On Semiprime Goldie Modules

0.23 MB·

by  Jaime Castro Pérez

On Semiprime Goldie Modules
6
1
Jaime Castro P´erez∗
0
2 Escuela de Ingenier´ıa y Ciencias, Instituto Tecnol´ologico y de

n Estudios Superiores de Monterrey
a Calle del Puente 222, Tlalpan, 14380, M´exico D.F., M´exico.
J
Mauricio Medina B´arcenas† Jos´e R´ıos Montes‡
3
1 Angel Zald´ıvar§

Instituto de Matem´aticas, Universidad Nacional
]
A Aut´onoma de M´exico
R Area de la Investigaci´on Cient´ıfica, Circuito Exterior, C.U.,
. 04510, M´exico D.F., M´exico.
h
t
a
January 15, 2016
m
[

1
v Abstract
6
ForanR-moduleM,projectiveinσ[M]andsatisfyingascendingchain
3
4 condition (ACC) on left annihilators, we introduce theconcept of Goldie
3 module. We also use the concept of semiprime module defined by Raggi
0 et. al. in [16]to givenecessary and sufficient conditions for an R-module
. M, to be a semiprime Goldie module. This theorem is a generalization
1
of Goldie’s theorem for semiprime left Goldie rings. Moreover, we prove
0
6 that M is a semiprime (prime) Goldie module if and only if the ring
1 S = EndR(M) is a semiprime (prime) right Goldie ring. Also, we study
: the case when M is a duomodule.
v
i Keywords: Prime module, Semiprime module, Goldie module, Essentially
X
compressible module, Duo module.
r
a 2010 Mathematics Subject Classification: 16D50, 16D80, 16P50,16P70.
Introduction
Goldie’s Theorem states that a ring R has a semisimple artinian classical left
quotient ring if and only if R is a semiprime ring with finite uniform dimension
andsatisfiesACC onleft annihilators. Wisbauer provesin([20],Theorem11.6)
∗[email protected],correspondingauthor
†[email protected]
‡[email protected]
§[email protected]
1
a versionofGoldie’s Theoreminterms ofmodules. For a retractableR-module
M with S = End (M) the following conditions are equivalent: 1. M is non
R
M-singular with finite uniform dimension and S is semiprime, 2. M is non
M-singular with finite uniform dimension and for every N ≤ M there exists
e
a monomorphism M → N, 3. End (M) is semisimple left artinian and it is
R
c
the classical left quotient ring of S, here M denotes the M-injective hull of M.
Also, in [8] the authors study when the encdomorphism ring of a semiprojective
module is a semiprime Goldie ring.
In this paper we give another generalization of Goldie’s Theorem. For this,
we use the product of submodules of a module M defined in [3] to say when
a module is a semiprime module. This product extends the product of left
ideals of a ring R, so R is a semiprime module (over itself) if and only if R is a
semiprime ring in the usual sense.
In orderto havea definition of Goldie Module such thatit extends the clas-
sicaldefinition of left Goldie ring,we introduce whatascending chaincondition
onleft annihilatorsmeans ona module. A leftannihilatorin M is a submodule
of the form A = Ker(f) for some X ⊆End (M). This definition with
X Tf∈X R
R=M is the usual concept of left annihilator.
The mainconceptofthis workis thatanR-module M is a Goldiemodule if
M satisfiesACConleftannihilatorsandhasfiniteuniformdimension. Weprove
some characterizations of semiprime Goldie modules (Theorem 2.8, Theorem
2.22 and Corollary 2.23) which generalize the Goldie’s Theorem and extends
the Theorem 11.6 of [20] and corollary 2.7 of [8].
We organize this paper in three sections. Section 1 proves several results
for semiprime modules. We also generalize Theorem 10.24 of [10] to semiprime
artinian modules.
Insection2weintroducetheconceptofGoldiemodules. Weprovethemain
Theorem of this paper and a characterization of semiprime Goldie modules.
We also obtain some examples of Goldie modules. We also prove that if M
has finitely many minimal prime submodules P ,...,P in M such that M/P
1 t i
(1 ≤ i≤ t) has finite uniform dimension, then M is Goldie module if and only
if eachM/P is Goldie module for(1≤i≤t). We also givea descriptionof the
i
submodule Z(N) with N ∈σ[M].
