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Journal of Algebra and Its ApplicationsVol. 6, No. 2 (2007) 337–353c© World Scientific Publishing Company

A GENERALIZATION OF BAER’S LOWERNILRADICAL FOR MODULES

MAHMOOD BEHBOODI

Department of Mathematical ScienceIsfahan University of Technology, Isfahan, Iran

and

Institute for Studies in Theoretical Physicsand Mathematics (IPM), Tehran, Iran

mbehbood@cc.iut.ac.ir

Received 28 November 2005Accepted 24 August 2006

Communicated by T. Y. Lam

Let M be a left R-module. A proper submodule P of M is called classical prime if forall ideals A,B ⊆ R and for all submodules N ⊆ M , ABN ⊆ P implies that AN ⊆ P orBN ⊆ P . We generalize the Baer–McCoy radical (or classical prime radical) for a module[denoted by cl.radR(M)] and Baer’s lower nilradical for a module [denoted by Nil∗(RM)].For a module RM , cl.radR(M) is defined to be the intersection of all classical primesubmodules of M and Nil∗(RM) is defined to be the set of all strongly nilpotent elementsof M (defined later). It is shown that, for any projective R-module M , cl.radR(M) =Nil∗(RM) and, for any module M over a left Artinian ring R, cl.radR(M) = Nil∗(RM) =Rad(M) = Jac(R)M . In particular, if R is a commutative Noetherian domain withdim(R) ≤ 1, then for any module M , we have cl.radR(M) = Nil∗(RM). We show thatover a left bounded prime left Goldie ring, the study of Baer–McCoy radicals of generalmodules reduces to that of torsion modules. Moreover, over an FBN prime ring R withdim(R) ≤ 1 (or over a commutative domain R with dim(R) ≤ 1), every semiprimesubmodule of any module is an intersection of classical prime submodules.

Keywords: Classical prime module; semiprime module; classical m-system set; lowernilradical; Baer–McCoy radical; strongly nilpotent; left Goldie ring.

Mathematics Subject Classification 2000: 16S38, 16D50, 16D60, 16N60

0. Introduction

All rings in this paper are associative with identity and modules are unitary leftmodules. Let R be a ring and M an R-module. If N is a submodule of M we writeN ≤ M . Also, we denote the classical Krull dimension of R by dim(R), and the leftannihilator of a factor module M/N of M by (N : M).

In the literature, there are many different generalizations of the notion of primetwo-sided ideals to left ideals and also to modules. For instance, a proper left ideal

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338 M. Behboodi

L of a ring R is called prime if, for any elements a and b in R such that aRb ⊆ L,either a ∈ L or b ∈ L. Prime left ideals have properties reminiscent of prime idealsin commutative rings. For example, Michler [13] and Koh [7] proved that the ring R

is left Noetherian if and only if every prime left ideal is finitely generated. Moreover,Smith [14] showed that if R is left Noetherian (or even if R has finite left Krulldimension) then a left R-module M is injective if and only if, for every essentialprime left ideal L of R and homomorphism ϕ: L → M , there exists a homomorphismθ: R → M such that θ|L = ϕ. Let us mention two other generalizations of the notionof prime two-sided ideals to modules and left ideals. Let M be a left R-module. IfM �= 0 and Ann(M) = Ann(N) for all nonzero submodules N of M then M is calleda prime module. Both Goodearl–Warfield [5] and McConnell–Robson [12], use thephrase “prime submodule” to mean “submodule that is prime” (for example, bythis notion every Z-submodule of M := Z is a prime submodule); but Dauns [4]and McCasland–Smith [11], use the phrase “prime submodule” for a submodule P

of M , to mean that M/P is a prime module (i.e., for every ideal A ⊆ R and everysubmodule N ⊆ M , if AN ⊆ P , then either N ⊆ P or AM ⊆ P ).

In this paper, we use a new generalization of the notion of prime two-sided idealsto modules. A proper submodule P of M is called a classical prime submodule ofM if, for all ideals A,B ⊆ R and every submodule N ⊆ M , if ABN ⊆ P , theneither AN ⊆ P or BN ⊆ P . This notion of classical prime submodule has beenextensively studied by this author [1] (see also [2], in which the notion of “weaklyprime submodule” is investigated). Also, a proper submodule P of M is calleda semiprime submodule of M if, for every ideal A ⊆ R and every submoduleN ⊆ M , if A2N ⊆ P , then AN ⊆ P . It is clear that a two-sided ideal I of anyring R, is a prime (resp., semiprime) ideal if and only if it is a classical prime(resp., semiprime) submodule of M = R. Therefore, in case M = R, where R isany commutative ring, classical prime (resp., semiprime) submodules coincide withprime (resp., semiprime) ideals, but we may have a left ideal L in a noncommutativering R such that it is a classical prime submodule of R but it is not a (Dauns)-primesubmodule (see [2, Example 3]). An R-module M is called a classical prime (resp.,semiprime) module if (0) � M is a classical prime (resp., semiprime) submodule.It is clear that for a submodule P � M , M/P is classical prime (resp., semiprime)if and only if P is a classical prime (resp., semiprime) submodule of M .

In ring theory, prime ideals are closely tied to m-system sets (a nonempty setS ⊆ R is said to be an m-system set if for each pair a, b in S, there exists r ∈ R

such that arb ∈ S). The complement of a prime ideal is an m-system, and, given anm-system set S, an ideal disjoint from S and maximal with respect to this propertyis always a prime ideal. Moreover, for an ideal I in a ring R, the set

√I := {s ∈ R |

every m-containing s meets I} equals the intersection of all the prime ideals con-taining I. In particular,

√I is a semiprime ideal in R (see for example [8, Chap. 4],

for more details). In Sec. 1, we extend these facts for modules. Specifically, wedefine classical prime radical of a submodule, and we obtain analogous results tothat of rings for modules. Moreover, we raise the following fundamental conjecture:in any module, every semiprime submodule is an intersection of classical prime

