COMPLETE HOMOMORPHISMS BETWEEN MODULE LATTICES

We examine the properties of certain mappings between the lattice L(R) of ideals of a commutative ring R and the lattice L(RM) of submodules of an R-module M , in particular considering when these mappings are complete homomorphisms of the lattices. We prove that the mapping λ from L(R) to L(RM) defined by λ(B) = BM for every ideal B of R is a complete homomorphism if M is a faithful multiplication module. A ring R is semiperfect (respectively, a finite direct sum of chain rings) if and only if this mapping λ : L(R) → L(RM) is a complete homomorphism for every simple (respectively, cyclic) R-module M . A Noetherian ring R is an Artinian principal ideal ring if and only if, for every R-module M , the mapping λ : L(R) → L(RM) is a complete homomorphism. Mathematics Subject Classification 2010: 06B23, 06B10, 16D10, 16D80


Introduction
In this paper we continue the discussion in [7] concerning mappings, in particular homomorphisms, between the lattice of ideals of a commutative ring and the lattice of submodules of a module over that ring.
A lattice L is called complete provided every non-empty subset S has a least upper bound ∨S and a greatest lower bound ∧S.Given complete lattices L and L we say that a mapping ϕ : L → L is a complete homomorphism provided ϕ(∨S) = ∨{ϕ(x) : x ∈ S} and ϕ(∧S) = ∧{ϕ(x) : x ∈ S}, for every non-empty subset S of L. A complete homomorphism which is a bijection (respectively, injection, surjection) will be called a complete isomorphism (respectively, complete monomorphism, complete epimorphism).The first result is standard and easy to prove.Lemma 1.1.The following statements are equivalent for a bijection ϕ from a complete lattice L to a complete lattice L .
(i) ϕ is a complete isomorphism.
Moreover, in this case the inverse mapping ϕ −1 : L → L is also a complete isomorphism.
An element x of a complete lattice L is called compact in case whenever x ≤ ∨S, for some non-empty subset S of L, there exists a finite subset F of S such that x ≤ ∨F .The next result is also easy to prove.Lemma 1.2.Let ϕ : L → L be a complete isomorphism from a complete lattice L to a complete lattice L and let x be a compact element of L. Then ϕ(x) is a compact element of L .
for all elements x, y, z in L. The next result is also well known and easy to prove.It states that a lattice is distributive if and only if its dual lattice is distributive.
for all x, y, z in L.
Throughout this note all rings will be commutative with identity and all modules will be unital.Let R be a ring and M be any R-module.Let L(R) denote the lattice of all ideals of the ring R and let L( R M ) denote the lattice of all submodules of the R-module M .In [7] we investigate the mapping λ : L(R) → L( R M ) defined by λ(B) = BM for every ideal B of R and the mapping µ : L( R M ) → L(R) defined by µ(N ) = (N : R M ) for every submodule N of M , where (N : R M ) denotes the set of elements r ∈ R such that rM ⊆ N .The module M is called a λ-module in [7] in case λ : L(R) → L( R M ) is a homomorphism.Similarly, in [7] the module M is called a µ-module if the above mapping µ is a homomorphism.For any unexplained terminology and notation, please see [7].
Note that the lattice L( R M ) is complete when we define for every non-empty collection S of submodules of M .In particular the lattice L(R) is complete.The module M will be called λ-complete in case the above mapping clear that every λ-complete module is a λ-module and every µ-complete module is a µ-module but, in each case, the converse is false in general, as we can easily show.
For example, let Z denote the ring of rational integers and let p be any prime in Z. Then the simple Z-module U = Z/Zp is a λ-module.Let q be any prime in Z other than p and let S denote the collection of ideals of Z of the form Zq n for all positive integers n.Then Then the Z-module V is a µ-module (see [7,Example 3.11]).However V contains an infinite collection T of proper submodules Thus the Z-module V is not µ-complete.
Proposition 1.4.Given any ring R and R-module M the following statements are equivalent.
Moreover, in this case M is a faithful R-module.Again let R be a ring and let M be an R-module.
will be denoted by λ.Note that if B is any ideal of the ring R/A then B = B/A for a unique ideal B of R containing A and hence In addition, the mapping µ : Let R be any ring.An R-module M is called a multiplication module in case for each submodule N of M there exists an ideal B of R such that N = BM .
Cyclic modules are multiplication modules as are projective ideals of R or ideals of R generated by idempotent elements (see [2]).We prove that for any ring R an R-module M is µ-complete if and only if M is a finitely generated multiplication module (Theorem 2.2).An easy consequence is that the mapping µ (respectively, λ) is a complete isomorphism if and only if M is a finitely generated faithful multiplication module (Corollary 2.4).
For any ring R, projective modules are λ-complete (Corollary 3.4) as are faithful multiplication modules (Theorem 3.6).We prove that a ring R is arithmetical if and only if every R-module is a λ-module (Theorem 4.6).The ring R is semiperfect if and only if every simple R-module is λ-complete (Theorem 4.2).On the other hand, R is a direct sum of chain rings if and only if every cyclic R-module M is λ-complete (Theorem 4.7).Note that we do not yet know which rings R have the property that every R-module is λ-complete.It is proved that a Noetherian ring R is an Artinian principal ideal ring if and only if every R-module is λ-complete (Theorem 4.12).

