MAPPINGS BETWEEN MODULE LATTICES

We examine the properties of certain mappings between the lattice of ideals of a commutative ring R and the lattice of submodules of an R-module M , in particular considering when these mappings are lattice homomorphisms. We prove that the mapping λ from the lattice of ideals of R to the lattice of submodules of M defined by λ(B) = BM for every ideal B of R is a (lattice) isomorphism if and only if M is a finitely generated faithful multiplication module. Moreover, for certain but not all rings R, there is an isomorphism from the lattice of ideals of R to the lattice of submodules of an R-module M if and only if the mapping λ is an isomorphism. Mathematics Subject Classification (2010): 06B99, 13F05, 13E05, 13C99


Introduction
This paper is concerned with mappings, in particular homomorphisms, between lattices. Let L and L be lattices. As usual, given a and b in L, the least upper bound and greatest lower bound of a and b are denoted by a ∨ b and a ∧ b, respectively. Given mappings ϕ and θ from the lattice L to the lattice L we define ϕ ≤ θ provided ϕ(a) ≤ θ(a) for all a ∈ L. Clearly, ϕ ≤ θ and θ ≤ ϕ together imply ϕ = θ. We begin with a very simple result. Lemma 1.1. Let L, L 1 and L 2 be lattices, let ϕ, ϕ 1 and ϕ 2 be mappings from L to L 1 and let θ, θ 1 and θ 2 be mappings from L 1 to L 2 such that ϕ 1 ≤ ϕ 2 , θ 1 ≤ θ 2 and θ(a) ≤ θ(b) for all a, b ∈ L 1 with a ≤ b. Then θϕ 1 ≤ θϕ 2 and θ 1 ϕ ≤ θ 2 ϕ.
The next result is absolutely standard and easy to prove.
Lemma 1.2. The following statements are equivalent for a bijection ϕ from a lattice L to a lattice L .
Moreover, in this case the inverse mapping ϕ −1 : L → L is also an isomorphism.
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. The collection of submodules of M form a lattice which we shall denote by L( R M ) with respect to the following definitions: In particular, we shall denote the lattice L( R R) of ideals of R by L(R). We shall be interested in mappings between L(R) and L( R M ). Let R be a ring and M an R-module. Let L and N be submodules of M . Then (L : R N ) will denote the set of elements r ∈ R such that rN ⊆ L. Note that (L : R N ) is an ideal of R. In particular, (0 : R M ) is the annihilator of M in R and we shall denote it simply by ann R (M ). As usual, M is called faithful in case ann R  Similarly, Lemma 1.1 gives that µ = 1µ ≤ (µλ)µ = µ(λµ) ≤ µ1 = µ, so that µ = µλµ. Lemma 1.3. With the above notation, the following statements are equivalent.
The proof of the next result is similar to the proof of Lemma 1.3 Lemma 1.4. With the above notation, the following statements are equivalent.
(i) λ is an injection.
Moreover, in this case M is faithful.
Corollary 1.5. With the above notation, the mapping λ is a bijection if and only if µ is a bijection. In this case λ and µ are inverses of each other.
Again let R be a ring and let M be an R-module. Let A = ann R (M ). By defining We shall prove that a domain R is Prüfer if and only if the above mapping λ : L(R) → L( R M ) is a (lattice) homomorphism for every (cyclic) R-module M (Theorem 2.3). Given a general ring R, if M is a faithful multiplication module then the mapping λ is a homomorphism (Theorem 2.12). Moreover, λ is an isomorphism if and only if M is a finitely generated faithful multiplication module and in this case the inverse of λ is µ which is also an isomorphism (Theorem 4.3). Furthermore, in case R is a semilocal Noetherian domain or a Dedekind domain then the lattices L(R) and L( R M ) are isomorphic if and only if the above mapping λ is an isomorphism (Corollary 5.4 and Theorem 5.5).