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A Generalization of Baer’s Lower Nilradical for Modules 339

submodules. For instance, in Sec. 3, we show that this conjecture is true for modulesover certain classes of rings like FBN prime ring R with dim(R) ≤ 1 and commu-tative domain R with dim(R) ≤ 1. In Sec. 2, we first extend the notion of stronglynilpotent element to modules, and for an R-module M , we define (Baer’s) lowernilradical of M [denoted by Nil∗(RM)] to be the set of all strongly nilpotent ele-ments of M . Also, the Baer–McCoy radical or classical prime radical of M [denotedby cl.radR(M)], is defined to be the intersection of all classical prime submodulesof M (note that, if M has no classical prime submodule, then cl.radR(M) := M).In the main results of Sec. 2, we show that for any projective R-module M , wehave cl.radR(M) = Nil∗(RM), and for any module M over a left Artinian ring R,we have cl.radR(M) = Nil∗(RM) = Rad(M) = Jac(R)M . This motivates us tothe following natural question: is cl.radR(M) = Nil∗(RM), for all modules M overa ring R? In addition to the special cases mentioned above, we see in Sec. 3 thatthe Baer–McCoy radical (classical prime radical) and Baer’s lower nilradical agreefor all modules over a commutative Noetherian domain with dim(R) ≤ 1. Also, weshow that over a left bounded prime left Goldie ring, the study of Baer–McCoyradicals of general modules reduces to that of torsion modules. In particular, for aleft bounded, left Noetherian prime ring R with dim(R) ≤ 1, the study of Baer–McCoy radicals of general modules reduces to that for finitely generated torsionmodules. Moreover, we determine the Baer–McCoy radical of any module over aleft bounded prime left Goldie ring R with dim(R) ≤ 1.

1. Classical Prime Radical of a Submodule

For an element a ∈ R, let us write (a) = RaR: this is the ideal generated by a in R.The following evident proposition offers several other characterizations of classicalprime submodules.

Proposition 1.1. Let M be an R-module. For a submodule P � M, the followingstatements are equivalent:

(1) P is classical prime;(2) for all a, b ∈ R and every m ∈ M, if (a)(b)m ⊆ P, then either (a)m ⊆ P or

(b)m ⊆ P ;(3) for all a, b ∈ R and every m ∈ M, if aRb(Rm) ⊆ P, then either aRm ⊆ P or

bRm ⊆ P ;(4) for all left ideals A,B ⊆ R and every m ∈ M, if AB(Rm) ⊆ P, then either

A(Rm) ⊆ P or B(Rm) ⊆ P ;(5) for all right ideals A,B ⊆ R and every m ∈ M, if ABm ⊆ P, then either

Am ⊆ P or Bm ⊆ P ;(6) for every 0 �= m̄ ∈ M/P, (0 : Rm̄) is a prime ideal;(7) {(0 : Rm̄) | 0 �= m̄ ∈ M/P} is a chain (linearly ordered set) of prime ideals;(8) (P : M) is a prime ideal, and {(0 : Rm̄) | 0 �= m̄ ∈ M/P} is a chain of prime

ideals.

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340 M. Behboodi

Also, the following evident proposition offers several other characterizations ofsemiprime submodules.

Proposition 1.2. Let M be an R-module. For a submodule P � M, the followingstatements are equivalent:

(1) P is semiprime;(2) for every a ∈ R and m ∈ M, if (a)2m ⊆ P, then (a)m ⊆ P ;(3) for every a ∈ R and m ∈ M, if aRa(Rm) ⊆ P, then aRm ⊆ P ;(4) for every left ideal A ⊆ R and every m ∈ M, if A2(Rm) ⊆ P, then A(Rm) ⊆ P ;(5) for every right ideal A ⊆ R and every m ∈ M, if A2m ⊆ P, then Am ⊆ P ;(6) for every 0 �= m̄ ∈ M/P, (0 : Rm̄) is a semiprime ideal;(7) (P : M) is a semiprime ideal, and for every 0 �= m̄ ∈ M/P, (0 : Rm̄) is also a

semiprime ideal.

Now we have to adapt the notion of an m-system set to modules.

Definition 1.3. Let R be a ring and M be an R-module. A nonempty set S ⊆M\{0} is called a classical m-system if, for all (left) ideals A, B ⊆ R, and forall submodules K, L ≤ M , if (K + AL) ∩ S �= ∅ and (K + BL) ∩ S �= ∅, then(K + ABL) ∩ S �= ∅.Corollary 1.4. Let M be an R-module. Then a submodule P � M is classicalprime if and only if M\P is a classical m-system.

Proof. (⇒). Suppose S = M\P . Let A, B be two ideals in R and K and L

be submodules of M such that (K + AL) ∩ S �= ∅ and (K + BL) ∩ S �= ∅. If(K +ABL)∩S = ∅ then ABL ⊆ P . Since P is classical prime, AL ⊆ P or BL ⊆ P .It follows that (K +AL)∩ S = ∅ or (K + BL)∩ S = ∅, a contradiction. Therefore,S is a classical m-system in M .(⇐). Let S = M\P be a classical m-system in M . Suppose ABL ⊆ P , where A,Bare ideals of R and L ≤ M . If AL �⊆ P and BL �⊆ P , then AL ∩ S �= ∅ andBL∩ S �= ∅. Thus, ABL∩ S �= ∅, a contradiction. Therefore, P is a classical primesubmodule of M .

Proposition 1.5. Let M be an R-module, P be a proper submodule of M, andS := M\P. Then the following statements are equivalent:

(1) P is classical prime;(2) S is a classical m-system;(3) for all left ideals A, B ⊆ R, and for every submodule L ≤ M, if AL ∩ S �= ∅

and BL ∩ S �= ∅, then ABL ∩ S �= ∅;(4) for all right ideals A, B ⊆ R, and for every submodule L ≤ M, if AL ∩ S �= ∅

and BL ∩ S �= ∅, then ABL ∩ S �= ∅;(5) for all ideals A,B ⊆ R, and for every m ∈ M, if Am∩ S �= ∅ and Bm∩ S �= ∅,

then ABm ∩ S �= ∅;

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A Generalization of Baer’s Lower Nilradical for Modules 341

(6) for all a, b ∈ R, and for all m ∈ M, if aRm ∩ S �= ∅ and bRm ∩ S �= ∅, thenaRb(Rm) ∩ S �= ∅.