Inthelastsectionweapplythepreviousresultstoduomoduleswhichextend
results for commutative rings. In [13] the authors say that they do not know a
duo module with a quotient not duo, in this section we show an example.
Throughout this paper R will be an associative ring with unit and R-Mod
will denote the category of unitary left R-modules. A submodule N of an R-
moduleM isdenotedbyN ≤M. IfN isapropersubmodulewewriteN <M.
We use N ≤ M for an essential submodule. Let M and X be R-modules. X
e
is said to be M-generated if there exists an epimorphism from a direct sum of
copies of M onto X. Every R-module X has a largestM-generatedsubmodule
called the trace of M in X, defined by trM(X)= {f(M)|f : M → X}. The
P
categoryσ[M]isdefinedasthesmallestfullsubcategoryofR-Modcontainingall
R-modulesX whichareisomorphictoasubmodule ofanM-generatedmodule.
2
A module N ∈σ[M] is called singular in σ[M] or M-singular, if there is an
exactsequenceinσ[M], 0→K →L→N →0withK ≤ L. TheclassS ofall
e
M-singular modules in σ[M] is closed under submodules, quotients and direct
sums. Therefore, any L∈σ[M] has a largest M-singular submodule
Z(L)=X{f(N)|N ∈S andf ∈HomR(N,L)}
L is called non M-singular if Z(L)=0.
Let M be an R-module. In [2] the annihilator in M of a class C of modules
is defined as Ann (C)= K, where
M TK∈Ω
Ω={K ≤M|thereexistsW∈C andf ∈Hom (M,W)withK=Ker(f)}
R
Also in [2], the author defines a product in the following way: Let N ≤M. For
eachmodule X,N·X =Ann (C)whereC isthe classofmodulesW suchthat
M
f(N)=0 for all f ∈Hom (M,W).
R
For an R-module M and K,L submodules of M, in [3] the product K L is
M
defined by K L = {f(K)|f ∈ Hom (M,L)}. Moreover, in [2] it is showed
M P R
that if M is projective in σ[M], and N ≤ M, then N ·X = N X for every
M
module X.
A nonzero R-module M is called monoform if for each submodule N of M
and each morphism f : N → M, f is either zero or a monomorphism. M
has enough monoforms if each nonzero submodule of M contains a monoform
submodule.
Let M-tors be the frame of all hereditary torsion theories on σ[M]. For a
family {M } of modules in σ[M], let χ({M }) the greatest element of M-tors
α α
for which all M are torsion free. Let ξ({M }) be the least element of M-tors
α α
for which all M are torsion. ξ({M }) and χ({M }) are called the hereditary
α α α
torsiontheory generatedby the family {M } and the hereditary torsiontheory
α
cogenerated by the same family. In particular, the greatest and least elements
in M-tors are denoted by χ and ξ respectively. If τ ∈M −tors, let Tτ, Fτ and
t denotethetorsionclass,thetorsionfreeclassandthepreradicalassociatedto
τ
τ, respectively. For details about concepts and terminology concerning torsion
theories in σ[M], see [19] and [20].
1 Semiprime Modules
Definition 1.1. Let M ∈R−Mod and K, L submodules of M. Put K L=
M
{f(K)|f ∈ Hom (M,L)}. For the properties of this product see [4] Propo-
P R
sition 1.3.
Definition 1.2. Let M ∈R−Mod. We say a fully invariant submodule N ≤
M is a prime submodule in M if for any fully invariant submodules K,L ≤ M
such that K L ≤ N, then K ≤ N or L ≤ N. We say M is a prime module if
M
0 is a prime submodule.
3
Proposition 1.3. Let M be projective in σ[M] and P a fully invariant sub-
module of M. The following conditions are equivalent:
1. P is prime in M.
2. For any submodules K, L of M containing P and such that K L ≤ P,
M
then K =P or L=P.
Proof. 1⇒2: By Proposition 1.11 of [4].
2⇒1: Suppose that K, L are submodules of M such that K L≤P.
M
We claimthat K (L+P)≤P. Since K L≤L∩P, by Proposition5.5of
M M
[2] K (L/L∩P)=0 so K (L+P/P)=0. Thus K (L+P)≤P.
