NEW ZEALAND JOURNAL OF MATHEMATICS Volume 31 (2002), 115-129

PROJECTIVE, FLAT AND MULTIPLICATION MODULES

M a j i d M . A l i a n d D a v i d J . S m i t h

(Received December 2001)

Abstract. In this note all rings are commutative rings with identity and all modules are unital. We consider the behaviour of projective, flat and multipli­cation modules under sums and tensor products. In particular, we prove that the tensor product of two multiplication modules is a multiplication module and under a certain condition the tensor product of a multiplication mod­ule with a projective (resp. flat) module is a projective (resp. flat) module.We investigate a theorem of P.F. Smith concerning the sum of multiplication modules and give a sufficient condition on a sum of a collection of modules to ensure that all these submodules are multiplication. We then apply our result to give an alternative proof of a result of El-Bast and Smith on external direct sums of multiplication modules.

Introduction

Let R be a commutative ring with identity and M a unital i?-module. Then M is called a multiplication module if every submodule N of M has the form I M for some ideal I of R [5]. Recall that an ideal A of R is called a multiplication ideal if for each ideal B of R with B C A there exists an ideal C of R such that B — AC. Thus multiplication ideals are multiplication modules. In particular, invertible ideals of R are multiplication modules. On the other hand cyclic modules are multiplication, and multiplication modules over local rings are cyclic, [4, Theorem 1] and [2, Theorem 2.1].

Let R be a ring and K and L submodules of an /2-module M . The residual of K by L is [K : L], the set of all x in R such that xL C K . The annihilator of M, denoted by ann(M), is [0 : M], and for any m & M the annihilator of m, denoted by ann(m), is [0 : Rm\. For an i?-module M, we put 6 (M ) = • ^1-

Let M be a multiplication module and N a submodule of M . There exists an ideal I of R such that N = IM . Note that I C [N : M] and hence N = I M C [N : M ]M C iV, so that N = [N : M ]M . Note also that K is a multiplication submodule of M if and only if N D K = [N : K] K for all submodules N of M .

In Section 2, we investigate the tensor product of modules. We show that the tensor product of two multiplication i2-modules is a multiplication i?-module, The­orem 2.1. We also prove that if M is a multiplication i?-module and P is a projec­tive (resp. flat) i?-module such that ann(M) C ann(P), then M ® P is a projective (resp. flat) A-module, Theorems 2.3 and 2.5. We apply Theorem 2.3 to gener­alize [18, Theorem 11] and show that multiplication modules whose annihilators are generated by an idempotent are in fact projective modules, Corollary 2.4. As

2000 A M S Mathematics Subject Classification: 13C13, 13A15, 15A69.K ey words and phrases: projective module, flat module, multiplication module, free module, ten­sor product, direct sum.

116 MAJID M. ALI AND DAVID J. SMITH

a consequence of Theorem 2.5, we give a generalization of [16, Theorem 2.2] and prove that if M is a multiplication i?-module such that ann(M) is a pure ideal, then M is a flat i?-module, Corollary 2.7.

Section 3 is concerned with the sum of modules. If N\(X G A) is a collection of submodules of an P-module M such that the sum of any two distinct members is multiplication, then need not be multiplication [18]. Theorem 3.1 givesseveral equivalent conditions under which this sum is multiplication, generalizing P.F. Smith’s theorems [18, Theorems 2 and 8 (i)].

We also consider the following question: Let R be a ring and N\(X G A) a collection of submodules of an i?-module M such that *s multiplication.Is at least one N\ a multiplication module? We show that the answer is no, and we give a sufficient condition on such a collection to ensure that all of these submodules are multiplication, Theorem 3.2. We apply Theorem 3.2 to give an alternative proof to a result of El-Bast and Smith [8 , Theorem 2.2].

All rings are commutative with identity and all modules are unital. For the basic concepts used, we refer the reader to [6], [9], [10], [12] and [13].

1. Preliminaries

P.F. Smith gave necessary and sufficient conditions for the sum of a collection of multiplication modules to be a multiplication module [18, Theorem 2]. We inves­tigate this theorem and prove a strengthened version using alternative methods.

Lemma 1.1. Let R be a ring and N\(A G A) a collection of submodules of an R-module M , let N = X^AeA and let A = ^2\eA[N\ : N]. Consider the following statements:

(i) N is a multiplication module.

(ii) H = A H for all submodules H of N .

(iii) R = A + ann(n) for all n G N .

(iv) [N : K] + ann(n) = X^agA^* : + ann(n) for all submodules K of M and all n G N.

(v) K H N — J^AeA K ^ A a for all submodules K of M .

Then (i) => (ii) <> (iii) <> (iv) => (v). If all N\ are multiplication modules, then the first four conditions are equivalent.

Proof, (i) => (ii). As N is multiplication, N\ = [N\ : N ]N for all A G A, and hence N = A N . Let H be any submodule of N . Then H = IN for some ideal I of R , and hence

H = IN = I (AN) = A (IN ) = AH.

