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Home , Kasch ring, Minimal ideal, Semiprimitive ring, Semisimple module

Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:20 https://doi.org/10.1007/s12044-018-0403-6

On a generalization of semisimple modules

NAZIM AGAYEV1,2,∗, CESIMú ÇELIKú 1,2 and TAHIREú ÖZEN1,2

1Computer Engineering Department, Fatih Sultan Mehmet Vakf University, Istanbul, Turkey 2Department of , Abant Izzetú Baysal University, Gölköy Kampüsü, Bolu, Turkey *Corresponding author. E-mail: [email protected]; [email protected]; [email protected]

MS received 28 January 2016; revised 14 May 2016; accepted 25 August 2016; published online 24 April 2018

Abstract. Let R be a with identity. A MR is called an r- if for any right I of R, MI is a direct summand of MR which is a generalization of semisimple and second modules. We investigate when an r-semisimple ring is semisimple and prove that a ring R with the number of nonzero proper ideals ≤4 and J(R) = 0isr-semisimple. Moreover, we prove that R is an r-semisimple ring if and only if it is a of simple rings and we investigate the structure of module whenever R is an r-semisimple ring.

Keywords. Semisimple modules; second modules; injective modules.

1991 Mathematics Subject Classification. 16U80.

1. Introduction Throughout, we consider associative rings with identity and all modules are unitary. For a module M over a ring R, we write MR to indicate that M is a right R-module. The and the of M are denoted by Soc(M) and J(M), respectively. The module M is called semisimple if Soc(M) = M. In section 2, we define and characterize r-semisimple modules and rings which are generalizations of semisimple and second modules and rings. A module MR is called a semisimple module if every submodule is a direct summand of MR. A right R-module M is second if M = 0 and annR(M) = annR(M/N) for every proper submodule N of M (see [9]). In [2], it is proved that a module MR is a second module if for every ideal I of R either MI = 0orMI = M. A module MR is called an r-semisimple module if for any right ideal I of R, MI is a direct summand of MR (or equivalently for every ideal I , MI is a direct summand of MR). It is clear that every semisimple and second modules are r-semisimple but the converse may not be true. If RR is r-semisimple, then it is semiprimitive, that is, J(R) = 0. Every direct summand and direct sum of r-semisimple modules are also r-semisimple. We investigate when an r-semisimple ring is semisimple and simple. Moreover, any module MR need not be r-semisimple although every proper

© Indian Academy of Sciences 1 20 Page 2 of 10 Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:20 submodule is r-semisimple. We show that a ring R with the number of the proper right ideals ≤4 and J(R) = 0isr-semisimple. In section 3, we investigate the structure of ring and modules whenever R is an r- semisimple ring. We obtain that R is an r-semisimple ring if and only if it is a direct sum of simple rings and we prove that every MR is r-semisimple and if RR is also self-injective and noetherian, then RR is semisimple . If R is an r-semisimple and right , then R is semisimple. For any unexplained terminology, we refer the reader to [1], [7] and [2].

2. Preliminaries In this section, we study some elementary properties of r-semisimple modules.

DEFINITION 2.1

Let M be a right R-module. M is called an r-semisimple module if for any right ideal I of R, MI is a direct summand of MR. The ring R is said to be an r-semisimple ring if RR is r-semisimple. It is clear that M is an r-semisimple right R module if and only if for any ideal I , MI is a direct summand of MR.

Let us give some examples of r-semisimple rings and modules:

(1) Clearly, every and semisimple ring (module) is an r-semisimple ring (module). Moreover, Zp∞ is an r-semisimple module, but not semisimple. (2) Second modules (see [2]) are r-semisimple, but the converse is not true. For example, M = Z2 ⊕ Z3 is an r-semisimple Z-module but is not second. (3) If R is both commutative and r-semisimple ring, then it is semisimple. (4) Let M be a torsion-free and r-semisimple module. Then M is second. Suppose, M is not second, then there exists a proper nonzero right submodule N of M and right ideal I such that MI ⊕ N = M. Then it implies NI = 0, which is a contradiction. (5) If M is a non-singular r-semisimple module on a uniform ring R, then M is second. If M is not second, then there exists a proper nonzero right submodule N and right ideal I such that MI ⊕ N = M. Then NI = 0 and so for any 0 = n ∈ N, I ≤ annR(n), but this contradicts the fact that M is non-singular. (6) Let R be a and r-semisimple ring. Then R is second and hence, it is simple. Moreover, xaR = xR for every minimal left ideal xR and any nonzero element a. Thus, if R is not a , then Soc(RR) = 0. (7) Let R be a uniform r-semisimple ring. Then R is simple.

