BAHMAN ASKARI, HADI ADIBAN and HAMID AGHA TAVALLAEE
Communicated by the former editorial board
Our aim in this paper is to find a lower bond for the number of elements of P−spectrum of module M over a ring R by having dimpM, and show that under some conditions the dimension of the weak multiplication moduleM over ringRis equal with the dimensionR. Also two properties for the 0−maximale submoduleN ofM over Dedekind domainRis investigated.
AMS 2010 Subject Classification: 16P70.
Key words: spectrum of module, prime submodule, weak multiplication module, Dedekind domain.
1. INTRODUCTION
Throughout this paper, all rings are commutative with identity and all modules are unitary. Let R be a ring and M a two sided R−module. A submoduleN of M is called prime if N 6=M and rm∈N (forr∈R, m∈M) implies that m∈N orrM ⊆N. In this case, if P = (N :M) ={t∈R|tM ⊆ N} we say thatN is aP−prime submodule ofM and it is easy to see thatP is a prime ideal ofR(see [4]). The collection of allP−prime submodules of an R−module M is called the P−spectrum of M and it is denoted by SpecPM. The collection of all prime submodules of M is called the spectrum of M and it denoted by SpecM. For any ring R, it is well known that if R 6= 0 then SpecR6=∅. This is not necessarily true for all modules. ([5], p. 3745].
LetKbe a prime submodule of theR−moduleM andnbe a non-negative integer. We say that K has height n if there exists a chain Kn ⊂ Kn−1 ⊂ ...K1 ⊂ K0 = K of prime submodules Ki (where 0 ≤ i ≤ n) of M, but no longer such chain. Otherwise, we say thatK has infinite height, the height ofK is denoted byhtK. The dimension of theR−moduleM,denoted bydimM or dimRM, is defined by sup{htN|N is prime submodule of M}, ifSpecM 6=∅.
Otherwise it is considered to be −1. Similarly dimPM = sup{htN|N is P− prime submodule of M}, if SpecPM 6= ∅. Otherwise, it is also considered to be−1.
MATH. REPORTS16(66),1(2014), 113–119
Let N be a prime submodule of the R−module M, we define cohtN = dimRMN = sup{htK|K is prime submodule of R−moduleMN} which is called the coheight of N. It is clear thatcoheightN is the largestn such that there is a chain N =N0 ⊂N1 ⊂...⊂Nn of prime submodules M. An R−module M is said to be a multiplication module, if every submodule of M is the form IM for some idealI of R. In this case it is easy to show that L= (L:M)M, whereL is a submodule ofM.
To prove what has been mentioned above, submodule can be substituted by prime submodule in the proof of Lemma 2.10.
In this note we prove that if the chain T1 ⊂T2 of P−prime submodules of M over a ring R is not saturated then there exist at least two P−prime submodules of M strictly betweenT1 and T2 (see Proposition 2.3).
In ([2], Corollary 2.11), it has been proven that if M is finitely generated R−module and P a prime ideal ofRsuch thatdimPM =n,then|SpecPM| ≥ 2n+1.The same as above mentioned conditions, we prove that|SpecPM| ≥3n (see Theorem 2.6).
In Theorem 2.7, we prove that if R is a Noetherian ring, P a maximal ideal of R, M a finitely generatedR−module and dimPM =n, then
|SpecPM| ≥Σn−1i=1rankki+2 ki
+n+ 2.
In Lemma 2.11, we prove that for every finitely generated weak multi- plication module M over a ring R that each the prime ideal of R contains Ann(M), then dimR=dimM.
In Theorem 2.13, we prove that if M is a finitely generated torsion free module over the Dedekind domainR andN is a 0−maximal submodule ofM, then cohtN ≥ 1. And if cohtN > 1, then there exists a 0−prime submodule N0 ofM such that htN < htN0.
