Study of Formal Triangular Matrix Rings

International Journal of Scientific & Engineering Research, Volume 5, Issue 12, December-2014 ISSN 2229-5518 21

STUDY OF FORMAL TRIANGULAR MATRIX RINGS

MRS Sunita Salunke SCOE Kharghar Navi Mumbai Email:[email protected]

Abstract—There has been a continuous study on formal triangular matrix rings. In this present study 2. Generalities on formal triangular matrix of formal triangular matrix rings few ring theoretic rings characteristic properties of formal triangular matrix rings have been studied in detail. Some definitive re- Proposition II.1. (1) The set of maximal right sults are verified on these rings concerning properties ideals of T is given by such as being respectively left Kasch, right mininjective, clean, potent or a ring of stable rank≤ n. The {(I ⊕ M) ⊕ K| either I = concepts of a strong left Kasch ring and a strong right A and K is a maximal right ideal of B mininjective ring are introduced and it is determined or I is a maximal right ideal of A and K = when the triangular matrix rings are respectively B} strong left Kasch or strong right mininjective (2) The set of minimal right ideals of T is the I.INTRODUCTION union of the two sets, {W ⊕ 0| W a simple submodule of (A ⊕ All the rings considered will be associative rings M) } and with identity, all the modules considered will be A {0 ⊕ K| with 0 the zero submodule of (A ⊕ unital modules. For any ring R the category of right M) and K a minimal right (resp left) R-modules will be denoted by Mod- A ideal of B satisfying KM = 0}. R(resp R-Mod). Let A and B be two given rings and M a left B IJSERright A bimodule. The formal trian- Proof. (1) Let W ⊕K with W ≤ (A⊕M)A,K ≤ A 0 gular matrix ring T = has its elements BB and KM ≤ W be a maximal right ideal of T . MB If K 6= B then choosing a maximal right ideal K0 a 0 0 0 formal matrices where a ∈ A, b ∈ B of BB such that K ≤ K , we see that (A⊕M)⊕K m b is a maximal right ideal of T , but since W ⊕ K is and m ∈ M with addition co-ordinate wise and 0 a 0 a0 0 maximal we get W ⊕ K = (A ⊕ M) ⊕ K , then multiplication given by . = 0 m b m0 b0 this implies that W = A ⊕ M and K = K ; that is aa0 0 I = A and K is a maximal right ideal of B. On the . The present is devoted to ma0 + bm0 bb0 other hand if K = B then 0 ⊕ KM ≤ W implies study of properties of Formal Triangular rings. that 0 ⊕ BM ≤ W ; that is W = I ⊕ M for some Specifically we characterize all maximal (resp min- I ≤ AA. The maximality of W ⊕ K implies that I imal) one sided ideals of T , use this to characterize is a maximal right ideal of A. Thus any maximal the jacobson radical J(T ) and the right (resp right ideal of T has to be either (A⊕M)⊕K with K a maximal right ideal of B or (I ⊕M)⊕B with left) socle socleTT (resp socT T ). We determine necessary and sufficient conditions for T to be I a maximal right ideal of A. semi-primary, left(or right) perfect, semi local etc. (2) Let W ⊕ K with W ≤ (A ⊕ M)A,K ≤ BB and KM ≤ W be the minimal right ideal of T . II.RING THEORETIC PROPERTIES OF T For any V ≤ W , V ⊕ 0 is a right ideal of T , (1) Left ideals of T are of the form I1 ⊕ I2 maximality of W ⊕ K implies that either K = 0 where and W is a simple submodule of (A ⊕ M)A or K I1 module and 0 ⊕ K for any minimal right ideal of where B with KM = 0 constitute minimal right ideal of J2 < BB,J1 5 (A ⊕ M)A and J2M ⊆ J1.[5] T . IJSER © 2014 http://www.ijser.org International Journal of Scientific & Engineering Research, Volume 5, Issue 12, December-2014 ISSN 2229-5518 22

