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On Radical Formula over Free Modules with TwoGenerators

Seil eken and Mustafa Alkan

Akdeniz University Department of Mathematics Antalya, Turkey

Abstract. In this paper, we study on the module M= RR over a commutative ring R with identity. After characterizing theradical of a submodule N ofM, we give some conditions for N to satisfy the radical formula. In particular, we show that RRsatisfy the radical formula ifR is an arithmetical ring.

Keywords: Prime submodule, Radical submodule, Radical FormulaPACS: 2010 MSC:13A18,13C13,13C99

INTRODUCTION

Throughout this paper, we consider R to be a commutative ring with identity and M to be an Rmodule with unitary.We denote the set of positive integers by Z+. A proper submodule N ofM is called a prime submodule if rm N forr R and m M implies that either m N or r P = (N : M). Then clearly P is a prime ideal but N does not need tobe prime whenever P is a prime ideal of R, which is the well known conjecture on prime submodules.

Let N be a submodule of M. The radical of N in M, denoted by RadMN, is defined to be the intersection of all

prime submodule of M containing N. If there is no prime submodule of containing N, then we put RadMN = M. Theenvelope submodule REM(N) ofN in M is a submodule ofM generated by the set

EM(N) = {rm : r R and m M such that rnm N for some n Z+}.It is clear that

(N : M)M REM(N) RadMN. In [9], McCasland and Moore proved that

(N : M)M = RadMN

for a submodule N of a finitely generated multiplication module M. In [10], they called that; i) N satisfy the radicalformula (s.t.r.f.) in M if RadMN = REM(N), ii) M satisfy the radical formula (s.t.r.f.) if every submodule of M s.t.r.f.in M, iii) R satisfies the radical formula (s.t.r.f.) if every Rmodule M s.t.r.f..

The question of what kinds of rings and modules s.t.r.f. has drawn the attention of many authors ([1],[2],[3], [4], [6],

[8], [13], and [14] ). In [6], Jenkins and Smith proved that Dedekind domains s.t.r.f. After that, in [8] Leung and Manproved that if R is a Noetherian domain with dimR 1 then R s.t.r.f. if and only if RR s.t.r.f. if and only if R is aDedekind domain. In [3], Azizi proved that every arithmetical ringR with dimR 1 satisfy the radical formula. Then in[4], Buyruk and PusatYlmaz proved that if R is Prfer domain then the Rmodule RR satisfies the radical formula.In this paper, we go on studying the module M= RR over a commutative ring R and our aim is to characterize whatkind of submodule of M satisfy the radical formula. We fix the following notations for the rest of this paper.

Unless stated otherwise, R denotes a commutative ring with identity. We use M to denote an Rmodule RR and Nto be a nonzero submodule of M generated by the set {(ai, bi) M: i }. Also we use the notation i j to indicatei j = aibj biaj and I=iRai +iRbi.

RADICAL SUBMODULE AND FORMULA

Let us start by characterizing (N : M). Then

Lemma 1 Let M and N be as above. Then i,jRi j (N : M) i,jRi j.

Proof Let i j = aibj biaj. Then for all i, j , we get

(aibj biaj)(1, 0) = (aibjbiaj, 0) = bj(ai, bi)bi(aj, bj)(aibj biaj)(0, 1) = (0, aibj biaj) = ai(aj, bj)aj(ai, bi)

Numerical Analysis and Applied Mathematics ICNAAM 2011

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and so i,jRi j (N : M).Let x (N:M). Then there exists a finite subset of such that x(1, 0) =i ti(ai, bi) and x(0, 1) =i ki(ai, bi)

for all i and ti, ki R . Then x =i tiai, x =i kibi, 0 =i tibi and 0 =i kiai. Thus we havex2 = i tikiaibi +ij,i=j tikjaibj0 =i tikiaibi +ij,i=j tjkiajbi

Therefore x2 =ij,i=j tjkii j i,j Ri j and so x i,j Ri j.Now we note that

(N : M) =

i,jRi j

I since i,jRi j I.

In [12], PusatYlmaz and Smith defined the submodule K(N, P) = {m M : cm N+ PM for c R\P}. Thenthey showed that it is the smallest Pprime submodule containing N and so RadMN={K(N, P) : P is a prime ideal}.Now we give a distinct characterization for K(N, P) and RadMN.

Lemma 2 Let M and N be as above and P be a prime ideal of R. Then K(N, P) = M orK(N, P) = PP or K(N, P) = {(m, n) M: for all i , mb inai P}.Proof Let P be a prime ideal of R. Assume (N : M) is not contained in P and c (N : M)\P. Then cM N and soM K(N, P).

