Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 42 (2001), No. 2, 557-573.

Finite and Infinite Collections of Multiplication Modules

Majid M. Ali David J. Smith

Department of Mathematics, University of Auckland, Auckland, New Zealand e-mail: majid@math.auckland.ac.nz smith@math.auckland.ac.nz

Abstract. All rings are commutative with identity and all modules are unitary. In this note we give some properties of a finite collection of submodules such that the sum of any two distinct members is multiplication, generalizing those which characterize arithmetical rings. Using these properties we are able to give a concise proof of Patrick Smith’s theorem stating conditions ensuring that the sum and intersection of a finite collection of multiplication submodules is a multiplication module. We give necessary and sufficient conditions for the intersection of a collection (not necessarily finite) of multiplication modules to be a multiplication module, generalizing Smith’s result. We also give sufficient conditions on the sum and intersection of a collection (not necessarily finite) for them to be multiplication. We apply D. D. Anderson’s new characterization of multiplication modules to investigate the residual of multiplication modules. MSC 2000: 13C05 (primary), 13A15 (secondary) Keywords: multiplication module, multiplication ideal, Prüfer domain, arithmetical ring, radical, direct sum, prime submodule, residual, torsion module

0. Introduction

Let R be a ring and M an R-module. For submodules K and L of M, the residual of K by L, denoted by [K : L], is the set of all x in R such that xL ⊆ K. An R-module M is called a multiplication module if for each submodule N of M there exists an ideal I of

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 2001 Heldermann Verlag

558 Majid M. Ali, David J. Smith: Finite and Infinite Collections of Multiplication Modules

R such that N = IM [3]. It is clear that every cyclic module is multiplication, and that a multiplication module over a local ring is cyclic [3]. Let N be a submodule of a multiplication module M. There exists an ideal I of R such that N = IM. Note that I ⊆ [N : M] and N = IM ⊆ [N : M]M ⊆ N so that N = [N : M]M. It follows that M is a multiplication R-module if and only if N = [N : M]M for all submodules N of M. An ideal A of R which is a multiplication module is called a multiplication ideal.

In Section 2 (Theorem 2.1) we establish several properties of a finite collection of submodules N_{i} of an R-module M which satisfy the condition that N_{i} + N_{j} is multiplication for all i < j. The interest in these properties lies in the fact that they generalize the standard characterization of arithmetical rings, see [6], [9] and [10]. Using these properties, we offer a short proof of Patrick Smith’s result giving conditions for the sum and intersection of a finite collection of multiplication modules to be a multplication module [18, Theorem 8]. Some examples will be given to highlight these properties by showing that they fail without the assumption of finiteness and that their converses are not true in general.

In Section 3 we give (see Theorem 3.2) necessary conditions for the intersection of a collection (not necessarily finite) of multiplication modules to be a multiplication module. We also give (see Theorems 3.6 and 4.2) conditions on the sum and intersection of an arbitrary collection of modules sufficient for them to be multliplication modules, generalizing Smith’s theorem.

In Section 4 we apply D.D. Anderson’s new characterization of multiplication modules, [2, Theorem 2.1], to obtain yet another characterization. In Theorem 4.3 we apply it to residuals of multiplication modules.

For the basic concepts we refer the reader to [5], [7], [11], [12] and [17].

1. Preliminaries

Let R be a ring and N_{i}(1 ≤ 1 ≤ n) a finite collection of submodules of an R-module M. Throughout this note we use the following notation:

S = n ; ∑_{l=1}N_{l}, N = n ; ⋂_{l=1}N_{l}, Ŝ_{i} = ∑_{j≠i}N_{j} and N̂_{i} = ⋂_{j≠i}N_{j}.

Lemma 1.1. Let R be a ring and N_{i}(1 ≤ i ≤ n) a finite collection of submodules of an R-module M such that

[N_{i} : N_{j}] + [N_{j} : N_{i}] = R for all i < j.

Then (i)  n ; ∑_{i=1}[Ŝ_{i} : S] = R,  (ii)  n ; ∑_{i=1}[N : N̂_{i}] = R.

Proof. (i) From the assumption we obtain n ; ∑_{i=1}∑_{j≠i}[N_{j} : N_{i}] = R. Obviously

∑_{j≠i}[N_{j} : N_{i}] ⊆ [Ŝ_{i} : N_{i}] for all 1 ≤ i ≤ n.

Hence n ; ∑_{i=1}[Ŝ_{i} : N_{i}] = R. But [Ŝ_{i} : N_{i}] = [Ŝ_{i} : S], and hence (i) is proved. The second part is

similar. □

Majid M. Ali, David J. Smith: Finite and Infinite Collections of Multiplication Modules 559

Compare the following result with [18, Proposition 4] and [7, Theorem 25.2]. Corollary 1.2. Let R be a ring and N_{i}(1 ≤ i ≤ n) a finite collection of submodules of an R-module M such that [N_{i} : N_{j}] + [N_{j} : N_{i}] = R for all i < j. Then

(i) [S : K] = n ; ∑_{i=1}[N_{i} : K] for every submodule K of M,

(ii) n ; ∑_{i=1}[N_{i} : S] = R,

(iii) [K : N] = n ; ∑_{i=1}[K : N_{i}] for every submodule K of M,

(iv) n ; ∑_{i=1}[N : N_{i}] = R,

(v) K ⋂ S = n ; ∑_{i=1}(K⋂N_{i}) for every submodule K of M, (vi) K + N = n ; ⋂_{i=1}(K + N_{i}) for every submodule K of M, (vii) IN = n ; ⋂_{l=1}IN_{l} for every ideal I of R. Proof. We prove (i) by induction on n. The result is true for n = 2 [18, Proposition 4]. Assume n > 2 and the result is true for n − 1, i.e. for every submodule K of M,

[Ŝ_{i} : K] = ∑_{j≠i}[N_{j} : K], for all 1 ≤ i ≤ n.