(ii) => (iii). Let n G N. Then Rn = An, and hence R = A + ann(n). (iii) => (ii) is obvious.

(iii) => (iv). Let K be any submodule of M . Let n G N . Clearly

^^[Ar> : K] + ann(n) C [N : K] + ann(n).a<ea

PR OJECTIVE, FLAT AND M ULTIPLICATION MODULES 117

Conversely, there exist a finite subset A' of A and elements x\ G [N\ : N ](A G A') and z G ann(n) such that 1 = J^xeA' x A + w G [N : K] + ann(n). Thenw = wi + W2 where W\K C N and W2 G ann(n), and hence

w ~ wix x + [w\z + W2) G : K] -I- ann(n)AeA' AeA

so that (iv) is satisfied, (iv) =>■ (iii) is clear by taking K = N. To complete the proof of the first assertion of the Lemma, it is enough to show that (iii) => (v). We do this locally. Thus we may assume that R is a local ring. If N = 0, the result is trivial. Thus we may suppose that N ^ 0. There exists 0 ^ n G N so that ann(n) 7 R , and hence A = R. There exists fi G A such that [N^ : N] = R and hence N = N^. This implies that

K H N = K n N ^ Q ^ K n N x C K n N ,AeA

so that K fl N = X^AeA N \K fl N\ , and the proof of the first part of the lemma is complete. For the last assertion, it is enough to show that (ii)=>- (i). We first prove that N\ = [iV* : N ]N for all A G A. We do this locally. Suppose that P is a maximal ideal of R. We discuss two cases.

Case 1: A C P. Then for all n G N , Rn = An C Pn, and hence (Rn)p = P p(R n)p , and by Nakayama’s Lemma we have that (Rn)p = Op, It follows that Np = Op, and hence (N\)p = Op for all A G A. Both sides of the equality N\ = [N\ : N ]N collapse to Op.

Case 2: A <£. P. There exists /i G A and p G P such that (1 — p)N C N^. It follows that Np = (NfL)p, and hence for all A G A,

[Nx - .N^ p C [N\ : (1 — p)N]p

= [ [ J V \ :jV ] :f l( l- p ) ] p

= [[ATa : N]p : (R(l - p))P] = [iVA : N]PQ {Nx -.N J p ,

so that [N\ : N^\p = [N\ : N]p. This implies that

[Nx :N ]p N p = [N x-.N ^p N p

= [N x :N f,]p(N fM)p = (N xn N ^ )p = {N\)p n {Nfj,)p = (N\)P f l Np = (N\)p.

From the two cases, we infer that the equation N\ = [N\ : N ]N is true locally and hence globally. Now assume that K is a submodule of M . Then by (v) of the lemma,

K n N = Y . K n N x = ’£ [ K :N X]N,,AeA AeA

= : N*\iN * : N ]N C [K : N }N C K n N,AeA

so that A 'f l iV = [K : N ]N , and hence N is a multiplication module. □

118 MAJID M. ALI AND DAVID J. SMITH

Note that the implication (v) => (iii) in the above Lemma is not true. Let R be a Priifer domain but not Noetherian. For example, take any non Noetherian valuation domain. There is a maximal ideal P of R which is not finitely generated. P is not multiplication, and hence 0(P) ^ R , [2, Theorem 2.1] and [3, Theorem 1]. However, for every non-zero ideal / of R, i n P = ^2peP In R p . The same example shows that (iii) does not imply (i), by taking Ni = N 2 = P.

The following consequence is straightforward.

Corollary 1.2. Let R be a ring and M an R-module. Then M is multiplication if and only if 9{M ) + ann(ra) = R for all m G M . Consequently, M is multiplication if and only if Rm = 9(M )m for all m G M .

Let M be an .R-module and let S = {ra,\|A G A} be a set of generators for M . Let F be the free i?-module generated by the set S' — {xa|A G A} and let n : F —> M be defined on S' by 7t(xa) = m\ and then extended to F linearly. Let K be the kernel of ir. Then the following exact sequence is called a presentation for M :

0 — » K — > F ^ M — > 0.The following characterizations of projective and flat modules can be found in [7] and [15].

Lemma 1.3. Let M an R-module. Then M is projective if and only if for each presentation

0 — > K — > F M — >0 for M , there exists an R-homomorphism 8 : F —> F such that (i) 7T o 9 = 7T and (ii) ker0 = ker7r.

Lemma 1.4. Let M be an R-module. Then M is flat if and only if for each presentation

0 — > K — > F ^ M — >0 and each u G K , there exists an R-homomorphism 9U : F —> F such that(i) 7T o 9U — tv and (ii) 9u{u) = 0.

We also need the next result about flat modules. The proof can be found in [7].