The Mapping λ
Let R be a ring and let M be an R-module. Let the mapping λ : L(R) → L( R M ) be as before, so that λ(B) = BM for every ideal B of R. It is clear that for all ideals B and C of R. Thus λ is a homomorphism if and only if for all ideals B and C of R. It will be convenient for us to call the module M a λ-module in case the above mapping λ is a homomorphism.
Lemma 2.1. The following statements are equivalent for an R-module M .
Proof. (i) ⇒ (ii) By induction on n, (iii) holds. Let G and H be any ideals of R. Then There exist a positive integer n and elements for all ideals G and H of R. Thus λ is a homomorphism and M is a λ-module.
Let R be a domain with field of fractions F . Let I be any non-zero ideal of R. Then I * is the R -submodule of F consisting of all elements f ∈ F such that f I ⊆ R. Note that I * I is an ideal of R. The ideal I is called invertible provided I * I = R (see, for example, [5, p. 67]). It is well known that I is an invertible ideal of R if and only if there exists an R-submodule X of F such that IX = R. The domain R is called Prüfer in case every non-zero finitely generated ideal of R is invertible. For a very comprehensive account of the properties of Prüfer domains see [5].
Lemma 2.2. Let R be a domain, let B and C be invertible ideals of R and let M be the R-module B + C. Then M is a λ-module only if M is also an invertible ideal of R.
Proof. We are given that Thus BC = (B ∩ C)(B + C). It follows that the ideal B + C is invertible. The next result is presumably known but we do not have a reference for it. Theorem 2.3. The following statements are equivalent for a (commutative) domain R.
Proof. (i) ⇒ (ii) Let M be any R-module. Let B and C be non-zero finitely generated ideals of R. Let D = B + C. Then D is a finitely generated ideal of R and hence D * D = R. Let G = D * B and H = D * C so that G and H are ideals of (iv) ⇒ (i) Let B, C and E be ideals of R. By (iv) and Lemma 2.1 we know that B(R/E)∩C(R/E) = (B ∩C)(R/E) and hence that (B +E)∩(C +E) = (B ∩C)+E. It follows that B ∩ (C + E) ⊆ (B ∩ C) + E and hence that B ∩ (C + E) ⊆ (B ∩ C) + (B ∩ E). Therefore B ∩ (C + E) = (B ∩ C) + (B ∩ E) for all ideals B, C and E of R. By [5,Theorem 25.2], R is a Prüfer domain.
(vi) ⇒ (i) By Lemma 2.2, every non-zero 2-generated ideal of R is invertible because every non-zero principal ideal is clearly invertible. By [5,Theorem 22.1], R is a Prüfer domain.
Note that if R is a domain which is not Prüfer then Theorem 2.3 shows that, despite the fact that the R-module R is a λ-module, there exists an ideal of R which is not a λ-module and there exists a homomorphic image of R which is not a λ-module. Thus in general the class of λ-modules is not closed under taking submodules and homomorphic images.
There is another class of rings for which every module is a λ-module. If R is a ring then an R-module M will be called a chain module provided the lattice L( R M ) is a chain, that is for any submodules N and L of M either N ⊆ L or L ⊆ N . The ring R will be called a chain ring if the R-module R is a chain module.
Proposition 2.4. Let R be any chain ring. Then every R-module is a λ-module.
Proof. Let M be any R-module. Let B and C be ideals of R. Without loss of generality we can suppose that B ⊆ C because R is a chain ring. Then BM ⊆ CM and we have: To find other examples of λ-modules we first prove a simple lemma.
Lemma 2.5. Let R be any ring. Then (a) Every direct summand of a λ-module is a λ-module.
(b) Every direct sum of λ-modules is also a λ-module.
Proof. (a) Let K be a direct summand of a λ-module M . Let B and C be any ideals of R. Then be any collection of λ-modules and let L = ⊕ i∈I L i . Given any ideals B and C of R we have: Corollary 2.6. Let R be any ring and let an R-module M = M 1 ⊕ M 2 be the direct sum of a projective submodule M 1 and a semisimple submodule M 2 . Then the module M is a λ-module.
