq i\leq n,\;1\leq j\leq\ell_{i}\\}=\mathbf{1}$ and
$[\mathbf{0},b_{ij}]\cong 2$. It follows from Proposition 5.5 that
$\mathcal{L}$ is Rickart. ∎

In general there might be indecomposable Artinian non Noetherian Rickart
lattices as the following example shows.

######  Example 5.9.

Consider the lattice $\mathcal{L}$ given by the following diagram.

$\textstyle{\mathbf{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\vdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{c_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{c_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{a\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{b\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathbf{0}}$

Then $\mathcal{L}$ is Artinian and indecomposable. Note that any nonzero
linear morphism $\varphi:\mathcal{L}\to\mathcal{L}$ must be an isomorphism.
Thus, $\mathcal{L}$ is Rickart.

Given a family of complete modular lattices $\\{\mathcal{L}_{i}\\}_{I}$, the
Cartesian product $\prod_{I}\mathcal{L}_{i}$ is a complete modular lattice
with supremum and infimimum pointwise. The cartesian product is defined as the
set of functions

$\prod_{I}\mathcal{L}_{i}=\left\\{f:I\to\bigcup_{I}\mathcal{L}_{i}\mid
f(i)\in\mathcal{L}_{i}\right\\}$

Then, $f\leq g$ if and only if $f(i)\leq g(i)$ for all $i\in I$, $(f\wedge
g)(i)=f(i)\wedge g(i)$ and $(f\vee g)(i)=f(i)\vee g(i)$. For every $i\in I$,
we have the canonical projections
$\pi_{i}:\prod_{I}\mathcal{L}_{i}\to\mathcal{L}_{i}$ defined as
$\pi_{i}(f)=f(i)$.

######  Proposition 5.10.

Let $\\{\mathcal{L}_{i}\\}_{I}$ be a family of complete modular lattices. Then
the projections $\pi_{i}:\prod_{I}\mathcal{L}_{i}\to\mathcal{L}_{i}$ are
linear morphisms.

###### Proof.

Fix $j\in I$, and consider $\pi_{j}$. Let $k_{j}\in\prod_{I}\mathcal{L}_{i}$
be the function defined as follows:

$k_{j}(i)=\begin{cases}\mathbf{0}\;\text{ if }\;i=j\\\ \mathbf{1}\;\text{ if
}\;i\neq j\end{cases}$

Then $\pi_{j}(k_{j})=k_{j}(j)=\mathbf{0}\in\mathcal{L}_{j}$. Also,

$\pi_{j}(f\vee k_{j})=(f\vee k_{j})(j)=f(j)\vee
k_{j}(j)=f(j)\vee\mathbf{0}=f(j)=\pi_{j}(f).$

For $a\in\mathcal{L}_{j}$, define

$f_{a}(i)=\begin{cases}a\;\text{ if }\;i=j\\\ \mathbf{1}\;\text{ if }\;i\neq
j\end{cases}$

Let
$\rho:\mathcal{L}_{j}\to[k_{j},\mathbf{1}]\subseteq\prod_{I}\mathcal{L}_{i}$
given by $\rho(a)=f_{a}$. It is clear that $\rho$ is a lattice morphism and
$k_{j}\leq f_{a}$ for all $a\in\mathcal{L}_{j}$. Consider $k_{j}\leq
g\leq\mathbf{1}$. Then,

$\rho\pi_{j}(g)=\rho(g(j))=f_{g(j)}$

and

$f_{g(j)}(i)=\begin{cases}g(j)\;\text{ if }\;i=j\\\ \mathbf{1}\;\text{ if
}\;i\neq j\end{cases}=g(i)$

Thus, $\rho\pi_{j}(g)=g$. Now consider $a\in\mathcal{L}_{j}$. Then,

$\pi_{j}\rho(a)=\pi_{j}(f_{a})=f_{a}(j)=a.$

Therefore $\pi_{j}$ induces an isomorphism of lattices
$\overline{\pi_{j}}:[k_{j},\mathbf{1}]\to\mathcal{L}_{j}$. Hence $\pi_{j}$ is
a linear morphism with kernel $\ker_{\pi_{j}}=k_{j}$. ∎

######  Definition 5.11.

Let $\mathcal{L}$ and $\mathcal{M}$ be lattices. It is said that $\mathcal{L}$
is $\mathcal{M}$-Rickart if $\ker_{\varphi}$ is complemented in $\mathcal{L}$
for every linear morphism $\varphi:\mathcal{L}\to\mathcal{M}$.