µ-complete modules
Let R be a ring and let M be an R-module.In this section we shall investigate µ-complete modules.We begin with the following basic result.The result follows.
Note that, given any ring R and R-module M , the mapping µ is not a surjection in case M is not a faithful R-module because in this case no submodule N of M has the property that (N : R M ) = 0.The next result characterizes µ-complete modules.
Theorem 2.2.Given any ring R, the following statements are equivalent for an (ii) M is a finitely generated multiplication module.Let r ∈ ( i∈I L i : R M ).Then rM is a finitely generated submodule of i∈I L i .
There exists a finite subset Thus ( i∈I L i : R M ) ⊆ i∈I (L i : R M ) and we have proved that ( i∈I L i : R M ) = i∈I (L i : R M ).By Lemma 2.1, M is µ-complete.
By the proof of (ii) ⇒ (i), the mapping µ is a complete isomorphism.Thus µ is a monomorphism.Given a ring R and an R-module M , note that Theorem 2.2 shows that whenever the mapping µ : L( R M ) → L(R) is a complete homomorphism then it is a monomorphism.This is not true if µ is merely a homomorphism (see, for example, [7, Example 3.11 and Proposition 3.12]).

Proof. By Theorem 2.2.
In contrast to Corollary 2.3 homomorphic images of λ-complete modules need not be λ-complete.For example, the Z-module Z is λ-complete but we have already noted that the simple Z-module Z/Zp is not λ-complete for every prime p in Z.
(Note that every homomorphic image of a λ-module over the ring Z is also a λ- (i) The mapping λ : L(R) → L( R M ) is a complete isomorphism.
(ii) The mapping µ : L( R M ) → L(R) is a complete isomorphism.
(iii) The R-module M is a finitely generated faithful multiplication module.
Proof.By Proposition 1.4 and Theorem 2.2.
Corollary 2.5.Let R be a ring and let M be any µ-complete R-module with A = ann R (M ).Then the (R/A)-module M is a λ-complete module.
Note that in general µ-complete modules are not λ-complete.For, let R be a domain that is not Prüfer.By [7,Theorem 2.3], there exists a cyclic R-module M which is not a λ-module and hence is not λ-complete.However, every cyclic module over any ring is a finitely generated multiplication module.

λ-complete modules
In contrast to the case of µ-complete modules, the situation for (non-faithful) λ-complete modules is more complex.We already know that simple modules over Z are not λ-complete although they are clearly finitely generated multiplication modules.First we prove an elementary result characterizing λ-complete modules.Proof.(a) Let K be a direct summand of a λ-complete module M .Let S be any non-empty collection of ideals of R. Then Given any non-empty collection S of ideals of R we have: Corollary 3.4.Given any ring R, every projective R-module is λ-complete.
Proof.Clearly the R-module R is λ-complete.Apply Lemma 3.3.
Recall the following result (see [2, Theorem 1.2] or [7, Lemma 2.10]).Lemma 3.5.Let R be any ring.Then an R-module M is a multiplication module if and only if for each maximal ideal P of R either (a) for each m in M there exists p in P such that (1 − p)m = 0, or (b) there exist x ∈ M and q ∈ P such that (1 − q)M ⊆ Rx.
We now strengthen [7, Theorem 2.12].Theorem 3.6.Let R be any ring.Then every faithful multiplication R-module is a λ-complete module.
Proof.Let M be a faithful multiplication R-module.Let S be any non-empty collection of ideals of R. Then (∩ B∈S B)M ⊆ ∩ B∈S (BM ).Suppose that there exists Then I is a proper ideal of R. Let P be a maximal ideal of R such that I ⊆ P .
Clearly (1 − p)m = 0 for some p ∈ P implies that 1 − p ∈ I, a contradiction.By Lemma 3.5 there exist x ∈ M and q ∈ P such that (1 − q)M ⊆ Rx.Note that for for some r B ∈ B for each ideal B in S. If B and C are ideals in S then (r B −r C )x = 0 and hence (1 (i) Every semisimple R-module is λ-complete.
(ii) Every simple R-module is λ-complete.
(iii) The ring R is semiperfect.
Proof.(i) ⇒ (ii) Clear.Next we investigate rings R with the property that every cyclic R-module is λ-complete.First we recall a result of Stephenson (see [9,Theorem 1.6]).
Lemma 4.3.The following statements are equivalent for a module M over a ring R.
(iii) Every 2-generated submodule of M is a µ-module.
(i) R is a field.
(iii) The R-module F is λ-complete.
(iii) ⇒ (i) Let B i (i ∈ I) denote the collection of all non-zero ideals of R. Then Lemma 3.1 gives that Thus ∩ i∈I B i = 0.It follows that R has non-zero socle and hence R = F .
Contrast the following result with Theorem 4.8.
Theorem 4.12.A Noetherian ring R has the property that every R-module is λ-complete if and only if R is an Artinian principal ideal ring.
Proof.Suppose first that every R-module is λ-complete.Let P be any prime ideal of R. By Lemma 4.10, every (R/P )-module is λ-complete and hence the domain

Other homomorphisms
In general there will be many complete homomorphisms ν : L(R) → L( R M ) for a given ring R and R-module M (see [7,Section 5]).Note the following result.
Proposition 5.1.Let R be a ring and let M be an R-module such that there exists a complete isomorphism ν : L(R) → L( R M ).Then M is a finitely generated R-module.

