On the lattice of conatural classes of linear modular lattices

Abstract

The collection of all cohereditary classes of modules over a ring R is a pseudocomplemented complete big lattice. The elements of its skeleton are the conatural classes of R-modules. In this paper we extend some results about cohereditary classes in R-Mod to the category $\mathcal {L_{M}}$ of linear modular lattices, which has as objects all complete modular lattices and as morphisms all linear morphisms. We introduce the big lattice of conatural classes in $\mathcal {L_{M}}$, and we obtain some results about it, paralleling the case of R-Mod and arriving at its being boolean. Finally, we prove some closure properties of conatural classes in $\mathcal {L_{M}}$.

1 Introduction

T. Albu and M. Iosif in [1] introduced the notion of a linear lattice morphism between two bounded modular lattices (which we shall define in the next section). Furthermore, they showed that the collection of all bounded modular lattices together with the collection of linear morphisms form a category, denoted by $\mathcal {L_{M}}$ and called the category of linear modular lattices. Write R-Mod for the class of all left modules over an associative ring R with 1. In [2, Example 0.2] the authors show that any morphism between two R-modules induces a linear morphism between their respective lattices of submodules; this in turn gives a functor ${\mathcal {F}}:R\text {-Mod} \longrightarrow \mathcal {L_{M}}$. This fact led us to study which primary results about closure properties for certain classes of modules in R-Mod can be extended to the corresponding classes of modular lattices in $\mathcal {L_M}$.

Of course, every complete lattice is bounded. In this paper we consider only complete lattices. For us, then, the objects of $\mathcal {L_M}$ are the complete modular lattices.

Recall that, in a lattice L with least element 0, for $a\in L$, an element b of L is called a (strong) pseudocomplement in L of a if and only if b is a maximal (resp., the largest) element of L with the property $a\wedge b=0$. Observe that any strong pseudocomplement in L of a is a pseudocomplement in L of a, and in fact, is the unique pseudocomplement in L of a. A lattice with 0 is said to be (strongly) pseudocomplemented if and only if each one of its elements has a (resp., strong) pseudocomplement in L. Moreover, recall that every lattice is a set, or at least cardinable (that is, bijectable with some set). A class which behaves like a lattice but is not necessarily cardinable is called a big lattice.

Following [4, Definition 13], let us say that a class ${\mathcal {C}}$ in R-Mod is a cohereditary class if and only if it is closed under homomorphic images (or equivalently, under isomorphisms and quotients). In that paper, A. Alvarado, H. Rincón and J. Ríos established that the collection of all non-empty cohereditary classes in R-Mod is a strongly pseudocomplemented complete big lattice, which they denoted by $R\text {-}quot$. According to [4, Definition 11], the skeleton of a pseudocomplemented (big) lattice L is the class of all pseudocomplements in L. With this in mind, the authors denoted as $R\text {-}conat$ the skeleton of $R\text {-}quot$, and named its elements conatural classes of R-modules. Moreover, in [3] they continue with the description of R-conat, and show some interesting closure properties.

The paper is organized as follows: Sect. 2 gives some preliminary concepts and definitions about the category $\mathcal {L_{M}}$ of all complete modular lattices. Section 3 extends to $\mathcal {L_{M}}$ the concept of cohereditary classes of R-modules for a ring R, and shows that one obtains a strongly pseudocomplemented big lattice. In Sect. 4 we give the lattice-theoretic counterpart of R-conat and we prove the extension of some fundamental results to the category $\mathcal {L_M}$. We end the section by giving some results on closure properties.

Let us say a word about the motivations behind, and the overall significance we see in, this paper. In module theory over a ring, there are several associated lattices whose properties are related to those of the ring. Among these lattices are lattices of module classes defined by closure properties. A module class is a torsion class when it is closed under quotients, extensions, and direct sums of its objects. A module class closed under submodules, injective hulls, and direct sums is called a natural class. The lattice of natural classes is denoted as $R\textit{-}nat$. Among other examples, we have the lattice of Wisbauer classes and the big lattice of Serre classes. The study of these lattices and of the relationships between them has been fruitful in investigating aspects of the underlying ring and has generated new concepts and techniques. Also, several research groups around the world have initiated the study of the $\mathcal {L_M}$ category.

Guided by the results for R-modules, analogous concepts about $\mathcal {L_M}$ can be explored. In particular, in previous works we have studied the big lattice of preradicals in $\mathcal {L_M}$ as well as the big lattice of natural classes in $\mathcal {L_M}$, and in this paper we introduce the big lattice of conatural classes in $\mathcal {L_M}$. Possible future avenues of research include trying to characterize conatural classes in terms of closure properties, or to describe atoms in this big lattice. One might also study some version in $\mathcal {L_M}$ of such lattices as that of hereditary torsion theories. The situation is sufficiently mature to study the consequences of comparisons between these new lattices. We should point out that the situations in R-Mod and in $\mathcal {L_M}$, although in some respects similar, in others turn out to be strikingly different. In our opinion, the situation for $\mathcal {L_M}$ is interesting in itself, and because it suggests aspects to study in R-Mod. Finally, it makes sense to ask the question of what categories admit of intuitive extensions of the notions of natural or conatural classes of objects, or maybe of preradicals.

Several results in this paper are dual to the corresponding ones in [6], but we prove them here for the reader’s convenience.

2 Preliminaries

We denote by ${\mathcal {L}}$ the class of all complete lattices. As usual, we denote by $0_{L}$ and by $1_{L}$ the least and the greatest elements of L, respectively. For a lattice L and elements $a\le b$ in L, we write

$$\begin{aligned} (b/a)=\{x\in L \mid a\le x \le b \}. \end{aligned}$$

An initial interval (resp. a quotient interval) of (b/a) is an interval (c/a) (resp. (b/c)) for some $c\in b/a$. Note that $(c/0_{L})$ (resp. $(1_{L}/c)$) is an initial (resp. quotient) interval of $L=1_{L}/0_{L}$ for any $c\in L$.

Throughout the work, when there is no ambiguity, we will refer to the lattice with only one element as 0.

The concept of a linear morphism summons the property of having a kernel $\ker (\eta )$ for every morphism $\eta :M\longrightarrow N$ in R-Mod, so that the First Isomorphism Theorem holds; this is:

$$\begin{aligned} M/\ker (\eta ) \cong \textrm{Im} \, \eta . \end{aligned}$$

Definition 2.1

[1, Definition 1.1]. Let $L,L'$ be any bounded lattices. Let us denote by $0_{L}$ and $1_{L}$ the least and the greatest elements of L and by $0_{L'}$ and $1_{L'}$ the corresponding elements of $L'$. A map $f:L\longrightarrow L'$ between the lattices L and $L'$ is called a linear morphism if there exist $n(f)\in L$, called a kernel of f, and $a'\in L'$ such that the following conditions are satisfied:

• a) $f(x)=f(x\vee n(f))$, $\forall x\in L$.

• b) f induces a lattice isomorphism ${\bar{f}}:\Big (1_{L}/n(f)\Big )\overset{\approx }{\longrightarrow }\ \Big (a'/0_{L'}\Big )$, such that

  $$\begin{aligned} {\bar{f}}(x)=f(x), \forall x\in (1_{L}/n(f)). \end{aligned}$$

Note that, given any linear morphism $f:L\longrightarrow L'$, one has that $f(L)=\Big (a'/0_{L'}\Big )$ where $a'=f(1_{L})$. More useful results and facts about linear morphisms can be found in [1, Sec. 1] and in [2, Sec. 0].

The class ${\mathcal {L}}$ of all complete lattices contains a subclass comprised by all the modular complete lattices, which will be denoted by ${\mathcal {M}}$.

Proposition 2.2

[1, Proposition 2.2 (1)]. The class ${\mathcal {M}}$ of all modular complete lattices becomes a category, called the category of linear modular lattices and denoted by $\mathcal {L_{M}}$, if for any $L,L'\in {\mathcal {M}}$ one takes as morphisms from L to $L'$ all the linear morphisms from L to $L'$.