J(A) 0 a 0 J(A) 0 Corollary II.1. [1] (a) J(T ) = and ∈ = J(T ). So MJ(B) m b MJ(B) and f is injective. soc(AA) 0 For any (a + J(A), b + J(B)) there exists (b) soc(TT ) = where soc(MA) soc(LB) a 0 a 0 ∈ T/J(T ) such that f = L = lB(M) 0 b m b (A + J(A), b + J(B)), and so f is surjective. Proof. T = A ⊕ M ⊕ B (ii) [Lifting Idempotents: If I is a non empty ( a). According to the definition subset of R then an idempotent of R/I or an J(R) = Intersection of maximal right ideals of idempotent modulo I is an element x ∈ R such that R. Now maximal right ideals of T are x2 − x ∈ I. Such an element is said to be lifted {I ⊕ M ⊕ B| I maximal right ideal of A} ∪ provided that there exists an idempotent y ∈ R {A ⊕ M ⊕ K| K maximal right ideal of B} so such that y − x ∈ I and we say that x has been J(T ) = J(A) ⊕ M ⊕ J(B) lifted to y.] J(A) 0 a 0 = . Let ∈ T be an idempotent mod MJ(B) m b (b) According to the definition, for any ring R J(T ) which can be lifted in T . This im- 2 soc(RR) = sum of all minimal right ideals a 0 a 0 A 0 plies that − = of R. For T = , by Prop II.1 the m b m b MB a2 − a 0 minimal right ideals of T are the union of two sets ma + bm − m b2 − b {W ⊕0| W a simple submodule of (A⊕M) }∪ A J(T ) a2 − a ∈ {0⊕K| K is a minimal right ideal of B satisfying which belongs to ; therefore J(A), b2 − b ∈ J(B) a KM = 0}. Now soc(T ) = P all minimal right ; that is is an idempotent T J(A) b J(B) ideals of T = P((W ⊕ 0) + (0 ⊕ K)) with W mod and is an idempotent mod . And a 0 and K as above and KM = 0. can be lifted in T implies that there m b Since K is minimal right ideal of B such a0 0 a 0 exists ∈ T such that − that KM = 0 which implies that K ⊆ lB(M). m0 b0 m b Therefore a0 0 a − a0 0 = ∈ J(T ) ; that soc(TT ) = soc(AA)⊕ soc(MA)⊕ soc(lB(M)) m0 b0 m − m0 b − b0 IJSER is a − a0 ∈ J(A) and b − b0 ∈ J(B), so a is lifted soc(A ) 0 = A to a0 and b is lifted to b0. soc(M ) soc(L ) A B Conversely, if a and b are idempotents mod J(A) where L = lB(M). and J(B) which can be lifted in A and B re- a 0 spectively to a0 and b0. Then is an m b Corollary II.2. (i) idempotent mod J(T ) for any m and it can be a 0 a0 0 +J(T ) (a+J(A), b+J(B)) is a ring lifted in T to . m b 0 b0 isomorphism of T/J(T ) with A/J(A) × B/J(B). (ii) Idempotents mod J(T ) can be lifted in T if and Corollary II.3. T is semilocal if and only if A and only if idempotents mod J(A) can be lifted in A B are semilocal. [12] and idempotents mod J(B) can be lifted in B . Proof. Suppose T is semilocal. This implies that Proof. (i) Define f : T/J(T ) → (A/J(A)) × every ideal of T/J(T ) is a direct summand of (B/J(B)) by T/J(T ). Now by Corollary II.2(i) we have a 0 T/J(T ) =∼ (A/J(A))×(B/J(B)) =∼ (A/J(A))⊕ f + J(T ) = (a + J(A), b + J(B)) m b (B/J(B)) and g :(A/J(A)) × (B/J(B)) → T/J(T ) by So every ideal of (A/J(A)) × (B/J(B)) is a 0 a direct summand of T/J(T ). The ideals of g(a + J(A), b + J(B)) = . m b (A/J(A))×(B/J(B)) are of the form I¯×J¯ where a 0 I and J are ideals of A and B respectively. So Now f + J(T ) = (J(A),J(B)) m b each I¯ × J¯ is a direct summand of T/J(T ); that this implies that is I¯ is a direct summand of T/J(T ) and J¯ is a (a + J(A), b + J(B)) = (J(A),J(B)) so a ∈ direct summand of T/J(T ) hence I¯ is a direct J(A), b ∈ J(B) summand of A/J(A) and J¯ is a direct summand of

