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Modules with the summand intersection propertyGeorge V. Wilson aa Department of Mathematics, University of Georgia, Athens, GA, 30602Published online: 27 Jun 2007.
To cite this article: George V. Wilson (1986): Modules with the summand intersection property, Communications inAlgebra, 14:1, 21-38
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COMMUNICATIONS IN ALGEBRA, 1 4 ( 1 ) , 21-38 (1986)
MODULES WITH THE
SUMMAND INTERSECTION PROPERTY
George V . W i l s o n Depar tment o f Ma themat i cs
U n i v e r s i t y o f Geo rg ia Athens, GA 30602
Kap lansky [?] showed t h a t i f F i s a f r e e module o v e r a
p r i n c i p a l i d e a l domain (P ID) , t h e n t h e i n t e r s e c t i o n o f any two
summands o f F i s a g a i n a summand. T h i s r e s u l t p rovoked Fuchs
[3] t o sugqes t t h e f o l l o w i n g p rob lem.
Prob lem 9: C h a r a c t e r i z e t h e a b e l i a n groups i n w h i c h t h e i n t e r s e c t i o n o f two d i r e c t summands i s a g a i n a summand.
I n t h i s pape r , we add ress t h e c o r r e s p o n d i n q q u e s t i o n f o r
modules o v e r a r i n g . I n s e c t i o n I, we d e v e l o p some t o o l s w h i c h
a p p l y t o any r i n g and use them t o d e s c r i b e i n j e c t i v e s w i t h S IP
o v e r n o e t h e r i a n r i n g s . S e c t i o n I 1 q i v e s a f a i r l y c o m p l e t e
d e s c r i p t i o n o f modules w i t h SIP o v e r a n o e t h e r i a n domain t h a t have
a nonze ro i n j e c t i v e submodule. S e c t i o n I 1 1 c o m p l e t e l y c l a s s i f i e s
t o r s i o n modules w i t h SIP o v e r Dedek ind domains. T h i s c l a s s i f i c a -
t i o n i s t h e n combined w i t h t h e r e s u l t s o f s e c t i o n I 1 t o g i v e a
Copyright O 1986 by Marcel Dekker, Inc.
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complete classification of modules with SIP that have a nonzero
injective submodule over a PID.
All rings considered here are associative with unity.
I. General Case
Definition: An R-module M has the summand intersection --
property (SIP) i f the intersection of two summands is apain a
summand. M has the strong summand intersection property (SSIP)
if the intersection of any number of summands is aqain a summand.
We begin by recordinq a fact which can be eas
Lemma 1 : If M has SIP (SSIP), then every summand ---
has SIP (resp. SSIP).
ily checked.
of M also
We now turn to our two main tools for the analysis of SIP
Proposition 1 : a) M has SIP if and only if for every pair of -
summands S and T with T T : M + S the projection map, the
kernel of the restricted map v I T is a summand.
b) If M has SIP and S c T is a summand of M , then
the kernel of any morphism from S to T is a summand.
Proof: a) Suppose M has SIP. For S' = ker .rr , we get -
M = S O S ' . Now ker 7 i I T = T n S ' is a summand. Conversely,
suppose that M has the stated property. Given S and T
summands of M , choose S' a complement to S and let o be
the projection to S' . We see that S n T = ker is a
summand.
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THE SUMMAND INTERSECTION PROPERTY
b ) By Lemma 1, S G T has SIP. Take any morphism o
mapping S t o T. I n S G T , we l o o k a t
Bo th S1 and S2 a r e summands o f S T w i t h complement U .
I t i s easy t o see t h a t S1 n S2 = C(k,O) / k E k e r 03 , so k e r a
must be a summand o f S .
P r o p o s i t i o n 2 : L e t M = G Mi be a d i r e c t sum o f f u l l y i n v a r i a n t
submodules Mi . The module M has SIP (SSIP) i f and o n l y i f
each Mi has SIP ( r e s p . SSIP).
P r o o f ; L e t S be any summand o f M. S i n c e each Mi i s f u l l y --
i n v a r i a n t , vSMi 5 S n Mi . A l s o each s E S can be w r i t t e n as
a f i n i t e sum s = Cmi w i t h mi E M i . So, s = n s = C v (m ) . S S i
We have S F G T ~ ( M ~ ) E G (S n M i ) E S , o r S = a (S O M i ) .
Now, g i v e n two summands S and T o f M ,
S ~ T = [ G ( S n M i ) ] fl [ G T n M i ] = G [ ( S n M i ) n ( T n M i ) ] .
S i n c e each Mi has SIP, t h i s i s a summand o f M . The p r o o f f o r
SSIP i s s i m i l a r .
Us ing P r o p o s i t i o n 1, we can c l a s s i f y some r i n g s i n t e rms o f
w h i c h modules have SIP.
