Sum-Essential Graphs of Modules
Jerzy Matczuk, Ali Majidinya

TL;DR
This paper introduces and studies the sum-essential graph of modules, exploring how module properties influence the graph's structure and properties, and analyzing subgraphs related to essential submodules.
Contribution
It defines the sum-essential graph of modules and investigates its properties and the relationship between module characteristics and graph structure.
Findings
Characterization of the sum-essential graph's properties
Analysis of the subgraph induced by non-essential submodules
Insights into the relationship between module properties and graph features
Abstract
The sum-essential graph of a left -module is a graph whose vertices are all nontrivial submodules of and two distinct submodules are adjacent iff their sum is an essential submodule of . Properties of the graph and its subgraph induced by vertices which are not essential as submodules of are investigated. The interplay between module properties of and properties of those graphs is studied.
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Sum-Essential Graphs of Modules
Jerzy Matczuk
Institute of Mathematics, Warsaw University, Banacha 2, 02-097 Warsaw, Poland
and
Ali Majidinya
Department of Mathematical Sciences, Salman Farsi University of Kazerun, Kazerun, Iran, P.O. Box 73175-457.
[email protected] and [email protected]
Abstract.
The sum-essential graph of a left -module is a graph whose vertices are all nontrivial submodules of and two distinct submodules are adjacent iff their sum is an essential submodule of . Properties of the graph and its subgraph induced by vertices which are not essential as submodules of are investigated. The interplay between module properties of and properties of those graphs is studied.
Key words and phrases:
sum-essential graph, essential submodule, uniform dimension
2010 Mathematics Subject Classification:
05C25, 05C40, 16D99.
Introduction
Throughout the paper a ring means a unital associative ring and a module is a left unitary module over a given ring .
There are many studies on various graphs associated to modules, rings and other algebraic structures. The aim of the paper is to investigate the interplay between module properties of a module and properties of its sum-essential and proper sum-essential graphs. This will result in classification of modules in terms of some specific properties of those graphs.
For a given module the *sum-essential graph *of is a simple graph (i.e. unweighted, undirected graph containing no graph loops or multiple edges) whose vertices are nontrivial submodules (i.e. different from 0 and ) and two distinct vertices are adjacent if and only if the sum of the corresponding submodules is an essential submodule of . We will also study the *proper sum-essential graph *of which is a subgraph of generated by vertices which, as submodules of are not essential.
The graph was introduced and studied by Amjadi in [5] in a very special case when is a commutative ring and . There are other graphs relating submodule structure of a given module. For example comaximal left ideal graphs, i.e. graphs whose vertices are nontrivial left ideals of a ring and two vertices are adjacent if and only if were considered in [4, 9, 10]. The intersection graph of a module, i.e. a simple graph whose vertices are nontrivial submodules and two distinct vertices are adjacent if the intersection of corresponding submodules is nonzero, was introduced and investigated in [8]. Its complement graph was also examined. Earlier those graphs were considered in the context of left ideals of a given ring (see [2, 3]). In [1] the authors considered the inclusion graph whose vertices were nontrivial left ideals of a ring.
The first, introductional section contains elementary observations needed later on. We also show that and are connected graphs of diameter not bigger than 3.
Section 2 concentrates on modules such that the associated graphs contains vertices of small degree. Theorems 2.1 and 2.3 show that if every vertex of () is of finite degree then the graph has only finitely many vertices. An information about the structure of such modules is also given. The remaining part of this section is mainly devoted to investigation of modules such that contains a vertex of degree 1. Theorem 2.13 offers necessary and sufficient condition for a submodule of to be of degree 1 as a vertex in . It also appears (see Theorem 2.18) that either all submodules of degree one in are simple or contains the unique largest submodule of degree 1, examples are presented.
Section 3 concentrates on modules whose graphs are complete, -regular and triangle-free or a tree. For example, Theorem 3.2 describes modules with complete graph . Theorem 3.11 and 3.12 give necessary and sufficient conditions for to be triangle-free and a tree, respectively. The girth of and is also described.
1. Preliminaries
If not specified otherwise denotes a left module over a given ring . We write if is an essential submodule of (i.e. if , for any nonzero submodule of ), stands for the *uniform dimension *of . For , denotes the annihilator of in . The reader can consult [6] and [7] for precise definitions and properties needed in the text.