2. DIMENSION AND SPECTRUM OF MODULES
Definition 2.1. Let T1 and T2 be both P−prime submodules of M such that T1 ⊂ T2. We say that the chain T1 ⊂ T2 is saturated if there dose not exist any P−prime submodule of M strictly betweenT1 and T2.
The following lemma is due to ([2], Lemma 2.4), we provide a more clear proof for it.
Lemma 2.2. Let R be a ring, P a prime ideal of R, K the quotient field of RP, M an R−module and T1 and T2 both be P−prime submodules of M such that T1 ⊂ T2. Then there exists a one to one correspondence between
the P−prime submodules of M that are strictly between T1 and T2 and the non-trivial subspace KTT2
1 over the fieldK.
Proof. It is easy to see that T is a P−prime submodule of M strictly between T1 and T2 if and only if TT
1 is a P−prime submodule of R−module
M
T1 strictly between TT1
1 and TT2
1 if and only if TT
1 is a 0R P
−prime submodule of
R
P−module MT
1 strictly between TT1
1 and TT2
1. Now, by ([5], Proposition 1), we have TT
1 is a 0R
P
−prime submodule of RP−module TM
1 strictly between TT1
1 and
T2
T1 if and only if KTT
1 is a subspace of KMT
1 strictly between 0 = KTT1
1 and KTT2
1. Set A= {T|T is a P−prime submodule of M and T1 ⊂T ⊂T2} and B = {V| V is a non-trivial subspace of KTT1
2 over K}. We define f : A → B by f(T) =KTT
1.It can be easily verified thatf is well defined, one to one and onto. The result now follows.
Proposition 2.3. Let M be an R−module,P a prime ideal of R andT1 and T2 both be P−prime submodules of M such that T1 ⊂ T2. If the chain T1 ⊂T2 is not saturated then there exist at least two prime submodules of M strictly between T1 and T2.
Proof. Let K be the quotient field of RP and {xi}i∈I be a basis for the vector space KTT2
1 over field K. Let T be a prime submodule of M, strictly between T1 and T2, then according to Lemma 2.2, there exists a non-trivial subspace of V =KTT
1 of vector spaceKTT2
1 with basis{xi}i∈I0 that 06=I0 $I. Let W be a subspace of KTT2
1 with basis {xi}i∈I\I0, then W ∩ V = 0 and consequently W 6= V. Again according to Lemma 2.2, for subspace W there exist a P−prime submoduleT0 of M strictly betweenT1 and T2 and T 6=T0. (Because ifT =T0 then W =V that is a contradiction)
Definition 2.4.LetM be anR−module andI an ideal ofR.A submodule K ofM will be called I−maximal if it satisfies the following two properties:
(i) (K :M) =I;
(ii) IfT is a submodule ofMcontainingKsuch that (T :M) =I,thenK=T. ([6, p. 1810)
Lemma 2.5 ([6], Lemma 3.2). Let R be a ring,P a prime ideal ofR and M an R−module. If N is an P−maximal submodule of M then N is prime in M.
Theorem 2.6. Let M be a finitely generated R−module and P a prime ideal of R. If dimPM =nthen |SpecPM| ≥3n.
Proof. LetdimPM =n=−1, then obviously|SpecPM| ≥ −3. Now let
n≥0 then there exists a chain ofP−prime submodules T1⊂T2 ⊂...⊂Tn+1.
By ([2], Corollary 2.11), |SpecPM| ≥ |A|= 2n+ 1 in which A is the set {T1, T2, ..., Tn+1, M1, M2, ...Mn}
and eachMi (1≤i≤n) is a P−maximal (consequentlyP−prime) submodule of M,Ti ⊂Mi,Ti+1*Mi and Mi*Ti+1.
We should be just find other (n−1)P−prime submodules ofM. Since for each i (1≤i≤n−1) the chain ofTi ⊂Ti+2 is not saturated then according to Proposition 2.3, there exists an P−prime submodule Ti+10 of M, that is, strictly betweenTi and Ti+2 andTi+10 6=Ti+1.