Proof. Clearly every simple module and the R-module R are λ-modules. Apply Lemma 2.5.
For any ring R, every semisimple R-module is a λ-module but the mapping λ is not a monomorphism (if R is not von Neumann regular) and seldom an epimorphism, as the following result shows. Proposition 2.7. Let R be any ring and let M be a semisimple R-module. Then the above mapping λ is a homomorphism. Moreover, (a) λ is a monomorphism only if R is von Neumann regular, and (b) λ is an epimorphism only if M is cyclic.
Proof. The first part follows by Corollary 2.6. Let M = ⊕ i∈I U i for some non-empty collection of simple R-modules U i (i ∈ I).
To prove (b), we suppose now that λ is an epimorphism. Suppose that I contains at least two elements.
Thus I is finite and M is a finite direct sum of non-isomorphic simple modules. In this case it is well known that M is cyclic.
Let Z denote the ring of rational integers. Let M denote the Z-module ⊕ p∈S (Z/Zp), for any infinite set S of primes in Z. Then Proposition 2.7 shows that the mapping λ : L(Z) → L( Z M ) is a homomorphism that is neither a monomorphism nor an epimorphism. The question remains, for a given ring R, which R-modules M are λ-modules. This is certainly the case if λ : L(R) → L( R M ) is a bijection (see Lemma 1.2) but note the following simple fact.
Example 2.8. Let R be any ring and let F be a free R-module of rank ≥ 2. Then the mapping λ : L(R) → L( R F ) is a monomorphism but not an epimorphism.
Proof. Given ideals B and C of R, λ(B) = λ(C) implies that BF = CF and hence B = C. It follows that λ is an injection. By Corollary 2.6 λ is a monomorphism. However, if F has a basis f i (i ∈ I) then, for each j ∈ I, Rf j = λ(B) for any ideal B of R. Thus λ is not an epimorphism.
Next we consider when the mapping λ is a surjection for a given module M . The module M is called a multiplication module in case λ is a surjection, in which case M satisfies the equivalent conditions of Lemma 1.3. In other words, M is a multiplication module if and only if for each submodule N of M there exists an ideal B of R such that N = BM . Multiplication modules have been extensively studied (see, for example, [1] - [4], [8] - [11]). Note the following simple fact about multiplication modules that is included for completeness. It is quite possible for the mapping λ to be a surjection but not an injection, as we shall see shortly. Note the following fact about multiplication modules which we shall require later.
Corollary 2.11. Let R be any ring. Then an R-module M is a finitely generated multiplication module if and only if for each maximal ideal Proof. Suppose that M is a finitely generated multiplication module. There exist a positive integer n and elements m i ∈ M (1 ≤ i ≤ n) such that M = Rm 1 + · · · + Rm n . Let P be any maximal ideal of R. By the lemma, either (a) or (b) in the lemma holds. Suppose that (a) holds. Then for each This proves the necessity. Conversely, suppose that the module M has the stated property. Suppose that the ideal x∈M (Rx : R M ) is proper and let Q be a maximal ideal of R such that x∈M (Rx : R M ) ⊆ Q.
By hypothesis, there exist u ∈ M, q ∈ Q such that (1 − q)M ⊆ Ru and hence 1 − q ∈ (Ru : R M ) ⊆ Q, a contradiction. Thus R = x∈M (Rx : R M ) and there exist a positive integer k and elements u i ∈ M (1 ≤ i ≤ k) such that R = (Ru 1 : R M ) + · · · + (Ru k : R M ). It follows that Hence M = Ru 1 + · · · + Ru k and M is finitely generated. By [4,Corollary 1.5] M is a multiplication module.
The next result shows that multiplication modules are λ-modules in certain cases.
Theorem 2.12. Let R be any ring. Then every faithful multiplication R-module is a λ-module.