M M M
On the other hand,
(K+P) (L+P)=K (L+P)+P (L+P)≤P
M M M
because P is fully invariant in M.
Then, by hypothesisK+P =P orL+P =P, hence K ≤P orL≤P.
Definition1.4. WesayafullyinvariantsubmoduleN ≤M isasemiprimesub-
module inM ifforanyfullyinvariantsubmoduleK ≤M suchthatK K ≤N,
M
then K ≤N. We saidM is a semiprime module if 0 is a semiprime submodule.
Lemma 1.5. Let M be projective in σ[M] and N a fully invariant submodule
of M. The following conditions are equivalent:
1. N is semiprime in M.
2. For any submodule K of M, K K ≤N implies K ≤M.
M
3. For any submodule K ≤ M containing N such that K K ≤ N, then
M
K =N.
Proof. 1 ⇒ 2 : Let K ≤ M such that K K ≤ N. Consider the submodule
M
K M ofM. ThisistheminimalfullyinvariantsubmoduleofM whichcontains
M
K and K X =(K M) X for every module X. Hence by Proposition 1.3 of
M M M
[4] we have that
K K =(K M) K ≤((K M) K) M)≤N M
M M M M M M M
Since N isfully invariantsubmodule ofM thenN M =N andby Proposition
M
5.5 of [2] (K M) (K M)=((K M) K) M)≤N. Since N is semiprime
M M M M M M
in M, K M ≤N. Hence K ≤N.
M
2⇒1: By definition.
1⇔3: Similar to the proof of Proposition 1.3.
In Remark 1.26 below, we give an example where the associativity of the
product (·) (·) is not true in general.
M
Definition 1.6. Let M ∈ R−Mod and N a fully invariant submodule of M.
We define the powers of N as:
4
1. N0 =0
2. N1 =N
3. Nm =N Nm−1
M
Lemma 1.7. Let M be projective in σ[M] and N semiprime in M. Let J be a
fully invariant submodule of M such that Jn ≤N then J ≤N.
Proof. By induction on n. If n=1 the result is clear.
Supposen>1andthePropositionisvalidforn−1. Wehavethat2n−2≥n
then
J2n−2 ≤N
so
(Jn−1)2 =Jn−1 Jn−1 ≤N
M
since N is semiprime Jn−1 ≤N then J ≤N.
Proposition 1.8. Let S :=End (M) and assume M generates all its submod-
R
ules. If N is a fully invariant submodule of M such that Hom (M,N) is a
R
prime (semiprime) ideal of S, then N is prime (semiprime) in M.
Proof. Let K and L be fully invariant submodules of M such that K L≤N.
M
Put I = Hom (M,L) and J = Hom (M,K). Let m ∈ M and f g ∈
R R P i i
IJ. Since g ∈ J and g (m) ∈ K then f (g (m)) ∈ K L ≤ N. Hence
i i P i i M
IJ ≤ Hom (M,N). Since Hom (M,N) is prime (semiprime) in S, then I ≤
R R
Hom (M,N) or J ≤ Hom (M,N). Hence trM(L) := Hom(M,L)M ≤ N or
R R
trM(K)≤N and since M generates all its submodules then L≤N or K ≤N.
Thus N is a prime (semiprime) submodule.
Next definition aper in [9]
Definition 1.9. A module M is retractable if Hom (M,N) 6= 0 for all 0 6=
R
N ≤M
Corollary 1.10. Let S := End (M) with M retractable. If S is a prime
R
(semiprime) ring then M is prime (semiprime).
Proof. Let K and L be fully invariant submodules of M such that K L = 0.
M
Since Hom (M,0) is a prime (semiprime) ideal of S then by the proof of 1.8,
R
trM(K)=0 o trM(L)=0. Since M is retractable, K =0 or L=0. Hence 0 is
prime (semiprime) in M. Thus M is prime (semiprime).
Proposition 1.11. Let M be projective in σ[M] and N a proper fully invariant
submodule of M. The following conditions are equivalent:
1. N is semiprime in M.
2. If m∈M is such that Rm Rm≤N, then m∈N.
M
3. N is an intersection of prime submodules.
5
Proof. 1⇒2: By Lemma 1.5.