Observe that two objects in $\mathcal {L_M}$ are isomorphic in $\mathcal {L_M}$ if and only if they are isomorphic as lattices.

3 Cohereditary classes of modular lattices

Definition 3.1

[2, Definition 2.1 (1)]. Let ${\mathcal {C}}$ be a class in ${\mathcal {L}}_{{\mathcal {M}}}$. We call ${\mathcal {C}}$ an abstract class if ${\mathcal {C}}$ is closed under isomorphisms, that is, if $L\in {\mathcal {C}}$ and $N\cong L$, then $N \in {\mathcal {C}}$.

Definition 3.2

[2, Definition 2.1 (2)]. Let ${\mathcal {C}}$ be a class in $\mathcal {L_M}$. We say that ${\mathcal {C}}$ is a cohereditary class if ${\mathcal {C}}$ is an abstract class and for any $L\in \mathcal {L_M}$ and any $a\le b \le c$ in L such that $(c/a)\in {\mathcal {C}}$, we have that $(c/b)\in {\mathcal {C}}$.

Recall that an element c of a complete lattice L is compact if for every $X\subseteq L$ with $c\le \bigvee X$ there is some finite $F\subseteq X$ such that $c\le \bigvee F$. A complete lattice K is said to be compact if $1_K$ is a compact element.

Example 3.3

The following classes of modular lattices are cohereditary:

• The class of all noetherian lattices.

• The class of all artinian lattices.

• The class of all lattices with at most two elements.

• The class of all compact complete lattices.

Remark 3.4

Let ${\mathcal {C}}$ be a cohereditary class in $\mathcal {L_{M}}$ and $L\in {\mathcal {C}}$. Then for any $c\in L$ we have that $\Big (1_{L}/c\Big )\in {\mathcal {C}}$.

Recall from [1, Proposition 2.2 (4)] that the epimorphisms in the category $\mathcal {L_{M}}$ are exactly the surjective linear morphisms.

Definition 3.5

Let ${\mathcal {C}}$ be a class in $\mathcal {L_{M}}$. We say that ${\mathcal {C}}$ is a class closed under epimorphisms if for any $L\in {\mathcal {C}}$ and any surjective linear morphism $\varphi :L\longrightarrow L'$ in $\mathcal {L_{M}}$, we have that $L'\in {\mathcal {C}}$.

Clearly, any class ${\mathcal {C}}$ closed under epimorphisms is an abstract class.

Proposition 3.6

Let ${\mathcal {C}}$ be a class in $\mathcal {L_{M}}$. Then ${\mathcal {C}}$ is a cohereditary class if, and only if, ${\mathcal {C}}$ is closed under epimorphisms.

Proof

$\Longrightarrow )$ Let ${\mathcal {C}}$ be a cohereditary class in $\mathcal {L_{M}}$, $L\in {\mathcal {C}}$ and let us take $\varphi :L\longrightarrow L'$ a surjective linear morphism. If we denote by $L=(1_{L}/0_{L})$, $L'=(1_{L'}/0_{L'})$, and $n(\varphi )$ the kernel of $\varphi$, then $\varphi$ induces a lattice isomorphism

$$\begin{aligned} {\bar{\varphi }}:\Big (1_{L}/n(\varphi )\Big )\longrightarrow \Big (1_{L'}/0_{L'}\Big ), \end{aligned}$$

where ${\bar{\varphi }}(x)=\varphi (x)$ for all $x\in \Big (1_{L}/n(\varphi )\Big )$.

Since ${\mathcal {C}}$ is a cohereditary class and $\Big (1_{L}/0_{L}\Big )=L\in {\mathcal {C}}$, then $\Big (1_{L}/n(\varphi )\Big )\in {\mathcal {C}}$. Therefore

$$\begin{aligned} L'=\Big (1_{L'}/0_{L'}\Big )\overset{{\bar{\varphi }}}{\cong } \Big (1_{L}/n(\varphi )\Big )\in {\mathcal {C}}, \end{aligned}$$

which implies that $L'\in {\mathcal {C}}$.

$(\Longleftarrow$ Let ${\mathcal {C}}$ be a class in $\mathcal {L_{M}}$ closed under linear epimorphisms and let $L\in \mathcal {L_{M}}$ with $a\le b \le c$ in L such that $\Big (c/a\Big )\in {\mathcal {C}}$. Note that the canonical morphism $\rho : \Big (c/a\Big ) \longrightarrow \Big (c/b\Big )$, with correspondence $x\longmapsto x\vee b$ is a linear surjective morphism with kernel $n(\rho )=b$, that is, $\rho$ is a linear epimorphism. Now, as ${\mathcal {C}}$ is a class closed under epimorphisms, $\Big (c/b\Big )\in {\mathcal {C}}$. Thus, ${\mathcal {C}}$ is a cohereditary class. $\square$

Remark 3.7

Any non-empty cohereditary class in $\mathcal {L_{M}}$ contains the lattice 0 (that is, the trivial lattice). Indeed, suppose that ${\mathcal {C}}$ is a non-empty cohereditary class and let $L\in {\mathcal {C}}$. Since the canonical morphism $L\overset{\rho }{\longrightarrow }\ 0$ is a surjective linear morphism, by the previous proposition it follows that $0\in {\mathcal {C}}$. (Also, $(1_L/1_L)$ is a quotient interval of L.)

Similar to [4], the union and the intersection of an arbitrary family of cohereditary classes in $\mathcal {L_{M}}$ is a cohereditary class.

Therefore, the collection of all non-empty cohereditary classes in $\mathcal {L_{M}}$, ordered by class inclusion, is a complete big lattice. Here, for any family $\{{\mathcal {C}}_{i}\}_{i\in I}$ of non-empty cohereditary classes, we have that

$$\begin{aligned} \curlywedge _{i\in I} \ {\mathcal {C}}_{i}= \bigcap _{i\in I} \ {\mathcal {C}}_{i} \text{ and } \curlyvee _{i\in I}\ {\mathcal {C}}_{i}= \bigcup _{i\in I} \ {\mathcal {C}}_{i}. \end{aligned}$$

Observe that the least element is $\{0\}$ and the greatest element is $\mathcal {L_{M}}$.

For any class ${\mathcal {C}}$ in $\mathcal {L_{M}}$, the smallest cohereditary class containing ${\mathcal {C}}$ is, of course, the intersection of all cohereditary classes containing ${\mathcal {C}}$. We will now describe it explicitly.

Proposition 3.8

Let ${\mathcal {C}}$ be a class in $\mathcal {L_{M}}$. Then

$$\begin{aligned} \xi _{\twoheadrightarrow }({\mathcal {C}})= \Big \{ T\in \mathcal {L_{M}} \mid T\cong \Big (1_{L}/b\Big ) \textit{ for some } L\in {\mathcal {C}}, b\in L\Big \} \end{aligned}$$

is the smallest cohereditary class that contains ${\mathcal {C}}$.

Proof

We will show first that $\xi _{\twoheadrightarrow }({\mathcal {C}})$ is cohereditary.

Clearly, $\xi _{\twoheadrightarrow }({\mathcal {C}})$ is an abstract class. Now, let $L\in \mathcal {L_{M}}$ and $a\le b \le c$ in L such that $\Big ( c/a\Big )\in \xi _{\twoheadrightarrow }({\mathcal {C}})$. If $\varphi :\Big (c/a\Big )\longrightarrow \Big (1_{L'}/b'\Big )$ is an isomorphism for some $L'\in {\mathcal {C}}$ and $b'\in L'$, then $\Big (c/b\Big )\cong \Big (1_{L'}/\varphi (b)\Big )$ with $b'\le \varphi (b)$, therefore $\Big (c/b\Big )\in \xi _{\twoheadrightarrow }({\mathcal {C}})$ and $\xi _{\twoheadrightarrow }({\mathcal {C}})$ is a cohereditary class.

