Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 41 (2000), No. 1, 257-266.

Quasi-Frobenius Modules

Adil G. Naoum Layla S. M. Al-Shalgy

Department of Mathematics, College of Science University of Baghdad, Baghdad, Iraq Department of Mathematics, College of Education (Iben Al-Haitham) University of Baghdad, Baghdad, Iraq

Abstract. Let R be a commutative ring with 1, and let M be a faithful R-module. We say that M is a quasi-Frobenius (in short QF) module if Hom_{R}(P, M) is either zero or a simple R-module for each simple R-module P. In this paper we give some charakterizations of QF modules and we study the relation between QF modules and multiplication modules.

Introduction

Let R and S be two rings with identities and let M be an R-S-bimodule. Following G. Azumaya [3], we say that M is a quasi-Frobenius (briefly QF) R-S-module if

(1) M is faithful on both R and S,

(2) Hom_{R}(P, M) (respectively Hom_{S}(Q, M)) is either zero or a simple S-module (respectively zero or a simple R-module) for each simple R-module P (respectively Smodule Q).

Equivalently, a faithful R-S-bimodule M is QF if and only if ann_{M}(I) (respectively ann_{M}(J)) is either zero or a simple S-module (respectively zero or a simple R-module) for each maximal ideal I of R (respectively J of S).

In this paper we will say that an R-module M is QF if M is a QF R-R-bimodule. On the other hand, a ring R is called a quasi-Frobenius ring if R is an Artinian left or right self-injective ring [5].

Let us observe that every QF ring is a QF R-module. However the converse is false, for example Z as a Z-module is QF but Z as a ring is not QF.

One of our main concerns in this paper is to generalize some of the basic properties of QF rings to modules in case R = S and R being a commutative ring with identity. We also study QF modules when S = End_{R}(M).

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix}2000 Heldermann Verlag

1. Preliminary results

We begin with the following simple remarks.

1.1. Remarks.

(1) If R is an integral domain then every torsion-free R-module is QF.

(2) If the ring R has no non-zero nilpotent elements, then R is a QF R-module.

Proof. (1) is obvious.

(2) Let P be a simple R-module and assume that Hom_{R}(P, R) ≠ 0, then R contains an ideal say I which is isomorphic to P. Since R has no non-zero nilpotent element, then one can show easily that I = eR for some idempotent e in R. Hence Hom_{R}(P, R) ≅ Hom_{R}(eR, R) ≅ eR is a simple R-module. Hence R is a QF R-module. □

A more interesting example of a QF module is given in the following theorem.

1.2. Theorem. Let R be a Dedekind domain and let K be the field of quotients of R. Let L be an R-submodule of K containing R such that L/R is a faithful R-module. Then L/R is a QF R-module.

Proof. Let P be a simple R-module, then P ≅ R/I for some maximal ideal I of R. Since R is a Dedekind domain, then I is an invertible ideal of R.

We show that (L ∩ I^{−1})/R is a simple R-module unless it is zero. Let J/R be an R-submodule of (L ∩ I^{−1})/R, then R ⊆ J ⊆ L ∩ I^{−1} and hence IR ⊆ IJ ⊆ I(L ∩ I^{−1}) which implies that I ⊆ IJ ⊆ R since L ∩ I^{−1} ⊆ I^{−1}. Therefore by the maximality of I, either IJ = I or IJ = R. Thus either J = R or J = I^{−1} and hence J = L ∩ I^{−1}.

The proof will be completed by showing that Hom_{R}(R/I, L/R) ≅ (L ∩ I^{−1})/R. So we define a map α : L ∩ I^{−1} → Hom_{R}(R/I, L/R) by α(x) = α_{x} for all x ∈ L ∩ I^{−1}, where α_{x} : R/I → L/R by α_{x}(r + I) = xr + R for all r ∈ R. One can check easily that α is a well-defined R-homomorphism. In fact α is an epimorphism. For if f ∈ Hom_{R}(R/I, L/R) then, since R/I is a cyclic R-module generated by 1 + I, we look at f(1 + I) as an element of L/R, let f(1 + I) = x + R for some x ∈ L. Then

I f(1 + I) = f(I + I) = f(I) = 0 = I(x + R).