######  Proposition 5.12.

Let $\mathcal{L}$ and $\mathcal{M}$ be complete modular lattices and suppose
$\mathcal{L}$ is $\mathcal{M}$-Rickart. If $a\in\mathcal{L}$ is a complement
and $x\in\mathcal{M}$, then $[\mathbf{0},a]$ is $[\mathbf{0},x]$-Rickart.

###### Proof.

It follows from Proposition 2.11 and the fact that the inclusion
$\iota:[\mathbf{0},x]\to\mathcal{M}$ is a linear morphism. ∎

######  Proposition 5.13.

Let $\mathcal{L}$ and $\mathcal{M}$ be complete modular lattices and
$\\{a_{i}\\}_{I}$ be a independent family of elements of $\mathcal{M}$ with
$\mathbf{1}=\bigvee_{i\in I}a_{i}$.

  1. (1)

If $\mathcal{L}$ has CIP, $\mathcal{L}$ is $\mathcal{M}$-Rickart if and only
if $\mathcal{L}$ is $[\mathbf{0},a_{i}]$-Rickart for all $i\in I$ with
$I=\\{1,...,n\\}$.

  2. (2)

If $\mathcal{L}$ is upper continuous and has SCIP, $\mathcal{L}$ is
$\mathcal{M}$-Rickart if and only if $\mathcal{L}$ is
$[\mathbf{0},a_{i}]$-Rickart for all $i\in I$ for any index set $I$.

###### Proof.

(1)$\Rightarrow$ It follows from Proposition 3.13.

$\Leftarrow$ Let $\varphi:\mathcal{L}\to\mathcal{M}$ be a linear morphism.
Then, we have the linear morphisms
$\pi_{a_{i}}\varphi:\mathcal{L}\to[\mathbf{0},a_{i}]$. By hypothesis,
$\ker_{\pi_{a_{i}}\varphi}$ is complemented in $\mathcal{L}$ for every $i\in
I$. We claim that $\ker_{\varphi}=\bigwedge_{I}\ker_{\pi_{a_{i}}\varphi}$. It
is clear that $\ker_{\varphi}\leq\bigwedge_{I}\ker_{\pi_{a_{i}}\varphi}$.
Suppose $I$ is finite. It follows from Proposition 2.14.(1) that

$\varphi\left(\bigwedge_{I}\ker_{\pi_{a_{i}}\varphi}\right)\leq\bigvee_{j\in
I}\pi_{j}\left(\varphi\left(\bigwedge_{I}\ker_{\pi_{a_{i}}\varphi}\right)\right)\leq\bigvee_{j\in
I}\pi_{a_{j}}\varphi(\ker_{\pi_{a_{j}}\varphi})=0.$

This proves the claim. Now, since $\mathcal{L}$ has the CIP, $\ker_{\varphi}$
is complemented in $\mathcal{L}$.

(2) The proof is similar using Proposition 2.14.(2). ∎

######  Proposition 5.14.

Let $\\{\mathcal{L}_{i}\\}_{I}$ be a family of complete modular lattices. If
$\prod_{I}\mathcal{L}_{i}$ is a Rickart lattice, then $\mathcal{L}_{i}$ is
$\mathcal{L}_{j}$-Rickart for all $i,j\in I$. If $I$ is finite, then the
converse is true.