2 ⇒3 : Since N is proper in M, let 0 6= m ∈M \N. Then Rm Rm (cid:2)
0 0M 0
N. Now, let 0 6= m ∈ Rm Rm but m ∈/ N Then Rm Rm (cid:2) N and
1 0M 0 1 1M 1
Rm Rm ≤ Rm Rm . We obtain a sequence of non-zero elements of M,
1M 1 0M 0
{m ,m ,...} such that m ∈/ N for all i and Rm Rm ≤Rm Rm .
0 1 i i+1M i+1 iM i
By Zorn’s Lemma there exists a fully invariant submodule P of M with
N ≤P, maximal with the property that m ∈/ P for all i .
i
WeclaimP isaprimesubmodule. LetK andLsubmodulesofM containing
P. Since P ≤ K and P ≤ L, then there exists m and m such that m ∈ K
i j i
and m ∈ L. Suppose i ≤ j, then Rm Rm ≤ K and by construction m ∈
j iM i j
Rm Rm and thus m ∈ K. If we put k = max{i,j}, then m ∈ K and
iM i j k
mk ∈ L. Hence, RmkMRmk ≤ KML, and so KML (cid:2) P. By Proposition 1.3,
P is prime in M.
3⇒1: It is clear.
Proposition 1.12. Let 06=M be a semiprime module and projective in σ[M].
Then M has minimal prime submodules in M.
Proof. By the proofPropositionof1.11, M has prime submodules. Let P ≤M
be a prime submodule. Consider Γ = {Q ≤ P|Q is prime}. This family
is not empty because P ∈ Γ. Let C = {Q } be a descending chain in Γ. Let
i
N,K ≤M befullyinvariantsubmodulesofM suchthatN K ≤ C. Suppose
M T
that N (cid:2) TC. Then there exists Qj such that N (cid:2) Qj and N (cid:2) Ql for all
Q ≤ Q . Therefore K ≤ Q for all Q ≤ Q , and since C is a chain then
l j l l j
K ≤ C. Therefore C ∈Γ. By Zorn’s Lemma Γ has minimal elements.
T T
Remark 1.13. Notice that if M is projective in σ[M] and M has prime sub-
modules in M, then M has minimal prime submodules.
Corollary 1.14. Let 0 6= M be a semiprime module and projective in σ[M].
Then
0=\{P ≤M|P isaminimalprimeinM}.
Proof. Letx∈ {P ≤M|P isaminimalprimeinM}andQ≤M beaprime
T
submodule in M. By Proposition1.12 there exists a minimal prime submodule
P such that P ≤Q then x∈Q and x is in the intersection of all primes in M.
By Proposition 1.11, x=0.
Lemma 1.15. Let M ∈ R−Mod and N a minimal submodule of M. Then
N2 =0 or N is a direct summand of M.
Proof. Suppose that N N 6= 0. Then there exists f : M → N such that
M
f(N) 6= 0. Since 0 6= f(M) ≤ N and N is a minimal submodule, f(M) = N.
On the other hand, Ker(f)∩N ≤ N, since f(N) 6= 0 then Ker(f)∩N = 0.
We have that (M/Ker(f)) ∼= N and since N is a minimal submodule Ker(f),
then is a maximal submodule of M. Thus Ker(f)⊕N =M.
Corollary 1.16. Let M be a retractable module. If N is a minimal submodule
in a semiprime module M, then N is a direct summand.
6
Proof. Since M is semiprime, N N 6=0.
M
Theorem 1.17. The following conditions are equivalent for a retractable R-
module M:
1. M is semisimple and left artinian.
2. M is semiprime and left artinian.
3. M is semiprime and satisfies DCC on cyclic submodules and direct sum-
mands.
Proof. 1⇒2: If M is semisimple then it is semiprime.
2⇒3: Since M is left artinian, then it satisfies DCC on cyclic submodules
and direct summands.
3 ⇒ 1 : Since M satisfies DCC on cyclic submodules, there exists K a
1
minimal submodule of M. By Corollary 1.16, M =K ⊕L . Now there exists
1 1
K a minimal submodule of L and L = K ⊕ L . With this process we
2 1 1 2 2
obtain a descending chain of direct summands, which by hypothesis it is finite
L ⊇L ⊇L ⊇...⊇L . SinceL issimpleandM =K ⊕K ⊕...⊕K ⊕L ,
1 2 3 m m 1 2 m m
then M is semisimple.