Therefore xI ⊆ R, that is x ∈ I^{−1} and hence x ∈ L∩I^{−1} and it is clear that f = α_{x}. Thus α is an epimorphism which implies that (L ∩ I^{−1})/ ker α ≅ Hom_{R}(R/I, L/R). Moreover, it is easily checked that ker α = R. Hence (L ∩ I^{−1})/R ≅ Hom_{R}(R/I, L/R). Therefore Hom_{R}(R/I, L/R) is a simple R-module which completes the proof. □

The following are special cases of 1.2.

1.3. Corollary. If R is a Dedekind domain and K is the field of quotients of R, then K/R is a QF R-module.

1.4. Corollary. Let p any prime number in Z, and let Q_{p} = \{\frac{a}{p^{n}} | a, n ∈ Z\}. Let Q_{p}/Z = Z_{p^{∞}}. Then Z_{p^{∞}} is a QF Z-module.

Proof. It is easily checked that Z_{p^{∞}} is a faithful Z-module. Hence by 1.2, Z_{p^{∞}} is a QF Z-module. □

Note that Z_{p^{∞}} in 1.4 is an Artinian but not Noetherian Z-module.

2. Some characterizations of quasi-Frobenius modules

Consider the following two characterizations for quasi-Frobenius rings.

(1) An Artinian ring R is a QF ring if and only if ann_{R}(ann_{R}(I)) = I for each simple ideal I of R, [5, Th. 3.4].

(2) An Artinian ring R is a QF ring if and only if every simple R-module is reflexive, [10, Th. 2.1].

We give in this section similar characterizations for QF modules. We start by the following theorem.

2.1. Theorem. Let M be a faithful R-module. Then M is a QF R-module if and only if ann_{M}(ann_{R}(U)) = U for each simple R-submodule U of M.

Proof. Assume that M is a QF R-module and let U be a simple R-submodule of M. Then U ≅ R/I for some maximal ideal I of R. Now ann_{M}(ann_{R}(U)) = ann_{M}(ann_{R}(R/I)) = ann_{M}(I). But M is a QF R-module, then ann_{M}(I) is a simple R-module unless it is zero. But ann_{M}(I) = ann_{M}(ann_{R}(U)) ⊇ U ≠ 0, therefore ann_{M}(I) ≠ 0 is a simple R-module containing U. Hence ann_{M}(I) = U.

For the converse, let P be a simple R-module such that Hom_{R}(P, M) ≠ 0. Then P ≅ R/I for some maximal ideal I of R. Then

Hom_{R}(P, M) ≅ ann_{M}(I) ≅ ann_{M}(ann_{R}(P)) = P.

Therefore Hom_{R}(P, M) is a simple R-module and hence M is a QF R-module. □

Next we recall that an R-module M is said to be fully stable if for each R-submodule N of M, f(N) ⊆ N for each R-homomorphism f : N → M, [1].

As a consequence of 2.1, we have

2.2. Corollary. Every faithful fully stable R-module is QF.

Proof. Let M be a faithful fully stable R-module. Then ann_{M}(ann_{R}(x)) = (x) for each x ∈ M by [1, Cor. 3.5]. In particular ann_{M}(ann_{R}(U)) = U for each simple R-submodule U of M. Therefore M is QF by 2.1. □

Note that the converse of 2.2 may not be true in general, for instance Z as a Z-module is QF but not fully stable since ann_{Z}(ann_{Z}(2)) = Z ≠ (2).

Recall that a ring R is called a self-injectice ring if for each ideal I of R and for each R-homomorphism f : I → R, there exists an element r ∈ R such that f(x) = rx for each x ∈ I, [2]. Hence we have 2.3. Corollary. Let R be a self-injective ring. Then R as an R-module is QF.

Proof. Since R is a self-injective ring, then it can be easily checked that R is a fully stable R-module, and hence the result follows by 2.2. □

Let M and X be two R-modules, X is called M-reflexive if the natural map ϕ : X → X^{∗∗} is an R-isomorphism where X^{∗} = Hom_{R}(X, M), and ϕ is defined by (ϕ(x))(f) = f(x) for all x ∈ X and f ∈ X^{∗}, see [8]. And recall that an R-module M is called distinguished if ann_{M}(I) ≠ 0 for each ideal I of R, see [3].

Using these concepts we extend in the following two theorems the second characterization of QF rings which is mentioned in the introduction.