Let be a simple graph. The vertex set of is denoted by , stands for the *degree *of , i.e. the cardinality of the set of all vertices which are adjacent to . The maximum and minimum degrees of the graph are the maximum and minimum degree of its vertices and are denoted by and , respectively. is a complete graph if every pair of distinct vertices of are adjacent, will stand for a complete graph with n\in\mbox{\mathbb{N}} vertices. The graph is -regular, if for every .
Let . We say that is a *universal vertex *of if is adjacent to all other vertices of and write if and are adjacent. The distance is the length of the shortest path from to if such path exists, otherwise . The *diameter *of is .
The *girth *of a graph , denoted by , is the length of a shortest cycle in . If has no cycles, then .
A subset is independent if no two vertices of are adjacent. For a positive integer , a *-partite *graph is a graph whose vertices can be partitioned into nonempty independent sets.
Let us recall that if is a left module over the ring , then:
- •
The *sum-essential graph *of is a simple graph such that consists of all nontrivial submodules of and two vertices are adjacent if and only if the sum of the corresponding submodules is an essential submodule of .
- •
The *proper sum-essential graph *of , denoted by , is a subgraph of induced by vertices which are not essential submodules of .
Let us remark that has exactly one vertex iff contains a submodule such that both and are simple modules. On the other hand, if the graph is not empty, then the module is not uniform and hence .
Henceforth we will assume that all considered modules have at least two proper submodules, i.e. . Notice that if is a nontrivial nonessential submodule of then , as a vertex of is adjacent to its complement and every vertex which is essential as a submodule of is a universal vertex of . Hence . Moreover, if contains a proper essential submodule (i.e. is not a semisimple module), then . The above remarks will be used freely in the text. Using the above it is easy to see that:
Proposition 1.1**.**
Suppose the module is not simple. Then is semisimple if and only if if and only if there exists a vertex of , such that .
Example 1.2**.**
Let be a direct sum of pairwise non isomorphic simple -modules, where is a finite set with . Then submodules of are in one to one correspondence with subsets of . Let and be a proper submodules such that . Then , where is a subset of different from . This shows that . In particular, maximal submodules correspond to vertices of of degree and minimal submodules to vertices of degree .
Notice that if and , then . The following example shows that this inequality can be sharp even when is a uniform submodule of .
Example 1.3**.**
Consider the -module and set A=\mbox{\mathbb{Z}}(1,0)=\mbox{\mathbb{Z}}_{8} and X=4\mbox{\mathbb{Z}}_{8}\subseteq A. Then is uniform and one can check that and .
Example 1.4**.**
Let , where ’s are simple -modules. Then:
(i)
(ii) is a complete graph of cardinality .
Indeed (i) is a direct consequence of [8, Lemma 1.1] (or can be easily seen directly) and (ii) is a consequence of (i).
Remark that Examples 1.2 and 1.4 apply when and is a finite product of division rings and is a two by two matrix ring over a division ring, respectively.
We close this section with the following observation:
Theorem 1.5**.**
* and are connected graphs of diameter not bigger than 3.*
Proof.
Suppose there are nontrivial submodules such that and are not adjacent. Let denote complements to and in , respectively.
If , then we have the path .
If , then we have the path . Notice all submodules appearing above are not essential, so the thesis follows. ∎
2. On vertices of finite degree
The aim of this section is to characterize modules such that all vertices of the proper sum-essential are of finite degree and to determine submodules of degree one. Let us begin the following theorem.
Theorem 2.1**.**
For a module the following conditions are equivalent:
- (1)
* has only finitely many submodules;* 2. (2)
* is finite;* 3. (3)
Every vertex of is of finite degree; 4. (4)
Either (i) contains a proper essential submodule of finite degree in or (ii) There exist simple submodules of such that and , for all .
Proof.
Implications are trivial.
. Suppose does not contain a proper essential submodule, i.e. is semisimple. Let be a direct sum of simple modules , where ranges over an index set . Pick . Since, for any proper subset of , is adjacent to , the set has finitely many subsets, so is finite, say .
Let us fix and set . Then, for any nonzero submodule of different from , we have . Since is finite, Example 1.4 shows that for all , i.e. (4)(ii) holds.