Proof. (1)⇔ (2) by Corollary 1.4.(2) ⇒ (3) ⇒ (4) ⇒ (5) ⇒ (6) is clear.(6) ⇒ (1). Suppose a, b ∈ R and m ∈ M such that aRb(Rm) ⊆ P . If a(Rm) �⊆ P

and b(Rm) �⊆ P , then a(Rm) ∩ S �= ∅ and b(Rm) ∩ S �= ∅. By our hypothesis,aRb(Rm) ∩ S �= ∅ i.e., aRb(Rm) �⊆ P , a contradiction. Therefore, aRm ⊆ P orbRm ⊆ P . Now by Proposition 1.1, P is a classical prime submodule.

Proposition 1.6. Let M be an R-module, S ⊆ M be a classical m-system, and P

be a submodule of M maximal with respect to the property that P is disjoint from S.

Then P is a classical prime submodule.

Proof. Suppose ABL ⊆ P , where A, B are ideals of R and L ≤ M . If AL �⊆ P

and BL �⊆ P , then by the maximal property of P , we have, (P + AL) ∩ S �= ∅ and(P + BL) ∩ S �= ∅. Thus (P + ABL) ∩ S �= ∅ and it follows that P ∩ S �= ∅, acontradiction. Thus, P must be a classical prime submodule.

Next we need a generalization of the notion of√

N for any submodule N of M .We adopt the following:

Definition 1.7. Let R be a ring and M be a R-module. For a submodule N of M ,if there is a classical prime submodule containing N , then we define

√N := {m ∈ M : every classical m-system containing m meets N}.

If there is no classical prime submodule containing N, then we put√

N = M .

Theorem 1.8. Let M be an R-module and N ≤ M. Then either√

N = M or√

N

equals the intersection of all the classical prime submodules of M containing N. Inparticular, if

√N �= M, then

√N is a semiprime submodule of M.

Proof. Suppose that√

N �= M . This means that

{P | P is a classical prime submodule of M and N ⊆ P} �= ∅.We first prove that

√N ⊆ {P | P is a classical prime submodule of Mand N ⊆ P}.

Let m ∈ √N and P be any classical prime submodule of M containing N . Consider

the classical m-system M\P . This classical m-system cannot contain m, for other-wise it meets N and hence also P . Therefore, we have m ∈ P . Conversely, assumem �∈ √

N . Then, by Definition 1.7, there exists a classical m-system S containingm which is disjoint from N . By Zorn’s Lemma, there exists a submodule P ⊇ N

which is maximal with respect to being disjoint from S. By Proposition 1.6, P is aclassical prime submodule of M , and we have m �∈ P , as desired.

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342 M. Behboodi

We recall the definition of the notion of n-system in a ring R. A nonempty setT ⊆ R is said to be an n-system set if for each a in T , there exists r ∈ R suchthat ara ∈ T (see for example [8, Chap. 4], for more details). The complement ofa semiprime ideal is an n-system set, and if T is an n-system in a ring R such thata ∈ T , then there exists an m-system S ⊆ T such that a ∈ S (see [8, Lemma 10.10]).Now we have to adapt the notion of an n-system set to modules.

Definition 1.9. Let M be an R-module. A nonempty set T ⊆ M\{0} is calleda classical n-system if, for every ideal A ⊆ R and all submodules K, L ≤ M , if(K + AL) ∩ T �= ∅, then (K + A2L) ∩ T �= ∅.

Lemma 1.10. Let M be an R-module. Then a submodule N � M is semiprime ifand only if M\N is a classical n-system.

Proof. (⇒). Let T = M\N . Suppose A is an ideal of R and K, L are submodulesof M such that (K +AL)∩T �= ∅. If (K +A2L)∩T = ∅ then K +A2L ⊆ N . SinceN is semiprime, K +AL ⊆ N . Thus, (K +AL)∩T = ∅, a contradiction. Therefore,T is a classical n-system in M .(⇐). Suppose that T = M\N is a classical n-system in M . Suppose A2L ⊆ N ,where A is an ideal of R and L ≤ M , but AL �⊆ N . It follows that AL∩ T �= ∅ andso A2L ∩ T �= ∅, a contradiction. Therefore, N is a semiprime submodule of M .

The proof of the next proposition is similar to the proof of Proposition 1.5.

Proposition 1.11. Let M be an R-module, P be a proper submodule of M, andlet T := M\P. Then the following statements are equivalent :

(1) P is semiprime;(2) T is a classical n-system;(3) for every left ideal A ⊆ R, and every submodule L ≤ M, if AL ∩ T �= ∅, then

A2L ∩ T �= ∅;(4) for every right ideal A ⊆ R, and every submodule L ≤ M, if AL ∩ T �= ∅, then

A2L ∩ T �= ∅;(5) for every ideal A ⊆ R, and every m ∈ M, if Am ∩ T �= ∅, then A2m ∩ T �= ∅;(6) for every a ∈ R, and every m ∈ M, if aRm ∩ T �= ∅, then (aR)2m ∩ T �= ∅;(7) for every a ∈ R, and every m ∈ M, if Ra(Rm)∩T �= ∅, then (Ra)2Rm∩T �= ∅.

We next explore a relationship between classical m-systems and classical n-systems and some interesting questions and consequences of this. It is clear thatevery classical m-system is a classical n-system, but not conversely. An interestingquestion is:

Question 1.12. If T is a classical n-system in a module M such that x ∈ T, doesthere exist a classical m-system S ⊆ T such that x ∈ S?

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Remark 1.13. Let R = Z[x] and F = R⊕R. Let f = (2, x) ∈ F and P = 2R+Rx

(which is a maximal ideal of R). In [6] it is shown that N = Pf is a semiprimesubmodule of F which is not an intersection of (Dauns)-prime submodules. But N

is an intersection of classical prime submodules of F (in fact, in [2] it is shown thatN is a classical prime submodule of F ). If the answer to Question 1.12 is yes, thenit can be shown that

√N = N for all semiprime submodules, and hence, that every

semiprime submodule is an intersection of classical prime submodules.