Suppose K is a submodule of M. Clearly n ; ∑_{l=1}[N_{l} : K] ⊆ [S : K]. By Lemma 1.1, there exists

x_{i} ∈ [Ŝ_{i} : S] such that 1 = n ; ∑_{i=1}x_{i}. Let z ∈ [S : K]. Then z = n ; ∑_{i=1}zx_{i} and zx_{i}K ⊆ x_{i}S ⊆

Ŝ_{i} for all 1 ≤ i ≤ n. It follows that

z ∈ n ; ∑_{i=1}[Ŝ_{i} : K] = n ; ∑_{i=1}[N_{i} : K]

so that [S : K] ⊆ n ; ∑_{i=1}[N_{i} : K], and (i) is proved. For (ii) take K = S in (i). (iii) is similar to (i). For (iv) take K = N in (iii). To prove (v), let K be any submodule of M. Clearly n ; ∑_{l=1}K⋂N_{l} ⊆ K⋂S. Using part (ii), there

exist x_{i} ∈ [N_{i} : S] such that 1 = n ; ∑_{i=1}x_{i}. Let u ∈ K ⋂ S. Then u = n ; ∑_{i=1}ux_{i} and ux_{i} ∈ K ⋂ N_{i}

for all 1 ≤ i ≤ n. Thus u ∈ n ; ∑_{i=1}K ⋂ N_{i} and hence K ⋂ S ⊆ n ; ∑_{i=1}K ⋂ N_{i}.

To prove (vi), assume that K is a submodule of M. Then clearly K + N ⊆ n ; ⋂_{l=1}(K + N_{l}). By

part (iv) there exist x_{i} ∈ [N : N_{i}] such that 1 = n ; ∑_{i=1}x_{i}. Let v ∈ n ; ⋂_{j=1}(K + N_{j}). Then v = k_{j} + n_{j}

for some k_{j} ∈ K, n_{j} ∈ N_{j} and all 1 ≤ j ≤ n. It follows that x_{j}v = x_{j}k_{j} + x_{j}n_{j} ∈ K + N, and

hence v = n ; ∑_{j=1}x_{j}v ∈ K + N so that n ; ⋂_{l=1}(K + N_{l}) ⊆ K + N and (vi) follows.

Finally, let I be an ideal of R. Clearly IN ⊆ n ; ⋂_{l=1}IN_{l}. Again by (vi), 1 = n ; ∑_{i=1}x_{i} for some

x_{i} ∈ [N : N_{i}]. Let y ∈ n ; ⋂_{r=1}IN_{r}. Then y ∈ IN_{r} for all 1 ≤ r ≤ n. Hence y = m ; ∑_{k=1}u_{rk}n_{rk} where

u_{rk} ∈ I and n_{rk} ∈ N_{r} for all 1 ≤ r ≤ n and all 1 ≤ k ≤ m. It follows that

y = n ; ∑_{r=1}yx_{r} = m ; ∑_{k=1} n ; ∑_{r=1}u_{rk}(n_{rk}x_{r}).

But n_{rk}x_{r} ∈ N and hence u_{rk}(n_{rk}x_{r}) ∈ IN for all 1 ≤ r ≤ n and all 1 ≤ k ≤ m. This implies that y ∈ IN and thus completes the proof of (vii). □

We observe that if R is a ring and N_{i}(1 ≤ i ≤ n) is a finite collection of finitely generated submodules of an R-module M such that N_{i} + N_{j} is multiplication for all i < j, then the conclusions of Corollary 1.2 are satisfied, see [16, Lemma 3.3] and [18, Corollary 3 of Theorem 1].

An R-module M is called a weak-cancellation module if, for all ideals I and J of R, if IM ⊆ JM then I ⊆ J + ann(M). It is clear that every cyclic module is weak-cancellation, from which it follows that finitely generated multiplication modules are weak-cancellation, see [4, Theorem 3.1] and [18, Corollary of Theorem 9]. However, the following holds for multiplication modules in general.

Lemma 1.3. Let R be a ring and M a multiplication R-module. Let I and J be ideals of R. Then IM ⊆ JM if and only if I ⊆ J + ann(m) for all m ∈ M.

Proof. Suppose first that I and J are ideals such that IM ⊆ JM. By [18, Theorem 9] either I ⊆ J + ann(M) and the result follows immediately, or M = [(J + ann(M)) : I]M. Let m ∈ M. Then Rm = AM for some ideal A of R. Now

Rm = AM = A[(J + ann(M)) : I]M
= [(J + ann(M)) : I]AM = [(J + ann(M)) : I]m,

and hence R = [(J + ann(M)) : I] + ann(m). Finally

I = RI = [(J + ann(M)) : I]I + ann(m)I
⊆ J + ann(M) + ann(m)I ⊆ J + ann(m),

as required. The converse is trivial. □

Corollary 1.4. Let R be a ring and N_{λ}(λ ∈ Λ) a collection of submodules of an R-module M, and let S = ∑_{λ∈Λ} N_{λ} be a multiplication module. Then

(i) ∑_{λ∈Λ}[N_{λP} : S_{P}] = R_{P} for every maximal ideal P of R. (ii) ∑_{λ∈Λ}[N_{λ} : S] + ann(a) = R for every a ∈ S. In particular, if K and L are submodules of an R-module M such that K +L is multiplication, then

(i) [K_{P} : L_{P}] + [L_{P} : K_{P}] = R_{P} for every maximal ideal P of R. (ii) [K : L] + [L : K] + ann(a) = R for every a ∈ K + L.

Proof. (i) As S is a multiplication module, we have N_{λ} = [N_{λ} : S]S for all λ ∈ Λ. Hence S = ∑_{λ∈Λ}[N_{λ} : S]S. Assume P is a maximal ideal of R. Then S_{P} is a weak-cancellation R_{P} -module. It follows that

R_{P} = ∑_{λ∈Λ}[N_{λ} : S]_{P} + ann(S_{P}) ⊆ ∑_{λ∈Λ}[N_{λP} : S_{P}] + ann(S_{P}).

But ann(S_{P}) ⊆ [N_{λP} : S_{P}] for all λ ∈ Λ. Thus (i) is proved. Part (ii) follows immediately by the preceding Lemma, since S = ∑_{λ∈Λ}[N_{λ} : S]S. □

We remark that Patrick Smith [18, Theorem 2] proved (ii) under the additional assumption that the submodules N_{λ} of M are multiplication.