Corollary 2.13. Let R be a ring and let M be any multiplication module with A = ann R (M ). Then the mapping λ : Proof. By Lemma 2.9 and Theorem 2.12.
Corollary 2.14. Let R be any ring and let an R-module M be a direct summand of a direct sum of faithful multiplication R-modules. Then M is a λ-module.
For any ring R, the R-module R ⊕ R is a faithful λ-module which is not a multiplication module and therefore the converse of Theorem 2.12 is false. Note further that in general if R is a domain which is not Prüfer then there exists a cyclic R-module which is not a λ-module (see Theorem 2.3). Clearly cyclic modules are multiplication modules. This shows that the modules in Theorem 2.12 need to be faithful. Moreover in Theorem 2.12, although λ is an epimorphism it need not be an isomorphism as the following result shows.
Proposition 2.15. Let R be a ring and let I be a proper ideal of R which is generated by idempotent elements such that ann R (I) = 0. Then the R-module I is a faithful multiplication module and the mapping λ : L(R) → L( R I) is an epimorphism but not a monomorphism.
Proof. Let J be any ideal of R with J ⊆ I. Let a ∈ J. Then a ∈ Re 1 + · · · + Re n for some positive integer n and idempotent elements e i (1 ≤ i ≤ n) of I. It is well known that there exists an idempotent e ∈ I such that Re 1 + · · · + Re n = Re. Then a = be for some b ∈ R and hence a = ae ∈ JI. It follows that J = JI. Hence the R-module I is a faithful multiplication module. By Theorem 2.12 λ is an epimorphism. However, λ(R) = RI = I 2 = λ(I), so that λ is not an injection.
To illustrate Proposition 2.15 we have the following example.
Example 2.16. Let a ring R = i∈I F i be the direct product of any infinite collection of fields F i (i ∈ I). Then R is a commutative von Neumann regular ring whose socle S = ⊕ i∈I F i . Moreover for any proper ideal B of R with S ⊆ B the R-module B is a faithful multiplication module such that the mapping λ : L(R) → L( R B) is an epimorphism but not a monomorphism.
Proof. It is clear that R is a commutative von Neumann regular ring. Let f = {f i } be any non-zero element of R where the ith component f i is an element of F i for each i ∈ I. For each j ∈ I let e (j) denote the element in R whose jth component is 1 and all of whose other components are 0. There exists k ∈ I such that f k = 0. Then e (k) f is a non-zero element of U k where U k is the ideal of R consisting of all elements {u i } in R such that u i = 0 for all i = k. Clearly U i is a simple submodule of R R for each i ∈ I and ⊕ i∈I U i is an essential submodule of R R so that S = ⊕ i∈I U i = ⊕ i∈I F i is the socle of R. Let B be any proper ideal of R with S ⊆ B. Then B is generated by idempotent elements and ann R (B) = 0. Apply Proposition 2.15.
Note that in Proposition 2.15 the ideal I is not finitely generated.

The Mapping µ
Given a ring R, in this section we shall consider the mapping µ from the lattice L( R M ) of submodules of an R-module M to the lattice L(R) of ideals of R defined by µ(N ) = (N : R M ) = ann R (M/N ). We shall say that M is a µ-module in case the mapping µ is a homomorphism. First note the following simple fact.  Clearly the result follows.
In contrast to Theorem 2.3, no (non-zero) ring R has the property that the mapping µ : L( R M ) → L(R) is a homomorphism for every R-module M , as the next result shows.  Proof. Let P be a maximal ideal of R such that M = P M . Note that M/P M is a non-zero semisimple module and hence contains a maximal submodule. Let L be a maximal submodule of M such that P M ⊆ L. Let m be any element of M with m / ∈ L. Clearly M = L + Rm. By Corollary 3.2 there exists an element p ∈ R such that pM ⊆ L and (1 − p)M ⊆ Rm. If p / ∈ P then R = P + Rp and hence M = P M + pM ⊆ L, a contradiction. Thus p ∈ P , as required.