Obviously Ti+10 for each 1 ≤ i ≤ n−1, is not P−maximal then non of Ti+10 (1≤i≤n−1) is same as Mi (1≤i≤n). ConsequentlyT20, T30, ..., Tn0 are other (n−1),P−prime submodules ofM, then |SpecPM| ≥(2n+ 1) + (n−1)
= 3n.
Recall that if R be an integral domain, the rank of R−module M is defined to be the maximal number of elements ofM linearly independent over R. (See [6])
Theorem 2.7. Let R be a Noetherian ring, P a maximal ideal of R, M a finitely generated R module. If dimPM =n for n≥2 andK1 ⊂K2 ⊂...⊂ Kn+1 be a chain ofP−prime submodules of M then
|SpecPM| ≥Σn−1i=1rankR P
Ki+2
Ki +n+ 2.
Proof. SinceRis a Noetherian ring andM is a finitely generatedR−module then by ([1], Proposition 6.5),M is Noetherian and from ([4], Theorem 6), each Ki for 1≤i≤n+ 1 is finitely generated. Now, since the chain Ki ⊂Ki+2 for each 1≤i≤n−1 is not saturated, then by ([2], Theorem 2.9, (ii)), Ki is an intersection m P−prime submodule ofM each of which is strictly betweenKi
and Ki+2,wherem=rankR P
Ki+2
Ki .
We claim that if T is an P−prime submodule of M strictly between Ki and Ki+2 then T is not strictly between Ki+1 and Ki+3. If not, then from Ki ⊂T ⊂Ki+2 and Ki+1 ⊂T ⊂Ki+3 it is concluded that Ki+1 ⊂T ⊂Ki+2
which is a contradiction. (Because according to hypothesis Ki+1 ⊂ Ki+2 is saturated).
Up to now, we found Σn−1i=1rankR P
Ki+2
Ki P−prime submodules ofM. Now, since the chainKi⊂Ki+1for each 1≤i≤nis saturated then according to ([2], Theorem 2.9), there exists a P−maximal (consequentlyP−prime) submodule
Mi of M such that Ki =Mi∩Ki+1 and Ki+1 6= Mi then M1, M2, ..., Mn are the other (n) P−prime submodule of M. Also, with adding K1 and Kn+1 to this set, we have that|SpecPM| ≥Σn−1i=1rankR
P
Ki+2
Ki +n+ 2.
We should note that Theorem 2.7 is justifying Theorem 2.6. Because rankR
P
Ki+2
Ki ≥2 for each 1≤i≤n−1. Consequently (Σn−1i=1rankR
P
Ki+2 Ki
) +n+ 2≥(Σn−1i=12) +n+ 2 = 2(n−1) +n+ 2 = 3n.
Lemma 2.8 ([6], Theorem 3.3). Let M be a finitely generated R−module, P a prime ideal of R and B a submodule of M. If (B :M) ⊆ P, then there exists a P−prime submoduleN of M containing B.
Definition 2.9. AR−module M is called a weak multiplication module if for each prime submoduleLofM there exists a idealI ofRsuch thatL=IM. In the following lemma we prove that if M is a weak multiplication module and L is a prime submodule ofM then (L:M)M =L.
Lemma 2.10. Let M be a weak multiplication module over a ring R and let L be a prime submodule of M. Then (L:M)M =L.
Proof. Obviously (L :M)M ⊂L for every submodule L of M. Suppose now that x ∈ L, since L is a prime submodule of M then there exist an ideal I of R such that L = IM. Thus, we have x = im, where i ∈ I and m ∈ M. Since iM ⊂ IM = L, so iM ⊂ L hence, i∈(L:M) and therefore x=im∈(L:M)M.ConsequentlyL⊂(L:M)M and the proof is complete.
Lemma 2.11. Let R be a ring and M be a finitely generated weak multi- plicationR−module. If each prime ideal of R contains Ann(M) thendimR= dimM.
Proof. Let P be a prime ideal of R then P contains Ann(M). Now, according to Lemma 2.8, there exists an P−prime submodule N containing submodule 0 of M, then SpecPM 6=∅.