. If is as in (4)(i), then has finitely many submodules. Suppose (4)(ii) holds and . Since is semisimple, it is enough to see that has only finitely many simple submodules. Let denote the set of all submodules isomorphic to and , where and . Set . If , then . Suppose . Then , for and any is a submodule of . By (4)(ii), so also is finite. This, in particular implies that is finite. Replacing by any , , in the above consideration we see that has finitely many simple submodules and (1) follows. ∎
For obtaining a similar result for the proper sum-essential graph we will need the following lemma:
Lemma 2.2**.**
Suppose that , for any . Then and contains only finitely many submodules.
Proof.
Suppose is not an empty graph. Let us observe first that is finite. Indeed in case would contain nonzero submodules such that , then would be adjacent to infinitely many vertices , for k\in\mbox{\mathbb{N}}. This is impossible as . Let be a uniform submodule of and a complement to in . Then, for any nonzero submodule of , , as . This shows that every nonzero uniform submodule contains only finitely many submodules, in particular, it contains a simple submodule. Hence , as . This means that every vertex of the graph is of finite degree, as has this property. Now, applying Theorem 2.1 to we get , i.e. the thesis holds. ∎
Theorem 2.3**.**
For a module the following conditions are equivalent:
- (1)
Every vertex of is of finite degree; 2. (2)
The graph is finite.
Proof.
In virtue of Lemma 2.2, it is enough to show that for every proper submodule of the set is finite. Let be a complement to in . Then as is a proper submodule of . Moreover is adjacent to any and follows, as .
The implication is clear. ∎
Lemma 2.4**.**
Let . If , then either is a simple module or there exists a simple submodule such that .
Proof.
Suppose is not simple and let be a nonzero proper submodule of . If is essential in , then clearly the second case holds.
Thus we may assume that is not essential in . Let be a complement to . If is a proper submodule of , then is adjacent to and . This yields , which is impossible. Thus . Then also and follows, a contradiction. This shows that if is not essential, then it is simple. ∎
Proposition 2.5**.**
For a submodule of a semisimple module the following conditions are equivalent:
- (1)
; 2. (2)
* is simple and has unique nonzero complement in ;* 3. (3)
* is simple, is not simple and has no other submodules isomorphic to .*
Proof.
Suppose . Then, by Lemma 2.4, is simple. It is also clear that the complement to in is unique. The reverse implication is obvious.
Let be the unique complement to . Let be any submodule of such that and a complement to in . Since is semisimple we get and follows as is unique. In case would be isomorphic to then both and the submodule , where is the isomorphism, would be contained in and . This is impossible, so does not contain submodules isomorphic to , i.e. the statement (3) holds.
Suppose is a submodule of satisfying (3). Let be complements to in and let denote the projection onto along . If , then as is simple. Then, as is semisimple, there exists a submodule such that . Thus is isomorphic to and follows, which is impossible. Hence and consequently and maximality of gives , i.e. (2) holds. ∎
As a direct application of the above proposition and Lemma 2.4 we obtain the following corollaries:
Corollary 2.6**.**
Let be a semisimple ring and a minimal left ideal of . Then if and only if is a two-sided ideal of . In particular, every minimal left ideal of is of degree 1 in if and only if is a direct product of division rings.
Corollary 2.7**.**
The sum-essential graph possesses a vertex of degree 1 if and only if either is a semisimple module possessing a proper simple submodule such that does not contain other submodules isomorphic to or is a chain module with the graph with two vertices (then any vertex is of degree 1).
Our next goal is to describe, in Theorem 2.13, modules such that contains a vertex of degree one. For this we need a series of lemmas.
Lemma 2.8**.**
Let be a proper uniform submodule of and . Suppose that any submodule of satisfying is contained in . Then:
- (1)
; 2. (2)
If , then is the unique complement to in ; 3. (3)
If , then is simple and a submodule is a complement to in if and only if is the complement to in . In particular, if is a complement to in , then .
Proof.
Let us notice that, due to our assumption, any complement of in is contained in the socle of . In particular (1) holds. The above also implies (2).
Suppose now that . Then is a simple module, as is uniform. Let be a complement to in and a complement to in . We know that is semisimple, so is also a complement of in (as ). Moreover, if , then , i.e. is a complement to in . This yields (3). ∎
In the following lemma we collect basic properties of vertices of degree one in .