2.4. Theorem. Let M be a distinguished QF R-module. Then every simple R-module is M-reflexive.

Proof. Let P be a simple R-module. Since M is distinguished, then P^{∗} =Hom_{R}(P, M)≠0 by [13]. But M is QF, therefore P^{∗} is a simple R-module. And again since M is distinguished and QF, then P^{∗∗} is a simple R-module. Thus both P and P^{∗∗} are simple Rmodules and ϕ : P → P^{∗∗} is a non-zero R-homomorphism, therefore ϕ is an R-isomorphism and hence P is M-reflexive. □

Note that the condition in 2.4 that M is distinguished, cannot be dropped, as it is shown in the following example.

The Z-module Z is QF. However, if P is any simple Z-module, then P ≅ Z_{p} for some prime number p, and

P^{∗} = Hom_{Z}(P, Z) ≅ Hom_{Z}(Z_{p}, Z) = 0

which implies that P^{∗∗} = 0 and hence P ≅ P̸^{∗∗}, that is, P is not Z-reflexive. Note that Z is not a distinguished Z-module.

Let A, B and M be any R-modules, a bilinear map α : A × B → M is called regular if α(a, b) = 0 for all a ∈ A implies b = 0 and α(a, b) = 0 for all b ∈ B implies a = 0, see [3].

The following theorem gives a converse of 2.4, under a certain condition.

2.5. Theorem. Let M be a faithful R-module. Assume that for each simple R-module P, P^{∗∗} ≅ P and P^{∗} contains a maximal submodule (where P^{∗} = Hom_{R}(P, M)). Then M is a QF R-module.

Proof. Let P be a simple R-module. We have to show that P^{∗} is simple. Note that ann_{P} (P^{∗}) is an R-submodule of P, it is either 0 or P because of the simplicity of P. If ann_{P} (P^{∗}) = P, then f(x) = 0 for all x ∈ P and all f ∈ P^{∗} and this implies that P^{∗} = 0 which is a contradiction since P^{∗∗} ≅ P by hypothesis. Hence ann_{P} (P^{∗}) = 0. Therefore it can be easily checked that the pairing (x, f) ↦ f(x) for all x ∈ P and all f ∈ P^{∗} is a regular bilinear map of P × P^{∗} into M. Now, let U be a maximal submodule of P^{∗}, then P^{∗}/U is a simple R-module. Let V = ann_{P} (U), then V ≅ ann_{P^{∗∗}}(U) since P ≅ P^{∗∗}. But ann_{P^{∗∗}}(U) ≅ Hom_{R}(P^{∗}/U, M) by [7, Prop. 23.12, p. 184]. Therefore V ≅ Hom_{R}(P^{∗}/U, M) = (P^{∗}/U)^{∗}. Since V is an R-submodule of P, then either V = 0 or V = P. If V = 0, then (P^{∗}/U)^{∗} = 0 and hence (P^{∗}/U)^{∗∗} = 0. But by hypothesis P^{∗}/U ≅ (P^{∗}/U)^{∗∗} and P^{∗}/U is simple, therefore a contradiction. Hence V = P. That is f(x) = 0 for all x ∈ P and all f ∈ U. Therefore f = 0 for all f ∈ U (by the regularity of the pairing (x, f) ↦ f(x) for all x ∈ P and all f ∈ P^{∗}). Thus U = 0 and hence P^{∗} is a simple R-module. Therefore M is a QF R-module. □

Note that the condition in the previous theorem, that P^{∗} contains a maximal submodule, holds for example in case P^{∗} is a finitely generated [9] or a projective [2] R-module.

3. Quasi-Frobenius modules and multiplication modules

Recall that an R-module M is said to be a multiplication module if for each submodule N of M there exists an ideal I of R such that N = IM, see [6]. In this section we investigate the relation between multiplication modules and quasi-Frobenius modules. We begin with the following proposition.

3.1. Proposition. Every faithful multiplication module over a QF ring is a QF module.

Proof. Let R be a QF ring and let M be a faithful multiplication R-module. Then according to [5], R is an Artinian ring and hence M is a cyclic R-module by [4, Prop. 4]. But M is faithful, therefore M ≅ R. Since R is a QF ring, then R is a QF R-module. Hence M is a QF R-module. □

We note that if we weaken the condition “R is a QF ring” in 3.1 to R being merely a Noetherian ring and use an extra condition on M we get that M is a QF module as in the following proposition.

3.2. Proposition. Let R be a Noetherian ring and let M be a faithful multiplication R-module such that for each simple R-module P, P^{∗∗} ≅ P, where P^{∗} = Hom_{R}(P, M). Then M is a QF R-module.