Now assume (N : M) P. Consider following two cases:i) Let iRai + Rbi P. Then PM contains N and since PM is a Pprime submodule of M, we get that

K(N, P) = P

P.

ii) Let iRai +Rbi is not contained in P. Then we can assume that a1 / P and consider the submoduleTP = {(m, n) M: for all i , mbinai P}.

It is clear that N Tp. Now we prove that TP is a prime submodule ofM.Let r (TP : M). Then we get r(0,1) TP and so r P. Hence (TP : M) = P since the the other inclusion is clear.Assume that rm TP where r R\P and m = (x,y) M. Then we get r(xbi yai) P and xbi yai P for all

i . Then (x,y) TP and so it is a Pprime submodule ofM.Let (m, n) TP. Then there is p P such that a1n = mb1 + p and so

a1(m, n) = (a1m, a1n) = (a1m, mb1) + (0,p) = m(a1, b1) + (0,p) N+ PMSince a1 / P, we get that (m, n) K(N, P) and so TP K(N, P). This completes the proof.

Theorem 3 Let M and N be as above. Then

RadMN = {(m, n) IM : for all i , mbinai

(N : M)}.In particular, for(a, b) N, RadMN R= {(x,y) M: aybx

(N : M)}

Proof Let K be the set of prime of ideals containing (N : M) and L = {P K : I P}. Then by Lemma 2,IM = PLK(N, P) and so we get

RadMN = PKK(N, P) = (

IM){PKL{(m, n) M: for all i , mbinai P}}Then RadMN = {(m, n)

IM : for all i , mbinai (N:M)PP}.

Let (m, n) RadMN. Take an element (a, b) N. Then we get that a = t1a1 + ... + tnan and b = t1b1 + ... + tnbn forsome ti R(i = 1, ..., n). Thenfor all i, weget that mbi = nai +pi for some pi

(N : M) and so mbiti = naiti +piti

and then

m(t1b1 + ... + tnbn) = n(t1a1 + ... + tnan) + p1t1 + ... + pntn

Therefore, mbna

(N : M) and so RadMN R= {(x,y) M: aybx

(N : M)}.

Theorem 4 Let N have the subset{(xj,yj) : j } such thatjRxj +Ryj = R. ThenRadMN =

(N : M)M+N = REM(N).

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Proof By the hypothesis, we may assume that is finite set. Then we get that 1 =j tjxj + sjyj for some tj, sj R.On the other hand, (xj,yj) = izji(ai, bi) for some zji R and so 1 = jizjitjai +zi jsjbi. Hence I = R and1 = i kiai + libi for some ki, li R. Then by Theorem 3, this gives us that RadMN ={(m, n) M : for all i ,mbinai

(N : M)}. Let (m, n) RadMN and so mbi = nai + ri for some ri

(N : M) and all i . It follows

that

m =i kiaim + li(bim) = i

kiaim + li(nai + ri) = i

kiaim + linai + liri

and

n =i ki(ain) + libin = i kimbi kiri + libin =i kimbi + libin kiri.Hence, we get that (m, n) = i(kimi + lin)(ai, bi) + ri(li,ki) N+

(N : M)M.

This completes the proof since

(N : M)M+NREM(N)RadMN.

Corollary 5 If N has the element(x,y) such that either x or y is invertible then N s.t.r.f. in M.

Theorem 6 If

I

(N : M). Then

RadMN =

(N : M)M= REM(N)

Proof Let I(N : M). Then I = (N : M) and so RadMN = {(m, n) IM : for all i , mbinai I}.Then since ai, bi

I for all i, we get that RadMN =

IM =

II. On the other hand, take rm IM where

r I and m M. Then we get rt (N : M) for t Z+ and so rtm N. Hence IMREM(N).

Theorem 7 Let N be a submodule of M with the element (x,tx) or(tx,x) for some t R and 0 = x R. If

(N : M)

is a prime ideal and

I=

Rx. Then N s.t.r.f. in M.