P r o p o s i t i o n 3 : a ) A l l p r o j e c t i v e R-modules have SIP i f and
o n l y i f R i s h e r e d i t a r y .
b ) The f o l l o w i n g a r e e q u i v a l e n t f o r a r i n g R :
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i) R i s sem is imp le .
i i ) A l l R-modules have SSIP.
i i i ) A l l R-modules have SIP.
i v ) A l l i n j e c t i v e R-modules have SIP.
P r o o f : a ) Suppose t h a t R i s h e r e d i t a r y and P i s any p r o -
j e c t i v e R-module w i t h summands P1 and P2 . F o r any morph ism
0: P1 + P i m o i s p r o j e c t i v e , so a s p l i t s . Thus, k e r a 2 ' i s a summand o f P1 and by P r o p o s i t i o n 1 P has SIP.
Conve rse l y , assume t h a t a l l p r o j e c t i v e s have SIP. L e t P
be any p r o j e c t i v e and N any submodule o f P . Choose a f r e e
module F and an ep imorph i sm a: F -. N . By assumpt ion , t h e
p r o j e c t i v e F C P has SIP, so by P r o p o s i t i o n 1 k e r a i s a
summand o f F . N = i m o i s i s o m o r p h i c t o a complement o f
k e r a and so i s p r o j e c t i v e . Thus, R i s h e r e d i t a r y .
b ) Tha t i i m p l i e s ii i m p l i e s iii i m p l i e s i v i s
t r i v i a l . So assume t h a t i v h o l d s . We show t h a t R i s semi-
s i m p l e by showing t h a t a l l modu les a r e i n j e c t i v e . F o r any module
M , t h e r e i s an i n j e c t i v e El and a monomorphism a, : M + El
L i k e w i s e , t h e r e i s a monomorphi sm a2: El/ im fi , + E2 f o r some
i n j e c t i v e module E2 . S i n c e El G E2 has SIP, M ;.. k e r o2 i s
a summand o f El , hence M i s i n j e c t i v e . We c o n c l u d e t h a t R ,
i s sem is imp le .
We n o t i c e t h a t p a r t a o f t h e p r e v i o u s p r o p o s i t i o n c o n t a i n s
t h e r e s u l t o f Kap lansky w h i c h was men t i oned a t t h e b e g i n n i n q o f
t h i s pape r .
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THE SUMMAND INTERSECTION PROPERTY
P r o p o s i t i o n 3 says t h a t when R i s n o t sem is imp le , t h e r e
a r e i n j e c t i v e R-modules t h a t do n o t have SIP. We c l o s e t h i s sec-
t i o n by s t u d y i n g t h o s e i n j e c t i v e 8 t h a t do have SIP. We assume
f o r t h e r e s t o f t h e s e c t i o n t h a t R i s a commutat ive , n o e t h e r i a n
r i n g . The i n j e c t i v e enve lope o f a module M w i l l be d e n o t e d
E(M) .
Lemma 2: L e t R be a commuta t i ve , n o e t h e r i a n r i n g . L e t El and E2
be indecomposab le i n j e c t i v e modules . I f El GE2 has SIP, t h e n e i t h e r
i ) Hom(E,,E2) = 0
o r i i ) El i s i s o m o r p h i c t o E2 and t h e r e i s some p r i m e i d e a l
A < R w i t h a n n ( x ) = A f o r e v e r y nonze ro x E El .
P roo f : Take 0 # u E Hom(E1 ,E2) . S i n c e k e r a i s a summand o f -- El w h i c h i s n o t a l l o f El , k e r u = 0 . Now i m 0 i s a non-
z e r o i n j e c t i v e submodule o f E2 , hence a summand o f E2 . S i n c e
E 2 i s indecomposab le , 0 i s o n t o and El i s i s o m o r p h i c t o E2.
I t rema ins t o show t h e c o n d i t i o n on a n n i h i l a t o r s . Suppose
x and y a r e nonze ro e lemen ts of E and t h a t t h e r e i s some 1
a E a n n ( x ) w i t h a f a n n ( y ) . De f i ne r: El -+ E2 b y
~ ( m ) = am) . We see t h a t x E k e r r , S O T i s n o t a mono-
morph i sm. A1 so y f k e r T , so T # 0 . Thus, k e r T i s n o t a
summand, c o n t r a d i c t i n g P r o p o s i t i o n 1 . Thus, ann x = ann y f o r
a l l nonze ro x,y E El . Tha t ann x i s p r i m e f o l l o w s i m m e d i a t e l y
f r o m @, Theorem 61.
P r o p o s i t i o n 4: L e t R be a commutat ive , n o e t h e r i a n r i n g . An
i n j e c t i v e R-module E has S IP i f and o n l y i f i t has SSIP.