2. Finite collections

The following theorem shows several properties of a finite collection of submodules N_{i}(1 ≤ i ≤ n) such that N_{i} + N_{j} is multiplication for all i < j. These properties generalize the characteristic properties of Prüfer domains (see for example [7, Theorem 25.2], [12, Theorem 6.6]) and of arithmetical rings (see [9, Lemma 2] and [10, Theorem 3]). Later we use these properties to give a concise proof of Smith’s theorem [18, Theorem 8].

Theorem 2.1. Let R be a ring and N_{i}(1 ≤ i ≤ n) a finite collection of submodules of an R-module M such that N_{i} + N_{j} is multiplication for all i < j. Then

(i) n ; ∑_{i=1}[Ŝ_{i} : S] + ann(a) = R for all a ∈ S,

(ii) n ; ∑_{i=1}[N : N̂_{i}] + ann(a) = R for all a ∈ S,

(iii) [S : K] = n ; ∑_{i=1}[N_{i} : K] (mod ann(a)) for all submodules K of M and all a ∈ S,

(iv) n ; ∑_{i=1}[N_{i} : S] + ann(a) = R for all a ∈ S,

(v) [K : N] = n ; ∑_{i=1}[K : N_{i}] (mod ann(a)) for all submodules K of M and all a ∈ S,

(vi) n ; ∑_{i=1}[N : N_{i}] + ann(a) = R for all a ∈ S, (vii) K ⋂ S = n ; ∑_{i=1}(K⋂N_{i}) for every submodule K of M,

(viii) K + N = n ; ⋂_{i=1}(K + N_{i}) for every submodule K of M,

(ix) IN = n ; ⋂_{l=1}IN_{l} for every ideal I of R.

Proof. As N_{i} + N_{j} is multiplication for all i < j, we infer from Corollary 1.4 that

[N_{i} : N_{j}] + [N_{j} : N_{i}] + ann(a) = R for all a ∈ N_{i} + N_{j}.

It follows that

n ; ∑_{i=1} n ; ∑_{j=1}[N_{i} : N_{j}] + [N_{j} : N_{i}] + ann(a) = R,

and hence by Lemma 1.1 we get that

n ; ∑_{i=1}[Ŝ_{i} : S] + ann(a) = R for all a ∈ N_{k} + N_{l}, k < l.

Let m ∈ S. Then m = ∑_{k<l}a_{kl} where a_{kl} ∈ N_{k} + N_{l}. Hence

R = n ; ∑_{i=1}[Ŝ_{i} : S] + ⋂_{k<l}ann(a_{kl}) = n ; ∑_{i=1}[Ŝ_{i} : S] + ann(∑_{k<l}Ra_{kl}) ⊆ n ; ∑_{i=1}[Ŝ_{i} : S] + ann(m),

so that

R = n ; ∑_{i=1}[Ŝ_{i} : S] + ann(m) for all m ∈ S.

(ii) is similar. For (iii), let K be a submodule of M. By induction it suffices to assume n = 2. By (i),

[N_{1} : N_{1} + N_{2}] + [N_{2} : N_{1} + N_{2}] + ann(a) = R for all a ∈ N_{1} + N_{2}.

Let a ∈ N_{1} + N_{2}. Clearly

[N_{1} : K] + [N_{2} : K] + ann(a) ⊆ [N_{1} + N_{2} : K] + ann(a).

Now let w ∈ [N_{1} + N_{2} : K] + ann(a). Then w = w_{1} + w_{2} where w_{1} ∈ [N_{1} + N_{2} : K] and w_{2} ∈ ann(a). Also, there exist x_{1}, x_{2}, z ∈ R such that x_{1} ∈ [N_{1} : N_{1}+N_{2}], x_{2} ∈ [N_{2} : N_{1}+N_{2}], z ∈ ann(a) and 1 = x_{1}+x_{2}+z. But w = w_{1}(x_{1}+x_{2}+z)+w_{2} and w_{1}x_{1}K ⊆ x_{1}(N_{1}+N_{2}) ⊆ N_{1}, and hence w_{1}x_{1} ∈ [N_{1} : K]. Similarly, w_{1}x_{2} ∈ [N_{2} : K]. Also, w_{1}z + w_{2} ∈ ann(a), and this shows that

[N_{1} + N_{2} : K] + ann(a) ⊆ [N_{1} : K] + [N_{2} : K] + ann(a).

For (iv), take K = S in (iii). (v) is similar to (iii), and for (vi), take K = N in (v). The last three parts of the theorem are true locally as seen by using Corollary 1.2 (parts (v), (vi), and (vii)) and Corollary 1.4, and hence they are true globally. □

Lüneburg [14, Theorem 3] proved that for ideals I, J of a domain R, if I + J is invertible, then (I + J)(I ⋂ J) = IJ. The following corollary extends this result.

Corollary 2.2. Let R be a ring and I, J ideals of R such that I + J is a multiplication ideal of R. Then

IJ = (I + J)(I⋂J).

Proof. As I + J is multiplication, we infer from Theorem 2.1 that

(I + J)(I⋂J) = (I + J)I⋂(I + J)J ⊇ JI⋂IJ = IJ.

The other inclusion is always satisfied. □

We apply Theorem 2.1 to give a concise proof of a theorem of P. Smith.

Theorem 2.3. [18, Theorem 8] Let R be a ring and N_{i}(1 ≤ i ≤ n) a collection of submodules of an R-module M such that N_{i} + N_{j} is multiplication for all i < j. Then:

(i) S is multiplication. (ii) If each N_{i} is multiplication, then N is multiplication.

Proof. (i) Let k ∈ {1, …, n}. It follows from Theorem 2.1(i) that

Ŝ_{k} = ( n ; ∑_{i=1}[Ŝ_{i} : S])Ŝ_{k}.

Then

Ŝ_{k} = [Ŝ_{k} : S]Ŝ_{k} + (∑_{i≠k}[Ŝ_{i} : S])Ŝ_{k} = [Ŝ_{k} : S]Ŝ_{k} + (∑_{i≠k}[Ŝ_{i} : Ŝ_{k}])Ŝ_{k}.