aλ 0 a B/J(B). Therefore we get A/J(A) and B/J(B) { | λ ∈ A}, we have uT = A+ are semsimple and so A and B are semilocal. Proof mλ 0 m 0 0 for the converse is on the similar lines. . bM bB Corollary II.4. T is semi-perfect if and only if A λ 0 Proof. For any ∈ T ; we have and B are semi-perfect. n µ a 0 λ 0 aλ 0 Proof. By Corollary II.3 we get T/J(T ) is semilo- = m b n µ mλ + bn bµ cal if and only if A/J(A) and B/J(B) are aλ 0 0 0 semilocal. Also from corollary II.2(ii) we get that = + . mλ 0 bn bµ idempotents mod J(T ) can be lifted in T if and a 0 0 only if idempotents mod J(A) can be lifted in A Hence uT ⊆ A + . and idempotents mod J(B) can be lifted in B. So m bM bB From T is semi-perfect if and only if A and B are semi- a 0 λ 0 aλ 0 = perfect. m b 0 0 mλ 0 a 0 0 0 0 0 = Corollary II.5. T is right (resp left) perfect if and m b n 0 bn 0 only if A and B are right(resp left) perfect. a 0 0 0 0 0 and = ; we see m b 0 µ 0 bµ Proof. Proof is clear by Corollary II.3. a that A ⊆ uT, m Corollary II.6. T is semi-primary if and only if A 0 0 0 0 and B are semi-primary. ⊆ uT and ⊆ uT . bM 0 0 bB Proof. Assume that A and B are semi-primary. a 0 Because of corollary II.3, to show that T is semi- Corollary II.7. Let u = ∈ T . Then uT primary we need only to show that J(T ) is nilpo- m b is a minimal right ideal of T if and only if one of tent. Since J(A) and J(B) are nilpotent we can the following is true. ﬁnd an integer k ≥ 1 with (J(A))k = 0 and (1) b = 0 and (a, m) generates a simple submodule (J(B))k = 0. of (A ⊕ M)A. ¯ J(AIJSER) 0 ¯ 0 0 If I = and I = , (2) a = 0, m = 0, bB is a minimal right ideal B 1 M 0 2 MJ(B) and bM = 0. then I¯1 and I¯2 are two sided ideals of T . Now since k+1 ¯ k+1 (J(A)) 0 Proof. From (2) of Proposition II.1 and Lemma I2 = k , It is easy to see that M(J(A)) 0 II.1 we see that uT is minimal right ideal of T if ¯ k+1 ¯ k+1 ¯ ¯ I1 = 0 = I2 . From J(T ) = I1 + I2, we and only if one of the following is valid. immediately conclude that J(T ) is nilpotent. a 0 0 (i) A + = W ⊕ 0 where W Now to prove the converse; again because of m bM bB Corollary II.3 we have only to show that J(T ) is is a simple submodule of (A ⊕ M)A. a 0 0 nilpotent implies that J(A) and J(B) are nilpotent. (ii) A + = 0 ⊕ K where K J(T )k = m bM bB (J(A))k 0 is a minimal right ideal of B satisfying KM = 0. M(J(A))k−1 + (J(B))k−1M (J(B))k Now (i) is valid if and only if b = 0 and (a, m) 0 0 generates a simple submodule of (A ⊕ M)A and = . 0 0 (ii) is valid if and only if a = 0 , m = 0 and bB is a minimal right ideal of B and bM = 0. We will now give an exact description of We take the analogue of Proposition II.1 is valid the principal right ideal uT of T where u = a 0 for left ideals of T. We state it without proof. and obtain necessary and sufﬁcient con- m b Proposition II.2. (1) The set of maximal left ditions for uT to be a minimal right ideal of T. ideals of T is given by a 0 {I ⊕ (M ⊕ K)| either I = Lemma II.1. Let u = ∈ T. Writing m b A and K is a maximal left ideal of B a or I is a maximal left ideal of A and K = B}. A for m (2) The set of minimal left ideals of T is the union IJSER © 2014 http://www.ijser.org International Journal of Scientific & Engineering Research, Volume 5, Issue 12, December-2014 ISSN 2229-5518 24