Corollary 3.5. Let R be a ring and let M be a µ-module over R such that M = P M for every maximal ideal P of R. Then the R-module M is a multiplication module. Moreover the mapping µ is a monomorphism.
We now aim to prove a partial converse of Corollary 3.5. It is proved in Theorem 2.3 that a domain is Prüfer if and only if every homomorphic image of a λ-module is a λ-module. Contrast this fact with the following result.  By Lemma 3.1 R R is a µ-module. Now apply Proposition 3.6.
We now characterize which finitely generated modules are µ-modules. Then B is a proper ideal of R and hence B ⊆ P for some maximal ideal P of R. Because M is a finitely generated multiplication module, Corollary 2.11 shows that there exist m ∈ M and p ∈ P such that (1 − p)M ⊆ Rm.
In particular, note that (1 − p)N and (1 − p)L are both submodules of the cyclic R-module Rm. Next, Because Rm is a µ-module by Corollary 3.7, we now see that   (iii) ⇒ (v) Let K be any finitely generated submodule of M . There exist a positive integer n and elements u i ∈ K (1 ≤ i ≤ n) such that K = Ru 1 + · · · + Ru n . If n = 1 then K is a multiplication module. Suppose that n ≥ 2. By induction on n the submodule L = Ru 1 + · · · + Ru n−1 is a multiplication module. Let P be any maximal ideal of R. By Corollary 2.11, there exist v ∈ L, p ∈ P such that (1 − p)L ⊆ Rv. Let u = u n and note that K = L + Ru. By hypothesis, R = (Ru : R Rv) + (Rv : R Ru) so that either (Ru : R Rv) P or (Rv : R Ru) P .
Suppose first that (Ru : R Rv) P . Then there exists p 1 ∈ P such that (1 − p 1 )v ∈ Ru. This implies that Now suppose that (Rv : R Ru) P . Then there exists p 2 ∈ P such that (1 − p 2 )u ∈ Rv. In this case, In case n ≥ 2 we have proved that for each maximal ideal P of R there exist an element w in K and an element q in P such that (1 − q)K ⊆ Rw. By Lemma 2.10 K is a multiplication module.
(i) ⇒ (iv) Given any finitely generated submodules N, L of M , the submodule N + L is also finitely generated and hence a µ-module. By Lemma 3.1, This completes the proof.
Next we give examples to show that the condition that the module M be finitely generated is necessary for both implications in Theorem 3.8.
Example 3.10. Let F be any field and let R denote the collection of all sequences {f 1 , f 2 , f 3 , . . . } of elements of F with the property that there exists a positive integer n (depending on the particular sequence) such that f n = f n+1 = f n+2 = . . . . Then R is a commutative von Neumann regular ring whose socle S is a multiplication module but not a µ-module.
Proof. It is not difficult to see that the commutative ring R is von Neumann regular. Next it is clear that S consists of all sequences {f 1 , f 2 , f 3 , . . . } of elements of F such that 0 = f n = f n+1 = f n+2 = . . . for some positive integer n. By Proposition 2.15, the module S is a faithful multiplication module. Let S 1 denote the subset of S consisting of all sequences {f 1 , f 2 , f 3 , . . . } such that f 2k = 0 (k ≥ 1) and let S 2 denote the subset of S consisting of all elements {f 1 , f 2 , f 3 , . . . } with f 2k−1 = 0 (k ≥ 1). Clearly S 1 and S 2 are submodules of the R-module S such that S = S 1 ⊕ S 2 , (S 1 : R S) = S 1 and (S 2 : R S) = S 2 . By Corollary 3.2 the R-module S is not a µ-module.
Example 3.11. Again let Z denote the ring of rational integers and let M denote the Prüfer p-group for any prime p in Z. Then the Z-module M is a µ-module but not a multiplication module.