Proof. M being a faithful multiplication R-module and R a Noetherian ring, then M is a Noetherian R-module, see [6]. Now, if P is any simple R-module, then P is cyclic and hence a finitely generated multiplication module. Moreover EM is a finitely generated submodule of M, where E = [ann M : ann P]. Therefore by [12, Th. 3.4], P^{∗} = Hom_{R}(P, M) is a finitely generated R-module. Thus P^{∗} contains maximal submodules and hence by 2.5, M is a QF R-module. □

Because of the fact that a faithful multiplication R-module M is Noetherian if and only if R is a Noetherian ring, see [6], the following is an immediate consequence of 3.2.

3.3. Corollary. If M is a Noetherian faithful multiplication R-module such that for each simple R-module P, P ≅ P^{∗∗}, then M is a QF R-module.

For our next result the following remark is needed.

Remark. IF R is any ring and M is a faithful R-module, then R is isomorphic to a subring of End_{R}(M).

Proof. Is obvious. □

And the following concept is also needed. Given an R-module M, a subring D of End_{R}(M) is said to be a dense subring of End_{R}(M), if given any finite set {x_{1}, x_{2}, …, x_{n}} of elements of M and any element f of End_{R}(M), there exists an element d of D such that f(x_{i}) = d(x_{i}) for all i = 1, 2, …, n, see [3].

Now we have the following proposition.

3.4. Proposition. If M is a QF R-module such that R is dense in End_{R}(M), then M is a QF End_{R}(M)-module.

Proof. Put S = End_{R}(M). Let U be a simple S-module such that Hom_{S}(U, M) ≠ 0. Then U can be considered as an S-submodule of M. And by the last remark U is an R-submodule of M. Since R is dense in S, then it can be easily seen that U is a simple R-module. Hence Hom_{R}(U, M) is a simple R-module because M is a QF R-module. Moreover if f ∈ Hom_{S}(U, M), x ∈ M and r ∈ R, then

f(r, x) = f(λ_{r}(x)) = λ_{r}(f(x)) = r(f(x))

(where λ_{r} : M → M is such that λ_{r}(x) = rx for all r ∈ R and x ∈ M). Hence f ∈ Hom_{R}(U, M) and thus Hom_{S}(U, M) ⊆ Hom_{R}(U, M). The simplicity of Hom_{R}(U, M) implies that Hom_{R}(U, M) = Hom_{S}(U, M). Therefore Hom_{S}(U, M) is a simple R-module and hence a simple S-module. Thus M is a QF S-module. □

As an immediate consequence of 3.4 we have the following

3.5. Corollary. Let M be a multiplication QF R-module. Then M is a QF S-module where S = End_{R}(M).

Proof. Since M is a faithful multiplication R-module, then by [11, Prop. 1.5], R is dense in S and hence the result follows by 3.4. □ Now, we need the following lemma.

3.6. Lemma. Let M be a finitely generated faithful multiplication R-module and let N be an R-submodule of M. Then N is a simple R-module if and only if there exists a simple ideal I of R such that N = IM.

Proof. Since M is a multiplication R-module, then there exists an ideal I of R such that N = IM, see [6].

Assume that N is simple. Let J be a non-zero ideal of R such that J ⊆ I, then JM ⊆ IM. But IM is a simple R-module and JM ≠ 0, hence JM = IM, then by [14, Cor. of Th. 9] J = I. Thus I is a simple ideal.

Conversely: Suppose that I is a simple ideal and let K be a non-zero R-submodule of N, then there exists a non-zero ideal J of R such that K = JM. Then JM ⊆ IM and according to [14, Cor. of Th. 9], this implies that J ⊆ I. Thus J = I and hence K = N which completes the proof. □

We are now ready to give the following result.

3.7. Theorem. Let M be a finitely generated faithful multiplication R-module. Then the following statements are equivalent:

(1) ann_{M}(ann_{R}(N)) = N for each simple R-submodule N of M,

(2) ann_{R}(ann_{R}(I)) = I for each simple ideal I of R.

Proof. Let I be a simple ideal of R, then N = IM is a simple R-module by 3.6, and

ann_{M}(ann_{R}(N)) = ann_{M}(ann_{R}(IM))

= ann_{M}(ann_{R}(I)) (since M is faithful)

⊇ ann_{R}(ann_{R}(I))M (since for any ideal J, ann_{M}(J) ⊇ ann_{R}(J)M ).