By induction suppose that Ŝ_{j} is multiplication for all j ∈ {1, …, n}. Then [Ŝ_{i} : Ŝ_{k}]Ŝ_{k} = [Ŝ_{k} : Ŝ_{i}]Ŝ_{i} = [Ŝ_{k} : S]Ŝ_{i} for all i ≠ k, and hence

Ŝ_{k} = [Ŝ_{k} : S]Ŝ_{k} + ∑_{i≠k}[Ŝ_{k} : S]Ŝ_{i} = [Ŝ_{k} : S]S.

Let K be any submodule of M. Then by Theorem 2.1 (vii) we have that

K⋂S = n ; ∑_{i=1}K⋂N_{i} ⊆ n ; ∑_{i=1}K⋂Ŝ_{i} = n ; ∑_{i=1}[K : Ŝ_{i}]Ŝ_{i} = n ; ∑_{i=1}[K : Ŝ_{i}][Ŝ_{i} : S]S ⊆ [K : S]S ⊆ K⋂S,

so that K ⋂ S = [K : S]S. This shows that S is a multiplication module. For the second part, let K be any submodule of M. By Theorem 2.1 (viii),

K⋂N = n ; ⋂_{i=1}K⋂N_{i} = n ; ⋂_{i=1}[K : N_{i}]N_{i} ⊆ n ; ⋂_{i=1}[K : N]N_{i} = [K : N]N ⊆ K⋂N,

so that K ⋂ N = [K : N]N, and N is a multiplication module. □

We prove two corollaries. The first is a generalization of [16, Corollary 3.4] and [18, Propositon 12], and the second shows that Theorem 2.1(vii) is true for an arbitrary collection of modules.

Corollary 2.4. Let R be a ring and N_{i}(1 ≤ i ≤ n) a finite collection of finitely generated multiplication submodules of an R-module M that can be generated by m_{i} elements respec-

tively, and let N = n ; ⋂_{i=1}N_{i}. If N_{i} + N_{j} is multiplication for all i < j, then N is a finitely

generated multiplication module that can be generated by n ; ∑_{i=1}m_{i} elements.

Proof. As N_{i} + N_{j} is multiplication, it follows from the remark made after Corollary 1.2

that n ; ∑_{i=1}[N : N_{i}] = R. Then there exist elements x_{i} ∈ [N : N_{i}] such that n ; ∑_{i=1}x_{i} = 1. It follows

that

N = n ; ∑_{i=1}x_{i}N ⊆ n ; ∑_{i=1}x_{i}N_{i} ⊆ N,

so that N = n ; ∑_{i=1}x_{i}N_{i} is a submodule of M generated by n ; ∑_{i=1}m_{i} elements. That N is mul-

tiplication follows by Theorem 2.3(ii). Alternatively, we may observe that x_{i}N ⊆ x_{i}N_{i} for

all i, and hence x_{i} ∈ [x_{i}N_{i} : N]. It follows that n ; ∑_{i=1}[x_{i}N_{i} : N] = R. But N = n ; ∑_{j=1}x_{j}N_{j},

and x_{i}N_{i} is multiplication [4, Corollary 1.4]. Thus by [18, Corollary 1 of Theorem 1], N is multiplication. □

Corollary 2.5. Let R be a ring and N_{λ}(λ ∈ Λ) a collection of submodules of an R-module M such that N_{λ} + N_{μ} is multiplication for all λ ≠ μ. Let S = ∑_{λ∈Λ}N_{λ}. Then K ⋂ S = ∑_{λ∈Λ}K⋂N_{λ}

for every submodule K of M. In particular, if N_{λ}(λ ∈ Λ) is a collection of multiplication modules such that

[N_{λ} : N_{μ}] + [N_{μ} : N_{λ}] = R for all λ ≠ μ,

then the result holds.

Proof. Let K be a submodule of M. Clearly ∑_{λ∈Λ}(K⋂N_{λ}) ⊆ K⋂S. For each x ∈ K ⋂ S there

exists a finite subset Λ_{x} of Λ such that x ∈ ∑_{λ∈Λ_{x}}N_{λ}. It follows that K ⋂ S ⊆ ∑ ; x∈K ⋂ S(K⋂ ∑_{λ∈Λ_{x}}N_{λ}).

As N_{λ} + N_{μ} is multiplication for all λ, μ ∈ Λ_{x} (λ ≠ μ), we infer from Theorem 2.1(vii) that K ⋂ ∑_{λ∈Λ_{x}}N_{λ} = ∑_{λ∈Λ_{x}}(K⋂N_{λ}). Therefore K ⋂ S ⊆ ∑_{λ∈Λ}K⋂N_{λ}, and the proof of the first

assertion is complete. The second assertion follows now by [18, Corollary 3 of Theorem 2]. □

Examples 1–3 below show that the conclusions of Theorem 2.1 (other than (vii)) are not valid without the finiteness assumption, and Example 4 shows, among other things, that their converses are not true.

Example 1. Let R = ℂ[[x]]. R is a Noetherian local domain and the unique maximal ideal of R is Rx. Let N_{k} = Rx^{k}. Then N_{k} is a multiplication ideal of R for each k ≥ 1 and so too is N_{k} + N_{n} for any positive integers k < n, because N_{k} + N_{n} = N_{k}. On the other hand N = ⋂_{k≥1}N_{k} = 0 and hence ∑_{l≥1}[N : N_{l}] = 0 ≠ R.

Example 2. Let R = [ℤ ℚ], the commutative ring of all matrices of the form ( \begin{matrix}n & r \\ 0 & n \end{matrix} )

with n ∈ ℤ, r ∈ ℚ.