Proof. If N and L are proper submodules of M then (N : R M ) = (L : R M ) = 0 by [7,Proposition 2.6]. Now Lemma 3.1 gives that M is a µ-module. On the other hand, if K is any proper non-zero submodule of M then K = BM for any ideal B of Z and hence M is not a multiplication module.
Recall that a module M over a general ring R is called hollow in case M is not the sum of two proper submodules. Next we generalise Example 3.11 Proposition 3.12. Let R be a domain which is not a field and let M be a non-zero injective R-module. Then (a) The above mapping λ is a homomorphism but is neither a monomorphism nor an epimorphism.
(b) The mapping µ is a homomorphism if and only if M is a hollow module. However µ is neither a monomorphism nor an epimorphism.
Proof. Note that M = BM = λ(B) for every non-zero ideal B of R and 0 = (N : R M ) = µ(N ) for every proper submodule N of M by [7,Proposition 2.6].
(a) For all ideals B and C of R we have (B ∩ C)M = BM ∩ CM . By Lemma 2.1 λ is a homomorphism. Let a be any non-zero element of R such that a is not a unit in R. Then λ(Ra) = M = λ(R) so that λ is not a monomorphism. For any maximal ideal P of R, P = 0 and hence M = P M . Thus M is not simple. If L is a non-zero proper submodule of M then L = λ(E) for any ideal E of R. Thus λ is not an epimorphism.
(b) It is clear that µ is a homomorphism if and only if whenever N and L are proper submodules of M then the submodule N + L is also proper, that is M is hollow. We have seen above that M is not simple and hence µ(L) = 0 = µ(0) for every non-zero proper submodule L of M . Thus µ is not a monomorphism. Moreover if a is as before we have Ra = µ(K) for every submodule K of M , so that µ is not an epimorphism.
The next result is an analogue of Theorem 2.3. It is an immediate consequence of Corollary 3.9 because domains with the property that every finitely generated ideal is a multiplication module are precisely Prüfer domains.
Theorem 3.13. The following statements are equivalent for a domain R.
Note that Theorem 3.13 shows that the class of µ-modules is not closed under submodules in general.
Lemma 3.14. Given any ring R, the following statements are equivalent for an R-module M . The next result should be compared with Proposition 2.4. Recall that a (nonzero) ring R is called local in case it contains precisely one maximal ideal.  Because R is local either 1 ∈ (Ru : R Rv) and v ∈ Ru or else 1 ∈ (Rv : R Ru) and u ∈ Rv ⊆ V . By the choice of u it follows that v ∈ Ru for all v ∈ V and hence V ⊆ Ru ⊆ U . We have proved that U ⊆ V or V ⊆ U for any submodules U and V of M . Thus M is a chain module.
We next consider semisimple modules.  Proof. We saw in Example 2.16 that S = ⊕ i∈I U i . For each i ∈ I let π i : S → U i denote the canonical projection. Let N and L be submodules of the R-module S such that N ∩ L = 0. It is rather easy to see that N ∩ U i = π i (N ) for all i ∈ I. Let I 1 = {i ∈ I : π i (N ) = 0}. Then N = ⊕ i∈I1 U i . Similarly, if I 2 = {i ∈ I : π i (L) = 0} then L = ⊕ i∈I2 U i . Because N ∩ L = 0, the set I 1 ∩ I 2 must be empty or, in other words, I 2 ⊆ I \ I 1 . Let f = {f i } ∈ R. Let g denote the element {g i } such that g i = f i (i ∈ I 2 ) and otherwise g i = 0. Then This proves that R = ann R (N ) + ann R (L). By Proposition 3.16 it follows that the R-module S is a µ-module.
We saw in Corollary 2.6 that every semisimple module is a λ-module. However Proposition 3.16 shows that if M is the semisimple Z-module ⊕ p∈Π (Z/Zp), where Π is the collection of all primes p in Z, then M is not a µ-module.

Finitely Generated Modules
Again R is a commutative ring and M a unital R-module. We begin this section with the following elementary result. Proof. Clear.
The next result can be found essentially in [4,Theorem 3.1].