Now suppose that ${\mathcal {D}}$ is a cohereditary class in $\mathcal {L_{M}}$ such that ${\mathcal {C}}\subseteq {\mathcal {D}}$. Since ${\mathcal {D}}$ is closed under quotient intervals and is an abstract class then $\xi _{\twoheadrightarrow }({\mathcal {C}})\subseteq {\mathcal {D}}$. $\square$

Proposition 3.9

Let ${\mathcal {C}}$ be a class in $\mathcal {L_{M}}$. Then

$$\begin{aligned} {\mathcal {A}}_{{\mathcal {C}}}= \Big \{ L \in \mathcal {L_{M}} \mid \forall a\in L, \ a\ne 1_{L}, \text { the quotient interval } \Big (1_{L}/a\Big )\notin {\mathcal {C}} \Big \} \end{aligned}$$

is a cohereditary class.

Proof

By Proposition 3.6, it suffices to show that ${\mathcal {A}}_{{\mathcal {C}}}$ is a class closed under epimorphisms. Let $L\in {\mathcal {A}}_{{\mathcal {C}}}$ and let $L\overset{\varphi }{\longrightarrow }\ L'$ be an epimorphism in $\mathcal {L_{M}}$. If $n(\varphi )$ denotes the kernel of $\varphi$, then $\varphi$ induces a lattice isomorphisms

$$\begin{aligned} \Big (1_{L}/n(\varphi )\Big )\overset{{\bar{\varphi }}}{\cong }\Big (1_{L'}/0_{L'}\Big ), \end{aligned}$$

such that ${\bar{\varphi }}(x)=\varphi (x)$ for all $x\in \Big (1_{L}/n(\varphi )\Big )$. Now, for every $a'\in L'$ with $a'\ne 1_{L'}$, the the quotient interval $\Big (1_{L'}/a'\Big )$ is not trivial. Moreover, since ${\bar{\varphi }}$ is a lattice isomorphism, then there exist $x\in \Big (1_{L}/n(\varphi )\Big )$, with $x\ne 1_{L}$, such that

$$\begin{aligned} \Big (1_{L}/x \Big )\overset{{\bar{\varphi }}}{\cong }\Big (1_{L'}/a'\Big ). \end{aligned}$$

This shows that every nontrivial quotient interval of $L'$ is isomorphic to a nontrivial quotient interval of L. As $L\in {\mathcal {A}}_{{\mathcal {C}}}$, no nontrivial quotient intervals of L belong to the class ${\mathcal {C}}$. Thus, it follows that $L'\in {\mathcal {A}}_{{\mathcal {C}}}$. $\square$

Proposition 3.10

Let ${\mathcal {C}}$ be a class in $\mathcal {L_{M}}$. Then

$$\begin{aligned} {\mathcal {C}}\cap {\mathcal {A}}_{{\mathcal {C}}}= \{0\}. \end{aligned}$$

Proof

Let $L\in {\mathcal {C}}\cap {\mathcal {A}}_{{\mathcal {C}}}$ then if $0_{L}\ne 1_{L}$ by the way in which ${\mathcal {A}}_{{\mathcal {C}}}$ was defined it would follow that $L=\Big (1_{L}/0_{L}\Big )\notin {\mathcal {C}}$ which contradicts the hypothesis. Therefore $0_{L}=1_{L}$ and $L=\{0\}$. $\square$

Proposition 3.11

Let ${\mathcal {C}}$ be a class in $\mathcal {L_{M}}$. If ${\mathcal {D}}$ is any cohereditary class in $\mathcal {L_{M}}$ such that ${\mathcal {C}}\cap {\mathcal {D}}=\{0\}$ then

$$\begin{aligned} {\mathcal {D}}\subseteq {\mathcal {A}}_{{\mathcal {C}}}. \end{aligned}$$

Proof

Let $L\in {\mathcal {D}}$. Since ${\mathcal {D}}$ is cohereditary, then ${\mathcal {D}}$ contains all nontrivial quotient intervals of the form $\Big (1_{L}/b\Big )$ with $b\in L$. Since ${\mathcal {C}}\cap {\mathcal {D}}=\{0\}$ it follows that $\Big (1_{L}/b\Big )\notin {\mathcal {C}}$ for all $b\in L$ with $b\ne 1_{L}$, which in turn implies that $L\in {\mathcal {A}}_{{\mathcal {C}}}$. s$\square$

We will denote by $\mathcal {L_{M}}\text {-}quot$ the big lattice of all non-empty cohereditary classes of modular complete lattices. By Proposition 3.9, Proposition 3.10, and Proposition 3.11 we have the following

Theorem 3.12

Let ${\mathcal {C}}\in \mathcal {L_{M}}\text {-}quot$. Then

$$\begin{aligned} {\mathcal {C}}^{\perp _{\twoheadrightarrow }}= \Big \{ L \in \mathcal {L_{M}} \mid \forall a\in L, \ a\ne 1_{L}, \text { the quotient interval } \Big (1_{L}/a\Big )\notin {\mathcal {C}} \Big \} \end{aligned}$$

is a strong pseudocomplement in $\mathcal {L_{M}}\text {-}quot$ of ${\mathcal {C}}$.

Corollary 3.13

$\mathcal {L_{M}}\text {-}quot$ is a strongly pseudocomplemented big lattice.

Appropriately generalized versions of the following proposition, corollary and remark hold for every strongly pseudocomplemented big lattice.

Proposition 3.14

• (i) Let ${\mathcal {C}}\in \mathcal {L_{M}}\text {-}quot$. Then

  $$\begin{aligned} {\mathcal {C}}\subseteq \Big ({\mathcal {C}}^{\perp _{\twoheadrightarrow }}\Big )^{\perp _{\twoheadrightarrow }}. \end{aligned}$$

• (ii) If ${\mathcal {A}}, {\mathcal {B}}\in \mathcal {L_{M}}\text {-}quot$ are such that ${\mathcal {A}}\subseteq {\mathcal {B}}$, then ${\mathcal {B}}^{\perp _{\twoheadrightarrow }}\subseteq {\mathcal {A}}^{\perp _{\twoheadrightarrow }}$.

Proof

Follows from the definition of strong pseudocomplements. $\square$

Corollary 3.15

Let ${\mathcal {C}}\in \mathcal {L_{M}}\text {-}quot$. Then

$$\begin{aligned} {\mathcal {C}}^{\perp _{\twoheadrightarrow }}=\Big (\big ({\mathcal {C}}^{\perp _{\twoheadrightarrow }}\big )^{\perp _{\twoheadrightarrow }}\Big )^{\perp _{\twoheadrightarrow }}. \end{aligned}$$

Consequently, ${\mathcal {C}}$ lies in the skeleton of $\mathcal {L_M}\text {-}quot$ if and only if

$$\begin{aligned} {\mathcal {C}}=\Big ({\mathcal {C}}^{\perp _{\twoheadrightarrow }}\Big )^{\perp _{\twoheadrightarrow }}. \end{aligned}$$

Remark 3.16

The skeleton of $\mathcal {L_M}\text {-}quot$ is closed under arbitrary intersections.

Proof

Let $\{{\mathcal {C}}_i\}_{i\in I}$ be a subset of the skeleton of $\mathcal {L_M}\text {-}quot$. By Proposition 3.14(i),

$$\begin{aligned} \bigcap _{i\in I}{\mathcal {C}}_i\subseteq \Bigg (\bigg (\bigcap _{i\in I}{\mathcal {C}}_i\bigg )^{\perp _{\twoheadrightarrow }}\Bigg )^{\perp _{\twoheadrightarrow }}. \end{aligned}$$

Now, for each $j\in I$, $\bigcap \nolimits _{i\in I}{\mathcal {C}}_i\subseteq {\mathcal {C}}_j$, so that, by Proposition 3.14(ii) and Corollary 3.15,

$$\begin{aligned} \Bigg (\bigg (\bigcap _{i\in I}{\mathcal {C}}_i\bigg )^{\perp _{\twoheadrightarrow }}\Bigg )^{\perp _{\twoheadrightarrow }}\subseteq \Big (\big ({\mathcal {C}}_j\big )^{\perp _{\twoheadrightarrow }}\Big )^{\perp _{\twoheadrightarrow }}={\mathcal {C}}_j. \end{aligned}$$

Then,

$$\begin{aligned} \Bigg (\bigg (\bigcap _{i\in I}{\mathcal {C}}_i\bigg )^{\perp _{\twoheadrightarrow }}\Bigg )^{\perp _{\twoheadrightarrow }}\subseteq \bigcap _{i\in I}{\mathcal {C}}_i, \end{aligned}$$

so the desired equality holds. $\square$

We make now an observation we shall use later on.