(i) For s ∈ ℚ, let N_{s} = R ( \begin{matrix}0 & s \\ 0 & 0 \end{matrix} ) = [0 ℤs]. N_{s} is a multiplication ideal of R (being

principal). If s, t ∈ ℚ, say s = \frac{u}{v}, t = \frac{x}{y} with gcd(u, v) = gcd(x, y) = 1, then let r = gcd(s, t) = \frac{1}{vy} gcd(u, x) gcd(v, y). (See [13] for properties of gcd in this sense.) Then for any s, t ∈ ℚ,

N_{s} +N_{t} is multiplication because N_{s} +N_{t} = N_{r}, r = gcd(s, t). But S = ∑_{s∈ℚ}N_{s} = [0 ℚ] which

is neither finitely generated nor multiplication, and ∑_{s∈ℚ}[N_{s} : S] ≠ R since [N_{s} : S] = [0 ℚ]

for all s ∈ ℚ.

(ii) For i ≥ 1, let N_{i} = R ( \begin{matrix}i & 0 \\ 0 & i \end{matrix} ) = [ℤi ℚ]. Then N_{i} is multiplication and so too is

N_{i} + N_{j} for all i, j ≥ 1 because N_{i} + N_{j} = N_{d} where d = gcd(i, j). But N = ⋂_{i≥1}N_{i} = [0 ℚ]

is not multiplication, and ∑_{j≥1}[N : N_{j}] = 0 ≠ R.

(iii) Let p be an odd prime integer and N_{p} = [ℤp ℚ]. Then ⋂_{p≠2}N_{p} = [0 ℚ], and hence

N_{2} + ⋂_{p≠2}N_{p} = N_{2}. But ⋂_{p≠2}(N_{2} + N_{p}) = [ℤ ℚ] = R. Also it is easy to verify that [N_{2} :

⋂_{p≠2}N_{p}] = R but ∑_{p≠2}[N_{2} : N_{p}] = N_{2}.

Example 3. [8, Example 31] Let R be a Prüfer domain which is not Noetherian. There is a maximal ideal P of R which is not finitely generated. P is not multiplication, and hence R ≠ θ(p) = ∑_{p∈P}[pR : P], [1, Theorem 1] and [2, Theorem 2.1]. However, aR + bR is

multiplication for all a, b, ∈ R.

Example 4. Let Q be the ring of all sequences of elements of ℤ_{2}, and put e_{n} = (0, 0, …, 1, 0, …). Let R = Q[x, y]. Then I = ∑_{n≥1}e_{n}R is a multiplication ideal of R

since it is generated by idempotents, [1] and [4]. Let I_{n} = (e_{1}, …, e_{n}, e_{n+1}x, e_{n+1}y)R. We show that I_{n} is not a multiplication ideal by showing that I_{n} is not locally principal, [1]. Let M_{n+1} = (1 − e_{n+1}, x, y)R. M_{n+1} is a maximal ideal of R (in fact R/M_{n+1} ≈ ℤ_{2}). For f = ∑_{i,j}f_{ij}x^{i}y^{j} ∈ R, define φ(f) = ∑(f_{ij})_{n+1}x^{i}y^{j}, where (f_{ij})_{k} is the k^{th} term of the sequence

f_{ij}. Then φ : R → ℤ_{2}[x, y] is a homomorphism, and ker φ = (1 − e_{n+1})R ⊂ M_{n+1}. Moreover, φ(M_{n+1}) = (x, y)ℤ_{2}[x, y]. So φ extends to a homomorphism φ : R_{M_{n+1}} → ℤ_{2}[x, y]_{(x,y)}, and φ(I_{n}) = (x, y)ℤ_{2}[x, y]_{(x,y)} is not principal, hence neither is I_{n}R_{M_{n+1}}. Now, I_{1} ⊂ I_{2} ⊂ ⋯, and I = ⋃_{n≥1}I_{n}. One may show directly that, as Corollary 1.4(ii) concludes, ∑_{n≥1}[I_{n} : I]+ann(a) = R

for all a ∈ I, a conclusion which does not follow from Smith’s theorem [18, Theorem 2]. It also shows that the converse of Corollary 1.4 is not true, because

[I_{i} : I_{j}] + [I_{j} : I_{i}] + ann(a) = R

for all i ≠ j and all a ∈ I_{i} + I_{j}, but I_{i} + I_{j} = I_{k} where k = max{i, j}, which is not multiplication. This example further shows that the converses of Theorem 2.1 and Corollary

2.2 are not true. For this purpose, let S = n ; ∑_{i=1}I_{i} and N = n ; ⋂_{i=1}I_{i}. Then S = I_{n} and N = I_{1} are

not multiplication. Also, for all i, j, I_{i} + I_{j} is not multiplication.

3. Infinite collections

Naoum and Hasan [16] gave a sufficient condition for the intersection of two multiplication modules to be multiplication. That result was generalized by Smith [18] to a finite collection of multiplication modules. Using different methods, we extend the results to intersections of arbitrary collections of modules. First, we need a lemma.

Lemma 3.1. Let R be a ring and N_{λ}(λ ∈ Λ) a collection of submodules of an R-module M and let N = ⋂_{λ∈Λ}N_{λ}. If ∑_{λ∈Λ}[N : N_{λ}] = R, then IN = ⋂_{λ∈Λ}IN_{λ} for every ideal I of R.

Proof. Let I be an ideal of R. Clearly IN ⊆ ⋂_{λ∈Λ}IN_{λ}. If the condition is satisfied, there

exist a finite subset Λ′ of Λ and elements x_{λ} ∈ [N : N_{λ}](λ ∈ Λ′) such that ∑_{λ∈Λ′}x_{λ} = 1. Let

w ∈ ⋂_{λ∈Λ}IN_{λ}. Then w ∈ IN_{λ} for all λ ∈ Λ′. For each λ ∈ Λ′, there exists a finite set Λ_{λ} such

that w = ∑_{μ∈Λ_{λ}}u_{μ}y_{λμ} where u_{μ} ∈ I and y_{λμ} ∈ N_{λ}. It follows that w = ∑_{λ∈Λ′} ∑_{μ∈Λ_{λ}}u_{μ}(x_{λ}y_{λμ}). But

x_{λ}y_{λμ} ∈ N for all λ ∈ Λ′ and all μ ∈ Λ_{λ}. Thus w ∈ IN, and the result follows. □

Theorem 3.2. Let R be a ring and N_{λ}(λ ∈ Λ) a collection of multiplication submodules of an R-module M and let N = ⋂_{λ∈Λ}N_{λ}. Then

∑_{λ∈Λ}[N : N_{λ}] + ann(a) = R for all a ∈ N

if and only if these conditions are satisfied:

(i) IN = ⋂_{λ∈Λ}IN_{λ} for every ideal I of R.