Lemma 1.5. Let M be aflat R-module with a presentation as above. Let Ui G K , 1 < i < n and let 9i be the corresponding homomorphisms given above. Then there exists a homomorphism 9 : F —» F such that(i) 7T o 9 = 7r and (ii) 9(ui) = 0 for all 1 < i < n.

2. Tensor Products

Let R be a ring. It is well known that if I and J are projective (resp. flat, multiplication) ideals of R , then IJ is projective (resp. flat, multiplication), [3] and [15]. If M is a multiplication .R-module and I is a multiplication ideal, then I M is a multiplication .R-module, [8 ]. If M is a finitely generated multiplication .R-module and I is a projective (resp. flat) ideal of R such that ann(M )fl/ = 0, then I M is a projective (resp. flat) .R-module [14]. It is also well-known [6 ] that the tensor product of two projective (resp. flat) i?-modules is a projective (resp. flat) .R-module. We now prove that the tensor product of two multiplication .R-modules is a multiplication .R-module. We also show that if M is a multiplication .R-module

PROJECTIVE, FLAT AND M ULTIPLICATION MODULES 119

and N is a projective (resp. flat) .R-module such that ann M C ann N , then the tensor product M <8 N is a projective (resp. flat) i?-module.Theorem 2.1. Let R be a ring and M i and M 2 multiplication R-modules. Then M i (8 M 2 is a multiplication R-module.

Proof. We first prove that 9(M i)9(M 2) C 9(M i <8 M2). Let mi E Mi and m2 £ M2. If x G [i?mi : Mi] and y G [i?m2 : M2], then xMi C i?mi and yM 2 C flm 2. It follows that

xy(M i <8 M 2) = xM i <8 yM 2 C Rrrii <8 Rm 2 = i?(mi <8 m2),and hence xy G [i?(mi (8) m2) : Mi <g) M 2]. This implies that [Rmi : Mi][Rrri2 : M2] C [i?(mi <8 m2) : Mi <8) M2] and hence 9(M i)9(M 2) C 0(Mi <8 M2). Next, since each of Mi and M2 is a multiplication .R-module, we infer from Corollary 1.2 that i?mi = 0(Mi)mi and Rni2 = 9 (M 2)m2■ Hence

R(mi (8 m 2) = Rrrii <8 Rm 2 = 0(M i)mi <8 9 (M 2)m 2

— 9(M i)9(M 2){mi <8 m2)C 0(Mi <8 M2)(mi <8 m2)

so that R(mi (8 m2) = 0(Mi <8 M 2)(mi <8 m2). From this we obtain that Rx = 9(M i ® M 2)x for all x G Mi <8 M2, and the result follows by Corollary 1.2. □

The next lemma should be compared with [16, Lemma 3.6].Lemma 2.2. Let R be a ring and M a multiplication R-module. Let {m A|A G A} be a set of generators for M . Then for each A G A, there exist a finite subset Aa of A and elements t^a G R with fi G Aa (depending on X) and a G A such that the following conditions are satisfied:

(i) rnx = Y , V I\neA\ J

(ii) t^ m p = t^pma for all a, ft G A and all fi E A\.

(iii) mA = ^

P roof. Let A G A. It follows by Corollary 1.2 that 9(M ) + ann(mA) = R • There exists a finite subset A* of A such that

^ [Rmp : M] + ann(mA) = R-

It follows by [16, Lemma 3.5] that there exist G [RmM : M], d^, d\ E R and w\ E ann(mA) such that

d» Cl» + dxWA = L

As G [Rmn : M], cmmM C i?mM, it follows that for all a E A there exists P such that — c Q.m. . Put t — d^c^^c^d. Then

(E dvcln mA + dxw2xm\ = mA, mgaa /

120 MAJID M. ALI AND DAVID J. SMITH

and hence m\ = ( ^ MeAAW j mA- This proves (i). Let cn,/? G A and E A\. Then

3 — d C ^ C f l — d u p 0 ^ 0 , 1 7 1 n

t[j,f3Tna -— d C ^C pTYlf — d C f C pTTl

and hence (ii) is satisfied. By (i) and (ii), it is now clear that

m x = ^ t^ m x = ^ 2 t xrrifj,.

□Theorem 2.3. Let R be a ring and M a multiplication R-module. If P is a projective R-module such that ann M C ann P, then M <g> P is a projective R-module.