Remark 3.17

Let ${\mathcal {C}}\in \mathcal {L_{M}}\text {-}quot$. Then

$$\begin{aligned} \Big ({\mathcal {C}}^{\perp _{\twoheadrightarrow }}\Big )^{\perp _{\twoheadrightarrow }}&= \left\{ L \in \mathcal {L_{M}} \,\bigg |\, \forall a\in L, \ a\ne 1_{L} \text { the quotient interval} \Big (1_{L}/a\Big )\notin {\mathcal {C}}^{\perp _{\twoheadrightarrow }} \right\} \\&= \Bigg \{ L \in \mathcal {L_{M}} \,\bigg |\, \forall a\in L,a\ne 1_{L}\, \text { there exists}\, c\in L \text { with}\, a \\&\le c < 1_{L}\, \text { such that}\, \Big (1_{L}/c\Big )\in {\mathcal {C}} \Bigg \}. \end{aligned}$$

4 Conatural classes in $\mathcal {L_{M}}$

Now that we have shown that $\mathcal {L_{M}}\text {-}quot$ is a strongly pseudocomplemented big lattice, we present the latticial counterpart of [3, Definition 1]:

Definition 4.1

We denote by $\mathcal {L_{M}}\text {-}conat$ the skeleton of $\mathcal {L_{M}}\text {-}quot$. The elements of $\mathcal {L_{M}}\text {-}conat$ will be called conatural classes.

We also present the lattice-theoretic counterpart of [4, Definition 22]:

Definition 4.2

Let ${\mathcal {C}}$ be a class in $\mathcal {L_{M}}$. We say that ${\mathcal {C}}$ satisfies the condition (CN) if the following holds:

Remark 4.3

Let ${\mathcal {C}}\in \mathcal {L_{M}}\text {-}quot$. Then, ${\mathcal {C}}$ having condition (CN) reduces to

Applying Remark 3.17, that can be rewritten as

$$\begin{aligned} \Big ({\mathcal {C}}^{\perp _{\twoheadrightarrow }}\Big )^{\perp _{\twoheadrightarrow }}\subseteq {\mathcal {C}}. \end{aligned}$$

Theorem 4.4

Let ${\mathcal {C}}$ be a class in $\mathcal {L_{M}}$. Then the following conditions are equivalent:

• (i) ${\mathcal {C}}\in \mathcal {L_{M}}\text {-}conat$,

• (ii) ${\mathcal {C}}$ satisfies (CN),

• (iii) ${\mathcal {C}}\in \mathcal {L_{M}}\text {-}quot$ and ${\mathcal {C}}=\Big ({\mathcal {C}}^{\perp _{\twoheadrightarrow }}\Big )^{\perp _{\twoheadrightarrow }}$.

Proof

(i) $\Longleftrightarrow$ (iii) By Corollary 3.15.

(ii) $\Longrightarrow$ (iii) It is straightforward to verify that ${\mathcal {C}}$, since it satisfies condition (CN), is necessarily abstract. We now show that it is in fact cohereditary. Let $L\in \mathcal {L_{M}}$ and $x\le y \le z$ in L such that $\Big (z/x\Big )\in {\mathcal {C}}$. We claim that $\Big (z/y\Big )\in {\mathcal {C}}$. Indeed, since ${\mathcal {C}}$ satisfies condition (CN), for the lattice $\Big (z/y\Big )\in \mathcal {L_{M}}$ and any $w\in \Big (z/y\Big )$ with $w\ne z$, by taking $M=\Big (z/x\Big )\in {\mathcal {C}}$ and $m=w\in \Big (z/x\Big )$ condition (CN) implies that $\Big (z/y\Big )\in {\mathcal {C}}$.

Therefore ${\mathcal {C}}\in \mathcal {L_{M}}\text {-}quot$. Also, as $0\in {\mathcal {C}}$, ${\mathcal {C}}\in \mathcal {L_{M}}\text {-}quot$ (being non-empty).

Now, by Proposition 3.14(i) we have that ${\mathcal {C}}\subseteq \Big ({\mathcal {C}}^{\perp _{\twoheadrightarrow }}\Big )^{\perp _{\twoheadrightarrow }}$. Remark 4.3 provides the reverse inclusion.

(iii) $\Longrightarrow$ (ii) By Remark 4.3. $\square$

In the following, we shall use Theorem 4.4 freely. Observe that, by appealing to the (CN) condition, we now have another way (independent of Remark 3.16) to prove that $\mathcal {L_M}\text {-}conat$ is closed under arbitrary intersections.

Remark 4.5

Let ${\mathcal {C}}$ be a class in $\mathcal {L_{M}}$. Set

$$\begin{aligned} \xi _{CN}({\mathcal {C}}){} & {} = \left\{ L \in \mathcal {L_{M}} \,\bigg |\, \forall a\in L, \text { with}\ a \right. \\{} & {} \ne 1_{L}, \text { there exists}\ b\in L, \text { with}\ a\le b < 1_{L},\\{} & {} \quad \left. \text { and there exist}\ M\in {\mathcal {C}} \text { and }m\in M \text { such that} \Big (1_{L}/b\Big )\cong \Big (1_{M}/m\Big ) \right\} \end{aligned}$$

Observe that, for non-empty ${\mathcal {C}}$,

$$\begin{aligned} \xi _{CN}({\mathcal {C}})=\Big (\big (\xi _{\twoheadrightarrow }({\mathcal {C}})\big )^{\perp _{\twoheadrightarrow }}\Big )^{\perp _{\twoheadrightarrow }}, \end{aligned}$$

whereas $\xi _{CN}(\emptyset )=\{0\}$. Thus, $\xi _{CN}({\mathcal {C}})$ is always a conatural class with ${\mathcal {C}}\subseteq \xi _{CN}({\mathcal {C}})$. Now, if ${\mathcal {D}}$ is a conatural class in $\mathcal {L_{M}}$ such that ${\mathcal {C}}\subseteq {\mathcal {D}}$, then it follows that $\xi _{CN}({\mathcal {C}})\subseteq {\mathcal {D}}$, because D satisfies condition (CN) (or, because, as $\xi _{\twoheadrightarrow }({\mathcal {C}})\subseteq {\mathcal {D}}$, we have, by Proposition 3.14(ii), that

$$\begin{aligned} \xi _{CN}({\mathcal {C}})=\Big (\big (\xi _{\twoheadrightarrow }({\mathcal {C}})\big )^{\perp _{\twoheadrightarrow }}\Big )^{\perp _{\twoheadrightarrow }}\subseteq \Big ({\mathcal {D}}^{\perp _{\twoheadrightarrow }}\Big )^{\perp _{\twoheadrightarrow }}={\mathcal {D}}). \end{aligned}$$

Therefore, $\xi _{CN}({\mathcal {C}})$ is the conatural class generated by ${\mathcal {C}}$.

Lastly, observe that, for an arbitrary class ${\mathcal {C}}$ in $\mathcal {L_M}$, it is plain that $\xi _{CN}({\mathcal {C}})\subseteq {\mathcal {C}}$ if and only if ${\mathcal {C}}$ has the (CN) condition.