of the two sets a1 ∈ rA(I) also Bb1 = 0 gives us bb1 = 0 for {0 ⊕ V | V a simple submodule of B(M ⊕ B)} ∪ all b ∈ B so b1 = 0. Since Ma1 is a simple {I ⊕ 0| with 0 the zero submodule of B(M ⊕ module and Ma1 + Bm1 ⊆ M ⊆ M ⊕ B, B) and I a minimal so Ma1 + Bm1 = 0 implies that left ideal of A satisfying MI = 0}. Ma1 = 0 and Bm1 = 0. Now Ma1 = 0 In particular, we see from (2) above that means a ∈ r (M) and Bm = 0 means 1 A 1 Soc(AH) 0 bm1 = 0 for all b ∈ B proving m1 = 0. Therefore Soc(T T ) = , where Soc(BM) Soc(BB) a1 0 t = where a1 ∈ rA(I) ∩ rA(M) and H = rA(M). 0 0 rA(I) ∩ rA(M) 0 III.LEFT KASCH NATURE,RIGHT hence r (I˜) = . T 0 0 MININJECTIVITY ˜ 3. Left (right) Kasch and strong left (strong Also rT (I) is a minimal right ideal of T if and right) Kasch rings. only if rA(I) ∩ rA(M) is minimal right deal of A. (2) r (K˜ ) = {t ∈ T |Kt˜ = 0}. Let T Deﬁnition III.1. A ring R is said to be left Kasch a1 0 t = ∈ rT (K˜ ), then K.t˜ = if rR(I) 6= 0 for every maximal left ideal I of R. m1 b1 A 0 a 0 Deﬁnition III.2. R will be called a strong left 0; that is 1 = 0, so Kasch ring if r (I) is a minimal right ideal of MK m1 b1 R Aa 0 R for every maximal left ideal I of R. 1 = 0. Thus we have Ma1 + Km1 Kb1 This section is devoted to studying conditions Aa1 = 0 and Kb1 = 0 which gives a1 = under which the triangular matrix ring T = 0 and b ∈ r (K). And also Ma + Km = 1 B 1 1 A 0 0 Km = 0; m ∈ r (K). is left Kasch (resp strong left Kasch). gives 1 that is 1 M There- MB 0 0 fore t = where m ∈ r (K) and m b 1 M r (I) = {r ∈ R| a.r = 0 for all a ∈ I}. 1 1 R 0 0 b1 ∈ rB(K). So rT (K˜ ) = . Lemma III.1. (1) Let I be any left ideal of A. rM (K) rB(K) For the left ideal rT (K˜ ) is a minimal right ideal of T if either I 0 I˜ = we have r (I˜) = rB(K) = 0 and rM (K) is a simple submodule MB T IJSERof MA or rB(K) is a minimal right ideal of B r (I) ∩ r (M) 0 A A . satisfying rB(K)M = 0 = rM (K) by using (2) of 0 0 Prop II.1 In particular rT (I˜) is a minimal left ideal of T if and only if Theorem III.1. (i) If T is left Kasch, so is A. rA(I) ∩ rA(M) is a minimal right ideal of A. Further rA(I) ∩ rA(M) 6= 0 for every maximal (2) Let K be any left ideal of B. For the left left ideal I of A. A 0 (ii) Suppose T is left Kasch and soc(BM) = 0. ideal K˜ = of T we have r (K˜ ) = MK T Then B is left Kasch. 0 0 (iii) Let A and B be left Kasch. Further assume . In this case rT (K˜ ) is a rM (K) rB(K) that rA(I) ∩ rA(M) 6= 0 for every maximal left minimal right ideal of T if and only if one of the ideal I of A. Then T is left Kasch. following is valid: Either Proof. (i) Assume T is left Kasch. Let I be (a) r (K) = 0 and r (K) is a simple submodule B M I 0 of M . Or any maximal ideal of A, then I˜ = A MB (b) r (K) is a minimal right ideal of B and B is a maximal left ideal of T and so r (I˜) 6= r (K)M = 0 = r (K). T B M 0. By using Lemma III.1 we get r (I˜) = T Proof. r (I˜) = {t ∈ T | I.t˜ = 0}. rA(I) ∩ rA(M) 0 (1) T 6= 0 and hence we get a 0 0 0 Let t = 1 . Now I.t˜ = 0 implies m1 b1 rA(I) ∩ rA(M) 6= 0 for every maximal left ideal I 0 a 0 I of A. that 1 = 0; that is MB m1 b1 (ii) Assume that T is left Kasch and let K be any Ia1 0 ˜ A 0 = 0. Now Ia1 = 0 maximal left ideal of B. Then K = Ma1 + Bm1 Bb1 MK implies that aa1 = 0 for all a ∈ I and so is a maximal left ideal of T. Hence rT (K˜ ) =