(ii) N is a multiplication module.

Proof. Suppose first that

∑_{λ∈Λ}[N : N_{λ}] + ann(a) = R for all a ∈ N.

We can prove (i) even without the assumption that the N_{λ} are multiplication. It suffices to prove it locally. Thus we may assume that R is a local ring. If N = 0, there is nothing to prove. Otherwise N ≠ 0, and hence there exists 0 ≠ a ∈ N so that ann(a) ≠ R. It follows that

∑_{λ∈Λ}[N : N_{λ}] = R,

and the result follows from Lemma 3.1. To prove (ii), let K be any submodule of M. Then

K⋂N = ⋂_{λ∈Λ}(K⋂N_{λ}) = ⋂_{λ∈Λ}[K : N_{λ}]N_{λ} ⊆ ⋂_{λ∈Λ}[K : N]N_{λ} = [K : N]N ⊆ K⋂N,

so that K ⋂ N = [K : N]N, and N is multiplication. Conversely, assume that (i) and (ii) are satisfied. As N_{λ} is multiplication for all λ ∈ Λ, we infer that

N = [N : N_{λ}]N_{λ} ⊆ (∑_{μ∈Λ}[N : N_{μ}])N_{λ} for all λ ∈ Λ.

Hence

N ⊆ ⋂_{λ∈Λ}(∑_{μ∈Λ}[N : N_{μ}])N_{λ} = (∑_{μ∈Λ}[N : N_{μ}])N.

By Lemma 1.3, the result follows. □

In the following example, condition (i) but not (ii) of Theorem 3.2 holds.

Example 5. Let R = Q[[x]], I = Rx. (R, I) is a discrete valuation ring. Let D = ℤ + I. Then I is a Prüfer domain [7, p.319]. Let N_{i} = Dp_{i} where p_{1} < p_{2} < ⋯ is the sequence of positive primes of ℤ. Let N = ⋂_{i≥1}Dp_{i}. Then N is not a finitely generated ideal of R and

hence it is not multiplication. One can easily verify that for all i, [N : N_{i}] = N and hence

∑_{i≥1}[N : N_{i}] ≠ D.

We mention three corollaries to Theorem 3.2. The first is an immediate consequence while the second and third give sufficient conditions for the radical of a module to be multiplication.

Corollary 3.3. Let R be a ring and N_{λ}(λ ∈ Λ) a collection of multiplication submodules of an R-module M and let N = ⋂_{λ∈Λ}N_{λ}. If ∑_{λ∈Λ}[N : N_{λ}] = R, then N is a multiplication module.

Let R be a ring and M an R-module. A submodule P of M is called a prime submodule if whenever rm ∈ P, for some m ∈ M, r ∈ R, then m ∈ P or r ∈ [P : M]. The M-radical of a submodule N of M is defined as the intersection of all prime submodules of M containing N, (see [15]).

Corollary 3.4. Let R be a ring and N a submodule of an R-module M such that

∑[rad N : P] = R,

where the sum is over all prime submodules P of M containing N.

(i) If every P is multiplication, then so too is rad N. (ii) If every P is finitely generated, then so too is rad N. (iii) If every P is faithful, then so too is rad N.

Proof. (i) Corollary 3.3. (ii) There exist a finite set of prime submodules {P_{1}, P_{2}, …, P_{n}} of M containing N and

elements x_{i} ∈ [rad N : P_{i}] such that n ; ∑_{i=1}x_{i} = 1. It follows that rad N = n ; ∑_{i=1}x_{i}P_{i}, and hence is

finitely generated.

(iii) As in (ii), 1 = n ; ∑_{i=1}x_{i} with x_{i} ∈ [rad N : P_{i}]. Let w ∈ ann(radN). Then wx_{i}P_{i} = 0, and

wx_{i} ∈ ann(P_{i}) = 0. But w = ∑wx_{i} = 0. So radN is faithful. □

The Jacobson radical of a module M over a ring is defined [11] to be the intersection of all maximal submodules of M.

Corollary 3.5. Let R be a semi-local ring with maximal ideals P_{1}, …, P_{n}. If each P_{i} is multiplication (i.e. principal), then J(R), the Jacobson radical of R is multiplication (i.e. principal). If M is a multiplication R-module (i.e. cyclic) with P_{i}M ≠ M, then J(M), the Jacobson radical of M is also multiplication (i.e. cyclic).

Proof. The first assertion is clear by Theorem 2.3(ii). On the other hand, by [4, Theorem 2.5], P_{i}M is a maximal submodule of M and by [4, Corollary 1.4], P_{i}M is multiplication. In fact P_{i}M are the only maximal submodules of M. For, if Q is any maximal submodule different from P_{i}M, then again by [4, Theorem 2.5] there exists a maximal ideal P ∈ {P_{1}, P_{2}, …, P_{n}} such that Q = P M. Hence P_{i}M ≠ P M and hence P_{i} ≠ P, a contradiction. It follows

that J(M) = n ; ⋂_{i=1}P_{i}M. Since P_{i}M + P_{j}M = M for all i ≠ j, the result is clear by Theorem

2.3(ii). □

Let R be a ring such that every maximal ideal P of R is multiplication and ∑_{P maximal}[J(R):P]=

R. Then by Corollary 3.3, J(R) is multiplication. Suppose that M is a multiplication module such that for all maximal ideals P of R, P M ≠ M. Then J(M) = ⋂_{P maximal}P M, and P M is a

multiplication submodule of M. Also it is easy to verify that ∑_{P maximal}[J(M) : P M] = R, and

hence again by Corollary 3.3, J(M) is multiplication.