We have not found any examples of a module M with a semiprime submodule N

for which√

N �= N . The lack of such counterexamples, together with the fact that√N = N (where N is a semiprime submodule of M) for modules M over certain

classes of rings like FBN prime ring R with dim(R) ≤ 1 and commutative domainR with dim(R) ≤ 1 (see Sec. 3), motivates the following fundamental conjecture:

Conjecture 1.14. If N is a semiprime submodule of a module M, then√

N = N.

Consequently, every semiprime submodule of M is an intersection of classical primesubmodules.

2. Classical Prime Radical and Baer’s Lower Nilradicalof a Module

It is well-known that the set of nilpotent elements of a commutative ring forms anideal and is equal to the intersection of all the prime ideals. This notion has beengeneralized by MaCasland and Moore to modules over a commutative ring (see [10]or [9], which used the notion of (Dauns)-prime submodule for prime radical of amodule). Unfortunately, not every module satisfies McCasland’s radical formula;see [6].

We recall the definition of the strongly nilpotent element in a ring. An elementa of a ring R is called strongly nilpotent if every sequence a1, a2, a3, . . . such thata1 = a and an+1 ∈ anRan (∀n) is eventually zero. Also, for any ring R, Nil∗(R) isthe smallest semiprime ideal in R, and is equal to the intersection of all the primeideals in R. This is called (Baer’s) lower nilradical or the Baer–McCoy radical ofR. It is a well-known result of Levitzki that the Nil∗(R) is precisely the set of allstrongly nilpotent elements of R (see, for example [8, p. 180]). Now we generalizethis notion to modules.

Definition 2.1. For any module M , we define cl.radR(M) =√

(0). This is calledBaer–McCoy radical or classical prime radical of M . Thus, if M has a classical primesubmodule, then cl.radR(M) is equal to the intersection of all the classical primesubmodules in M but, if M has no classical prime submodule, then cl.radR(M)=M .

Definition 2.2. An element m of an R-module M is called nilpotent if m =∑ri=1 aimi for some ai ∈ R, mi ∈ M and r ∈ N, such that ai

kmi = 0 (1 ≤ i ≤ r)for some k ∈ N, and m is called strongly nilpotent if m =

∑ri=1 aimi for some

ai ∈ R, mi ∈ M and r ∈ N, such that for every i (1 ≤ i ≤ r) and every sequence

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344 M. Behboodi

ai1, ai2, ai3, . . . where ai1 = ai and ain+1 ∈ ainRain (∀n), we have aikRmi = 0 forsome k ∈ N.

It is clear that every strongly nilpotent element of a module M is a nilpotentelement but the converse is not true (see the following example).

Example 2.3. Let

R ={[

a b

c d

] ∣∣∣∣ a, b, c, d ∈ Q}

, M ={[

0 x

0 y

] ∣∣∣∣ x, y ∈ Q}

.

Then by formal matrix multiplication, M becomes a left R-module. Let

r1 =[

0 10 0

], r2 =

[0 01 0

]∈ R, m =

[0 10 1

]∈ M.

One can easily see that m = r1m + r2m and r12m = r2

2m = 0, and hence m

is a nilpotent element of M . We claim that m is not a strongly nilpotent elementof M , for if not, then m =

∑ri=1 aimi for some ai ∈ R, mi ∈ M and r ∈ N,

such that for every i (1 ≤ i ≤ r) and every sequence ai1, ai2, ai3, . . . whereai1 = ai and ain+1 ∈ ainRain (∀n), we have aikRmi = 0 for some k ∈ N. Withoutloss of generality we can assume that a1m1 �= 0. Thus we can take a sequencea11, a12, a13, . . . where a11 = a1 and a1n+1 ∈ a1nRa1n (∀n), such that a1n �= 0,for all n ≥ 1 (since R is a prime ring, a1n �= 0 implies that a1nRa1n �= 0). Thusa1kRm1 = 0 for some k ∈ N. Thus Ra1kRm1 = 0, and since R is a simple ring, itfollows that m1 = 0, a contradiction.

Definition 2.4. For any module M , we define Nil∗(RM) to be the set of all stronglynilpotent elements of M . It is easy to check that Nil∗(RM) is a submodule of M .This is called (Baer’s) lower nilradical of M .

Proposition 2.5. Let M be an R-module and I be an ideal of R such that I ⊆Ann(M). Then

cl.radR(M) = cl.radR/I(M), and Nil∗(RM) = Nil∗(R/IM).

Proof. Elementary.

Lemma 2.6. Let R be a ring and M be an R-module. Then

Nil∗(R)M ⊆ Nil∗(RM) ⊆ cl.radR(M).

Proof. Let m ∈ M . It is clear that for every strongly nilpotent element a of R,am is a strongly nilpotent element in M , and so Nil∗(R)M ⊆ Nil∗(RM). Supposem ∈ Nil∗(RM) but m �∈ cl.radR(M). Then m =

∑ri=1 aimi for some ai ∈ R, mi ∈

M and r ∈ N, such that for every 1 ≤ i ≤ r and every sequence ai1, ai2, ai3, . . .

where ai1 = ai and ain+1 ∈ ainRain (∀n), we have aikRmi = 0 for some k ∈ N.Without loss of generality we can assume that a1m1 �∈ cl.radR(M). Thus, there

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exists a classical m-system S such that a1m1 ∈ S, so that Ra1(Rm1) ∩ S �= ∅.Thus, (Ra1)2(Rm1) ∩ S �= ∅. Assume r1 = s1 = 1, a11 = a1 and r2a12s2m1 ∈(Ra1)2(Rm1)∩ S, where r2, s2 ∈ R and a12 = a11t1a11 for some t1 ∈ R. Therefore,(Ra12)Rm1 ∩ S �= ∅ and so (Ra12)2Rm1 ∩ S �= ∅. Thus, there exists r3a13s3m1 ∈(Ra12)2(Rm1) ∩ S, where r3, s3 ∈ R and a13 = a12t2a12 for some t2 ∈ R. We canrepeat this argument to get sequences {rn}n∈N, {sn}n∈N and {a1n}n∈N in R, wherea11 = a1 and a1n+1 ∈ a1nRa1n (∀n), such that rna1nsnm1 ∈ S for all n ≥ 1. Nowby our hypothesis a1kRm1 = 0 for some k ∈ N, and so rka1kskm1 = 0 ∈ S, acontradiction.