Also, for ${\mathcal {C}}\in \mathcal {L_M}\text {-}quot$, it holds that

$$\begin{aligned} \xi _{CN}({\mathcal {C}})=\Big ({\mathcal {C}}^{\perp _{\twoheadrightarrow }}\Big )^{\perp _{\twoheadrightarrow }}. \end{aligned}$$

It is clear by now that $\mathcal {L_{M}}{} \textit{-}conat$, ordered by class inclusion, is a complete big lattice, where $\{0\}$ and $\mathcal {L_M}$ are respectively the least and greatest elements, infima are intersections, and the supremum of $\{{\mathcal {C}}_i\}_{i\in I}\subseteq \mathcal {L_{M}}{} \textit{-}conat$ is

$$\begin{aligned} \xi _{CN}\Bigg (\bigcup _{i\in I}{\mathcal {C}}_i\Bigg ). \end{aligned}$$

For ${\mathcal {C}}\in \mathcal {L_{M}}{} \textit{-}conat$, write ${\mathcal {C}}^{\perp _{CN}}$ for ${\mathcal {C}}^{\perp _{\twoheadrightarrow }}$.

Recall that a module M is called a coatomic module if each of its proper submodules is included in a maximal submodule. We now present the lattice-theoretic counterpart of this concept.

Definition 4.6

We say that an element m in a lattice L with greatest element $1_L$ is a coatom if m is maximal in $m\ne 1_L$. We say that L is a coatomic lattice if for all $a\in L$ with $a\ne 1_{L}$ there exists a coatom $m\in L$ such that $a\le m$.

Recall that a lattice is said to be a simple lattice if and only if it has precisely two elements. Of course, there is, up to isomorphism, only one simple lattice.

Example 4.7

Let L be the simple lattice in $\mathcal {L_{M}}$. Then

• (i)
  $$\begin{aligned} \xi _{CN}(L)&= \left\{ M\in \mathcal {L_{M}} \,\bigg |\, \forall a\in M, \text {with } a\ne 1_{M}, \text {there exists}\ b\in M, \text {with } a \right. \\&\left. \le b < 1_{M}, \text {such that}\ \Big (1_{L}/b\Big ) \cong L \right\} \\&=\Big \{ M\in \mathcal {L_{M}} \mid M \ \text{ is } \text{ a } \text{ coatomic } \text{ lattice } \Big \} \end{aligned}$$
• (ii)
  $$\begin{aligned} \Big (\xi _{CN}(L) \Big )^{\perp _{CN}} =\Big \{ N\in \mathcal {L_{M}} \mid N \text{ does } \text{ not } \text{ have } \text{ coatoms } \Big \}. \end{aligned}$$

• (iii) $\Big (\xi _{CN}(L) \Big )^{\perp _{CN}}$ is a complement of $\xi _{CN}(L)$ in $\mathcal {L_{M}}\text {-}conat$.

Proof

(i) Straightforward from Remark 4.5.

(ii) By Remark 4.5 and Corollary 3.15,

$$\begin{aligned} \Big (\xi _{CN}(L) \Big )^{\perp _{CN}}&=\Big (\big ((\xi _{\twoheadrightarrow }(L))^{\perp _{\twoheadrightarrow }}\big )^{\perp _{\twoheadrightarrow }}\Big )^{\perp _{CN}}\\&=\Big (\big ((\xi _{\twoheadrightarrow }(L))^{\perp _{\twoheadrightarrow }}\big )^{\perp _{\twoheadrightarrow }}\Big )^{\perp _{\twoheadrightarrow }}\\&=(\xi _{\twoheadrightarrow }(L))^{\perp _{\twoheadrightarrow }}\\&=\Big \{ N\in \mathcal {L_{M}} \mid N \text{ is } \text{ trivial } \text{ or } \text{ simple } \Big \}^{\perp _{\twoheadrightarrow }}\\ {}&=\Big \{ N\in \mathcal {L_{M}} \mid N \text{ does } \text{ not } \text{ have } \text{ coatoms } \Big \}. \end{aligned}$$

(iii) It suffices to show that the conatural class generated by $\xi _{CN}(L) \cup \Big (\xi _{CN}(L) \Big )^{\perp _{CN}}$ is all of $\mathcal {L_{M}}$. Let $N\in \mathcal {L_{M}}$ and $\Big (1_{N}/n\Big )$ be a quotient interval of N with $n\ne 1_{N}$. If $\Big (1_{N}/n\Big )$ has a coatom then $\Big (1_{N}/n\Big )$ has a nonzero quotient interval isomorphic to $L\in \xi _{CN}(L)$, but if $\Big (1_{N}/n\Big )$ has no coatoms, we have that $\Big (1_{N}/n\Big )\in \Big (\xi _{CN}(L) \Big )^{\perp _{CN}}$. Therefore, in one case or the other we have that $\Big (1_{N}/n\Big )$ has a nonzero quotient interval in $\xi _{CN}(L) \cup \Big (\xi _{CN}(L) \Big )^{\perp _{CN}}$, and thus, by Remark 4.5, $N\in \xi _{CN}\Big (\xi _{CN}(L) \cup \Big (\xi _{CN}(L) \Big )^{\perp _{CN}}\Big )$. $\square$

Next we present the lattice-theoretic counterpart of [3, Definition 4].

Definition 4.8

We say that the lattices L and $L'$ in $\mathcal {L_{M}}$ share quotients, if there exists $a\in L$, with $a\ne 1_{L}$, and $a'\in L'$, with $a'\ne 1_{L'}$, such that

$$\begin{aligned} \Big (1_{L}/a\Big ) \cong \Big (1_{L'}/a'\Big ). \end{aligned}$$

We say that the classes ${\mathcal {C}}$ and ${\mathcal {D}}$ in $\mathcal {L_{M}}$ share quotients, if there exists $L\in {\mathcal {C}}$ and $M\in {\mathcal {D}}$ such that L and M share quotients.

Proposition 4.9

The classes ${\mathcal {C}}$ and ${\mathcal {D}}$ in $\mathcal {L_{M}}$ share quotients if, and only if,

$$\begin{aligned} \xi _{CN}({\mathcal {C}})\cap \xi _{CN}({\mathcal {D}})\ne \{0\}. \end{aligned}$$

Proof

$\Longrightarrow )$ Clear.

$\Longleftarrow )$ Let $0\ne L\in \xi _{CN}({\mathcal {C}})\cap \xi _{CN}({\mathcal {D}})$. Since $L\in \xi _{CN}({\mathcal {C}})$ and $0_{L}<1_L$, there exist $a\in L$ with $0_{L}\le a<1_{L}$ and $C\in {\mathcal {C}}$ with $c\in C$ such that $\Big (1_{L}/a\Big ) \cong \Big (1_C/c\Big )$, say through an isomorphism $f:\Big (1_{L}/a\Big ) \rightarrow \Big (1_C/c\Big )$.

As $L\in \xi _{CN}({\mathcal {D}})$ and $a<1_L$, there exist $b\in L$ with $a\le b<1_{L}$ and $D\in {\mathcal {D}}$ with $d\in D$ such that $\Big (1_{L}/b\Big ) \cong \Big (1_D/d\Big )$.

Of course, in C, $c\le f(b)<1_C$, so that $\Big (1_C/f(b)\Big )$ is a nonzero quotient interval of $C\in {\mathcal {C}}$. And, clearly, $\Big (1_C/f(b)\Big )\cong \Big (1_{L}/b\Big ) \cong \Big (1_D/d\Big )$. $\square$

The following definition is the lattice-theoretic counterpart of [3, Definition 11]:

Definition 4.10

We say that $L\in \mathcal {L_{M}}$ is q-atomic if $\xi _{CN}(L)$ is an atom in $\mathcal {L_{M}}\text {-}conat$.

Proposition 4.11

Let $L\in \mathcal {L_{M}}$, $L\ne 0$. Then L is q-atomic if, and only if, for any $a,b\in L$ with $a,b\ne 1_{L}$, we have that $\Big (1_{L}/a\Big )$ and $\Big (1_{L}/b\Big )$ share quotients.

Proof

$\Longrightarrow )$ Suppose that $0\ne L\in \mathcal {L_{M}}$ is q-atomic. Let $a,b\in L$ with $a,b\ne 1_{L}$. Now, $0\ne \Big (1_{L}/a\Big ),\Big (1_{L}/b\Big )\in \xi _{CN}(L)$.