Let N0 ∈ SpecPM. Then, since M is a weak multiplication module and N0 = (N0 :M)M =P M we haveSpecPM ={P M}.
Now let dimR=n and P1 ⊂P2 ⊂...⊂Pn+1 be a chain of prime ideals of R, then PiM ∈SpecpiM and PiM ⊆Pi+1M for each 1≤i≤n+ 1.
If PiM = PjM for 1 ≤ i, j ≤ n+ 1, then Pi = (PiM : M) = (PjM : M) =Pj that is a contradiction. SoP1M ⊂P2M ⊂...⊂Pn+1M is a chain of prime submodules of M and hence, dimR≤dimM.
Conversely, sinceR6= 0 so there exists at least a prime ideal P of R and since SpecPM = {P M} hence, SpecM 6= ∅. Consequently, dimM ≥ 0. Let P1M ⊂ P2M ⊂... ⊂Pn+1M be a chain of prime submodules of M, then we have P1 ⊆P2 ⊆...⊆Pn+1. If Pi =Pj for 1≤i, j≤n+ 1, thenPiM =PjM that is a contradiction (because we assume that PiM ⊂ Pi+1M). So P1 ⊂ P2⊂...⊂Pn+1 is a chain of prime ideals of Rand hence,dimM ≤dimR.
Lemma2.12 ([2], Proposition 2.13). LetM be a R−module, then for each prime ideal P of R, we have dimPM =rankR
P
M P M −1.
Theorem 2.13. Let R be a Dedekind domain andM be a finitely gener- ated torsion-free R−module. If N is a 0−maximal submodule of M then:
(i) cohtN ≥1;
(ii) If cohtN > 1 then there exists a 0−prime submodule N0 of M such that htN < htN0.
Proof. Before we start going the work, we make a remark:
Since 0 is a prime ideal of R and N is a 0−maximale submodule ofM so, according to Lemma 2.5,N is a 0−prime submodule ofM. Now, from ([3], Theorem 3.1) we have
(2.1) htN +cohtN =dimM =rankM,
also by Lemma 2.12, we have
(2.2) dim0M =rankR
0
M
0M −1 =rankM−1.
Therefore, from (2.1) and (2.2) we can conclude that:
(2.3) htN +cohtN =dim0M+ 1.
If cohtN = 0,then htN =htN +cohtN = dim0M + 1. So, htN −1 = dim0M and consequentlyhtN > dim0M that is a contradiction. (Because N is an 0−prime submodule of M).
If cohtN > 1, then from (2.3) we can conclude that dim0M = htN + cohtN −1> htN+ 1−1 =htN. Hence,dim0M > htN and so there exists a 0−prime submodule N0 of M such that dim0M =htN0 > htN.
REFERENCES
[1] M.F. Atiyah and I.G. Macdonald, Introduction to commutative algebra. University of Oxford, 1969.
[2] A. Azizi,Intersection of prime submodules and dimension of modules. Acta Math. Sci- entia. 25(2005), 385–394.
[3] A. Azizi and H. Sharif,On prime submodules. Honam Math. J.21(1999), 1–12.
[4] C.P. Lu, Prime submodule of modules. Comment. Math. Univ. St. Pauli 33 (1984), 61–69.
[5] C.P. Lu,Spectra of modules. Comm. Algebra23(1995), 3741–3752.
[6] R.L. MacCasland and M.E. Moore,Prime submodules. Comm. Algebra20(1992), 1803–
1817.
Received 17 December 2011 Islamic Azad University, Young Researchers and Elite Club,
Ghorveh Branch, Ghorveh, Iran
bahman askari2003@yahoo.com Islamic Azad University, Department of Mathematics,
Dezful Branch, Dezful, Iran hadiadiban@gmail.com Islamic Azad University, Department of Mathematics,
Tehran-North Branch, Tehran, Iran tavallaee@iust.ac.ir