The following example shows that in general, the first inclusion in Lemma 2.6,is strict inclusion.

Example 2.7. Let R = Z and M = Zp∞ ⊕ Z for some prime number p. Wesee easily that cl.rad(M) = Zp∞ and Nil∗(R)M = 0. But for any m =

(rpt , 0

) ∈Zp∞ ⊕ Z, we can write m = p

(r

pt+1 , 0)

such that pt+1(

rpt+1 , 0

)= 0, and it follows

that Zp∞ ⊆ Nil∗(RM). Therefore, Nil∗(R)M � Nil∗(RM) = cl.radR(M).

The prime radical of M is defined to be the intersection of all (Dauns)-primesubmodules of M . As in [6], we denote the prime radical of M by rad(M). LetR = Z[x] and F = R ⊕ R. Let f = (2, x) ∈ F and P = 2R + Rx (which is amaximal ideal of R). If M = F/N , then Nil∗(RM) = 0 but rad(M) = Rf/N �= 0(see [6, p. 3600]). On the other hand, N ≤ F is a classical prime submodule and socl.radR(M) = Nil∗(RM) = 0 (see [2, Remark 4]). However, we have not found anyexample where cl.radR(M) �= Nil∗(RM). Thus, we have the following fundamentalquestion.

Question 2.8. Let M be an R-module. Is cl.radR(M) = Nil∗(RM)?

Remark 2.9. We shall see later in this section that the answer to Question 2.8 is“yes” for every projective R-module M and for every module over a left Artinianring R (see Theorems 2.14, 2.15). Also, in Theorem 3.14 of Sec. 3, we show thatthe answer to Question 2.8 is “yes” for modules over a commutative Noetheriandomain R with dim(R) ≤ 1.

Proposition 2.10. Let R be a ring. Then

Nil∗(R) = Nil∗(RR) = cl.radR(R).

Proof. By Lemma 2.6, Nil∗(R) ⊆ Nil∗(RR) ⊆ cl.radR(R). However, for any ring R,the set of all prime ideals of R is contained in the set of all classical prime submod-ules of RR, and hence cl.radR(R) ⊆ Nil∗(R). Therefore, cl.radR(R) = Nil∗(RR).

Lemma 2.11. Let M be an R-module and N be a submodule of M . Then

cl.radR(N) ⊆ cl.radR(M).

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Proof. Let P be any classical prime submodule of M . If N ⊆ P , then cl.radR(N) ⊆P . If N �⊆ P , then it is easy to check that N ∩ P is a classical prime submoduleof N , and hence cl.radR(N) ⊆ N ∩ P ⊆ P . Thus, in any case, cl.radR(N) ⊆ P . Itfollows that cl.radR(N) ⊆ cl.radR(M).

Lemma 2.12. Let M be an R-module such that M =⊕

Λ Mλ is a direct sum ofsubmodules Mλ (λ ∈ Λ). Then

cl.radR(M) =⊕Λ

cl.radR(Mλ).

Proof. By Lemma 2.11, cl.radR(Mλ) ⊆ cl.radR(M) for all λ ∈ Λ, so that⊕

Λ

cl.radR(Mλ) ⊆ cl.radR(M). Now let m �∈ ⊕Λ cl.radR(Mλ), for some m ∈ M . Then

there exists µ ∈ Λ such that πµ(m) �∈ cl.radR(Mµ), where πµ: M → Mµ denotesthe canonical projection. Thus there exists a classical prime submodule Nµ of Mµ

such that πµ(m) �∈ Nµ. Let K = Nµ

⊕(⊕

λ�=µ Mλ). It is easy to check that K is aclassical prime submodule of M and m �∈ K. Thus m �∈ cl.radR(M). It follows thatcl.radR(M) ⊆ ⊕

Λ cl.radR(Mλ).

Lemma 2.13. Let M be an R-module such that cl.radR(M) = Nil∗(RM). Thencl.radR(N) = Nil∗(RN) for every direct summand N of M .

Proof. Suppose that M = N⊕

K for some submodules K of M . By Lemma 2.6,Nil∗(RN) ⊆ cl.radR(N). Let m ∈ cl.radR(N). By Lemma 2.11, m ∈ cl.radR(M).By hypothesis, m =

∑ri=1 aimi for some ai ∈ R, mi ∈ M and r ∈ N, such

that for every 1 ≤ i ≤ r and every sequence ai1, ai2, ai3, . . . where ai1 = ai

and ain+1 ∈ ainRain (∀n), we have aikRmi = 0 for some k ∈ N. For each 1 ≤i ≤ r, there exist elements xi ∈ N , yi ∈ K such that mi = xi + yi. Clearly,m = a1x1 + a2x2 + · · · + anxn, and aikRxi = 0, for some k ∈ N, (1 ≤ i ≤ n). Thusm ∈ Nil∗(RN). It follows that cl.radR(N) ⊆ Nil∗(RN).

Theorem 2.14. Let R be a ring, and let M be a projective R-module. Then

cl.radR(M) = Nil∗(RM).

Proof. There exists a free R-module F such that M is a direct summand of F .There exists an index set Λ and cyclic submodules Fλ (λ ∈ Λ) of F such thatF =

⊕Λ Fλ. By Lemma 2.12, cl.radR(F ) =

⊕Λ cl.radR(Fλ), and by Proposi-

tion 2.10 and Lemma 2.11 cl.radR(Fλ) = Nil∗(RFλ) ⊆ Nil∗(RF ), for all λ ∈ Λ.Hence cl.radR(F ) = Nil∗(RF ). Now apply Lemma 2.13.