Lemma 2.2. Let R be an r-semisimple ring. Then R is a .

Proof. R is an r-semisimple ring, so J(R) is a direct summand of R. Since J(R) is a small submodule of R,wehaveJ(R) = 0. So, R is semiprimitive. 

Example 2.3. The inverse of Lemma 2.2 is not true. Let R = Z. Then J(Z) = 0, but Z is not an r-semisimple ring. Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:20 Page 3 of 10 20

COROLLARY 2.4

If R is an artinian and r-semisimple ring, then R is semisimple.

Proof. It follows from Proposition 15.16 in [1]. 

Lemma 2.5. Let R be an r-semisimple ring and has at least three maximal right ideals. If R has a maximal and minimal right ideal I , then RI = R.

Proof. Let I, I1, I2 be three different maximal right ideals of R. Assume that RI = R. Then RI = I , hence I is a two sided ideal of R. Since I is both maximal and , I ∩ I1 = I ∩ I2 = 0. Therefore I1 I = I2 I = 0 and I ⊆ RI = (I1 + I2)I ⊆ I1 I + I2 I = 0. But this is a contradiction. So RI = R. 

COROLLARY 2.6

Let R be an r-semisimple ring and every proper right ideal be maximal. If R has at least three right ideals, then R is simple.

Lemma 2.7. Let R be r-semisimple and for every proper right ideal I, RI = R. Then R is semisimple.

Proof. Let I be any proper right ideal of R. Then I cannot be an essential ideal of R.By Lemma 1 in [8], either I is a direct summand or there exists a proper essential right ideal L of R containing I which implies that RL = R. Then I is a direct summand of R, whence R is semisimple. 

Lemma 2.8. Let R be r-semisimple and u · dim(RR)<∞. Then Soc(RR) is a direct summand of R.

Proof. We can assume u ·dim(RR) ≤ 2. Then Soc(RR) = I1 ⊕ I2 where all Ii are minimal right ideals for i = 1, 2. Since J(R) = 0, there exists a maximal right ideal M1 of R such that M1 ∩ I1 = 0 and so M1 ⊕ I1 = R. Therefore Soc(M1) is simple and J(M1) = 0. Then there exists a maximal right submodule M2 of M1 such that Soc(M1) ∩ M2 = 0 and Soc(M1) ⊕ M2 = M1. Thus R = I1 ⊕ Soc(M1) ⊕ M2 = Soc(RR) ⊕ M2. 

Lemma 2.9. Infinitely many of r-semisimple modules is r-semisimple if every ideal of R is a finitely generated left ideal. Moreover, a direct sum of r-semisimple modules is r-semisimple too.

Proof. See [1], page 174. 

Example 2.10. Let R = Z2×Z3×···×Zp×···; p is prime (ring with an identity element). Then R is not semisimple. Since every commutative r-semisimple ring is semisimple, R can not be an r-semisimple ring. But, clearly, R is an r-semisimple Z-module. Moreover, if R = Z2 ⊕ Z3 ⊕···⊕Zp ⊕···; p is prime (which is a ring without identity), then it is an r-semisimple Z-module, too. 20 Page 4 of 10 Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:20

Example 2.11. Let M = ZZ. Then E(M) = Q. It is clear that E(M) is an r-semisimple Z- module. But M is notr-semisimple. This example shows that, submodule of ar-semisimple module need not be r-semisimple. Moreover, it is not extension closed which follows from the exact sequence 0 → Z2 → Z4 → Z2 → 0.

The following example shows that there exists a simple nonreduced r-semisimple ring with zero divisors and with only three both maximal and minimal right ideals, where RI = R for all nonzero right ideals I .

Example 2.12. Let R = M2(Z2). R has zero divisors. The proper right ideals of R are       ab 00 I = |a, b ∈ Z , I = |a, b ∈ Z 1 00 2 2 ab 2    ab I = |a, b ∈ Z . 3 ab 2

All I1, I2, I3 are maximal and minimal.

PROPOSITION 2.13

Let R be a reduced r-semisimple but not domain ring. Then R is not simple and has a nonzero proper ideal being a direct summand.

Proof. Since R is not domain, there exists at least one a ∈ R such that r(a) = 0. Since R is reduced r(a) = l(a) = 0. Assume that RaR = R. Then 1R = ri asi where ri , si ∈ R. Since there exists at least one 0 = x ∈ r(a), x = 1R x = ri asi x = 0for 2 (asi x) = 0. This is a contradiction. Therefore, RaR = R. Then R is not simple. Since R is r-semisimple, RaR is a direct summand. 