Now, assume (1), then ann_{M}(ann_{R}(N)) = N and N = IM ⊇ ann_{R}(ann_{R}(I))M. Hence I ⊇ ann_{R}(ann_{R}(I)) by [14, Cor. of Th. 9]. But ann_{R}(ann_{R}(I)) ⊇ I, and thus (2) follows. Conversely: Assume (2), and let N be a simple R-submodule of M, then by 3.6, there exists a simple ideal I of R such that N = IM. By (2), ann_{R}(ann_{R}(I)) = I. Then

ann_{M}(ann_{R}(N)) = ann_{M}(ann_{R}(IM))

= ann_{M}(ann_{R}(I)) (since M is faithful) ⊇ ann_{R}(ann_{R}(I))M = IM = N.

But N ⊆ ann_{R}(ann_{R}(N)), therefore (1) follows. □

The following are some consequences of 3.7.

3.8. Corollary. A finitely generated faithful multiplication R-module is QF if and only if ann_{R}(ann_{R}(I)) = I for each simple ideal I of R.

Proof. Is obvious by 2.1 and 3.7. □

3.9. Corollary. If R is a self-injective ring and M is a finitely generated faithful multiplication R-module, then M is a QF R-module.

Proof. Since R is a self-injective ring, then by [1, Prop. 3.4], ann_{R}(ann_{R}(I)) = I for each cyclic ideal I of R, in particular for each simple ideal I of R. Therefore M is a QF R-module by 3.8. □

We end this section by the following example.

3.10. Example. Let X be an infinite set and let R = P^{X} be the power set of X. For all A, B ∈ R, define A + B = A ∪ B/A ∩ B and AB = A ∩ B. Then R is a Boolean ring.

Let I be an ideal of R generated by all singletons in X. In fact I is the set of all finite subsets of X (I is not finitely generated). I is a pure ideal and hence I is a multiplication ideal, that is a multiplication R-module. Note that the simple R-modules are generated by singletons and each simple R-module contains only two elements. Let P be a simple Rmodule. Clearly I contains a copy of P and hence Hom_{R}(P, I) ≠ 0. Moreover Hom_{R}(P, I) is simple. In fact Hom_{R}(P, I) ≅ P since I contains a unique copy of P. Hence I is a QF R-module.

4. Quasi-Frobenius bimodules

Let R and S be two commutative rings with identities. We consider in this section QF R-S-modules.

Because of the fact that the endomorphism ring of a multiplication module is a commutative ring, see [11], we start with the following proposition.

4.1. Proposition. Let M be a multiplication QF R-module and let S = End_{R}(M). Then M is a QF R-S-module.

Proof. Since M is a multiplication QF R-module, then by 3.5, M is a QF S-module. Let I be a maximal ideal of R such that ann_{M}(I) ≠ 0, then ann_{M}(I) is a simple R-module. Note that ann_{M}(I) is an R-submodule of M, therefore it is an S-submodule of M (since R is dense in S because M is a faithful multiplication R-module, see [11, Th. 1.5]). In fact ann_{M}(I) is a simple S-submodule of M, for if U is an S-submodule of ann_{M}(I), then U is an R-submodule of ann_{M}(I) (since R is dense in S), and hence either U = 0 or U = ann_{M}(I). Thus ann_{M}(I) is a simple S-module.

Now, let L be a maximal ideal of S such that ann_{M}(L) ≠ 0, ann_{M}(L) is a simple S-submodule. Let V be an R-submodule of ann_{M}(L), then rx ∈ V for all r ∈ R and x ∈ V. Let f ∈ S, then there exists t ∈ R such that f(x) = tx (since R is dense in S). Therefore f(x) ∈ V for all f ∈ S and x ∈ V. Hence V is an S-submodule of ann_{M}(L), which implies that either V = 0 or V = ann_{M}(L). Hence ann_{M}(L) is a simple R-module and this completes the proof. □

Now, we consider the following concept:

Let M be an R-S-module. A left R-module A and a right S-module B are said to form an orthogonal pair with respect to M, if there exists a regular bilinear map of A × B into M.

Next we give the following two lemmas:

4.2. Lemma. Let M be a QF R-S-module, let X be an R-module and Y be an S-module which form an orthogonal pair with respect to M. Then:

(1) If X is simple, then X^{∗} ≅ Y.

(2) If Y is simple, then X ≅ Y^{∗}.