Let M be an R-module. Because every maximal submodule of M is a classicalprime submodule, it is clear that cl.radR(M) is contained in the Jacobson radicalRad(M) of M . Now we have the following result.

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A Generalization of Baer’s Lower Nilradical for Modules 347

Theorem 2.15. Let M be a module over a left Artinian ring R. Then

cl.radR(M) = Nil∗(RM) = Rad(M) = Jac(R)M.

Proof. Since every maximal submodule of M is a classical prime submodule,cl.radR(M) ⊆ Rad(M). By [12, 31.5], Rad(M) = Jac(R)M . On the other hand,Jac(R) = Nil∗(R) and by Lemma 2.6, Nil∗(R)M ⊆ Nil∗(RM) ⊆ cl.rad(RM). There-fore cl.radR(M) = Nil∗(RM) = Rad(M) = Jac(R)M .

3. On the Classical Prime Radical of Modules Over a LeftBounded, Prime Left Goldie Ring

An essential (or large) submodule of a module M is any submodule N which hasnonzero intersection with every nonzero submodule of M . We write N ≤e M todenote this situation. Clearly, if I is a nonzero ideal in a prime ring R, then I isboth an essential left ideal and an essential right ideal of R. Given a left moduleM over a ring R, the set

Z(M) = {m ∈ M | Im = 0 for some I ≤e RR} = {m ∈ M | Ann(m) ≤e RR}is a submodule of M (see for example [5, Lemma 3.25]). The submodule Z(M)described above, is called the singular submodule of M . If Z(M) = M , then M iscalled singular module, while if Z(M) = 0, then M is called nonsingular module.For example, for a commutative domain R the nonsingular R-modules are exactlythe torsion-free R-modules.

A regular element in a ring R is any element x ∈ R such that �.Ann(x) = 0and r.Ann(x) = 0 (i.e., x is neither a left nor right zero divisor of R). Let M be aleft module over a ring R, and let

T (M) = {m ∈ M |am = 0 for some regular element a ∈ R}.In general T (M) is not a submodule of M (see [5, Exercise 6C]). But T (M) = Z(M),for all left modules M over a semiprime left Goldie ring R (see [5, Proposition 6.9]).Moreover, if R is a ring such that T (M) = Z(M) for all left R-modules M , then R

is a semiprime left Goldie ring (see [5, Exercise 6D]). Thus, for a module M over asemiprime Goldie ring R, the submodule T (M) is called the torsion submodule of M .Also, M is called a torsion module if T (M) = M , and M is called a torsion-freemodule if T (M) = 0. Now we have the following definition.

Definition 3.1. Let R be a semiprime left Goldie ring, and M be an R-module.We define

S(M) := {m ∈ M | aRm = 0 for some regular element a ∈ R}.One can check that S(M) is a submodule of M . Since S(M) ⊆ T (M), S(M) is calledthe strongly torsion submodule of M . Also, M is called a strongly torsion module ifS(M) = M , and M is called a strongly torsion-free module if S(M) = 0. It is clearthat if R is a commutative domain, then S(M) = T (M), for all R-modules M .

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Let I be a left ideal in a semiprime left Goldie ring R. By Goldie Theo-rem [5, Proposition 5.9], I is an essential left ideal of R if and only if I containsa regular element. Hence every nonzero ideal of a prime left Goldie ring R has aregular element (since every nonzero ideal of a prime ring is an essential left (right)ideal). Thus, we have the following lemma.

Lemma 3.2. Let R be a prime left Goldie ring, and M be an R-module. Then

S(M) = {m ∈ M |Ann(Rm) �= 0}.We recall that a ring R is left bounded if every essential left ideal of R contains

an ideal which is essential as a left ideal. For instance, every commutative ring isleft bounded, as is every semisimple ring (since a semisimple ring has no properessential left ideals). Note that a prime ring R is left bounded if and only if everyessential left ideal of R contains a nonzero ideal (see for example [5, p. 132]). Also,a ring R is left fully bounded provided every prime factor ring of R is left bounded.By way of example, note that every commutative ring is of course fully bounded. Itis also the case that every PI-ring (ring with polynomial identity) is fully bounded.

A priori, a left fully bounded ring need not be left bounded. However, in aleft fully bounded left Noetherian ring, it can be shown that all factor rings areleft bounded (see for example [5, Exercise 8F]). A left (right) FBN ring is any left(right) fully bounded left (right) Noetherian ring. An FBN ring is any left andright FBN ring.

Proposition 3.3. Let R be a left bounded prime left Goldie ring, and M be anR-module. Then S(M) = T (M).

Proof. In view of Definition 3.1, it is suffices to show that T (M) ⊆ S(M). Letm ∈ T (M). By [5, Proposition 6.9], Ann(m) is an essential left ideal of R. Since R

is left bounded, Ann(m) contains a nonzero ideal J of R. Thus J(Rm) = Jm = 0and so Ann(Rm) �= 0. Now by Lemma 3.2, T (M) ⊆ S(M).

Lemma 3.4. Let R be a left bounded prime left Goldie ring, and M be a torsion-free R-module. Then M is a classical prime module.

Proof. Let aRb(Rm) = 0, where a, b ∈ R and 0 �= m ∈ M . If aRb �= 0, then byLemma 3.2 and Proposition 3.3, m = 0, a contradiction. Thus aRb = 0 and sinceR is a prime ring, a = 0 or b = 0. It follows that either aRm = 0 or bRm = 0. ThusM is a classical prime module.

Proposition 3.5. Let R be a left bounded prime left Goldie ring, and M be anR-module with torsion submodule T (M). Then

(i) PM = M or PM is a classical prime submodule of M for all maximal idealsP of R.

(ii) T (M) = M or T (M) is a classical prime submodule of M .

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Proof. (i) is evident.(ii) Suppose T (M) �= M . By [5, Proposition 3.29], M/T (M) is a torsion-free R-module. Now by Lemma 3.4, M/T (M) is a classical prime module i.e., T (M) is aclassical prime submodule of M .