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P r o o f : C l e a r l y , a module w i t h SSIP has SIP. So assume t h a t E --
has SIP. E decomposes i n t o a d i r e c t sum o f indecomposab le i n -
j e c t i v e modules . We g roup t h e s e i n t o components Ci a l l o f whose
summands a r e e q u a l , i . e . E = @ C . and Ci = jEi 1
where f o r
each i , Ci has a l l t h e indecomposab le summands o f E ( i n t h e
g i v e n d e c o m p o s i t i o n ) w h i c h a r e i s o m o r p h i c t o jEi . By Lemma 2,
t h e C i l s a r e f u l l y i n v a r i a n t , so by P r o p o s i t i o n 2 i t s u f f i c e s
t o show t h a t each Ci has SSIP.
Each Ci i s a sum o f i s o m o r p h i c indecomposab le summands.
By Lemma 3 e i t h e r Ci i s indecomposab le , i n w h i c h case i t has
SSIP, o r i t i s a d i r e c t sum o f c o p i e s o f some indecomposab le
i n j e c t i v e module I = E(R/A) where A i s a p r i m e i d e a l and
a n n ( x ) = A f o r e v e r y x E I . We see t h a t I i s a t o r s i o n -
f r e e , i n j e c t i v e module o v e r t h e domain R I A , so I i s i somor -
p h i c t o t h e q u o t i e n t f i e l d o f R I A . Thus, Ci i s a v e c t o r
space o v e r t h i s f i e l d . I n t h i s case t o o , Ci has SSIP.
11. Modules w i t h a nonze ro i n j e c t i v e summand
F o r n o e t h e r i a n r i n g s R , a1 l R-modules have a u n i q u e maxi -
ma1 i n j e c t i v e summand i f and o n l y i f R i s h e r e d i t a r y . However,
o v e r any n o e t h e r i a n r i n g , modules w i t h SIP have a u n i q u e such
summand.
P r o p o s i t i o n 5 : L e t M be a module o v e r t h e n o e t h e r i a n r i n g R .
I f M has SIP, t h e n M has a u n i q u e maximal i n j e c t i v e summand.
P r o o f : By Z o r n ' s lemma, choose a maximal i ndependen t s e t CE,} --
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THE SUMMAND INTERSECTION PROPERTY 2 7
o f indecomposable, i n j e c t i v e submodules o f M . Since R i s
noe ther ian , E = G E i s i n j e c t i v e and so a d i r e c t summand o f M. a
We c l a i m t h a t E con ta ins a l l i n j e c t i v e submodules o f M . L e t
I be an i n j e c t i v e submodule o f PI . Since M has SIP, E f l I
i s a summand o f I , say I = (E n I ) G I' . I f I' # 0 , we
would ge t a s t r i c t l y l a r g e r s e t o f indecomposable, i n j e c t i v e
summands. Thus, I' = 0 and 1 5 E .
For t h e r e s t o f t h e s e c t i o n we w i l l s tudy modules o v e r a
n o e t h e r i a n domain D w i t h f i e l d o f q u o t i e n t s Q . Wi th t h i s
a d d i t i o n a l r e s t r i c t i o n , we w i l l be a b l e t o s t reng then t h e d e s c r i p -
t i o n o f i n j e c t i v e s w i t h SIP. I n t u r n , t h i s s t r o n g e r v e r s i o n w i l l
be used t o show t h a t when a module w i t h SIP has a nonzero maximal
i n j e c t i v e submodule, t h i s summand e s s e n t i a l l y determines t h e whole
module.
P r o p o s i t i o n 6 : L e t D be a n o e t h e r i a n domain. For an i n j e c t i v e
D-module, E , t h e f o l l o w i n g a r e e q u i v a l e n t .
a ) E has SSIP.
b ) E has SIP,
c ) e i t h e r i ) E i s t o r s i o n - f r e e
o r i i ) E i s t o r s i o n and f o r any two d i s t i n c t , indecomposable summands I and J o f E , Hom(1,J) = 0 .
Proo f : P r o p o s i t i o n 4 shows t h a t a i s e q u i v a l e n t t o b. To
show t h a t b i m p l i e s c, we beg in by showing t h a t E i s n o t mixed.
For any nonzero i d e a l A o f D , Hom(Q,E(D/A)) # 0 because t h e
map D + D/A + E(D/A) extends t o Q . Thus, no module w i t h SIP
can have Q 6 E(D/A) as a summand. E i s e i t h e r t o r s i o n o r
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28 WILSON
t o r s i o n - f r e e . Suppose t h a t E i s t o r s i o n w i t h indecomposab le
summands I and J . Lemma 2 shows t h a t i f Hom(1,J) f 0 , t h e n
I i s i s o m o r p h i c t o J and a n n ( x ) = a n n ( y ) f o r e v e r y x,y E I . We show t h a t t h e r e a r e e lemen ts i n I w i t h d i s t i n c t a n n i h i l a t o r s .