Lemma 2.14. Let M be an r-semisimple module. Then every direct summand of M is also an r-semisimple module.

Proof. Let M = M1 ⊕ M2 and I be any right ideal of R. Since M is r-semisimple, MI is a direct summand of M. So, MI = M1 I ⊕ M2 I and for some N ≤ M, M = MI ⊕ N = M1 I ⊕ M2 I ⊕ N. It means that M1 = M1 ∩ M = M1 I ⊕ (M1 ∩ (M2 I ⊕ N)) and M1 is r-semisimple. 

Lemma 2.15. Let M and N be two right R-modules such that M =∼ N. If M is r- semisimple, then N is also an r-semisimple R-module.

Proof. Let f : M → N be an R-isomorphism. Then f (M) = N. Since for any right ideal I of R, M = MI ⊕ K for some submodule K of M, f (M) = f (M)I + f (K ) = N.Let y ∈f (M)I ∩ f (K ). Theny = f (mi )ri = f (z) where mi ∈ M, ri ∈ I, z ∈ K .So f ( mi ri − z) = 0, thus mi ri = z ∈ MI ∩ K = 0. Then N = NI ⊕ f (K ), hence N is also r-semisimple.  Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:20 Page 5 of 10 20

COROLLARY 2.16

Let MR be a torsion free module and R be an r-semisimple ring. Then for all x ∈ M, xR is an r-semisimple R-module.

Proof. Since M is torsion free, for any x ∈ M, xR =∼ R and so, by Lemma 2.15, xRis an r-semisimple R-module. 

PROPOSITION 2.17

Let R be a ring such that it has more than one nonzero proper right ideal and all of them are maximal. Then R is an r-semisimple ring.

Proof. Let T1 and T2 be maximal right ideals. Then they are also minimal ideals. Thus R = T1 ⊕ T2 and so R is r-semisimple. 

COROLLARY 2.18

Let R be a ring with two nonzero proper right ideals such that J(R) = 0. Then R is an r-semisimple ring.

Proof. Let I1, I2 be two proper right ideals of R. Since J(R) = 0, both of them should be maximal and I1 ∩ I2 = 0. By Proposition 2.17, R is an r-semisimple ring. 

Now, let us give some characterization of r-semisimple rings by its proper right ideals. Let R be a ring. (i) Suppose R has only one nonzero proper right ideal, then J(R) = 0, so R is not an r-semisimple ring by Lemma 2.2. (ii) Suppose R has two nonzero proper right ideals M1 and M2 where J(R) = 0. By Corollary 2.18, R is r-semisimple. (iii) Suppose R has three nonzero proper right ideals Mi , 1 ≤ i ≤ 3. Then R is r-semisimple as follows: (a) If all three right ideals are maximal, then R is r-semisimple by Proposition 2.17. (b) If there are only two proper maximal right ideals, then two cases may arise: (1) Both two maximal right ideals include another third proper right ideal. Then J(R) = 0 and R is not r-semisimple. (2) Only one of two maximal right ideals includes the third right ideal. We are going to show that, in this case, there is a contradiction. For instance, assume M2  M3 and M1, M3 are maximal. Then M1 ⊕ M2 = R and M1 ⊕ M3 = R. Let a ∈ M3 \ M2. We may write a = m1 + m2 ∈ R where m1 ∈ M1, m2 ∈ M2. Then a − m2 = m1 ∈ M3 ∩ M1 = 0 and a = m2 which is a contradiction. (c) If there is only one proper maximal right ideal, then J(R) = 0 and R is not r-semisimiple.

(iv) If R has only four nonzero proper right ideals M1, M2,M3 and M4 such that J(R) = 0, then it is r-semisimple as follows: 20 Page 6 of 10 Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:20

(a) If all four right ideals are maximal then R is r-semisimple by Lemma 2.17. (b) If there are only three proper nonzero maximal right ideals M1, M2 and M3, then three cases may arise:

(1) All three maximal ideals include M4. Then J(R) = 0 and R is not r- semisimple. (2) Only two maximal ideals include M4. Then M3 ⊕ M4 = R and so R is r-semisimple. (3) Only M1 includes M4. Then M3 ⊕ M4 = R and so R is r-semisimple.

(c) If there are only two nonzero proper maximal right ideals M1 and M2, then three cases may arise:

(1) M3 and M4 are not comparable, then four cases may arise:

(i1) M3 and M4 ≤ M1. Then M2 ⊕ M3 = R and so R is r-semisimple. i2) M3, M4 ≤ M1 and M4 ≤ M2. Then J(R) = 0 and R is not r-semisimple. (i3) M3, M4 ≤ M1 and M3, M4 ≤ M2. Then M3 ⊕ M4 = M1 = M2.Thisis a contradiction. (i4) M3 ≤ M1 and M4 ≤ M2. Then M3 ⊕ M4 = R and so R is r-semisimple.