Proposition 3.6. Let R be a left bounded prime left Goldie ring, and M be anR-module. Let P be a proper submodule of T (M). Then P is a classical primesubmodule of T (M) if and only if it is a classical prime submodule of M .

Proof. Suppose P is a classical prime submodule of T (M). If T (M) = M , then weare through. Thus, we assume that T (M) �= M and so by Proposition 3.5, T (M)is a classical prime submodule of M . Now if IJK ⊆ P , where K ≤ M and I, J

are ideals of R, then IK ⊆ T (M) or JK ⊆ T (M). Clearly, if I = 0 or J = 0,then IK ⊆ P or JK ⊆ P . Now let IJ �= 0. Since R is a prime left Goldie ring,there is a nonzero regular element c ∈ IJ . But cK ⊆ IJK ⊆ T (M), so for allk ∈ K, ck ∈ T (M). Thus by Lemma 3.2 and Proposition 3.3, for each k ∈ K,there exists a nonzero ideal Ik of R such that Ikck = 0. Since R is a prime ring,Ikc = IkRc �= 0, for all k ∈ K. Since R is left bounded, Ikc contains a nonzero ideal,for all k ∈ K. It follows that Ann(Rk) �= 0, for all k ∈ K. Now by Lemma 3.2 andProposition 3.3, K ⊆ T (M). Thus IJK ⊆ P , where K ≤ T (M) and I, J are idealsof R. Since P is a classical prime submodule of T (M), IK ⊆ P or JK ⊆ P . Itfollows that P is a classical prime submodule of M . The converse is evident.

The following corollary shows that for a left bounded prime left Goldie ringR, the study of classical prime radicals of general R-modules reduces to that fortorsion modules.

Corollary 3.7. Let R be a left bounded prime left Goldie ring, and let M be anR-module. Then

cl.radR(M) = cl.radR(T (M)).

Proof. By Lemma 2.11, cl.radR(T (M)) ⊆ cl.radR(M). Therefore, by Proposi-tion 3.6, we have cl.radR(M) = cl.radR(T (M)).

Lemma 3.8. Let R be a left bounded prime left Goldie ring, and let M be anR-module. Then

cl.radR(M) ⊆⋂

{PT (M) | P is a maximal ideal of R}.

Proof. By Corollary 3.7 and Proposition 3.5(i).

Proposition 3.9. Let R be a left bounded prime left Goldie ring such thatdim(R) ≤ 1, and let M be an R-module. Then

cl.radR(M) =⋂

{PT (M) | P is a maximal ideal of R}.

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Proof. If dim(R) = 0, then R is a simple ring, and hence every R-module is aclassical prime module. Thus, it follows that cl.radR(M) = 0, for all R-modulesM , and so we are through. We assume that dim(R) = 1. We begin by showingthat every classical prime submodule of T (M) contains a submodule of the formPT (M) for some maximal ideal P of R. Suppose N is any classical prime submoduleof T (M). Let T̄ = T (M)/N . Then T̄ is a classical prime module, and hence byProposition 1.1, P = (0 : T̄ ) is a prime ideal and Γ := {(0 : Rx̄) | 0 �= x̄ ∈ T̄} isa chain of prime ideals of R. By Lemma 3.2 and Proposition 3.3, Ann(Rx) �= 0,for every x ∈ T (M). It follows that (0 : Rx̄) �= 0, for all x̄ ∈ T̄ . Thus Γ is achain of nonzero prime ideals of R. Since dim(R) = 1, Γ is singleton, and hence(0 : T̄ ) = P = (0 : Rx̄), for all 0 �= x̄ ∈ T̄ . Thus P is a maximal ideal of R suchthat PT (M) ⊆ N . By Corollary 3.7 and Lemma 3.8, we conclude that

cl.radR(M) ⊆⋂

{PT (M) | P is a maximal ideal of R}

⊆⋂

{N | N is a classical prime submodule of T (M)}= cl.radR(T (M)) = cl.radR(M),

and hence equality holds.

The following proposition shows that in some instances, the study of classicalprime radicals of general modules reduces to that for finitely generated (strongly)torsion modules.

Proposition 3.10. Let R be a left bounded prime left Goldie ring that satisfies theascending chain condition (ACC) on ideals. If dim(R) ≤ 1, then for an R-moduleM we have

cl.radR(M) =⋃

cl.radR(L),

where the union is taken over all finitely generated torsion submodules L of M .

Proof. If dim(R) = 0, then R is a simple ring, and hence every R-module is aclassical prime module. This follows that cl.radR(M) = 0, for all R-modules M , andso we are through. Thus, we assume that dim(R) = 1. Also by Corollary 3.7, we mayassume that M is a torsion R-module. By Lemma 2.11, cl.radR(L) ⊆ cl.radR(M)for all submodules L of M . Assume, m ∈ cl.radR(M). By Proposition 3.3, M isstrongly torsion, and hence there exists a nonzero ideal I of R such that Im = 0.Since R satisfy the ACC on ideals, R/I also satisfy the ACC on ideals, and henceby [8, p. 180, Exercise 15], R/I has only finitely many minimal prime ideals. Sincedim(R) = 1 and I �= 0, dim(R/I) = 0. This implies that there exists a positiveinteger n such that P1, . . . ,Pn are all maximal ideals of R containing the ideal I.By Proposition 3.5, for each i (1 ≤ i ≤ n), m ∈ PiM , and hence m ∈ PiLi, forsome finitely generated submodule Li of M . Let L = L1 + · · · + Ln. Then L is afinitely generated torsion submodule of M and m ∈ PiL, for all i (1 ≤ i ≤ n).

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Now let P be any maximal ideal of R. Suppose that P �= Pi, for all i (1 ≤ i ≤ n).Then I + P = R, and hence

Rm = (I + P)m ⊆ Im + Pm = Pm ⊆ PL.

Therefore

m ∈⋂

{PL | P is a maximal ideal of R}.Now by Proposition 3.9,

m ∈⋂

{PL | P is a maximal ideal of R} = cl.radR(L).

Thus, we have

cl.radR(M) ⊆⋃

cl.radR(L),

where the union is taken over all finitely generated torsion submodules L of M .