Lemma 2.9**.**
Let . If , then:
- (1)
* is a uniform module;* 2. (2)
, for every vertex ; 3. (3)
The complement to in is a semisimple module; 4. (4)
Let be a submodule of such that and , then .
Proof.
Suppose and let be the complement to in . Then, for any nonzero submodule of , . Since and , follows. This easily yields (1) and (2).
(3) If is a proper essential submodule of , then , as . Therefore, as , has no proper essential submodules so is semisimple and (3) holds.
(4) By (1), is uniform. Let be a submodule of satisfying assumptions of (4) and be a simple submodule of . If , then .
Now suppose that . If also then would not be essential in and would be adjacent to two different vertices, and the complement of which contains . Hence and follows. ∎
Lemmas 2.8 and 2.9 give immediately the following
Corollary 2.10**.**
Suppose is a vertex of degree one. Let be the unique complement to in and . Then:
- (1)
If , for a submodule of , then ; 2. (2)
* is the unique complement to in ;* 3. (3)
* iff *(in this case ); 4. (4)
* iff **(*in this case );
Proof.
(1) is a direct consequence of Lemma 2.9(3). This statement yields (2) if . For , (2) is just Lemma 2.8(3). Now it is easy to complete the proof of the corollary. ∎
The statement (2) of Lemma 2.9 yields also:
Corollary 2.11**.**
Suppose that the graph has only finitely many vertices of degree 1, then every such vertex contains a simple submodule of .
In the sequel we will need yet another lemma.
Lemma 2.12**.**
Suppose is a nontrivial uniform submodule of a left -module with a nonzero complement , where . Suppose there exist a simple submodule , and such that for any , if then . Then .
Proof.
If has two different complements, then the result is clear. Thus let us assume that is the unique complement to in . Since is uniform, and follows.
Let be as in the formulation of the lemma and . Notice that if then , so also , i.e. . This shows that the map defined by is a well-defined epimorphism of -modules. Moreover, since gives , is an isomorphism. Observe that and hence, as is simple, we have . Thus , where . Notice that is not essential in . Otherwise, and then , for some and . So . Thus and by assumption follows. Now, from , we get , a contradiction. Therefore is not essential in and the essentiality of in shows that is adjacent with two distinct vertices and . So . ∎
Now we are ready to prove the following theorem:
Theorem 2.13**.**
Let a left -module and .
The following conditions are equivalent:
- (1)
; 2. (2)
- (i)
* is uniform and the complement to in is unique and is semisimple;* 2. (ii)
For every simple submodule , and there exists an element such that and ; 3. (3)
- (i)
* is uniform and the complement to in is unique and is semisimple;* 2. (ii)
If and , for a submodule of , then ; 4. (4)
For any submodule of the following conditions hold:
- (i)
If then ; 2. (ii)
* and , then ;* 3. (iii)
* is uniform and has unique complement in .*
Proof.
Suppose . Then part (i) holds by Corollary 2.10 and part (ii) is a direct consequence of Lemma 2.12.
Let be a simple submodule of . If , then the condition (2)(ii) implies that , for any and . Hence . This shows that , for every simple submodule , and .
Let be a submodule of such that and . If then, as is uniform, , so . Suppose and pick . Then , for some suitable and . If then and follows. If , then . Thus in any case and (ii) follows.
Let be a complement to in . By (3)(i), is unique and semisimple. Let be a submodule of such that . If , then as is semisimple. Thus, to prove the implication, it is enough to show that if , then has to be essential in . To this end, suppose . Then, as the complement is unique, . Therefore, by (3)(ii), . Notice also that the condition (3)(i) guarantees that satisfies assumptions of Lemma 2.8. Let .
If (i.e. ). Then the sum is direct. By (i), is uniform, hence . This implies , as .
If then Lemma 2.8(3) implies that simple and is the unique complement to in . Then and follows, as . This completes the proof of (1).
We have already noticed that (3)(i) implies that satisfies the assumptions of Lemma 2.8. Then the equivalence is a direct consequence of this lemma. ∎
Remark 2.14*.*
Using Proposition 2.5, one can see that the condition (4)(iii) in the above theorem is equivalent to
(iii’) is uniform and has no other submodules isomorphic to .
The above remark together with Theorem 2.13 and Proposition 2.5 yield:
Remark 2.15*.*
Suppose . Then is only adjacent to , where .