(2) M4 ≤ M3 ≤ M1. Then M2 ⊕ M4 = R and so R is r-semisimple. (3) M4 ≤ M3 ≤ M1 and M3 ≤ M2. Then J(R) = 0 and R is not r-semisimple. (d) If there are only one proper maximal right ideal, then J(R) = 0 and R is not r-semisimple.

Lemma 2.19. Let R be a ring with Soc(RR) = 0 and J(R) = 0. If every non-essential maximal right ideal of R is r-semisimple, then so is R.

Proof. Assume that every maximal right ideal is essential in R. Then Soc(RR) ⊆ J(R),so Soc(RR) = J(R) = 0. It implies that there exists a right ideal M such that M is maximal but not essential. Thus there exists IR ≤ R such that M ∩ I = 0 and R = M ⊕ I . Since I is a simple right ideal, I is r-semisimple. Therefore R is r-semisimple. 

There exists some non r-semisimple modules such that every submodule is r-semisimple such as Z4 on Z and the example of Teply on (see Lemma2in[5]).

3. r-Semisimple rings In this section we investigate the structure of ring and modules whenever R is r-semisimple.

Lemma 3.1. Let R be an r-semisimple ring. Then for any ideal I of R, l(I ) = r(I ) = Ann(I ).

Proof. Since R is r-semisimple, R is semiprimitive. Il(I ))2 = 0 implies Il(I ) = 0, so l(I ) ⊆ r(I ). Similarly (r(I )I )2 = 0 implies r(I ) ⊆ l(I ). Thus l(I ) = r(I ) = Ann(I ).  Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:20 Page 7 of 10 20

Lemma 3.2. Let R be an r-semisimple ring and I be a right ideal of R. Then there exists a right ideal K of R such that K = eR, R = RI ⊕ K, l(RI) = Ann(RI) = eR, r(eR) = RI, lr(RI) = rl(RI) = R, where e2 = e.

Proof. Since R is r-semisimple, R = RI ⊕ K for some right ideal K of R. Since RI is an ideal, RI = fRfor some f ∈ R. Thus, K = (1 − f )R.NowK ⊆ l(RI) implies l(RI) = l(RI) ∩ R = K ⊕ (l(RI) ∩ RI) = K since l(RI) ∩ RI = 0. Similarly, we may find r(K ) = RI. By Lemma 3.1, lr(RI) = rl(RI) = RI. 

PROPOSITION 3.3

For any ring R, the following are equivalent: (i) R is an r-semisimple. (ii) For every right ideal I of R, there exists an idempotent e ∈ R such that l(RI) = eR, r(eR) = RI and 1 − e ∈ RI.

Proof.

(i) ⇒ (ii). By Lemma 3.2. (ii) ⇒ (i). Since J(R) is an ideal of R, there exists e = e2 ∈ R such that 1−e ∈ RJ(R) = J(R). Then e is invertible, so e = 1. By (ii), r(eR) = r(R) = 0 = RJ(R) = J(R).Now, let I be any right ideal of R. By (ii), there exists f 2 = f ∈ R such that 1 − f ∈ RI. Since J(R) = 0, RI ∩ fR= 0. This implies that R = RI ⊕ fR. So, R is r-semisimple. 

Lemma 3.4. Let R be an r-semisimple ring. Then for every P,Ann(P) is a simple ring.

Proof. By Lemma 3.2, R = P ⊕ Ann(P). Assume that Ann(P) has a proper ideal L which is also an ideal of R. Since Ann(P) is also an r-semisimple module, Ann(P) = L ⊕ Ann(L). Then there exist 0 = a ∈ L − Ann(L) and 0 = b ∈ Ann(L) − L such that aRb = 0 ⊆ P. Since P is a prime ideal, either a ∈ P or b ∈ P. This is a contradiction. Thus Ann(P) is simple. 

Theorem 3.5. R is r-semisimple if and only if it is a direct sum of simple rings.  Proof. Let S = {Ann(P) : P is a prime ideal of R}. Since S is an ideal of R, by Lemma 3.2, R = S ⊕ Ann(S). Since Ann(S) ⊆∩Ann(Ann(P)) =∩P = 0 and J(R) contains the prime radical by [3], Ann(S) = 0. Thus R is a direct sum of simple rings by Lemma 3.4. The converse implication is obvious since R (which has an identity) is a finite direct sum of simple rings. 