Example 6. Contrary to what happens in the finite case, in Example 2(ii) and Example 5 an intersection of multiplication modules is not multplication. On the other hand, let R = k[x, y], k an infinite field. Then R has an infinite number of maximal ideals M_{λ}, λ ∈ Λ. The Jacobson radical J = ⋂_{λ∈Λ}M_{λ} = 0 is a multiplication ideal. Also M_{λ} + M_{μ} = R, a

multiplication ideal for all λ ≠ μ, but the M_{λ} are not all multiplication ideals. Thus an intersection of modules may be multiplication even if the components are not. In each of these examples, however, the sum of any two distinct modules is multiplication.

The next theorem establishes necessary and sufficient conditions for the intersection of a collection (not necessarily finite) of multiplication submodules to be a multiplication module. It is a generalization of [18, Theorem 8 (ii)].

Theorem 3.6. Let R be a ring and N_{λ}(λ ∈ Λ) a collection of submodules of an R-module M. Let N = ⋂_{λ∈Λ}N_{λ} and S = ∑_{λ∈Λ}N_{λ}. Suppose that N_{λ} + N_{μ} is a multiplication module for all

λ ≠ μ. Let A = ∑_{λ∈Λ}[N : N_{λ}], and suppose that A + ann(n) = R for all n ∈ S. Then N is

multiplication if and only if N_{λ} is multiplication for all λ ∈ Λ.

Proof. Suppose first that N is multiplication. Let λ ∈ Λ and K a submodule of M such that K ⊆ N_{λ}. Clearly [K : N_{λ}]N_{λ} ⊆ K. Let x ∈ K and set

H = {r ∈ R | rx ∈ [K : N_{λ}]N_{λ}}.

If H ≠ R, then there exists a maximal ideal P of R such that H ⊆ P. We discuss two cases. Case 1: A ⊆ P. As A + ann(n) = R for all n ∈ S, we have N_{λ} + N_{μ} = A(N_{λ} + N_{μ}) for all μ ≠ λ. Then

N_{λ} + N_{μ} = A(N_{λ} + N_{μ}) ⊆ P (N_{λ} + N_{μ}) ⊆ N_{λ} + N_{μ}

so that N_{λ} + N_{μ} = P (N_{λ} + N_{μ}). Also N_{λ} + N_{μ} is multiplication for all μ ≠ λ. Thus Rx = I(N_{λ} + N_{μ}) for some ideal I of R and hence

Rx = I(N_{λ} + N_{μ}) = IP (N_{λ} + N_{μ}) = P x.

There exists p ∈ P such that (1 − p)x = 0, and hence 1 − p ∈ H ⊆ P, a contradiction. Case 2: A ⊈ P. Then

[N : N_{λ}] + ∑_{μ≠λ}[N : N_{μ}] ⊈ P.

If [N : N_{λ}] ⊈ P, there exists q ∈ P such that (1 − q)N_{λ} ⊆ N, and hence (1 − q)K ⊆ N. Since N is multiplication, there exists an ideal J of R such that (1 − q)K = JN. Now

(1 − q)JN_{λ} = J(1 − q)N_{λ} ⊆ JN = (1 − q)K ⊆ K

so that (1 − q)J ⊆ [K : N_{λ}]. It follows that

(1 − q)^{2}x ∈ (1 − q)^{2}K = (1 − q)JN ⊆ (1 − q)JN_{λ} ⊆ [K : N_{λ}]N_{λ}.

Hence (1−q)^{2} ∈ H ⊆ P, a contradiction. Finally, if ∑_{μ≠λ}[N : N_{μ}] ⊈ P, then there exists μ ≠ λ,

such that [N : N_{μ}] ⊈ P, and hence there exists q′ ∈ P such that (1 − q′)N_{μ} ⊆ N ⊆ N_{λ}. Then (1 − q′)(N_{λ} + N_{μ}) ⊆ N_{λ}. Next

K = [K : (N_{λ} + N_{μ})](N_{λ} + N_{μ}) ⊆ [K : N_{λ}](N_{λ} + N_{μ}),

and hence

(1 − q′)x ∈ (1 − q′)K ⊆ [K : N_{λ}](1 − q′)(N_{λ} + N_{μ}) ⊆ [K : N_{λ}]N_{λ}.

But this gives (1 − q′) ∈ H ⊆ P, a contradiction. Thus H = R and x ∈ [K : N_{λ}]N_{λ} so that K = [K : N_{λ}]N_{λ} and this proves that N_{λ} is a multiplication module. The converse follows by Theorem 3.2. □

4. Applications of Anderson’s new characterization of multiplication modules

D. D. Anderson [2, Theorem 2.1] has proved that a submodule A of an R-module M is multiplication if and only if for each maximal ideal P of R with A_{P} ≠ 0_{P}, A_{P} is cyclic and [N : A]_{P} = [N_{P} : A_{P}] for each submodule N of M. We use this new characterization to obtain two further characterizations of multiplication modules which we then apply to investigate the residual of multiplication modules. We further illustrate uses of the characterization by providing alternative proofs of some of our results in Sections 2 and 3.

Proposition 4.1. Let R be a ring and A a submodule of an R-module M. The following conditions are equivalent:

(i) A is a multiplication module. (ii) For each maximal ideal P of R with A_{P} ≠ 0_{P} , there exists a multiplication submodule

B containing A and an element p ∈ R such that (1 − p)B ⊆ A. (iii) For each maximal ideal P of R with A_{P} ≠ 0_{P} , there exists a multiplication submodule

B of A and an element p ∈ P such that (1 − p)A ⊆ B.