Proof. (1) Let α : X × Y → M be a regular bilinear map. For each y ∈ Y, define α_{y} : X → M by α_{y}(x) = α(x, y) for all x ∈ X. It can be easily seen that α_{y} is a welldefined R-homomorphism, and hence α_{y} ∈ X^{∗}. Define λ : Y → X^{∗} such that λ(y) = α_{y} for all y ∈ Y. Clearly λ is an S-homomorphism. Moreover, if λ(y) = 0 for some y ∈ Y, then α_{y} = 0 and therefore α(x, y) = 0 for all x ∈ X which implies that y = 0 by the regularity of α. Therefore Y is isomorphic to an S-submodule of X^{∗}. But M is a QF R-S-module and X^{∗} ≠ 0 (since α_{y} ∈ X^{∗}), hence X^{∗} is a simple S-module, therefore X^{∗} ≅ Y. Similarly for (2). □

4.3. Lemma. Let M be a QF R-S-module. Let U be a simple R-submodule of M and V be a simple S-submodule of M. Then:

(1) S/ ann_{S}(V) is a simple S-module.

(2) R/ ann_{R}(V) is a simple R-module.

Proof. (1) Define λ : (S/ ann_{S}(U)) × U → M by λ(a + ann_{S}(U), x) = xa for all a ∈ S and all x ∈ U. It can easily be checked that λ is a regular bilinear map. Therefore S/ ann_{S}(U) and U form an orthogonal pair with respect to M. Hence 4.2 implies that U^{∗} ≅ S/ ann_{S}(U) and thus U^{∗} ≠ 0. But M is a QF R-S-module, then U^{∗} is a simple S-module. Therefore S/ ann_{S}(U) is a simple S-module. Similarly for (2). □

Finally, we have the following proposition.

4.4. Proposition. Let M be a QF R-S-module. Then ann_{M}(ann_{S}(U)) = U for each simple R-submodule U of M and ann_{M}(ann_{R}(V)) = V for each simple S-submodule V of M.

Proof. Let U be a simple R-submodule of M. Then ann_{M}(ann_{S}(U)) ≅ Hom_{S}(S/ann_{S}(U), M) by [1]. But by 4.3, S/ ann_{S}(U) is a simple S-module and since M is a QF R-S-module, then Hom_{S}(S/ ann_{S}(U), M) is a simple R-module and since it contains U, it is equal to U. The second part is proved similarly. □

References

[1] Abbas, M. S.: On Fully Stable Modules. Ph.D. Thesis, University of Baghdad 1990.

[2] Anderson F. W.; Fuller, K. R.: Rings and Categories of Modules. Springer-Verlag, New York Heidelberg Berlin 1973.

[3] Azumaya, G.: A duality Theory for Injective Modules (Theory of Quasi-Frobenius Modules). Amer. J. Math. 81 (1959), 249–287.

[4] Barnard, A.: Multiplication Modules. J. Algebra 71 (1981), 174–178.

[5] Dieudonné, J.: Remarks on Quasi-Frobenius Rings. Illinois J. Math. 2 (1958), 346– 354.

[6] El-Bast, Z. A.; Smith, P. F.: Multiplication Modules. Comm. Algebra 16 (1988), 755–779.

[7] Faith, C.: Algebra II: Ring Theory. Springer Verlag, Berlin Heidelberg New York [8] Huger, G.; Zimmermann, M.: Quasi-Frobenius Moduln. Arch. Math. (Basel) 24

(1973), 379–386.

[9] Kasch, F.: Modules and Rings. Academic Press, New York and London 1982.

[10] Morita, K.; Tachikawa, H.: Character Modules, Submodules of a Free Module and Quasi-Frobenius Rings. Math. Z. 65 (1956), 414–428.

[11] Naoum, A. G.: On the Ring of Endomorphisms of a Multiplication Module. Period. Math. Hungar. 29 (1994), 277–284.

[12] Naoum, A. G.; Al-Hashimi, B.; Kider, J. R.: On the Module of Homomorphisms of Finitely Generated Multiplication Modules I. Period. Math. Hungar. 22 (1991), 97–105.

[13] Naoum, A. G.; Al-Shalgy, L. S. M.: Distinguished Modules. To appear.

[14] Smith, P. F.: Some Remarks on Multiplication Modules. Arch. Math. (Basel) 50 (1988), 223–235.

Received March 10, 1997