Proof Let (x,xt) N. Take (m, n) RadMN. It follows that (m, n) Rby Theorem 3 and so nxmxt= x(nmt) (N : M). Then we have two cases:

i) Ifx

(N : M), then by Theorem 6, it follows that RadMN =

(N : M)M= REM(N).

ii) If (nmt) = p

(N : M) then (m, n) = (m, mt+ p) = (m, mt) + (0,p) I(1,t) +

(N : M)M. Now it is

enough to show

I(1,t) REM(N) for the complete proof. Take y

I and so yl = kx for l Z+ and k R. Itfollows that y

l

(1,t) = k(x,xt) N and so I(1,t)REM(N).Recall that a ring R is said to be an arithmetical ring, if for all ideals L,J and K of R we have L + (JK) =

(L +J) (L + K), or equivalently for each prime ideal P of R, any two ideals of the ring RP are comparable.Obviously Prfer domains and Dedekind domains are arithmetical. In [4, Theorem 2.4], for a Prfer domain R it is

proved that RR is s.t.r.f. By using Theorem 7, we obtain the following corollary which gives us a generalization ofTheorem 2.4 in [4].

Theorem 8 Let R be a arithmetical ring and N be a submodule of M . Then N s.t.r.f. in M.

Proof We can assume that any two ideals ofR is comparable by [7, Proposition 1.8] and so take an element of RadMN

as (m, mt). Then miRai +Rbi and so for n, l Z+, we have that ml Ra1 +Rb1 + ...+Ran +Rbn. Since everyideal of R is comparable, we get that ml Ry where y {a1, b1, ..., an, bn} and so ml = ry for rR. Now we get twocases either y

(N : M) or not.

Now assume that y (N : M) then ml

= ry (N : M) and m (N : M). Then m(1,t) (N : M)MREM(N).

Ify /

(N : M), then again we have two subcases i)(y, ky) N or ii)(ky,y) N for kR.i) Let (y, ky) N. Since (m, mt) R, we get that y(mtmk)

(N : M). Then mtmk= p

(N : M) and so

(m, mt) = (m, mk) + (0,p). On the other hand, by the proof of Theorem 7, we get that m(1, k) REM(N) and so(m, mk) + (0,p) REM(N) +

(N : M)M= REM(N).

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ii) If(ky,y) N then by the same argument (i), the proof is completed.

Theorem 9 Let R be a domain and(N : M) = 0. Then for a fixed t ,RadMN = {(m, n)

IM : mbt = nat}.

In particular, if I = R then RadMN = N is a prime submodule of M.

Proof Let R be a domain. Then

(N : M) = 0 and so i j = aibj biaj = 0 for all i, j . Then by Theorem 3, wehave that RadMN = {(m, n) IM : mbi = nai for all i }.

Let (m, n) IM and mbt = nat for a fixed t . Since (N : M) = 0, we get that it = 0 for all i and sobiat = aibt. Then for all i , we get that natai = mbtai = mbiat and so nai = mbi. Therefore, RadMN = {(m, n)

IM : mbt = nat}.In particular, ifI = R, then we get that RadMN = N = REM(N) by Theorem 4 and also clearly N is prime.

Corollary 10 Let R be a domain and N be a submodule of M with the element(x,tx) for some t R and 0 = x R If(N : M) = 0 and

I =

Rx, then RadMN =

I(1,t).

Proof Take (m, n) RadMN. It follows that (m, n) Rby Theorem 3 and so nx = mxt and so n = mt. Then we havethat (m, n) I(1,t) and so RadMN =

I(1,t).

Theorem 11 Let R be a one dimensional domain and either (x,xt) N ((tx,x) N) for x,tR or (N : M) = 0. ThenN s.t.r.f. in M.

Proof LetM be a maximal ideal of R. It is enough to show that NM s.t.r.f. in MM.

Case i) assume that (N: M) = 0 and so 0 = (N: M)M (NM :RM MM). Since RM is a local one dimensional ring,we get that

(NM :RM MM) is a maximal ideal of RM and it follows that

iRMai +RMbi is equal to either

(NM :RM MM) or RM.

LetiRMai +RMbi =

(NM :RM MM), then by Theorem 6, we get that

RadMNM =

(NM :RM MM)MM and so NM s.t.r.f. in MM.

LetiRMai +RMbi = RM then by Theorem 4, NM s.t.r.f in MM.

Case ii) now assume that (N : M) = 0 and (x,xt) N. We note that for any a, b R, Ra +Rb is equal to Ry forall y Ra +Rb or R ifR is a local one dimensional ring. Then it follows that either iRMai +RMbi =

RMx or

iRMai +RMbi = RM and so by using Theorem 7 or Theorem 4 respectively, we get that NM s.t.r.f. in MM.

Acknowledgement: The author was supported by the Scientific Research Project Administration of Akdeniz

University.

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