Take any nonzero x E I and nonze ro a E a n n ( x ) . S i n c e I i s
d i v i s i b l e , t h e r e i s some y E I w i t h a y = x . C l e a r l y ,
a n n ( y ) f a n n ( x ) , so I J does n o t have SIP. T h i s c o n t r a -
d i c t i o n shows t h a t Hom(1,J) = 0 . F i n a l l y , suppose t h a t c h o l d s . I f E i s t o r s i o n - f r e e ,
t h e n i t i s a Q - v e c t o r space and c l e a r l y has SSIP. I f E i s as i n
c, ii, t h e n E decomposes i n t o a d i r e c t sum o f f u l l y i n v a r i a n t
indecomposab le modules E = G E S i n c e indecomposab les have i '
SSIP, P r o p o s i t i o n 2 shows t h a t E has SSIP.
To s t a t e t h e n e x t r e s u l t s we need t h e n o t i o n o f q u a s i -
i somorph ism. F o r t o r s i o n - f r e e modules A and B , a q u a s i -
homomorphism f r o m A t o B i s a homomorphism cp: E(A) + E(B)
such t h a t f o r some d E D , dcp(A) 5 B . A quasi-homomorphism cp
i s a q u a s i - i s o m o r p h i s m i f i t i s an i somorph i sm o f E (A ) w i t h
E(B) and cp-' i s a quasi-homomorphism. F o r A and B o f
f i n i t e r a n k , A i s q u a s i - i s o m o r p h i c t o B i f and o n l y i f A i s
i s o m o r p h i c t o a submodule o f B and B i s i s o m o r p h i c t o a sub-
module o f A . F o r f u r t h e r i n f o r m a t i o n , t h e r e a d e r i s r e f e r r e d
t o [3 5921 w h i c h t r e a t s t h e case o f a b e l i a n g roups e x t e n s i v e l y
and t o [ 2 ] w h i c h t r e a t s modules o v e r any domain.
Lemma 3: L e t D be a n o e t h e r i a n domain w i t h f i e l d o f q u o t i e n t s Q. --
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THE SUMMAND INTERSECTION PROPERTY 2 9
a) L e t A and B be rank one, reduced, t o r s i o n f r e e D-modules.
If C! G A G B has SIP, then A i s quasi - isomorphic t o B .
b ) L e t {Aili EIN be a c o l l e c t i o n o f quas i - i somorph ic ,
rank one, reduced, t o r s i o n - f r e e D-modules. W r i t i n g A = G A i '
Q G A does n o t have S I P .
P roo f : a ) We i d e n t i f y A and B w i t h submodules o f Q . S ince
A i s n o t d i v i s i b l e , we can choose d E D w i t h l / d F A .
De f ine a map 0: A G B -+Q by o ( r , s ) = d r - s .
k e r a = { ( r , d r ) : r € A , d r E B } . By P r o p o s i t i o n 1, k e r o i s a
summand o f A 6 B , say A G B = k e r o G X . k e r a has rank one
as does X . Obvious ly , A i s n o t con ta ined i n k e r a , so vx
t h e p r o j e c t i o n t o X sends A i s o m o r p h i c a l l y t o a submodule o f
X . A lso, ( 0 , l ) E A G B can be w r i t t e n
( 0 , l ) = (c,dc) G ( - c , l - d c ) E k e r o X . We cannot have dc = 1 ,
s i n c e l / d E A . Thus, 1 - dc f 0 and nBX f 0 . Now .ir .rr B X
sends A i s o m o r p h i c a l l y t o a submodule o f B . Symmet r i ca l l y ,
B i s isomorphic t o a submodule o f A , so A and B a r e quas i -
isomorphic .
b ) Again, we i d e n t i f y t h e Ai w i t h submodules o f Q .
Replac ing Ai w i t h an isomorphic copy as needed, we can assume
t h a t A < A < A . . . < Q . Since A1 i s n o t d i v i s i b l e , t h e r e 1 - 2 - 3 - i s some d C D w i t h dA1 f A1 . I t f o l l o w s e a s i l y t h a t
dAi # Ai f o r a l l t h e i ' s . Now d e f i n e a map o f rom A = G A i
t o Q by a ( o ) = a/di f o r a E Ai. C l e a r l y , d ( i m a) = im o ,
so o does n o t s p l i t . Thus, Q G A does n o t have SIP.
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We a r e now i n a p o s i t i o n t o p r o v e t h e ma in r e s u l t o f t h i s
s e c t i o n .