Of course it may happen that the complement to a uniform nonessential submodule of is unique but . Indeed, let R=\mbox{\mathbb{Z}} and M=\mbox{\mathbb{Z}}\oplus\mbox{\mathbb{Z}}_{2}. Then has unique complement (equal to ) but is adjoint to any submodule , k\in\mbox{\mathbb{N}}. Thus .
The following example and proposition offer various possible interrelations between submodules of degree one in a module .
Example 2.16**.**
- (1)
Let , and . Then if and only if U\subseteq R(1,0)=\mbox{\mathbb{Z}}. In this case . 2. (2)
Let p,q\in\mbox{\mathbb{N}} be prime numbers, R=\mbox{\mathbb{Z}} and M=\mbox{\mathbb{Z}}_{p\infty}\oplus\mbox{\mathbb{Z}}_{q}, where \mbox{\mathbb{Z}}_{p\infty}=\mbox{\mathbb{Z}}[\frac{1}{p}]/\mbox{\mathbb{Z}} is the -quasicyclic group. Then soc(M)=\mbox{\mathbb{Z}}_{p}\oplus\mbox{\mathbb{Z}}_{q}. If , then \deg_{\mathcal{P}_{R}(M)}(\mbox{\mathbb{Z}}_{q})=\infty and a submodule is of degree 1 if and only if U\subseteq\mbox{\mathbb{Z}}_{p\infty}. If , then has no vertices of degree one. 3. (3)
Let be the -module \mathbb{Z}_{4}\oplus\mbox{\mathbb{Z}}_{3} . Then soc(M)=2\mbox{\mathbb{Z}}_{4}\oplus\mbox{\mathbb{Z}}_{3}, \mbox{\mathbb{Z}}_{4} and 2\mbox{\mathbb{Z}}_{4} are the only submodules of degree one, while \deg_{\mathcal{P}_{\mathbb{Z}}(M)}(\mbox{\mathbb{Z}}_{3})=2. 4. (4)
Let P\subseteq\mbox{\mathbb{N}} denote the set of primes. Then any simple submodule of the -module \bigoplus_{p\in P}\mbox{\mathbb{Z}}_{p} is of degree 1.
Proposition 2.17**.**
Suppose there exist two distinct submodules and of a module such that . Then:
- (1)
If then and are simple non isomorphic modules; 2. (2)
If then ; 3. (3)
If then are simple non isomorphic and .
Proof.
(1) Suppose . Then, by Theorem 2.13(4) applied to and , we see that and are simple modules. Moreover, due to Remark 2.14, .
(3) is an immediate consequence of Remark 2.15.
(2) Suppose . If and are simple, the thesis is clear. Suppose is not simple and let . Then, by (3) is not essential in . Moreover , and have the same unique complement described in Remark 2.15.
Let be a vertex of , which is adjacent to . Applying Theorem 2.13 to and we obtain . Hence, as , also and gives , i.e. . ∎
As we have seen earlier, submodules of which are of degree one as elements of are uniform. By the above lemma, if and are such submodules having nonzero intersection then is a uniform submodule of . Notice that, in general, a sum of two uniform submodules having nonzero intersection is not always uniform.
We close this section with the following theorem:
Theorem 2.18**.**
Let be a module containing a submodule of degree 1 in . Then either all submodules of degree 1 are simple or contains the unique largest submodule of degree 1.
Proof.
Suppose contains a not simple submodule such that . Let with . Then, by Proposition 2.17, we know that and . Let and be the complement to in . Then, by Remark 2.15, is the unique complement to in .
Let be a uniform submodule such that . We claim is a complement to in . Clearly . If , then there exists a simple module such that . Then , as is uniform, a contradiction. This shows that . Notice also that if then and follows, as . This proves the claim.
Let , Notice that to prove the theorem it is enough to show only that every chain in is bounded, then the thesis is a consequence of Zorn’s Lemma and Proposition 2.17(2). To this end, let be a chain in and . Then is uniform and the above shows that is the unique complement to in . Let be a simple submodule of such that . Let and . Then , for some index . Since we may apply Theorem 2.13(2)(ii) to find such that and . The above shows that satisfies Theorem 2.13(2) and proves that , i.e. . This yields the thesis. ∎
Let us notice that in case is a finite graph, the above theorem is a direct consequence of Proposition 2.17.