PROPOSITION 3.6

Let R be an r-semisimple ring. Then every injective R-module M is r-semisimple.

Proof. Let I be a right ideal of R. Then RI is a direct summand of R,soR = RI ⊕ J for some right ideal J of R.Let f : I → MI be any homomorphism and i : I → R 20 Page 8 of 10 Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:20

= RI ⊕ J, j : MI → M be injection maps. Since M is an injective module, there exists an R-homomorphism β : R = RI ⊕ J → M such that β |I = jf. Define β : R → MI by β(x + y) = β(x), where x ∈ RI, y ∈ J. It is clear thatβ is an R−homomorphism. Since x = ri ai ; ri ∈ R, ai ∈ I, β(x + y) = β(x) = β( ri ai ) = β(ri )ai ∈ MI and βi = f . Thus MI is injective and so MI is a direct summand of M. This implies that M is an r-semisimple module. 

Every injective Z-module is r-semisimple. But Z is not r-semisimple. Thus we can give the following theorem without proof.

Theorem 3.7. Consider the following statements: (i) Every projective R-module is r-semisimple. (ii) Every free R-module is r-semisimple. (iii) R is an r-semisimple ring. (iv) Every injective R-module is r-semisimple. Then (i) ⇔ (ii) ⇔ (iii) ⇒ (iv). In general, (iv) ⇒ (iii) does not hold.

The following examples show that not all injective modules are r-semisimple.

Example 3.8. Let R be a commutative ring Z4. Then R is self-injective but is not an r-semisimple R-module.

Example 3.9. Let R be a 2 × 2 upper triangular matrix on division ring D. Then M2(D)R is an injective envelope of RR.LetEi, j be a matrix such that if i = j, then aij = 1, otherwise 0. Then M2(D)E1,2 R is not a direct summand of M2(D)R. Thus M2(D)R is injective on noncommutative ring R, but it is not r-semisimple.

Theorem 3.10 (Corollary 7.41 in [7]). The following are equivalent for a ring R: (i) R is right self injective. (ii) R is right weakly continuous and R ⊕ R is C S as a right module. (iii) R is right C2 ring and R ⊕ R is C S as a right module.

COROLLARY 3.11

If an r-semisimple ring R is self-injective, then it is a regular ring. Then every finitely generated right ideals are r-semisimple.

Proof. Since R is r-semisimple, J(R) = 0. By Theorem 3.10 R is a regular ring. 

The converse of the last corollary may not be true. For example, infinite Boolean rings are regular without having to be semisimple. But commutative r-semisimple rings are semisimple.

COROLLARY 3.12

If R is an r-semisimple, self-injective and right , then R is semisimple. Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:20 Page 9 of 10 20

For r-semisimple ring, it is not necessary to be self-injective. The first A1(k) is an r-semisimple and Notherian ring, but not semisimple.

Theorem 3.13 (Example 8.15 in [6]). For any ring R, the following are equivalent: (i) R is semisimple. (ii) R is von Neumann regular and right Kasch. (iii) R is Jacobson semisimple (i.e. J(R) = 0) and right Kasch. (iv) R is semiprime and right Kasch.

We can give the following results of the above theorem.

COROLLARY 3.14

Let R be an r-semisimple and right Kasch ring. Then R is semisimple.

Theorem 3.15. Let R be an r-semisimple ring and T be any maximal right ideal of R. Then either T is an idempotent or direct summand of R.

Proof. Let T be a maximal right ideal of R. Then R = RT ⊕ J for some right ideal J of R and T ≤ RT. Since T is the maximal right ideal in R, either T = RT or RT = R.IfT = RT, then T ≤⊕ R. Assume RT = R. Then T (RT) = TR= T . Hence T (RT) = (TR)T = T 2 implies T = T 2. 

COROLLARY 3.16

Let R be an r-semisimple ring such that for any maximal right ideal T , T 2 = T . Then R is a right Kasch ring.

Proof. By Theorem 3.15, T = RT ≤⊕ R for any maximal right ideal T of R.SoR is right Kasch. 

Example 3.17. Let K (x) be a field of rational functions over a field K and

α : K (x) → K (x), α(x) = x2 be an . Clearly,

Im(α) = K (x2).

Let R bethematrixringoftheform    α(a) b R = | a, b ∈ K (x) . 0 a

Then   0 K (x) I = 00 is the unique nontrivial right ideal of R. It is clear that R is right Kasch but not r-semisimple. 20 Page 10 of 10 Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:20

Acknowledgement The authors thank the referee and Prof. Christian Lomp for their valuable suggestions.

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Communicating Editor: B Sury