P r o p o s i t i o n 7 : L e t D be a n o e t h e r i a n domain and l e t M be a
D-module w i t h SIP. L e t I f 0 be i t s maximal i n j e c t i v e summand
w i t h complement I , M = I I .
a ) I f I i s t o r s i o n , a l l o f M i s t o r s i o n .
b ) I f I i s t o r s i o n - f r e e , t h e n t h e t o r s i o n submodule T
o f M i s a summand and M = I G T GU, where U i s a f i n i t e
d i r e c t sum o f q u a s i - i s o m o r p h i c , r a n k one modules .
P r o o f : a ) Suppose t h a t A i s an i d e a l o f D and t h a t
E = E(D/A) i s an indecomposab le summand o f I . The n a t u r a l map
D + D/A -+ E ex tends t o a nonzero , non isomorph ism T E Hom(Q,E) .
I f I ' i s n o t t o r s i o n , t h e n I ' / t o r s ( I 1 ) i s t o r s i o n - f r e e .
Thus, t h e r e a r e nonze ro morphisms o E Hom(1' ,()) . S i n c e Q i s
d i v i s i b l e , m u l t i p l i c a t i o n by any q E Q d e t e r m i n e s a n automor-
ph ism m o f Q . F o r a s u i t a b l e q , im(m a) i s n o t c o n t a i n e d 9 9
i n k e r -r . Thus, T m n i s a nonze ro morph ism f r o m I ' t o E . 9
B u t T m o c a n n o t s p l i t , because o a n n i h i l a t e s t o r s i o n . T h i s 9
would g i v e t h a t E G I ' does n o t have SIP, c o n t r a d i c t i n g o u r
assumpt ion. Thus, I' must be t o r s i o n .
b ) I f I ' i s t o r s i o n , t h e r e i s n o t h i n g t o show. So
assume t h a t I' i s n o t t o r s i o n and x E I ' i s t o r s i o n - f r e e .
The submodule <x> g e n e r a t e d by x i s i s o m o r p h i c t o D and so
embeds i n Q . S i n c e Q i s i n j e c t i v e , t h i s map ex tends t o a non-
z e r o m a p a f r o m I ' t o Q . S i n c e I' GQ has SIP, o s p l i t s
and I , I I a - c Q i s i s o m o r p h i c t o a summand o f I ' . By Z o r n ' s
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lemma, choose a maximal independent s e t Csi 1 o f rank one,
t o r s i o n - f r e e summands o f I ' . L e t U = G Si. C l e a r l y , U G T F I ' .
We c l a i m t h a t U C T = 1 ' . Say n o t . Take any x f U E T and l e t
r;: I' + I ' / U G T be t h e q u o t i e n t map. I f <x> n ( U G T ) = 0 ,
then nx i s a t o r s i o n - f r e e element. The embedding o f a x > i n t o
Q extends t o a nonzero map -r f rom I 1 / U G T t o Q . Now TTT
must s p l i t , g i v i n g a rank one, t o r s i o n - f r e e summand S o f I ' .
Since a r a n n i h i l a t e s U T , S 0 U = 0 c o n t r a d i c t i n g t h e
maximal i t y o f CSi 3 . Therefore, we must have dx E U G T f o r
some d E D . We can f i n d a nonzero a E D w i t h a d x E U .
Since U i s pure, ad x = ad w f o r some w E U . We ge t
ad(x - w) = 0 so x d i f f e r s f rom w by an element o f T . We
conclude t h a t x E U T and t h a t I' = U G i . Since I G U
must have SIP, Lemma 3 shows t h a t a l l t h e rank one summands o f U
must be quas i - i somorph ic and t h a t U must have f i n i t e rank .
Wi th some a d d i t i o n a l r e s t r i c t i o n s on the domain D , we can
complete t h e d e s c r i p t i o n g i v e n i n P r o p o s i t i o n 7 o f t h e modules
w i t h SIP t h a t have a nonzero i n j e c t i v e submodule. We d e l a y t h e
a n a l y s i s o f t h e t o r s i o n submodule u n t i l t h e n e x t s e c t i o n . We
f i n i s h t h i s s e c t i o n w i t h a r e s u l t on t h e t o r s i o n - f r e e p a r t , when
D i s a PID. I n t h i s case, quasi - isomorphic rank one modules a r e
isomorphic , so t h e U o f P r o p o s i t i o n 7 i s a homogeneous,
comp le te ly decomposable module o f f i n i t e rank.
P r o p o s i t i o n 8: L e t D be a PID. L e t I be a t o r s i o n - f r e e i n -
j e c t i v e module and U be a homogeneous, comp le te ly decomposabl e
module o f f i n i t e rank. The module M = I G U has SSIP.