Every left Noetherian ring is left Goldie (see for example [5, p. 88]). Thus, wehave the following corollary.

Corollary 3.11. Let R be a left bounded, left Noetherian prime ring withdim(R)≤ 1, and let M be an R-module. Then

cl.radR(M) =⋃

cl.radR(L),

where the union is taken over all finitely generated torsion submodules L of M .

Now we are in a position to show that the fundamental conjecture given in Sec. 1is true for modules over an FBN prime ring with dim(R) ≤ 1.

Theorem 3.12. Let R be an FBN prime ring with dim(R) ≤ 1, and let M be anR-module. Let N be a proper submodule of M . Then the following statements areequivalent:

(1) N is a semiprime submodule of M ;(2)

√N = N ;

(3) N is an intersection of classical prime submodules.

Proof. (1) ⇒ (2). Let N be a semiprime submodule of M . Then M̄ := M/N isa semiprime R-module. It suffices to show that cl.radR(M̄) = 0. Since every left(right) Noetherian ring is left (right) Goldie, by Proposition 3.10, we may assumethat M̄ is a finitely generated torsion R-module. If dim(R) = 0, then R is a simplering and so M̄ is a classical prime module, and hence cl.radR(M̄) = 0. Thus, weassume that dim(R) = 1. Let M̄ = Rm1 + Rm2 + · · · + Rmn. By Proposition 3.3,M̄ is a strongly torsion module (since M̄ is a torsion module), and hence for each i

(1 ≤ i ≤ n), we have Ii = Ann(Rmi) �= (0). It follows that I = Ann(M̄) �= 0. SinceM̄ is a semiprime module, by Proposition 1.2, I is a semiprime ideal of R. Since R

is left and right bounded, by [5, p. 132], every factor ring of R is also left and right

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352 M. Behboodi

bounded, and hence R̄ = R/I is an FBN semiprime ring with dim(R̄) = 0. Sinceevery prime ideal P̄ of R̄ is a minimal prime, by [5, Proposition 6.3], P̄ containsno regular elements, i.e., every regular element of R̄ is unit. Thus M̄ is a finitelygenerated torsion-free R̄-module. By Gentile, Levy Theorem [5, Proposition 6.19],M̄ can embedded in a finitely generated free left R̄-module. Let M̄ ≤ R̄n. Since R̄

is a semiprime ring, cl.radR̄(R̄n) = 0. Thus by Proposition 2.5 and Lemma 2.11,we have cl.radR(M̄) = cl.radR̄(M̄) ⊆ cl.radR̄(R̄n) = 0.

(2) ⇒ (3) by Theorem 1.8.(3) ⇒ (1) is evident.

Let M be an R-module over a commutative ring R. In [3, Theorem 2.3], it isshown that, every proper submodule of M is an intersection of maximal submodulesif and only if for each 0 �= m ∈ M , R/Ann(m) is a regular (von Neumann) ring. Byapplying this fact, we can show that for a commutative domain R with dim(R) ≤ 1the Theorem 3.12 above is true without the assumption that R is Noetherian.

Theorem 3.13. Let R be a commutative domain with dim(R) ≤ 1, and let M bean R-module. Then for any submodule N � M , the following are equivalent :

(1) N is a semiprime submodule of M ;(2)

√N = N ;

(3) N is an intersection of classical prime submodules.

Proof. (1) ⇒ (2). Let N be a semiprime submodule of M . Then M̄ := M/N is asemiprime R-module. It suffices to show that cl.radR(M̄) = (0). By Corollary 3.3,we may assume that M̄ is a torsion R-module. If dim(R) = 0, then R is a field andM̄ is a vector space over R. Since M̄ is a torsion R-module, M̄ = (0) and thereforecl.radR(M̄) = (0). Now we assume that dim(R) = 1. For each 0 �= m ∈ M̄ , wehave Ann(m) �= (0) and it is a semiprime ideal of R. Therefore the ring R/Ann(m)is reduced (i.e., R/Ann(m) has no nonzero nilpotent elements). Moreover, everyprime ideal of R/Ann(m) is a maximal (because, dim(R) = 1 and Ann(m) �= (0)).By [8, Exercise 15, p. 69], for each 0 �= m ∈ M̄ , R/Ann(m) is von Neumannregular. Now by [3, Theorem 2.3], every proper submodule of M̄ is an intersectionof maximal submodules of M̄ , and hence the zero submodule of M̄ is an intersectionof maximal submodules of M̄ . Since every maximal submodule of M̄ is a classicalprime submodule, cl.radR(M̄) = (0).

(2) ⇒ (3), by Theorem 1.8.(3) ⇒ (1), is evident.

We conclude this paper with the following theorem, which shows that the answerto Question 2.8 is “yes” for modules over a commutative Noetherian domain R withdim(R) ≤ 1.

Theorem 3.14. Let R be a commutative Noetherian domain with dim(R) ≤ 1 andM be an R-module. Then cl.radR(M) = Nil∗(RM).

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April 16, 2007 19:47 WSPC/171-JAA 00228

A Generalization of Baer’s Lower Nilradical for Modules 353

Proof. By Proposition 3.10, cl.radR(M) =⋃

cl.radR(L), where the union is takenover all finitely generated torsion submodules L of M . Suppose L = Rm1+· · ·+Rmn

is a finitely generated torsion submodule of M . Let Ii = Ann(mi), for all 1 ≤ i ≤ n.Then I = Ann(L) = I1I2 · · · In �= 0. Thus, every prime ideal of R/I is maxi-mal (i.e., R/I is Artinian) and L is an R/I-module. Therefore, by Theorem 2.15,cl.radR/I(L) = NiL∗(R/IL). Now, by Proposition 2.5, we have

cl.radR(L) = cl.radR/I(L) = NiL∗(R/IL) = NiL∗(RL) ⊆ NiL∗(RM).

Therefore, cl.radR(M) = NiL∗(RM).

Acknowledgments

This work was partially supported by IUT (CEAMA), and in part supported by agrant from IPM (No. 83130018). The author wishes to express his deepest gratitudeto the referee for his/her valuable comments.

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