3. Complete, -regular and triangle-free graphs and
Let us begin this section with the following easy observation:
Lemma 3.1**.**
Let be a universal vertex. If , then is a simple module.
Proof.
Notice that if then can not be adjacent to any of its submodules. Since is a universal vertex, the thesis follows. ∎
The following theorem gives a characterization of modules with a complete sum-essential graph.
Theorem 3.2**.**
Let be a nonzero nonsimple module. Then, is a complete graph if and only if one of the following conditions hold:
- (1)
* is a uniform module;* 2. (2)
- (i)
Every nonzero nonessential submodule of is simple; 2. (ii)
, where are simple submodules.
In particular, if is semisimple, then is a complete graph if and only if is direct sum of two simple modules.
Proof.
Suppose is not uniform. Then Lemma 3.1 shows that (2)(i) holds. Now (2)(ii) is a consequence of (i) and the assumption that every vertex is universal. ∎
Let us notice that every universal vertex of is also universal in and hence a nonempty graph is complete if and only if is such. Thus the above theorem gives the following corollary:
Corollary 3.3**.**
Suppose is not uniform. Then, is a complete graph if and only if the following conditions hold:
- (i)
Every nonzero nonessential submodule of is simple; 2. (ii)
, where are simple submodules.
If does not contain proper essential submodules (i.e. if is semisimple) something more can be said. The last statement of the following corollary is a consequence of Example 1.4.
Corollary 3.4**.**
For a semisimple module , the following conditions are equivalent:
- (1)
* is a complete graph;* 2. (2)
* contains a universal vertex;* 3. (3)
* for some simple modules and .*
In this case is a complete graph and .
Corollary 3.5**.**
Let be a semisimple ring with a nontrivial left ideal. Then is a complete graph if and only if or , where , and are division rings.
In Theorem 3.2 a characterization of modules with a complete sum-essential graph was given. The following theorem shows that -regular sum-essential graph have to be complete.
Theorem 3.6**.**
Let be a nonzero nonsimple module and k\in\mbox{\mathbb{N}}. The following conditions are equivalent:
- (1)
The graph is -regular; 2. (2)
* is a complete graph and .*
Proof.
Let be a -regular. Suppose contains a proper essential submodule . Then is adjacent to any . Then and is a complete graph.
Assume now that does not have proper essential submodule, i.e. is semisimple. Let be a simple submodule of . Then is the number of maximal submodules of not containing . Let be such maximal submodules. Then is adjacent to any , and to arbitrary proper submodule of containing . The fact that is -regular implies that has no proper submodules containing . Thus the semisimple module for suitable simple submodule of . Now the result is a consequence of Example 1.4.
The implication is clear. ∎
The following theorem presents the structure of modules having triangle-free proper sum-essential graphs.
Theorem 3.7**.**
Let be a module such that . The following conditions are equivalent:
- (1)
The graph is triangle-free; 2. (2)
Either is a direct sum of two non-isomorphic simple modules or is a chain module with exactly two proper submodules. 3. (3)
* the complete graph with two vertices.*
Proof.
Suppose is a triangle-free. If would contain an essential direct sum of nonzero submodules, then would form a triangle, a contradiction. Hence .
If , then any two vertices are adjacent, so has to be a chain module with two vertices only.
Suppose and let be a uniform submodule and its complement in . If would contain a proper nonzero submodule then and would form a triangle. Thus and are simple and . If would be a proper submodule, then we would have a triangle with vertices . Therefore in this case. Moreover and are not isomorphic, as three distinct simple submodules would form a triangle. This completes the proof.
Implications are clear. ∎
Corollary 3.8**.**
Let be a nonzero nonsimple module. Then the girth of the graph is either or .
Proof.
If contains a triangle then . If is triangle-free then, by Theorem 3.7, or is a single point and . ∎
Remark that bipartite graphs are triangle-free, so the above corollary applies in this case.
Definition 3.9**.**
We say that submodules and of a module are strongly disjoint if there are no nonzero isomorphic submodules of such that and .
The following lemma offers some other characterizations of strongly disjoint submodules.
Lemma 3.10**.**
For submodules and of a left -module the following conditions are equivalent:
- (1)
* and are strongly disjoint;* 2. (2)
For any nonzero elements and , ; 3. (3)
- (i)
; 2. (ii)
If is a nonzero submodule of , then or .