Lemma 4.2. Let R be any ring and let M be an R-module with annihilator A in R. If M is finitely generated then M = P M for every maximal ideal P of R with A ⊆ P . Moreover, the converse holds in case M is a multiplication module.
Proof. Suppose first that M is finitely generated. Suppose further that M = P M for some maximal ideal P of R. Then the usual determinant argument gives that (1 − p)M = 0 for some p ∈ P . Thus 1 − p ∈ A and hence A P . Conversely, now suppose that M is a multiplication module such that M = P M for every maximal ideal P of R with A ⊆ P . We have to prove that M is finitely generated. Suppose that R = m∈M (Rm : R M ). Then there exists a maximal ideal Q of R such that   Proof. (i) ⇔ (ii) By Corollary 1.5.
(i) ⇒ (iii) Because λ is an isomorphism, λ is a bijection. In particular, λ is a surjection and hence M is a multiplication module. Next Lemma 1.4 shows that µλ = 1 because λ is an injection. Therefore, for each ideal B of R, This proves (iii). (iv) ⇒ (i) By [4,Theorem 3.1], if C and D are ideals of R such that λ(C) = λ(D) then CM = DM and hence C = D. Thus λ is an injection. But M being a multiplication module gives that λ is a surjection. Being a bijection, λ is an isomorphism by Lemma 1.2. Proof. Suppose that M is a finitely generated multiplication module. Then the (R/A)-module M is a finitely generated faithful multiplication module (Lemma 2.9) and the mapping λ is an isomorphism by Theorem 4.3. Conversely, suppose that λ is an isomorphism. By Theorem 4.3, the (R/A)-module M is a finitely generated multiplication module. It follows that the R-module M is a finitely generated multiplication module by Lemma 2.9. Proof. Suppose that M is a (non-zero) faithful multiplication module. By [4,Lemma 4.3 and Theorem 4.4] M is finitely generated. Now apply Theorem 4.3.

Isomorphisms of Lattices
Let R be a ring and M an R-module. Suppose that ρ : L(R) → L(R) is a lattice isomorphism (in particular, ρ could be the lattice isomorphism induced from an automorphism of the ring R) and suppose that σ : L( R M ) → L( R M ) is a lattice isomorphism(in particular, σ could be the lattice isomorphism induced from an Risomorphism from M to M ). Then σλρ : L(R) → L( R M ) is a mapping which is an isomorphism if and only if λ is an isomorphism. Thus in general there can be many mappings from L(R) to L( R M ) each of which may or may not be an isomorphism. Now let R be a ring which is not Hopfian, that is there exists a ring epimorphism ν : R → R which is not a monomorphism. Let A denote the non-zero kernel of ν. Then R ∼ = R/A so that L(R) is isomorphic to L(R/A) which in turn is isomorphic to L( R (R/A)). However in this case λ : L(R) → L( R (R/A)) is not an isomorphism because R/A is not a faithful R-module. For a specific example, let R denote the polynomial ring F [x 1 , x 2 , x 3 , . . . ] in indeterminates x i (i ≥ 1) over a field F . Let n be any positive integer. The linear mapping ν from the F -vector space R to R defined by ν(x i ) = 0 if 1 ≤ i ≤ n and ν(x i ) = i − n if i ≥ n + 1 induces a ring epimorphism with non-zero kernel A n = Rx 1 + · · · + Rx n .
This leads us to ask the following question. As a contribution towards the answer to this question we offer some modest results. We look at Noetherian rings and modules. Any Noetherian module has Krull dimension (see, for example, [6,Lemma 6.2.3]). For the definition and properties of Krull dimension see [6,Chapter 6]. We shall denote the Krull dimension of the ring R by kdim(R) and of an R-module M by kdim( R M ), if either exists.
Lemma 5.2. Let R be a Noetherian domain and let M be an R-module such that the lattices L(R) and L( R M ) are isomorphic. Then M is a finitely generated faithful R-module.