Proof. Let {ra,\|A G A} be a set of generators for M . Let F\ be the free .R-module generated by {xa|A G A} and define 7Ti : F\ —» M by 7Ti(xa) = m\ for all A G A. Define 9\ : F\ —> F\ by 9\{x\) = J2u£Ax where A\ and t^\ are as defined inLemma 2.2. It follows that for all A G A,

(ttl o 91 )(xx) = 7ri(6>i(arA)) = 7Ti I t^\x^ = ^\neAx ) fJ-& aa

= y : tfiXm/i = mx = 7ri(rcA)./x€:Aa

Hence 7Ti o = 7Ti .Assume that P is generated by {pk\k G / } . Let P2 be the free P module gener­

ated by {yk\k G I} . Consider the exact sequence

0 — * K 2 — ► P2 P — + 0,

where 7r2 : P2 —» P is defined by 7r2 ( fc) = Pk f°r all k E I, and iC2 = ker7r2. It follows by Lemma 1.3 that there exists an P homomorphism 92 : P2 —> P2 such that (i) 7r2 o 92 = 7r2 and (ii) ker7r2 = ker#2. The P-module M 0 P is generated by the set {m^ <8>Pfc|A G A, A: G / } . Let F = F\ ® F2 and consider the exact sequence

0 — > K — > F M 0 P — >0,where 7r = 7Ti ® 7r2. Define 9 : F —> F by 9 = 9\ 0 02. Then 7t(xa <8> yk) — tti(^a) ® 7T2(2/fc) and 9(x\ <g> y^) = 9i{x\) <g> 92{yk) for all A G A and all k E I. To prove that M ® P is a projective P module, we show that 9 satisfies the conditions of Lemma 1.3. Let k E I and let 92(yk) = Ylh=i rikVi- Then for all A G A,

(7r o 9)(xx <g> yk) = tt(91 (xx) <^92(yk)) = tt1 (91 (xx)) 0 7T2 (92{yk))

= tti(^a) 0 n2(yk) = n(x\ <g) yfc),i.e. 7ro9 — 7r. For the second condition of Lemma 1.3, we obtain from [10, Theorem 7.7] that ker7r is generated by elements x ® y where either x E ker7ri or y E ker7r2. We discuss two cases.

PROJECTIVE, FLAT AND M ULTIPLICATION MODULES 121

Case 1: y E ker it2 - Then y E ker 02, and hence #2(2/) = 0- It follows that

0(ar ® y) = 6*i (rc) ® 0 = 0 ,

and hence x ® i/E ker0.

Case 2: ar E ker7Ti. There exists a finite subset A' of A such that x — X^aeA' wax a with wa E R. Hence

0 = 7Ti(ar) = 7Ti I ^ 2 W°cXoc I = ^ 2 Wa7rl(Xa) = ^ Wam a- VaG A' / a£A' a£A'

Next,

0(x <gi y) = (9i(ar) <g> 02 (y ) = 0 i ( ^ ti)a i a J ® ^ 2 ( y )vaGA'

S ® “ S ® 02( y ) ( * )^aGA' /i£Aa J

where Aa and t^a are as defined in Lemma 2.2. Since YlaeA' we havefor all (3 E A,

I X I W* tw ml3 = ^ 2 Wat ^ m a = 0 .VaG A' / aGA' q £A'

Hence XlaeA' £ ann(m^), and therefore X)«gA' t awa £ annM C ann P. It

follows that tn<xwaP = 0 for all p E P, and hence ^ £MaWa7r2(y) = 0. SoaGA'

EaGA' ^ « w«y G ker 7T2 = ker 2, and

0 = 02 I X t^away j = X t^awa62(y).VaG A' / aGA'

Hence from (*) we have that 6 (x <S> y) = 0, and this completes the proof of the theorem. □

Although not all projective /2-modules are multiplication, it is well known [19] that every projective ideal is multiplication. Moreover, it is proved [15] that if I is a projective ideal then ann I = ann L for some idempotent ideal L of R. More is known for the finitely generated case. For example, it is shown [16] that a finitely generated ideal / of R is projective if and only if I is multiplication and ann I = Re for some idempotent element e. This result has been extended for modules [18]: If M is a finitely generated multiplication A-module such that ann M = Re for some idempotent e, then M is projective. The next consequence of Theorem 2.3 is a generalization of this last result to multiplication modules (not necessarily finitely generated).

Corollary 2.4. Let R be a ring and M a multiplication R-module such that ann M = Re for some idempotent e. Then M is projective.

122 MAJID M. ALI AND DAVID J. SMITH

Proof. As ann M = Re = ann(l — e) and R( 1 — e) is a projective ideal of R , we infer from Theorem 2.3 that M 0 R( 1 — e) is a projective i? module. But R = Re-\- R( 1 — e). Thus

M ® R = M ® R e + M<g> R( 1 — e).Finally M ® Re = eM ® R = 0<8>R = 0, and M <g> R = M . It follows that M = M CS> i?(l — e), and hence M is projective. □

Theorem 2.5. Let R be a ring and M a multiplication R-module. If P is a flat R-module such that ann M C ann P , then M <g> P is a flat R-module.

Proof. Assume that Fi, F2 and F are as in the proof of Theorem 2.3. Also assume that 7Ti, 7t2, 7t and 9\ are as defined before. Let x ® y be a generator in ker7r. To show that M eg) P is flat, we need to show that there exists an P-homomorphism

'• F —> F such that (i) 7r o 9x®y = 7r and (ii) Qx®y (x <S> y) — 0. We discuss two cases.