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P r o o f : L e t CSi} be any c o l l e c t i o n o f summands. S i n c e M i s
t o r s i o n - f r e e , K = n Si i s p u r e [3 526 f a c t e l . IK = I n K i s a
p u r e subgroup o f I , hence i s d i v i s i b l e and a summand. W r i t e
K = IK G U K . By [4, - Lemma 61, i t s u f f i c e s t o show t h a t K / I K i s
a summand o f M / I K , so we can ssume t h a t IK = 0 . A g a i n u s i n g
[4, Lemma 61, i t s u f f i c e s t o show t h a t ( K + I ) / I i s a summand o f
M / I . B u t ( K + I ) / I i s a p u r e subgroup o f M / I = U , hence i s a
summand by [3, 86.81. T h e r e f o r e , K i s a summand o f M .
Of cou rse , t h e above r e s u l t i n c l u d e s t h e case I = 0 . We
c o n c l u d e t h i s s e c t i o n by g e n e r a l i z i n g t h i s case . Our r e s u l t i s
based on a lemma o f Botha and Grabe, w h i c h does a l l t h e work .
These a u t h o r s c o n s i d e r e d t h e c l a s s @ o f a l l f i n i t e r a n k ,
t o r s i o n - f r e e modules whose endomorphism r i n g i s a PID. T h i s
c l a s s i n c l u d e s a l l t o r s i o n - f r e e modules o f r a n k one. The p a r t o f
t h e i r work w h i c h i s u s e f u l h e r e i s t h e f o l l o w i n g lemma.
Lemma 4 : [I] L e t A be a f i n i t e d i r e c t sum o f c o p i e s o f some -- -
C E @ . The k e r n e l o f each endomorphism o f A i s a summand o f
A , w h i c h i s i s o m o r p h i c t o a d i r e c t sum o f c o p i e s o f C .
C o r o l l a r y : A has SSIP. --
P r o o f : By P r o p o s i t i o n 1, A has SIP. S i n c e A has f i n i t e r a n k ,
i t has SSIP.
Remark: I n f i n i t e r a n k , homogeneous, c o m p l e t e l y decomposable
modules need n o t have SSIP. F o r example, a f r e e a b e l i a n g roup o f
t h e power o f t h e con t i nuum has SIP, b u t n o t SSIP.
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THE SUMMAND INTERSECTION PROPERTY
111. T o r s i o n
Hav ing d e t e r m i n e d t h e i n j e c t i v e p a r t o f a module w i t h SIP,
we t u r n t o t h e reduced p a r t and, i n p a r t i c u l a r , t h e t o r s i o n .
Throughout t h i s s e c t i o n , D w i l l d e n o t e a Dedek ind domain w i t h
f i e l d o f q u o t i e n t s Q . I n t h i s case, a t o r s i o n module T i s
u n i q u e l y a d i r e c t sum T = TB o f Q -p r imary submodules, one
f o r each p r i m e i d e a l 8 . We n o t e t h a t Tg i s u n i q u e l y d i v i s i b l e
by each d f B .
Lemma 5 : L e t D be a Dedek ind domain and l e t T = GTG be a
t o r s i o n module. T has SIP (SSIP) i f and on
B , t h e B-component TQ has S IP ( r e s p . SSIP
P r o o f : The TB a r e f u l l y i n v a r i a n t , so t h e --
i m m e d i a t e l y f r o m Lemma 1 and P r o p o s i t i o n 2.
P r o p o s i t i o n 9: L e t D be a Dedek ind domain --
l y i f f o r e v e r y p r i m e
) .
lemma f o l l o w s
a n d l e t C b e a
reduced module w i t h SIP. F o r each p r i m e i d e a l 8 , t h e @ - t o r s i o n
Cg i s a d i r e c t summand o f C w h i c h i s e i t h e r e l e m e n t a r y o r c y c l i c .
P r o o f : Cg i s a module o v e r t h e PID DQ, so we can use some o f --
t h e s t a n d a r d r e s u l t s on P I D ' s . I n p a r t i c u l a r , QDQ = ( p ) f o r
some p r i m e e lemen t p . Not e v e r y e lemen t o f o r d e r p i n CQ
can have i n f i n i t e h e i g h t ( a t p ) , s i n c e t h a t wou ld i m p l y t h a t CB
i s d i v i s i b l e 14, Lemma 81.