###### Proof.

Fix $j,k\in I$. Let $\varphi:\mathcal{L}_{k}\to\mathcal{L}_{j}$ be a linear
morphism. Define $\psi:\prod_{I}\mathcal{L}_{i}\to\prod_{I}\mathcal{L}_{i}$ as

$\psi(f)(i)=\begin{cases}\varphi(f(k))\;\text{ if }\;i=j\\\ \mathbf{0}\;\text{
if }\;i\neq j.\end{cases}$

It is not difficult to see that $\psi$ is a linear morphism with

$\ker_{\psi}(i)=\begin{cases}\ker_{\varphi}\;\text{ if }\;i=k\\\
\mathbf{1}\;\text{ if }\;i\neq k.\end{cases}$

By hypothesis, $\ker_{\psi}$ has a complement $f$ in
$\prod_{I}\mathcal{L}_{i}$. Then $\mathbf{0}=(\ker_{\psi}\wedge
f)(k)=\ker_{\psi}(k)\wedge f(k)=\ker_{\varphi}\wedge f(k)$ and
$\mathbf{1}=(\ker_{\psi}\vee f)(k)=\ker_{\psi}(k)\vee f(k)=\ker_{\varphi}\vee
f(k)$ in $\mathcal{L}_{k}$. Thus, $\ker_{\varphi}$ is a complement.

Now assume $I=\\{1,...,n\\}$. Suppose that $\mathcal{L}_{i}$ is an
$\mathcal{L}_{j}$-Rickart lattice for all $1\leq i,j\leq n$. Let
$\varphi:\prod_{i=1}^{n}\mathcal{L}_{i}\to\prod_{i=1}^{n}\mathcal{L}_{i}$ be a
linear morphism. For each $1\leq j\leq n$, consider the inclusions
$\iota_{j}:\mathcal{L}_{j}\to\prod_{i=1}^{n}\mathcal{L}_{i}$ given by

$\iota_{j}(a)(i)=\begin{cases}a\;\text{ if }\;i=j\\\ \mathbf{0}\;\text{ if
}\;i\neq j.\end{cases}$

Then we have a linear morphism
$\pi_{i}\varphi\iota_{j}:\mathcal{L}_{j}\to\mathcal{L}_{i}$ for all $1\leq
i,j\leq n$. Note that if $(a_{1},...,a_{n})\in\prod_{i=1}^{n}\mathcal{L}_{i}$
then $(a_{1},...,a_{n})=\bigvee_{i=1}^{n}\iota_{i}(a_{i})$. Since $\varphi$ is
a linear morphism
$\varphi(a_{1},...,a_{n})=\varphi(\bigvee_{i=1}^{n}\iota_{i}(a_{i}))=\bigvee_{i=1}^{n}\varphi\iota_{i}(a_{i})$
by [1, Proposition 1.3]. Hence

$\varphi(a_{1},...,a_{n})=\left(\bigvee_{i=1}^{n}\pi_{1}\varphi\iota_{i}(a_{i}),...,\bigvee_{i=1}^{n}\pi_{n}\varphi\iota_{i}(a_{i})\right).$

Thus, $\varphi(a_{1},...,a_{n})=\mathbf{0}$ if and only if
$\pi_{j}\varphi\iota_{i}(a_{i})=\mathbf{0}$ for all $1\leq i,j\leq n$. Let
$k_{ij}=\ker_{\pi_{j}\varphi\iota_{i}}$ and let
$k_{i}=\bigwedge_{j=1}^{n}=k_{ij}$ for $1\leq i,j\leq n$. It is clear that
$\varphi(k_{1},...,k_{n})=\mathbf{0}$. Now, if
$\varphi(a_{1},...,a_{n})=\mathbf{0}$, then
$a_{i}\leq\bigwedge_{j=1}^{n}=k_{ij}=k_{i}$. Thus,
$\ker_{\varphi}=(k_{1},...,k_{n})$. By Proposition 3.24, $\mathcal{L}_{i}$ has
the CIP for all $1\leq i\leq n$. Therefore each $k_{i}$ has a complement
$x_{i}$ in $\mathcal{L}_{i}$. Thus $(x_{1},...,x_{n})$ is a complement for
$\ker_{\varphi}$ in $\prod_{i=1}^{n}\mathcal{L}_{i}$. ∎

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