Case 1: y G ker7r2. As P is flat, it follows by Lemma 1.4 that there exists an P-homomorphism 9y : F2 —»► F2 such that (i) 7t2 o 0y = 7r2 and (ii) 9y (y) = 0. Define 9x(g)y : F —> F by 0*®,, = 0i <8> 0y. Then

7T 0 0x ®y = ( 7 T 1 <8> 7T2 ) o ( 6 * 1 <g) 0y) = ( T T i O 9\) <8> ( 7 T 2 O 0 y ) = 7 T i ® 7T2 = 7T .

As 0y(y) = 0 , 9x®y(x ® y ) = 0.Case 2: x G ker7Ti. There exists a finite subset A' of A such that

x = A' wax a, where wa G R. It follows that

= I WaXa ) = ^2VccE A' J a(zA'

di {x) = 61 f Wax a 1 = 2 w a6l ( x a ) = Watl-\a£A' / a£A' a€A' n£Aa

0 = 7Ti(x

and

neAcwhere Aa and t^a are as defined in Lemma 2.2. As we have seen in the proof of Theorem 2.3, E « ga' Wat^a G ann M C annP, and hence Y2aeA' watnaP = 0 f°r all p G P. This implies 7r2 ( ^ aeA/ wat^ay) = 0, i.e. ^ aGA/ watm y G ker7r2. We infer from Lemma 1.5 that there exists an P-homomorphism 6 : F2 —* F2 such that(i) 7r2o0 = tt2 and (ii) 9 (J^aeA' w<*t»<*y) = 0, and hence X q6A' wat^aV G ker 0. Define : F -> F by 9x®y = <S> 0. As 0(y) G F2, put 0(y) = ^ ”=1 riVi- Hence

0x®y(z ® y) = 9x(x) <8> 9(y) = I ^ S w<*tw x v I ® ^ 2 nVl\a^A ' fxeAa J 1=1

n

= ^ 2 ^ 2 '5 2 w°‘tv<*ri(x n ® y i) . (*) « e A' fie Aa 1 = 1

Now,

0 = 0 £ Wat^ y I = Watna9(y) - ^ 2 (**)V qiGA' / a g A ' a £ A ' Z=1

PR O JECTIVE, FLAT AND M ULTIPLICATION MODULES 123

But F2 is a free i?-module. Thus for all 1 < / < n, the coefficient of yi in (**) is zero. Hence

n

Y . y ^ Wat pgr I = 0 .aGA' 1=1

It follows from (*) that 0x®y (x <g) y) = 0. Also,

7T O 6 x ® y — (7Ti <g> 7T2) O ( 6*1 (8 ) 0 ) = (7Ti O 6>i) (8 ) (7T2 O 0 ) = 7Ti <g> 7r2 = 7T,

and this finishes the proof of the theorem. □

Following [17, p. 100], a submodule N of an P-module M is called a pure submodule if AT n aM = aN for all a G R. In particular, an ideal I of R is pure if Rx — Ix for every x G I. It is known that every pure ideal of a ring R is either locally zero or locally R. For, take M to be any maximal ideal of R. If I C M , then for all x G 7, ( R x ) m = I m ( R x ) m Q M m ( R x ) m ? and hence (R x ) m = M m ( R x ) m > and by Nakayama’s Lemma, (R x ) m = 0m , and hence I m = 0m - On the other hand, if / ^ M , then I fl (R \ M ) ^ 0 , and hence I m — R m -

It is well known that if I is a finitely generated multiplication ideal with pure annihilator, then I is a flat ideal [16, Theorem 2.2]. We generalize this result to multiplication modules (not necessarily finitely generated). We first give a lemma.

Lemma 2.6. Let R be a ring and M a multiplication R-module such that ann M is a finitely generated pure ideal. Then M is a flat module.

Proof. Let ann M = Y^i=\ R ai- Then for all i, there exists yi € ann M such that ai( 1 - yi) = 0. Put

1 - y = (1 - 2/ i ) ( i - s f e ) - - - ( l - yn)-

Then y € ann M, and a ( 1 — y) = 0 . It follows that £ ann(l — y) for all i, and hence ann M C ann(l — y). On the other hand, as y M = 0, we infer that M = (1 — y )M and hence ann(l — y )Q annM . This gives that annM = ann(l — y) and that y is idempotent. Now, ann(l — y) is a pure ideal, and by [16, Theorem2.2] i?(l — y) is a flat ideal of R. It follows by Theorem 2.5 that M <g> R(1 — y) is a flat module. But

M <g> R — M <8) Ry -+ M <g) R (l — y)

and M (g) R = M and M <g> Ry = y M <g) R = 0, thus M = M (g> R( 1 — y), and M is a flat module. □

Corollary 2.7. Let R be a ring and M a multiplication R-module. Suppose that ann M is a pure ideal. Then M is a flat module.