Case 1 : Assume t h a t t h e r e i s a n e lemen t t E Cg o f o r d e r p
n w i t h h ( t ) = n 0 < n < . Take y E Cg w i t h p y = t . Now
(<y> + # c ) / Q ~ c i s bounded and p u r e as a submodule o f t h e Dy
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WILSON 3 4
module C/@C . T h e r e f o r e , [3, 27 .51 (<y> + Q ~ c ) / Q ~ c i s a
summand o f C/@C . A l s o <y> fl = 0 , so <y> i s a summand
o f C [$, Lemma 61, say C = <y> 6 X . We c l a i m t h a t X can
have no Q - t o r s i o n . I f x E X has a n n ( x ) = B , t h e n t h e r e i s a
map a f r o m <y> t o X w h i c h sends y t o x . S i n c e k e r a
i s n o t a summand, t h i s i s a c o n t r a d i c t i o n . We c o n c l u d e t h a t
Cg = <y> w h i c h i s a summand.
Case 2: Assume t h a t a l l t E Cg o f o r d e r p have h e i g h t 0 o r
t h i s
CQ
CQ
We c l a i m t h a t Cg i s an e l e m e n t a r y summand o f C . To show
we n o t e t h a t a n ascend ing u n i o n o f e l e m e n t a r y summands o f
s p u r e [3, 526 f a c t f ] and e l e m e n t a r y and so a summand o f
3, 27.51. By Z o r n ' s lemma, choose a maximal e l e m e n t a r y -.
summand B o f Cg. We w i s h t o show t h a t B = C g . W r i t e
Cg = B X and assume t h e c o n t r a r y . As above, n o t a l l e l emen ts
o f o r d e r p i n X can have i n f i n i t e h e i g h t . By assumpt ion ,
t h e r e i s no e lemen t o f o r d e r p w i t h nonze ro f i n i t e h e i g h t , so
t h e r e i s some x E X w i t h o r d e r p and h e i g h t z e r o . B u t now,
t h e c y c l i c module <x> i s a p u r e and bounded submodule, and so a
summand o f X . We see t h a t B 6 <x> i s a l a r g e r e l e m e n t a r y
summand o f Cg, c o n t r a d i c t i n g o u r c h o i c e o f B . T h i s shows
Cg = B . Now, (Cg + BC)/BC i s a summand o f C /QC and
C Q n 8 C = 0 , so Cg i s a summand o f C .
C o r o l l a r y : Cg has SSIP.
P r o o f : I f CQ i s e lemen ta ry , i t i s a v e c t o r space o v e r D/&
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THE SUMMAND INTERSECTION PROPERTY
Before go ing on, we remark t h a t t h e t o r s i o n need n o t be a
summand o f a module w i t h SIP. For example, t h e 72-module i'I 22 P '
t h e p roduc t be ing taken over any i n f i n i t e s e t o f ( d i s t i n c t ) pr imes,
has SSIP, b u t i t s t o r s i o n G 72 i s n o t a summand. P
For a module w i t h SIP, we can now w r i t e M = E 6 C&. 6 XG
where E i s d i v i s i b l e , Cg i s B -p r imary reduced and XQ i s
some reduced complement. The p o s s i b l e E ' s and Cgls have been
determined. However, t h e r e can be some i n t e r a c t i o n between t h e
summands, so t h a t n o t a l l E G CB combinat ions occur . We now
determine which ones do have SIP.
P r o p o s i t i o n 10: L e t D be a Dedekind domain. L e t E be d i v i s i b l e
and CQ be a reduced Q-primary module. The f o l l o w i n g a r e equ iva len t .
a ) E G Cg has SSIP.
b ) E G Cg has SIP.
c ) i ) b o t h E and Cg have SIP
and i i ) i f E(DIB) i s a summand o f E, CQ i s e lementary.
P roo f : a o b v i o u s l y i m p l i e s b .
Suppose t h a t b ho lds . By Lemma 1, bo th E and Cg have
SIP. To see t h a t c o n d i t i o n ii i s a l s o r e q u i r e d , assume t h a t
E(Dl.8) i s a summand o f E and t h a t CG i s n o t e lementary. Then
C = <y> i s c y c l i c o f o r d e r pk f o r some 1 < k < m . Choose 8
z E soc(E(D/B)) . There i s a map u f rom <y> t o E(D/B)
sending y t o Z . k e r u = <py> i s n o t a summand, c o n t r a d i c t i n g
P r o p o s i t i o n 1 .