(
iii) The mapping µ : L( R/A M ) → L(R/A) is a complete isomorphism.(iv) The mapping λ : L(R/A) → L( R/A M ) is a complete isomorphism.Moreover in this case the mapping µ : L( R M ) → L(R) is a monomorphism.Proof.(i) ⇒ (ii) Let T denote the collection of all cyclic submodules of the µcomplete module M .Then M = N ∈T N .By Lemma 2.1, R = (M : R M ) = ( N ∈T N : R M ) = N ∈T (N : R M ), and hence R = (Rm 1 : R M ) + • • • + (Rm n : R M ) for some positive integer n and elements other words, M is finitely generated.By[7,   Theorem 3.8], M is also a multiplication module.(ii)⇒ (i) Suppose that M is a finitely generated multiplication module.By [7, Lemma 3.1 and Theorem 3.8] and induction,(K 1 + • • • + K n : R M ) = (K 1 : R M ) + • • • + (K n : R M ),for every positive integer n and submodulesK i (1 ≤ i ≤ n).Let L i (i ∈ I)be any non-empty collection of submodules of M .Clearly, i∈I (L i : R M ) ⊆ ( i∈I L i : R M ).

module by [ 7 ,Corollary 2 . 4 .
Theorem 2.3].)Given a ring R, the following statements are equivalent for an R-module M .

Lemma 3 . 1 .
Let R be a ring.Then an R-module M is λ-complete if and only if (∩ B∈S B)M = ∩ B∈S (BM ) for every non-empty collection S of ideals of R. Proof.Let S be any non-empty collection of ideals of R. Then λ(∨S) = ( B∈S B)M = B∈S (BM ) = ∨{λ(B) : B ∈ S}.In addition, λ(∧S) = (∩ B∈S B)M and ∧{λ(B) : B ∈ S} = ∩ B∈S (BM ).The result follows.Corollary 3.2.Let A be any ideal of a ring R. Then the R-module R/A is λcomplete if and only if ∩ B∈S (A+B) = A+(∩ B∈S B) for every non-empty collection S of ideals of R. Proof.Apply Lemma 3.1 to the module M = R/A.Lemma 3.3.Let R be any ring.Then (a) Every direct summand of a λ-complete module is λ-complete.(b) Every direct sum of λ-complete modules is also λ-complete.
a contradiction.Thus ∩ B∈S (BM ) = (∩ B∈S B)M for every non-empty subset S of ideals of R. By Lemma 3.1 M is λ-complete.Proof.Suppose first that U is λ-complete.Let S denote the collection of ideals B of R such that R = P +B.By Corollary 3.2 R = P +C where C = ∩ B∈S B. Clearly C is a supplement of P in R R. Conversely, suppose that P has a supplement G in R R. Let T be any non-empty collection of ideals of R. Then P + (∩ D∈T D) = P = ∩ D∈T (P + D), unless D P for all D ∈ T .Now suppose that D P for all D ∈ T .Let D ∈ T .Then R = P + G = P + D implies that R = P + (D ∩ G) and hence G = D ∩ G ⊆ D. It follows that R = P + G ⊆ P + (∩ D∈T D) ⊆ ∩ D∈T (P + D) ⊆ R. Thus in any case P + (∩ D∈T D) = ∩ D∈T (P + D).By Corollary 3.2, the R-module U is λ-complete.Theorem 4.2.The following statements are equivalent for a ring R.

Corollary 4 . 4 .
The following statements are equivalent for a module M over a ring R.(i) The lattice L( R M ) is distributive.

R/P
is a field by Lemma 4.11.Thus every prime ideal of R is maximal.By[5,   Theorem 4.6], the ring R is Artinian.Next, by Theorem 4.6 every ideal of R is a multiplication module and hence, by [2, Corollary 2.9] every ideal of R is principal.Thus R is a principal ideal ring.Conversely, suppose that R is an Artinian principal ideal ring.Let M be any R-module.Let S be any non-empty collection of ideals of R. Because R is Artinian, there exists a finite subset S of S such that ∩ B∈S B = ∩ B∈S B. Noting that R is a principal ideal ring and so every ideal of R is a multiplication module, Theorem 4.6 and [7, Lemma 2.1] together give that (∩ B∈S B)M = ∩ B∈S (BM ).Thus,∩ B∈S (BM ) ⊆ ∩ B∈S (BM ) = (∩ B∈S B)M = (∩ B∈S B)M,and hence (∩ B∈S B)M = ∩ B∈S (BM ).By Lemma 3.1 the R-module M is λcomplete.