Proof. Assume P is a maximal ideal of R. By [8 , Theorem 1.2] either M p = Op in which case it is obvious that M p is a flat Z?p- module, or there exist p € P and m G M such that (1 —p)M C Rm. It follows that (1 — p)Mann(m) = 0, and hence (1 — p)ann(m) C ann M . This implies that

ann(m)p C (ann M )p C ann (M p) C ann(i?ra)p = ann(m)p,

so that ann(Mp) = ann(M )p , see also [3, Theorem 2 .1 ]. As ann M is a pure ideal of R, we have that ann (M p) is a pure ideal of Rp. Since every pure ideal is idempotent and hence multiplication, we infer that ann (M p) is a principal ideal of

124 MAJID M. ALI AND DAVID J. SMITH

Rp. By Lemma 2.6, M p is again a flat R p -module. Hence, M is locally flat and therefore M is flat [6 ]. □

We make three observations. First, as a finitely generated pure ideal is a prin­cipal ideal generated by an idempotent, Corollary 2.4 shows that a multiplication .R-module M whose annihilator is a finitely generated pure ideal is projective. But every projective module is flat, [6], [7], [13]. Thus M is a flat module. This gives a different proof of Lemma 2.6. Second, from the proof of Corollary 2.7 it is obvi­ous that a multiplication module whose annihilator is an idempotent ideal is a flat module. Third, Q is a flat Z-module, but is not multiplication, which shows that the converse of Corollary 2.7 is not true in general.

Our final result of this chapter may be compared with [14, Theorems 2.1 and2.2]. The proof is now straightforward.

Corollary 2.8. Let R be a ring and M a multiplication R-module. If I is a projective (resp. flat) ideal of R such that annMPl/ = 0, then M ® I is a projective (resp. flat) R-module.

3. Sums of Multiplication Modules

Let Z be the ring of integers and p a prime number. Consider the Priifer p-group Zpoo as a Z-module. Note that Zp can be identified with a submodule of Zp2. In fact, pZp2 = Zp. More generally, for all m : k > 1 with m > k, Zpk can be identified with a submodule of Zpm. Hence

Zp C Zp2 C • • • C Zpoo

and Zpoo = (J Zpfc. These Zpk are the only nonzero submodules of Zpoo, and each is fc>i

a cyclic (hence multiplication) submodule. Also, for all m, k > 1, 7Lpk + Z pm = Zpr, where r = max{fc, m}, so that the sum of any two submodules of Zpoo is multipli­cation. We note however that Zpoo is neither finitely generated nor multiplication. The following theorem gives several equivalent conditions for the sum of a collection of submodules in which the sum of any two distinct members is multiplication to be a multiplication module. It generalizes [18, Theorems 2 and 8 (i)].

Theorem 3.1. Let R be a ring and M an R-module. Let N\(A £ A) be a col­lection of submodules of M such that N\ + Ntl is multiplication for all \ ^ pi. Let N = and A = Y lx eA ^ x ■ N]. Then the following conditions ar-eequivalent:

(i) A + ann(n) = R for all n G N .

(ii) N\ = [Nx : N ]N for all A e A.(iii) For every finite sum H of two or more modules N\, H is multiplication and

H = [H : N ]N .

(iv) N is multiplication.

Proof, (i) = > (ii). Let A, ^ G A with A ^ H- By Lemma 1.1, N\ + = A(N\ + Nfj,). Let A G A and x G N\. Suppose that

K = { r e R\rx G [Nx : A ]A }.

PR OJECTIVE, FLAT AND M ULTIPLICATION MODULES 125

Assume that K ^ R. Then there exists a maximal ideal P of R such that K C P. We discuss two cases.

Case 1: AC. P. Then for all /i ^ A, N\ + — A(N\ + N^) C P(N\ + N^) C N\ + N^ so that N\ + ATM = P(N\ + N^). As N\ + iVM is multiplication, we infer that Rx = I(N\ + ) for some ideal I of R, and hence

Rx = I(N\ + A g = IP(N\ + N J = P (I(N X + iVM)) = Px.

There exists p G P such that (1 — p)x = 0 . It follows that (1 — p) G K C P, a contradiction.

Case 2: A P. There exists fi G A such that [A^ : AT] P, and hence there exists q G P such that (1 — q)N C TV . It follows that (1 — q)N\ Q N\ + NM and hence there exists an ideal J of R such that (1 — q)N\ = J(N\ + N ^). Next

(1 - g) JN = J( 1 - g)AT C J(iVA + A g = (1 - g)iVA C Nx

so that (1 — q)J C [N\ : AT], This implies that

(1 - g f x € (1 - </)2 N A = (1 - 9 )J(ATa + N „ ) C (1 - g )JJV C [jV* : JVJiV.

But this implies (1 — q)2 G K C P, a contradiction. Thus K — R and x G [AT\ : Ar]A', that is A ^ C [ATa : AT] AT. Since the other inclusion is always true, N\ = [N\ : AT]N.