Thus, since $\xi _{CN}(L)$ is an atom in $\mathcal {L_{M}}{} \textit{-}conat$ it follows that

$$\begin{aligned} \xi _{CN}\Big (1_{L}/a\Big )=\xi _{CN}(L)=\xi _{CN}\Big (1_{L}/b\Big ). \end{aligned}$$

So $\Big (1_{L}/a\Big )\in \xi _{CN}\Big (1_{L}/b\Big )$. Then, every nonzero quotient interval of $\Big (1_{L}/a\Big )$ (in particular $\Big (1_{L}/a\Big )$itself) shares quotients with $\Big (1_{L}/b\Big )$.

$\Longleftarrow )$ Suppose that for any $a,b\in L$, with $a,b\ne 1_{L}$, we have that $\Big (1_{L}/a\Big )$ and $\Big (1_{L}/b\Big )$ share quotients. In $\mathcal {L_{M}}{} \textit{-}conat$, let

$$\begin{aligned} \{0\}\ne {\mathcal {K}}\subseteq \xi _{CN}(L). \end{aligned}$$

Take some $0\ne K\in {\mathcal {K}}$. Then $K\in \xi _{CN}(L)$, so that there are $k\in K$ with $0_K\le k<1_K$ and $b\in L$ such that $\Big (1_{K}/k\Big )\cong \Big (1_{L}/b\Big )$. Let us verify that $L\in \xi _{CN}(K)$. Let then $a\in L$ with $b<1_L$. By hypothesis, $\Big (1_{L}/a\Big )$ shares quotients with $\Big (1_{L}/b\Big )$, thus also with $\Big (1_{K}/k\Big )$ and therefore with K. It follows that

$$\begin{aligned} \xi _{CN}(L)\subseteq \xi _{CN}(K)\subseteq {\mathcal {K}}, \end{aligned}$$

so that $\xi _{CN}(L)={\mathcal {K}}$. $\square$

We will now show that $\mathcal {L_{M}}\text {-}conat$ is a boolean big lattice. For this, we recall

Theorem 4.12

[7, Ch. III, Proposition 4.4]. A lattice L is boolean if and only if for each $a\in L$ exists a unique complement $a'$ of a in L, and $a \wedge b = 0$ holds if and only if $b\leqslant a'$.

Now, by Corollary 3.13, $\mathcal {L_{M}}\text {-}quot$ is a strongly pseudocomplemented big lattice. Thus, in order to show that $\mathcal {L_{M}}\text {-}conat$ is a boolean big lattice, it suffices, by Theorem 4.12, to prove that any strong pseudocomplement in $\mathcal {L_{M}}\text {-}quot$ is a unique complement in $\mathcal {L_{M}}\text {-}conat$.

Theorem 4.13

Let ${\mathcal {C}}\in \mathcal {L_{M}} \text {-}conat$. Then ${\mathcal {C}}^{\perp _{CN}}$ is a complement of ${\mathcal {C}}$ in $\mathcal {L_{M}} \text {-}conat$.

Proof

It suffices to show that $\mathcal {L_M}={\mathcal {C}} \vee ({\mathcal {C}}^{\perp _{CN}})=\xi _{CN}\big ({\mathcal {C}} \cup ({\mathcal {C}}^{\perp _{\twoheadrightarrow }}) \big )$. Let then $L\in \mathcal {L_{M}}$, and let $(1_L/a)$ be a nonzero quotient interval of L. Now, either $(1_L/a)$ has a nonzero quotient interval in ${\mathcal {C}}$ or it does not. In the latter case, $(1_L/a)$, which is a nonzero quotient interval of itself, lies in ${\mathcal {C}}^{\perp _{\twoheadrightarrow }}$. Either way, $(1_L/a)$ has a nonzero quotient interval in ${\mathcal {C}} \cup ({\mathcal {C}}^{\perp _{\twoheadrightarrow }})$. It follows that $L\in \xi _{CN}\big ({\mathcal {C}} \cup ({\mathcal {C}}^{\perp _{\twoheadrightarrow }}) \big )$. Therefore,

$$\begin{aligned} \mathcal {L_{M}} \subseteq \xi _{CN}\big ({\mathcal {C}} \cup ({\mathcal {C}}^{\perp _{\twoheadrightarrow }}) \big ). \\ \end{aligned}$$

$\square$

Theorem 4.14

Let ${\mathcal {C}}\in \mathcal {L_{M}}\text {-}conat$. If ${\mathcal {D}}$ is a complement of ${\mathcal {C}}$ in $\mathcal {L_{M}}\text {-}conat$ then

$$\begin{aligned} {\mathcal {D}}={\mathcal {C}}^{\perp _{CN}}. \end{aligned}$$

Proof

Let ${\mathcal {D}}$ be a complement of ${\mathcal {C}}$ in $\mathcal {L_{M}}\text {-}conat$. Since

$$\begin{aligned} \lbrace 0\rbrace ={\mathcal {C}}\wedge {\mathcal {D}}={\mathcal {C}}\cap {\mathcal {D}}, \end{aligned}$$

it follows that ${\mathcal {D}}\subseteq {\mathcal {C}}^{\perp _{\twoheadrightarrow }}$. Our task now is to prove the reverse inclusion. Let then $L\in {\mathcal {C}}^{\perp _{\twoheadrightarrow }}$ and any $a\in L$ with $a\ne 1_L$. As

$$\begin{aligned} L\in \mathcal {L_{M}}={\mathcal {C}}\vee {\mathcal {D}}=\xi _{CN}({\mathcal {C}}\cup {\mathcal {D}}), \end{aligned}$$

there exists $b\in L$, with $a\leqslant b<1_L$, such that the quotient interval $(1_L/b)\in {\mathcal {C}}\cup {\mathcal {D}}$. As $L\in {\mathcal {C}}^{\perp _{\twoheadrightarrow }}$ and $(1_L/b)$ is a nonzero quotient interval of L, it follows that $(1_L/b)$ cannot lie in ${\mathcal {C}}$. Therefore, $(1_L/b)\in {\mathcal {D}}$, and thus, because of condition (CN), $L\in {\mathcal {D}}$. Hence,

$$\begin{aligned} {\mathcal {D}}={\mathcal {C}}^{\perp _{\twoheadrightarrow }}={\mathcal {C}}^{\perp _{CN}}. \\ \end{aligned}$$

$\square$

Theorem 4.15

$\mathcal {L_{M}}\text {-}conat$ is a boolean big lattice.

4.1 Some closure properties

In this subsection we prove some closure properties for conatural classes in ${\mathcal {L}}_{{\mathcal {M}}}$. For this purpose, we will start recalling the following

Definition 4.16

Let ${\mathcal {C}}$ be a class in ${\mathcal {L}}_{{\mathcal {M}}}$. We say that ${\mathcal {C}}$ is closed under extensions if for any $L\in {\mathcal {L}}_{{\mathcal {M}}}$ and $a\le b \le c$ in L such that $(b/a),(c/b) \in {\mathcal {C}}$, we have that $(c/a)\in {\mathcal {C}}$.

Proposition 4.17

Let ${\mathcal {C}}$ be a conatural class in $\mathcal {L_{M}}$. Then ${\mathcal {C}}$ is closed under extensions.

Proof

Let ${\mathcal {C}}\in \mathcal {L_{M}}\text {-}conat$ and let $L\in \mathcal {L_M}$. Suppose that $a\le b \le c$ in L are such that $(b/a), (c/b)\in {\mathcal {C}}$. We will show that $(c/a)\in {\mathcal {C}}$.

Let us take $x\in (c/a)$ with $x\ne c$. Since $x\vee b \le c$, we have the following two cases:

• (a) If $x\vee b\ne c$, then $x\le x\vee b<c$ and $\Big (c/x\vee b\Big )$ is a quotient interval of $(c/b)\in {\mathcal {C}}$.

• (b) If $x\vee b =c$, then by modularity we have that

  $$\begin{aligned} (c/x)=\Big (x\vee b/x\Big )\cong \Big (b/x\wedge b\Big )\subseteq (b/a). \end{aligned}$$

  This implies that (c/x) is isomorphic to a quotient interval of $(b/a)\in {\mathcal {C}}$.