Now, if M is semisimple and satisfies DCC on direct summands then M is
artinian.
Definition 1.18. Let M ∈R−Mod and N ≤M. We say N is an annihilator
submodule if N =Ann (K) for some 06=K ≤M.
M
Lemma 1.19. Let M be semiprime and projective in σ[M]. Let N,L≤M. If
L N =0, then N L=0 and L∩N =0.
M M
Proof. Since L N =0, then
M
0=N (L N) L=(N L) (N L).
M M M M M M
Hence N L=0 .
M
Now, since L∩N ≤L and L∩N ≤N, then
(L∩N) (L∩N)≤L N =0.
M M
Thus L∩N =0
Corollary 1.20. Let M be semiprime and projective in σ[M]. If N ≤M, then
N Ann (N)=0.
M M
Proposition 1.21. Let M be semiprime and projective in σ[M] and N ≤ M.
Then N is an annihilator submodule if and only if N =Ann (Ann (N))
M M
7
Proof. ⇒:By Lemma 1.20N ≤Ann (Ann (N)). Thereis K ≤M suchthat
M M
N =Ann (K), hence
M
K N =K Ann (K)=0
M M M
and thus K ≤Ann (N). Therefore,
M
Ann (Ann (N))≤Ann (K)=N
M M M
It follows that N =Ann (Ann (N)).
M M
⇐: By definition of annihilator submodule.
Proposition 1.22. Let M be semiprime and N ≤ M. Then, Ann (N) is
M
the unique pseudocomplement fully invariant of N. Moreover, N Ann (N)
L M
intersects all fully invariant submodules of M.
Proof. Let L≤M be a fully invariant pseudocomplement of N in M. Then
L N ≤L∩N =0
M
Thus L≤Ann (N). Observe that
M
(Ann (N)∩N) (Ann (N)∩N)≤(Ann (N)∩N) N =0
M M M M M
Since M is semiprime, Ann (N)∩N =0. Thus L=Ann (N).
M M
Lemma 1.23. Let M be a semiprime module and N ≤ M. Let S be the
set of all minimal prime submodules of M which do not contain N. Then
Ann (N)= {P|P ∈S}.
M T
Proof. Put K = {P|P ∈ S}. Any element in K ∩N is in the intersection of
T
all minimal prime submodules of M which is zero. Then K ∩N = 0. Since K
is fully invariant in M, K N ≤ K∩N = 0. Thus, K ≤ Ann (N). Now, let
M M
P ∈S. Since AnnM(N)MN =0≤P and N (cid:2)P, then AnnM(N)≤K.
Lemma 1.24. Let M be projective in σ[M]. If M is semiprime then M is
retractable.
Proof. Let N ≤ M and suppose Hom (M,N) = 0. Then Ann (N) = M.
R M
So M N = 0 but N N ⊆ M N = 0. Since M is semprime then N = 0 by
M M M
Lemma 1.5.
Proposition 1.25. Let M be projective in σ[M] and semiprime. The following
conditions are equivalent for N ≤M:
1. N is a maximal annihilator submodule.
2. N is an annihilator submodule and is a minimal prime submodule.
3. N is prime in M and N is an annihilator submodule.
8
Proof. 1 ⇒ 2 : Let K ≤ M such that N = Ann (K). Let L,H ≤ M be fully
M
invariant submodules of M such that LMH ≤ N. Assume H (cid:2) N. Then 0 6=
H K. Hence Ann (K) ≤ Ann (H K), but since Ann (K) is a maximal
M M M M M
annihilator submodule, then Ann (K)=Ann (H K).
M M M
As M is projective in σ[M], by Proposition 5.5 of [2], we have that
L (H (H K))=(L H) (H K)≤N (H K)=0
M M M M M M M M
Now, since H (H K)≤H K, then
M M M
Ann (K)=Ann (H K)≤Ann (H (H K))
M M M M M M
Therefore Ann (H K)=Ann (H (H K)). Thus L≤Ann (K)=N.