(ii) ==>• (iii). Let A' be a finite subset of A, and let H = J^agA' The fact that H is multiplication follows from [1, Theorem 2.3]. As N\ = [N\ : N ]N for all A G A', we have

H = £ ^ = ( Z : ) N Q : N N -AgA' \AgA' /

But [H : N ]N C H. Thus = [tf : AT]AT.(iii) ==>• (iv). Let i f be any submodule of M . For all x G K fl A”, there exists

a finite subset Ax of A such that x G X]agax ^a = Nx. It follows that

K n iV = X P z c X K n iV ,.xGifrw xeKnN

Nx is multiplication and Nx = [ATX : AT] A/". Hence

K n N C ^ [X : ATX]ATX C ^ [if : NX][NX : N]N C [X : AT] AT C i f f l iVxGXnyv xG/cniV

so that K D N = [K : N ]N and N is multiplication.(iv) =>■ (i) follows by Lemma 1.1.This finishes the proof of the theorem. □

If R is a ring and N\(\ G A) a collection of submodules of an P-module M such that £ a g a -^a is multiplication, must all of N\ (or at least some of them) be multiplication modules? The answer is no, as the following example shows: Let R be a ring with distinct maximal ideals M\ and M 2. Then R — M\ + M 2 is a multiplication ideal but neither M\ nor M 2 is multiplication. (See for example [11, Example 38]).

We show that the answer is yes under an additional condition.

126 MAJID M. ALI AND DAVID J. SMITH

Theorem 3.2. Let R be a ring and N x(X G A) a collection of submodules of an R-module M such that N = X^agA multiplication. If, moreover, N x nis multiplication for all X, fi G A with A ^ /x, then all submodules N\(X G A) are multiplication modules.

Proof. Let A = '^2Xe/i[Nx ■ N], By Lemma 1.1, N = A N . Let A 6 A. Let K C N\. Clearly [K : N X]NX C K . For the reverse inclusion, let y G K . Set

H = { r e R\ry G [K : NX]NX}.

Suppose that H ^ R. There exists a maximal ideal P of R such that H C P.

Case 1: A C P. Then N = A N C P N C N so that N = PAT. Next, Ry = IN for some ideal I of R and hence

Ry = IN = /FAT = P (IN ) = Py.

There exists p G F such that (1 — p)y = 0 and hence 1— p G F C P , a contradiction.

Case 2: A <£. P. Then

[JVa : N] + : N ] £ P .

If [N\ : N ] % P , there exists q G F such that (1 — q)N C N\. Now

K = [K : TV] AT C [X : N X]N,

and hence

(1 - q)y 6 (1 - q)K c [K : JVA](1 - q)N C [K : ATa]ATa.

It follows that 1 — q G H C P , also a contradiction. Finally if X ^ a ^ m ■ - 1 £ there exists /iG A and g' G P such that (1 — </)AT C A^. It follows that (1 — g') Aa C Aa H N and hence (1 — C Aa H ATm. But Aa fl ATM is multiplication. Thus (1 — q')K = J(N\ n N^) for some ideal J of R. Next

(1 - q')JNx = J( 1 - q')Nx C J (Nx fl N^) = (1 — 9') t f C K

so that (1 — q')J C [A : A a ] , and hence

(1 - </)2«/ 6 (1 - q ' f K = (1 - <{)J(Nx 0 N„) C (1 - C [K : JVA]JVA.

This gives a contradiction since (1 — q' ) 2 G H C P. Therefore H = R and x G [K : Aa]Na so that K = [K : NX]NX. This shows that N x is a multipli­cation module. □

It is well-known that if A and B are F-modules, then A ® B is projective if and only if A and B are projective, [10]. The same is not true for multiplication modules, for example, R is a multiplication F-module but R(B R is not. We apply Theorem 3.2 to give an alternative proof to [8 , Theorem 2.2] characterizing modules which are direct sums of multiplication modules.

If N x(X G A) is a non-empty collection of F-modules and A G A, we let N x = ® Nn and N = ® Nx. Note N = Nx ® Nx for all A G A. We iden-

Ag Atify the N x with their natural injections in N.

PR O JECTIVE, FLAT AND M ULTIPLICATION MODULES 127

Theorem 3.3. Let R be a ring and Nx(X E A) a collection of R-modules. ThenN = © N\ is multiplication if and only if:

AeA

(i) N\ is multiplication for all A G A.(ii) For each A G A, there exists an ideal A x of R such that A XNX = Nx and

A\N\ = 0 .