In any case, (c/x) either lies in ${\mathcal {C}}$ or has a nonzero quotient interval in ${\mathcal {C}}$. Since this argument is valid for all $x\in (c/a)$ with $x\ne c$, the condition (CN) implies that $(c/a)\in {\mathcal {C}}$. $\square$

We recall that in any lattice L with greatest element $1_l$, we say that $s\in L$ is superfluous if for any $a\in L$ with $a\ne 1_{L}$ we have that $s\vee a \ne 1_{L}$.

In the category $\mathcal {L_M}$. an epimorphism $\varphi :L\rightarrow L'$ is said to be a superfluous epimorphism if its kernel is a superfluous element in L.

We say that a class ${\mathcal {C}}$ in $\mathcal {L_{M}}$ is closed under superfluous epimorphisms if whenever $L\in \mathcal {L_M}$, $L'\in {\mathcal {C}}$ and $\varphi :l\rightarrow L'$ is a superfluous epimorphism it happens that $L\in {\mathcal {C}}$.

Proposition 4.18

Let ${\mathcal {C}}$ be a conatural class in $\mathcal {L_M}$. Then ${\mathcal {C}}$ is closed under superfluous epimorphisms.

Proof

Let ${\mathcal {C}}\in \mathcal {L_{M}}\text {-}conat$ and $L'\in {\mathcal {C}}$. Suppose that $\varphi :L\longrightarrow L'$ is a superfluous epimorphism with kernel $n(\varphi )$. Let us note first that

$$\begin{aligned} \Big (1_{L}/n(\varphi )\Big ) \overset{{\bar{\varphi }}}{\cong } \Big (1_{L'}/0_{L'}\Big )=L'. \end{aligned}$$

Let now $a\in L$ with $a\ne 1_{L}$. Let us define $b=n(\varphi )\vee a$. Observe that b lies in the interval $\Big (1_{L}/n(\varphi )\Big )$ and that $b \ne 1_{L}$ as $n(\varphi )$ is superfluous in L. Thus, it follows that

$$\begin{aligned} 0\ne \Big (1_{L}/b\Big )\overset{{\bar{\varphi }}}{\cong } \Big (1_{L'}/\varphi (b)\Big ). \end{aligned}$$

Note that $\Big (1_{L'}/\varphi (b)\Big )$ is a quotient interval of the lattice $L'\in {\mathcal {C}}$. Thus $\Big (1_{L}/b\Big )$ is a nonzero quotient interval in ${\mathcal {C}}$ of $\Big (1_{L}/a\Big )$. Since the class ${\mathcal {C}}$ is a conatural class, it satisfies condition (CN). Therefore, we conclude that $L\in {\mathcal {C}}$. $\square$

Recall that a non-empty subset A of nonzero elements of a complete lattice L is said to be independent if and only if for every $a\in A$, $a\wedge \bigvee \big (A\backslash \{a\}\big )=0_L$.

Definition 4.19

[6, Definition 4.12]. Let ${\mathcal {C}}$ be a class in ${\mathcal {L}}_{{\mathcal {M}}}$. We call ${\mathcal {C}}$ a class closed under (independent) joins over lattices in $\mathcal {L_{M}}$ if ${\mathcal {C}}$ is an abstract class and for any $K\in {\mathcal {L}}_{{\mathcal {M}}}$ and any (independent) subset A of K, with $(a/0_{K})\in {\mathcal {C}}$ for all $a\in A$, one has that

$$\begin{aligned} \Bigg ( \left( \bigvee A \right) /0_{K}\Bigg )\in {\mathcal {C}}. \end{aligned}$$

Example 4.20

Consider the class ${\mathcal {C}} \in {\mathcal {L}}_{{\mathcal {M}}}$ of all semiartinian lattices, i.e. the class of all complete modular lattices for which every quotient interval has an atom, then by [5, Proposition 5.4] if L is a lattice in ${\mathcal {L}}_{{\mathcal {M}}}$ and A a subset of L such that $(a/0_{L})$ is semiartinian for all $a\in A$ one has that $\Big ( (\bigvee A )/0_{K}\Big )$ is semiartinian. Then ${\mathcal {C}}$ is closed under joins over lattices in ${\mathcal {L}}_{{\mathcal {M}}}$.

Definition 4.19 can be extended in the following way. Let ${\mathcal {C}}$ and ${\mathcal {D}}$ be two classes of lattices in $\mathcal {L_{M}}$. We say that ${\mathcal {C}}$ is closed under (independent) joins over lattices in ${\mathcal {D}}$ if ${\mathcal {C}}$ is abstract and for each $L\in {\mathcal {D}}$ and each (independent) family $A\subseteq L$ with $(a / 0_L) \in {\mathcal {C}}$ for all $a\in A$, it holds that $\Big ( (\bigvee A )/0_L\Big ) \in {\mathcal {C}}$. As a first example, we show that any conatural class is closed under independent joins over lattices in the class ${\mathcal {M}}_d$ of all distributive complete lattices.

Proposition 4.21

Any conatural class ${\mathcal {C}}$ is closed under independent joins over lattices in ${\mathcal {M}}_d$.

Proof

We will show that, for any $K\in {\mathcal {M}}_{d}$ and any independent family $\{a_{i}\}_{i\in I}\subseteq K$ with $(a_{i}/0_{K})\in {\mathcal {C}}$ for all $i\in I$, one has that

$$\begin{aligned} \Bigg ( \left( \bigvee _{i\in I} a_{i}\right) /0_{K}\Bigg ) \in {\mathcal {C}}. \end{aligned}$$

It suffices to show that given any $x\in \left( \bigvee \nolimits _{i\in I} a_{i}/0_K\right)$ with $x< \bigvee \nolimits _{i\in I} a_{i}$, the quotient interval $\left( \bigvee \nolimits _{i\in I} a_{i}/ x\right)$ has a nonzero quotient interval in ${\mathcal {C}}$.

Let $x\in \big (\bigvee \nolimits _{i\in I} a_{i}/0_K\big )$ such that $x< \bigvee \nolimits _{i\in I} a_{i}$. Then, there exists $i_{0}\in I$ such that $a_{i_{0}} \not \le x$. Indeed, if the latter did not hold, we would have that $a_{i}\le x$ for all $i\in I$, so that $\bigvee \nolimits _{i\in I} a_{i} \le x$, a contradiction on the condition on x. Therefore, $x\wedge a_{i_0}<a_{i_0}$.

Fig. 1

Lattice-theoretic relations in an independent family

Further, we have the diagram of Figure 1, from which, by modularity, we obtain the isomorphism

$$\begin{aligned} \big ( a_{i_{0}}/0_{K} \big ) \overset{\varphi }{\cong } \Bigg (\bigvee _{i\in I} a_{i}/ \bigvee _{i\ne i_{0}} a_{i} \Bigg ). \end{aligned}$$

Note that the isomorphism $\varphi : \big ( a_{i_{0}}/0_{K} \big ) \longrightarrow \Big (\bigvee \nolimits _{i\in I} a_{i}/ \bigvee \nolimits _{i\ne i_{0}} a_{i} \Big )$ sends $z\longmapsto z\vee \Big ( \bigvee \nolimits _{i\ne i_{0}} a_{i} \Big )$. With this in mind, we have that the nonzero quotient interval $\big ( a_{i_{0}}/x\wedge a_{i_{0}} \big )$ of $\big ( a_{i_{0}}/0_{K} \big )$ is isomorphic to

$$\begin{aligned} \varphi \left( a_{i_{0}}/ x \wedge a_{i_{0}} \right)&= \left( \varphi (a_{i_{0}})/\varphi (x \wedge a_{i_0})\right) \\&=\left( \bigg ( a_{i_{0}}\vee \bigvee \limits _{i\ne i_{0}} a_{i} \bigg ) / \big ( ( x\wedge a_{i_{0}} ) \vee \bigvee \limits _{i\ne i_{0}} a_{i} \big ) \right) \\&=\left( \bigg (\bigvee \limits _{i\in I} a_{i} \bigg ) / \bigg ( x \vee \bigvee \limits _{i\ne i_{0}} a_{i} \bigg )\right) , \end{aligned}$$

where the last equality follows because K is a distributive lattice:

$$\begin{aligned} ( x\wedge a_{i_0} ) \vee \bigvee \limits _{i\ne i_0} a_{i}&= \left( x\vee \bigvee \limits _{i\ne i_0} a_{i} \Big ) \wedge \Big ( a_{i_0} \vee \bigvee \limits _{i\ne i_0} a_{i} \right) \\&=\left( x\vee \bigvee \limits _{i\ne i_0} a_{i} \Big ) \wedge \Big ( \bigvee \limits _{i\in I} a_{i} \right) = x\vee \bigvee \limits _{i\ne i_0} a_{i}. \end{aligned}$$