M M M M M M
Now, let P ≤ M be a prime submodule of M such that P < N. We have
that N K = 0 ≤ P. So K ≤ P < N. Hence K K = 0. Thus K = 0, but M
M M
is semiprime, a contradiction. It follows that N is a minimal prime submodule
of M.
2⇒3: By hypothesis.
3⇒1: Suppose N <K with K an annihilator submodule. Then
Ann (K) K =0≤N
M M
Since N is prime in M, then Ann (K) ≤ N < K. By Proposition 1.22
M
Ann (K)∩K =0,henceAnn (K)=0. SinceK isanannihilatorsubmodule,
M M
by Proposition 1.21, K =Ann (Ann (K))=Ann (0)=M.
M M M
Remark 1.26. Following the notation of Example 1.12 of [4] Let R = Z ⋊
2
(Z ⊕Z ). ThisringhasonlyonemaximalidealI andithasthreesimpleideals:
2 2
J , J , J , which are isomorphic. Then, the lattice of ideals of R has the form
1 2 3
R
•
I
•
❁
✂ ❁
✂ ❁
✂ ❁
✂ ❁
✂ ❁
✂ ❁
✂ ❁
✂
J•1 J•2 J•3
❁
❁❁❁❁❁❁ ✂✂✂✂✂✂
❁ ✂
❁ ✂
0 ✂
•
Moreover, R is artinian and R-Mod has only one simple module up to iso-
morphism. Let S be a simple module. By Theorem 2.13 of [15], the lattice of
fully invariant submodules of E(S) has tree maximal submodules N, L and K,
and it has the form
9
E(S)
•
❄
⑧ ❄
⑧ ❄
⑧ ❄
⑧ ❄
⑧ ❄
⑧ ❄
⑧ ❄
⑧ ❄
⑧
K L N
• • •
❄
❄ ⑧
❄ ⑧
❄ ⑧
❄ ⑧
❄ ⑧
❄ ⑧
❄ ⑧
❄ ⑧
S ⑧
•
0
•
Put M =E(S). Since K∩L=S and K L≤K∩L, then K L≤S. On
M M
the other hand consider the composition
M π // M/N ∼= //S i //L
f = i◦π where π is the natural projection and i is the inclusion. Then,
f(K)=S andS ≤KML. Thus,KML=S. NoticethatKML≤N butK (cid:2)N
and L (cid:2) N. Hence N is not prime in M. Analogously, we prove that neither
K nor L are prime in M. We also note that K K =S. Moreover, π(K)=S,
M
so K S =S. In the same way L S =S and N S =S
M M M
Let g : M → K be a non zero morphism. If Ker(g)∩S = 0 then g is a
monomorphism, a contradiction. So Ker(g)∩S = S. Thus S K = 0 and
M
Ann (K) = S. Analogously Ann (L) = S = Ann (N) = Ann (S). Since
M M M M
S S ≤ S K, S S = 0. Thus M is not semiprime. Hence, S is a maximal
M M M
annihilator submodule of M which is not prime because K K =S. With this
M
wecanseethatassociativityisnottrueingeneral,becauseL (K S)=L S =
M M M
S and (L K) S =S S =0. Notice that, in this example Hom (M,H)6=0
M M M R
forallH ∈σ[M]inparticularM isretractable,butM isnotprojectiveinσ[M].
Proposition 1.27. Let M be projective in σ[M] and semiprime. For N ≤M,
if N = Ann (U) with U ≤ M a uniform submodule, then N is a maximal
M
annihilator in M.
Proof. Suppose that N < K with K an annihilator submodule in M. Since
N = Ann (U) by Proposition 1.22, K ∩U 6= 0. By hypothesis U is uniform
M
and thus K∩U ≤ U. Then
e
(K∩U)⊕Ann (U)≤ U ⊕Ann (U)
M e M
Now,notice that if L≤ M, by Proposition1.22 (U⊕Ann (U))∩L6=0. So
FI M
((K∩U)⊕Ann (U))∩L6=0. Therefore, K∩L6=0 andK intersects all fully
M
invariant submodules of M . Since K∩Ann (K)=0 and Ann (K)≤ M,
M M FI
then Ann (K)=0. Thus, K =Ann (Ann (K))=Ann (0)=M.
M M M M
10