Proof. Suppose first that N is multiplication. Then 0 = N\ f l N\ D N\ f l A^ D 0, so that N \ n = 0 which is a multiplication module for all A, fi G A, A ^ fi. From Theorem 3.2 it follows that N x is multiplication for all A G A. Let A G A. Then N\ = A\N = A\N\ + A xN xjo r some ideal A\ of R. Hence A\N\ Q N X C\ Nx = 0 so that N\ = A\N\ and A XNX = 0. Conversely, suppose the modules N x(X G A) satisfy (i) and (ii). Then

Aa C ann(TVA) C [Nx : Nx] = [Nx : N],

and henceNx C A XN C [Nx : N }N C Nx,

so that Nx = [Nx : N ]N for all A G A.Put A = E X£A[NX : AT]. Then N = A N , and hence Nx = [A a : N ]N —

[ATa : N ]A N = A N X for all A G A. But Nx is multiplication for all A G A. Thus, by Lemma 1.1, R = A + ann(x) for all x e N x(X E A). Let n G N . There exist a finite subset A' of A and elements x x G N x(X G A') such that n = X^AeA' x *' ^ follows that

R — A + P| ann(^A) = A + annAeA'

I R xa I C A + ann(n) \AeA' J

so that R = A + ann(n). By Lemma 1.1, N is multiplication. This concludes the proof of the theorem. □

The next result should be compared with [8 , Corollary 2.3].Corollary 3.4. Let R be a ring N x(X G A) a collection of finitely generatedR-modules. Then N = © N x is multiplication if and only if

AeA(i) Nx is multiplication for all A G A.

(ii) a n n (^ ) + a n n (^ ) = R for all A G A.Proof. Suppose that N is multiplication. Then (i) follows by Theorem 3.3. Let A G A. As Nx D N x = 0, we infer from [1, Theorem 2.1] that

R = ann ( Nx fl N x j = ann(A^A) + ann (j^xj + ann(a:)

for all x G N and in particular for all x E Nx. As N x is finitely generated,R = &nn(Nx) + a n n (^ ) + f] ann(x), and (ii) follows. Conversely, suppose the

x £ N xmodules Nx(X E A) satisfy (i) and (ii). Let A G A. Then

R C ann(^A) + [Nx : Nx\ = ann{Nx) + [Nx : iV] C ann(A^A) + : N] C RneA

so that R = ann(Nx) + X^ eAt-^M • N] and by [18, Theorem 2 Corollary 2], N is multiplication. □

128 MAJID M. ALI AND DAVID J. SMITH

We close with the following property of the direct sum of a finite set of multipli­cation modules.

Corollary 3.5. Let R be a ring and N{(1 < i < n) a finite collection of R-modules. Then TV = ©TV; is multiplication if and only if:(i) TV, is multiplication for all i € {1 , . . . , n}.

(ii) ann(TV*) + ann(TV,) + ann(a) = R for all a G TV, + Nj and all i < j.

Proof. The sufficiency is immediate by Theorem 3.3 and Corollary 3.4. Con­versely, suppose the modules Ni satisfy (i) and (ii). Let i < j . Then [Ni : TV,-] + [Nj : TV,]+ann(a) = R and hence Ra = ([Ni : Nj] + [Nj : Ni])Ra for all a G TVj+TV,-. Summing over all a G Ni, we get

Ni = [Ni : Nj]Ni + [Nj : TV,] TV, = [TV, : TV ,-]^ + TV, n N3

= [Ni : Nj]Ni + [Ni : TV,] AT, C [Ni : Nt + N3}(Ni + N3) C Nt,

so thatNi = [Ni-.Ni + Nj^N i + Nj).

Similarly,Nj = [Nj : Ni + N j^N i + Nj).

Let K be a submodule of Ni + Nj. It follows from Lemma 1.1 thatK n (Ni + Nj) = K D N i + K H Nj = [K : TV,]TV, + [K : Nj]N3

= [K '.N iftN izN i + N j^N i + Nj)

+ [K : Nj][Nj : TV, + N3](Nt + TV,)

C [K : TV, + TVj](TVi + TV,) C K fl (TV, + TV,),

so that[i^ : TV, + TV,-](TV, + TV,) = K fl (TV, + TV,),

and hence Ni + TV,- is multiplication. It follows from [1, Theorem 2.3] that TV is multiplication, and the proof is complete. □

In particular the case when A is finite in Corollary 3.4 (or the modules TVj are finitely generated in Corollary 3.5) is of interest. If TV* (1 < i < n) is a finite collection of finitely generated i?-modules, then TV = ©TV; is multiplication if and only if all Ni are multiplication and ann(TV ) + ann(TV,) = R for all i < j . In this case, ann(TV) is multiplication if and only if ann(TV ) are all multiplication, see [18, Theorem 8 (ii)] and [1, Theorem 2.3]. Also, using [1 , Corollary 2.4], it is easily verified that if ann(TV ) are finitely generated, then so too is ann(TV).

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1057-1061.

Majid M. AliDepartment of Mathematics Sultan Qaboos University Muscat O M AN

mali@squ.edu.om

David J. SmithDepartment of MathematicsUniversity of AucklandAucklandN E W ZEALAND

smith@math.auckland.ac.nz