Finally, since $\big ( a_{i_{0}}/0_K \big )\in {\mathcal {C}}$ and ${\mathcal {C}}$ is cohereditary, it follows that

$$\begin{aligned} \left( \bigvee \limits _{i\in I} a_{i} / (x \vee \bigvee \limits _{i\ne i_0} a_{i}) \right) \cong \left( a_{i_{0}} / x\wedge a_{i_{0}} \right) \in {\mathcal {C}}\backslash \{0\}, \end{aligned}$$

which shows that the quotient interval $\Big ( \bigvee \nolimits _{i\in I}a_{i} / x \Big )$ has a nonzero quotient interval in ${\mathcal {C}}$. $\square$

Proposition 4.22

Any conatural class ${\mathcal {C}}$ in $\mathcal {L_{M}}$ is closed under joins over noetherian lattices in $\mathcal {L_{M}}$.

Proof

Let $N\in \mathcal {L_{M}}$ be a noetherian lattice and $A\subseteq N$ such that $(a/0_{N})\in {\mathcal {C}}$ for all $a\in A$. We will show that

$$\begin{aligned} \left( \left( \bigvee A \right) /0_{N}\right) \in {\mathcal {C}}. \end{aligned}$$

To do so, we shall prove that any nonzero quotient interval of $\Big ((\bigvee A )/0_{N}\Big )$ has a nonzero quotient interval in ${\mathcal {C}}$.

Let $x\in \Big ((\bigvee A )/0_{N}\Big )$ with $x< \bigvee A$. Then, there exists some $a_1\in A$ such that $a_1\not \le x$. Thus, by modularity, we have that

$$\begin{aligned} 0 \ne \left( x\vee a_1/x \right) \cong \left( a_1/a_1\wedge x \right) . \end{aligned}$$

Note that since $\left( a_1/0_{N}\right) \in {\mathcal {C}}$ and ${\mathcal {C}}$ is cohereditary, $\Big ( a_1/a_1\wedge x \Big )\in {\mathcal {C}}$. Thus, if $(x\vee a_1)=\bigvee A$, then we are done with the proof, since

$$\begin{aligned} \left( (\bigvee A )/x\right) = \left( x\vee a_1/x \right) \cong \left( a_1/a_1\wedge x \right) \in {\mathcal {C}}. \end{aligned}$$

Otherwise, if $x\vee a_1< \bigvee A$, then there exists $a_2\in A$ such that $a_2\not \le (x\vee a_1)$. Again, by modularity, we have that

$$\begin{aligned} 0\ne \left( x\vee a_1 \vee a_2/x\vee a_1 \right) \cong \left( a_2/(x\vee a_1)\wedge a_2 \right) \in {\mathcal {C}}. \end{aligned}$$

In case that $x\vee a_1 \vee a_2=\bigvee A$, we are done. Otherwise, there exists $a_3\in A$ such that $a_3\not \le x\vee a_1\vee a_2$. Note that this construction leads to a sequence of the form

$$\begin{aligned} x<x\vee a_1<x\vee a_1 \vee a_2<x\vee a_1\vee a_2\vee a_3<\cdots . \end{aligned}$$

Since by hypothesis the lattice N is noetherian, the above sequence has to stop at an element of the form $x\vee a_1 \vee \cdots \vee a_k$. This in turn implies that $x\vee a_1 \vee \cdots \vee a_k=\bigvee A$ and thus, $\Big ((\bigvee A)/x\Big )$ has the nonzero quotient interval

$$\begin{aligned} 0&\ne \Big ((\bigvee A)/x\vee a_1 \vee \cdots \vee a_{k-1}\Big )\\&= \Big ( x\vee a_1\vee \cdots \vee a_k/x\vee a_1 \vee \cdots \vee a_{k-1} \Big )\\&\cong \Big ( a_k/(x\vee a_1 \vee \cdots \vee a_{k-1})\wedge a_k \Big ) \in {\mathcal {C}}. \\ \end{aligned}$$

$\square$

Similarly to Definition 4.19, we can consider classes ${\mathcal {C}}$ in ${\mathcal {L}}_{{\mathcal {M}}}$ which are closed under (independent) joins of atoms.

Definition 4.23

Let ${\mathcal {C}}$ be a class in ${\mathcal {L}}_{{\mathcal {M}}}$. We call ${\mathcal {C}}$ a class closed under (independent) joins of atoms over ${\mathcal {D}} \subseteq \mathcal {L_{M}}$ if ${\mathcal {C}}$ is an abstract class and for any $K\in {\mathcal {D}}$ and any (independent) subset ${\mathcal {A}}$ of atoms of K, with $(a/0_{K})\in {\mathcal {C}}$ for all $a\in A$, one has that

$$\begin{aligned} \left( \left( \bigvee {\mathcal {A}} \right) /0_{K}\right) \in {\mathcal {C}}. \end{aligned}$$

Example 4.24

Consider the class ${\mathcal {C}}$ in ${\mathcal {L}}_{{\mathcal {M}}}$ of lattices for which every nonzero initial interval is infinite, called torsion-free lattices in [5], and let $L \in {\mathcal {L}}_{{\mathcal {M}}}$. If $a\in L$ is an atom, then $(a/0_{L})$ is not in ${\mathcal {C}}$. Therefore, vacuously, ${\mathcal {C}}$ is closed under joins of atoms over all ${\mathcal {L}}_{{\mathcal {M}}}$.

Recall that a complete lattice L is said to be upper semicontinuous (or just upper continuous if for every $a\in L$ and upper directed $D\subseteq L$, $a\wedge (\bigvee D)=\bigvee \nolimits _{d\in D}(a\wedge d)$.

Also, a complete lattice L is called semiatomic if $1_L$ is a join of atoms in L.

Proposition 4.25

Let ${\mathcal {C}}$ be a conatural class in $\mathcal {L_M}$. Then, ${\mathcal {C}}$ is closed under joins of atoms over the class of upper semicontinuous lattices in ${\mathcal {L}}_{{\mathcal {M}}}$.

Proof

Let $L\in \mathcal {L_{M}}$ be an upper semicontinuous lattice, ${\mathcal {A}}$ a set of atoms in L such that $(a/0_{L})\in {\mathcal {C}}$ for all $a\in {\mathcal {A}}$, and $x\in \big ((\bigvee {\mathcal {A}} )/0_{L}\big )$ with $x< \bigvee {\mathcal {A}}$. Since $\big ((\bigvee {\mathcal {A}} )/0_{L}\big )$ is modular, upper semicontinuous, and semiatomic, [5, Lemma 6.12] provides a maximal $m\in \big ((\bigvee {\mathcal {A}} )/0_{L}\big )$ such that $x\le m$. Take $a\in {\mathcal {A}}$. Then,

$$\begin{aligned} \bigg (\left( \bigvee {\mathcal {A}} \right) /m\bigg )\cong (a/0_L)\in {\mathcal {C}}, \end{aligned}$$

which implies that $\big ((\bigvee {\mathcal {A}} )/x\big )$ has a nonzero quotient interval in ${\mathcal {C}}$. $\square$