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36 WILSON
To prove t h a t c i m p l i e s a , we examine t h e rema in ing
p o s s i b i l i t i e s . Note t h a t E and CQ b o t h have SSIP. I f E i s
t o r s i o n - f r e e , bo th E and CQ a r e f u l l y i n v a r i a n t , so by Propo-
s i t i o n 2 E G Cff has SSIP. I f E i s a summand o f Q/D , then
E G CG i s t o r s i o n , so i t s u f f i c e s t o check t h a t f o r each pr ime
2 , (E GCQ)2 has SSIP. Obvious ly , we need o n l y l o o k a t
(E G Cy) = EQ Cq. I f Eff = 0 t h e r e i s n o t h i n g t o show, so
assume t h a t EQ = E(D/ff) (and t h a t CQ i s e lementa ry ) . L e t
?'Si Ii I be a c o l l e c t i o n o f summands o f E(D/B) 6 Cg . C l e a r l y ,
t h e o n l y elements o f E(D/ff) CG which a r e d i v i s i b l e by p E D
a r e o f t h e form (e,O) . Since a l l Si a r e pure, t h i s means t h a t
e i t h e r each Si c o n t a i n s E(D/B) o r i s e lementary. I f a l l Si
c o n t a i n E(D/ff) , then w r i t i n g Si = E(D/Q) G B we g e t i
Ti Si = E(D/B) G n Bi i s a summand. I f some Si i s e lementary, I I then n S i s a submodule o f t h e DIG v e c t o r space Si , hence
r i 1
i s a summand o f Si .
C o r o l l a r y : If M has SIP, i t s t o r s i o n submodule has SSIP. A-
C o r o l l a r y : L e t M = CE MG be a t o r s i o n module. M has SIP i f
and o n l y i f each MQ i s e i t h e r elementary, E(D/ff) C elementary
o r c y c l i c .
P r o p o s i t i o n 7 showed t h a t t o r s i o n i n E t h e d i v i s i b l e p a r t
of M = E C makes M t o r s i o n . Tors ion i n t h e reduced p a r t
a l s o p laces severe r e s t r i c t i o n s on C , and so on M .
P r o p o s i t i o n 11 : L e t D be a Dedeki nd domai n . C = Cg G Xg has
SIP (SSIP) i f and o n l y i f
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THE SUMMAND INTERSECTION PROPERTY
i ) b o t h CQ and XQ have SIP (SSIP)
and i i ) i f C@ f 0 , t h e n XB i s B - d i v i s i b l e .
P r o o f : Assume t h a t C has SIP(SS1P). By Lemma 1, b o t h CQ and
Xg have SIP (SS IP ) . Assume t h a t CB f 0 and t h a t XB i s n o t
B - d i v i s i b l e . Tank any y E CQ w i t h @ = 0 . The c y c l i c
module < y > i s i s o m o r p h i c t o D/B . X@ f BXQ so t h e r e a r e non-
z e t o maps 8: X@+ <y> . Ker e c a n n o t be a summand l e s t we have
XQ = k e r 8 6 i m 8 w h i c h has @ - t o r s i o n . Thus, CQ f 0
i m p l i e s XG i s B - d i v i s i b l e .
Conve rse l y , i f ii h o l d s , t h e n CQ and XQ a r e f u l l y
t o g e t h e r w i t h i i m p l i e s t h a t XB G C@ has
C o r o l l a r y :
M has a d
P r o o f : I f
i n v a r i a n t . T h i s
SIP (SS IP ) .
L e t D be a PID and M be a D-module w i t h SIP. I f
i v i s i b l e submodule, t h e n M has SSIP.
M i s t o r s i o n , t h i s i s a c o r o l l a r y t o P r o p o s i t i o n 10.
I f t h e d i v i s i b l e submodule o f M i s t o r s i o n - f r e e , t h e n b y
P r o p o s i t i o n 7, M = I G T G U . By P r o p o s i t i o n 8 and 10, I G U
and T have SSIP. Of c o u r s e , T i s f u l l y i n v a r i a n t . By
P r o p o s i t i o n 11, I G U i s f u l l y i n v a r i a n t a l s o . Now, P r o p o s i t i o n
2 shows t h a t M has SSIP.
Acknowledgement
I wou ld l i k e t o t h a n k Dr . L. Fuchs. H i s h e l p f u l comments
g r e a t l y improved t h e o r i g i n a l p r e s e n t a t i o n o f t h e s e r e s u l t s .
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REFERENCES
WILSON 1
[l] J.D. Botha and P. J. Grabe, On t o r s i o n - f r e e a b e l i a n groups whose endomorphism r i n g s a r e p r i n c i p a l i d e a l domains, Comm. i n A l g . (11 ) 1 2 (1983), 1343-1354.
[ 2 ] M.C.R. B u t l e r , A c l a s s o f t o r s i o n - f r e e a b e l f i n i t e r a n k , Proc. London Math. Soc. ( 3 ) 15 680-698.
[3] L. Fuchs, " I n f i n i t e Abel i a n Groups", two vo Press, New York, 1970.
i a n groups o f (1 965 ) ,
1 umes , Academic
[ 4 ] I. Kaplansky, " I n f i n i t e Abel i a n Groups", U n i v . o f M i c h i g a n Press, Ann A r b o r , 1969.
[5 1 , "Commutat ive R ings " , U n i v . o f Chicago Press, Chicago, 1974.
Received: July 1984 Revised: May 1985
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