Proof.
The implication is clear as any cyclic -module isomorphic to .
Suppose (2) holds. Then clearly . Let be a nonzero submodule of and , with and . If one of and is equal to zero, then the result holds. If both elements are nonzero we can use (2) to find an element which belongs to exactly one of the sets and . Then either or and the result follows.
Let and be isomorphic submodules. Let us consider the submodule , where is a given isomorphism. Let and be such that . If , then and follows. Similarly, if , then yields . Therefore, by (3), , i.e. and are strongly disjoint. ∎
Theorem 3.11**.**
Let be a module with nonempty graph . The following conditions are equivalent:
- (1)
* is triangle-free;* 2. (2)
- (i)
* (so any nonzero, nonessential submodule is uniform);* 2. (ii)
If and , then and are strongly disjoint. 3. (3)
If and , then and are strongly disjoint.
Proof.
(1)(2) Assume is triangle-free. By assumption is not uniform. The argument used at the beginning of the proof of Theorem 3.7 shows that , i.e. holds.
Let and . Let = and be the complement to in . If would be nonzero, then would be a proper nonessential submodule of and we would have a triangle with vertices (as ). This is impossible, so . Let be a uniform nonzero submodule of . Using (2)(i) we see that or , as otherwise would form a triangle. Now (2)(ii) is a consequence of Lemma 3.10.
The implication is clear.
Let be such that and . In particular, by assumption, we have . Let . Notice that , as otherwise and (3) would imply that either or , which is impossible. This shows that is triangle free. ∎
Theorem 3.12**.**
Let be a module with . Then the following conditions are equivalent:
- (1)
* is a tree;* 2. (2)
If and , then and are strongly disjoint and one of and is a simple module; 3. (3)
* is a star graph with the center , for a simple submodule of .*
Proof.
(1)(2) Every tree is triangle-free, so the first part of the statement (2) is given by Theorem 3.11. Let be such that . In case we would have proper nonzero submodules and of and , respectively, we would have a cycle in a tree. Hence one of , has to be a simple module, i.e. (2) holds.
Suppose (2) holds. Then contains a simple submodule, call it . Theorem 3.11 shows that is triangle free and . In particular has at most two different simple submodules.
If then, by (2), is a star graph with center .
Suppose is not simple. Since , where are the only simple submodules of . In particular, any contains one of or . The statement (2) implies that one of the simples, say , does not have proper essential extensions in . Therefore, if , then and is a star graph with center .
The implication is clear. ∎
Let us remark that the proof of the implication gives the module structure of modules with being a tree. The -modules \mbox{\mathbb{Z}}\oplus\mbox{\mathbb{Z}}_{q} or \mbox{\mathbb{Z}}_{p^{n}}\oplus\mbox{\mathbb{Z}}_{q} are examples of such modules, where p,q,n\in\mbox{\mathbb{N}} and are prime. We have seen in Corollary 3.8 that the girth of is equal to 3 when finite. In case of the graph we have the following:
Corollary 3.13**.**
Suppose is a nonempty graph. Then .
Proof.
Assume is a triangle-free graph, i.e. . If is not a tree then, using Theorems 3.12 and 3.11 we can pick two strongly disjoint uniform submodules and in such that and neither nor is simple. Let , be proper nonzero submodules of and , respectively. Then form a cycle, so . This yields the thesis. ∎
The girth of can take any value as described above. If , then . If consider the group as a -module, where p,q\in\mbox{\mathbb{N}} are different primes and . Then any nonessential submodule of is contained either in \mbox{\mathbb{Z}}_{p^{m}} or \mbox{\mathbb{Z}}_{q^{n}} and Theorem 3.11 (or direct argument) shows that is triangle free. If , then . If , then .
We close the paper with the following result:
Proposition 3.14**.**
Suppose that the module has exactly maximal submodules. The following conditions are equivalent:
- (1)
* is -partite graph;* 2. (2)
* is a semisimple module.*
Proof.
Let be the maximal submodules of , and be the complement to in .
In case would be nonzero then would generate a complete subgraph of . Since is -partite graph, this is impossible and follows. This yields the thesis.
Suppose is a semisimple module. For , let us define and for . Then, using semisimplicity of it is easy to see that form -partitioning subsets of . ∎
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