Proof. (i) ⇒ (ii) and (i) ⇒ (iii) are obvious by [2, Theorem 2.1 Part 5] and by taking A = B, B = Ra for some a ∈ A respectively. (ii) ⇒ (i): Let P be a maximal ideal of R such that A_{P} ≠ 0_{P} . There exists a multiplication submodule B of M containing A and an element p ∈ P such that (1 − p)B ⊆ A. It follows that A_{P} = B_{P} and for each submodule N of M,

[N_{P} : A_{P}] = [N_{P} : B_{P}] = [N : B]_{P} ⊆ [N : A]_{P} ⊆ [N_{P} : A_{P}],

and by [2, Theorem 2.1], A is multiplication. (iii) ⇒ (i): Let P be a maximal ideal of R such that A_{P} ≠ 0_{P} . There exists a multiplication submodule B of A and an element p ∈ P such that (1 − p)A ⊆ B. Hence A_{P} = B_{P} and for each submodule N of M,

[N : B]_{P} ⊆ [N : (1 − p)A]_{P} = [[N : A] : (1 − p)R]_{P} = [[N : A]_{P} : (1 − p)R_{P}] = [N : A]_{P} ⊆ [N : B]_{P} ,

and therefore [N : A]_{P} = [N : B]_{P} = [N_{P} : B_{P}] = [N_{P} : A_{P}], and again by [2, Theorem 2.1], A is multiplication. □

In Example 4 at the end of Section 2, we showed that a sum of modules may be multiplication even if no summand is multiplication. It is known [4, Theorem 2.2] that if a multiplication module S is a direct sum of submodules, then all of the summands are multiplication. We show that the same conclusion follows if the assumption of directness is weakened by assuming only that the intersection of distinct summands is multiplication.

Theorem 4.2. Let R be a ring and N_{λ}(λ ∈ Λ) a collection of submodules of an R-module M and let S = ∑_{λ∈Λ}N_{λ}. If S is multiplication and N_{λ} ∩ N_{μ} is multiplication for all λ ≠ μ, then

all N_{λ} are multiplication.

Proof. Let P be a maximal ideal of R. Let λ ∈ Λ, and suppose that N_{λP} ≠ 0_{P} . Then S_{P} ≠ 0_{P} , and hence for some a ∈ S, ann(a) ⊆ P. By Corollary 1.4, ∑_{μ∈Λ}[N_{μ} : S] ⊈ P, and

hence there exist μ ∈ Λ and p ∈ P such that (1 − p)S ⊆ N_{μ}. Let K be a submodule of M. If λ = μ, then

[K_{P} : N_{λP}] = [K_{P} : S_{P}] = [K : S]_{P} ⊆ [K : N_{λ}]_{P} ⊆ [K_{P} : N_{λP}],

and by [2, Theorem 2.1], N_{λ} is multiplication. Otherwise, λ ≠ μ, and hence (1 − p)N_{λ} ⊆ N_{λ} ∩ N_{μ}. It follows that (N_{λ} ∩ N_{μ})_{P} = N_{λP} , and hence

[K_{P} : N_{λP}] = [K_{P} : (N_{λ} ∩ N_{μ})_{P}] = [K : (N_{λ} ∩ N_{μ})]_{P} ⊆ [K : (1 − p)N_{λ}]_{P} = [K : N_{λ}]_{P} ⊆ [K_{P} : N_{λP}],

and again by [2, Theorem 2.1], N_{λ} is multiplication. □

We remark that this same method applying Anderson’s new characterization of multiplication modules may be used for example to give alternative proofs for our results 2.3, 3.2 and 3.6.

Naoum and Hasan [16, Theorem 2.5] proved that if R is an arithmetical ring and A and B are finitely generated ideals in R such that ann(B) is finitely generated, then [A : B] is finitely generated and hence a multiplication ideal. Patrick Smith [18, Theorem 10] showed that if M is a finitely generated faithful multiplication module, then N is a multiplication submodule of M if and only if [N : M] is multiplication. Compare Smith’s theorem with the following in which Anderson’s characterization of multiplication modules is applied to give an alternative proof. An R-module M is torsion if ann(m) ≠ 0 for all m ∈ M, otherwise it is called non-torsion [17]. Clearly every non-torsion module is faithful.

Theorem 4.3. Let R be a ring and B a non-torsion multiplication submodule of an R-mo-dule M. Then

(i) I = [IB : B] for every ideal I of R, (ii) a submodule A of B is multiplication if and only if [A : B] is a multiplication ideal of

R, (iii) IB is a multiplication module if and only if I is a multiplication ideal of R, (iv) a submodule A of B is non-torsion if and only if [A : B] is non-torsion.

Proof. (i) Let P be a maximal ideal of R. Since B is non-torsion, B_{P} ≠ 0_{P} . Since B is multiplication, it follows that IB = [IB : B]B, and from [2, Theorem 2.1] that ann(B_{P}) = ann(B)_{P} = 0_{P} . Hence, I_{P} B_{P} = [IB : B]_{P} B_{P} . Since B_{P} is cyclic, we conclude from [18, Corollary to Theorem 9] that I_{P} + ann(B_{P}) = [IB : B]_{P} + ann(B_{P}). But ann(B_{P}) = 0_{P} , so I_{P} = [IB : B]_{P} . Since the equality holds locally, it holds globally. (ii) Assume A is a multiplication submodule of B, and suppose P is a maximal ideal of R with [A : B]_{P} ≠ 0_{P} . Since B is non-torsion, we infer that B_{P} ≠ 0_{P} and [A : B]_{P} = [A_{P} : B_{P}]. Hence, A_{P} ≠ 0_{P} . It is easy to show that [A_{P} : B_{P}] is multiplication, and hence [A : B]_{P} is principal. Let I be an ideal of R. Since B is non-torsion and multiplication, it is easy to show that [IB : A] = [I : [A : B]]. Since A is multiplication,

[I_{P} : [A : B]_{P}] = [I_{P} : [A_{P} : B_{P}]] = [I_{P} B_{P} : A_{P}] = [IB : A]_{P} = [I : [A : B]]_{P} .

It follows from [2, Theorem 2.1] that [A : B] is multiplication. The converse follows by [4, Corollary 1.4]. (iii) follows from (i) and (ii), (iv) is routine. □

Finally, we observe that it follows easily from Lemma 1.3 that every non-torsion multiplication module over any ring R is a cancellation module and hence is finitely generated. On the other hand if R is an integral domain, then any faithful multiplication R-module M is non-torsion. In fact, since M is multiplication, M = Mθ(M), so that θ(M) ≠ 0. For some m ∈ M, [Rm : M] ≠ 0. Hence rM ⊆ Rm for some 0 ≠ r ∈ R, so that ann(m) ⊆ ann(rM) = 0.

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Received February 7, 2000; revised version November 23, 2000