0 0 6= 0. Since rM (K) ⊂ all the maximal left ideals of T, we see that T is rM (K) rB(K) strong left Kasch. Soc(BM) we get rM (K) = 0. Hence rB(K) 6= 0, (ii) Let I be any maximal left ideal of A. Then showing that B is left Kasch. I˜ is a maximal left ideal of T and rT (I˜) = (iii) Assume that A and B are left Kasch and that r (I) ∩ r (M) 0 A A , since T is strong left rA(I) ∩ rA(M) 6= 0 for every maximal left ideal 0 0 ˜ I of A. For any maximal left ideal I of T of Kasch rA(M) ∩ rA(I) is a minimal right ideal I 0 the form I˜ = , the equality r (I˜) = of A. We have Soc(AA) ⊆ rA(M) and rA(I) ⊆ MB T Soc(AA) we get rA(I) ∩ rA(M) = rA(I). On rA(I) ∩ rA(M) 0 shows that rT (I˜) 6= 0. the other hand if rA(M) is essential in AA and 0 0 r (I) ⊆ Soc(A ) so we get r (I) ∩ r (M) = ˜ ˜ A A A A For the maximal left ideal K of the form K = r (I), thus r (I) is a minimal right ideal of A. A 0 A A with K any maximal left ideal of This shows that A is a strong left Kasch ring. MK 0 0 Let K be any maximal left ideal of B. The as- B the equality r (K˜ ) = T r (K) r (K) sumption that B is left kasch implies that rB(K) 6= M B A 0 shows that r (K˜ ) 6= 0 since r (K) 6= 0. This 0. Since K˜ = is a maximal left ideal T B MK T proves that is left Kasch. of T and T is strong left Kasch, it follows that 0 0 We remark that in case r (M) is an essential r (K˜ ) = is a minimal right A T r (K) r (K) right ideal of A, from (iii) above we see that A, B M B ideal of T. Since rB(K) 6= 0 from Prop II.1 we left Kasch implies that T is left Kasch. see that rB(K) is a minimal right ideal of B and rM (K) = 0 = rB(K)M. Theorem III.2. (i) Assume that A and B are strong left Kasch and that r (M) is an essential A There is an obvious deﬁnition of a right (resp right ideal of A. Then T is strong left Kasch strong right) Kasch ring. A ring R is right (resp provided r (K) = r (K)M = 0 for all maximal M B strong right ) Kasch ring if l (I) 6= 0 (resp l (I) left ideals K of B. R R is a minimal left ideal of R) for every maximal (ii) Let B be left Kasch. Assume that either right ideal of R. We state the following analogues Soc( A) ⊆ r (M) or that r (M) is essential A AIJSERA for Theorems III.1 and III.2. in AA and rA(I) ⊆ Soc(AA) for every maximal left ideal I of A. Then T strong left Kasch implies Theorem III.3. (i) If T is right Kasch so is B. that A and B are strong left Kasch and further Further lB(K) ∩ lB(M) 6= 0 for every maximal rM (K) = rB(K)M = 0 for all maximal left ideals right ideal K of B. K of B. (ii) Suppose that T is right Kasch and Soc(MA) = 0. Then A is right Kasch. Proof. (i) Assume that A, B are strong left Kasch, (iii) Let A and B be right Kasch. Further assume r (M) is an essential right ideal of A and A that l (K) ∩ l (M) 6= 0 for every maximal right r (K) = r (K)M = 0 for all maximal left ideals B B M B ideal K of B. Then T is right Kasch. K of B. When I is a maximal left ideal of A, since A is strong left Kasch rA(I) is a minimal Proof. (i) let K be any maximal right ideal of B right ideal of A. Since r (M) is essential right A 0 A then K˜ = is a maximal right ideal of ideal of A we get rA(M) ∩ rA(I) = rA(I). From MK Lemma III.1 we see that rT (I˜) is a minimal right T. Since T is right kasch we have I 0 l (K˜ ) 6= 0. Now to ﬁnd l (K˜ ), let ideal of T where I˜ = . Also, when T T MB a1 0 t = ∈ lT (K˜ ). This K is a maximal left ideal of B, from the strong m1 b1 left Kasch nature of B; we see that r (K) is a a 0 A 0 B implies that 1 = minimal right ideal of B. From (2) of Lemma III.1 m1 b1 MK ˜ 0 0 a1A 0 we have rT (K) = , where = 0. Now a1A = 0 rM (K) rB(K) m A + b M b K 1 1 1 A 0 implies a1 = 0 , b1K = 0 implies K˜ = . The assumption rM (K) = MK b1 ∈ lB(K) and m1A + b1M = 0 implies rB(K)M = 0 shows that rT (K˜ ) is a minimal right that m1A = 0 and b1M = 0. So m1A = 0 ideal of T. Since ideals of the form I˜ or K˜ give gives m1 = 0 and b1M = 0 gives b1 ∈ lB(M).

0 0 Therefore lT (K˜ ) = . so lB(K)∩lB(M) is a minimal left ideal of B and 0 lB(K) ∩ lB(M) l (K˜ ) is minimal left ideal of T showing that T ˜ T If T is right Kasch then lT (K) 6= 0 so is strong right Kasch. lB(K) ∩ lB(M) 6= 0 for every maximal right ideal (ii) Let K be any maximal right ideal of B, then ˜ K and so lB(K) 6= 0. Thus B is right Kasch. A 0 K˜ = is a maximal right ideal of (ii) Let I be any maximal right ideal of A, then MK ˜ I 0 0 0 I = is a maximal right ideal of T. T. Now l (K˜ ) = , since MB T 0 l (K) ∩ l (M) B B lA(I) 0 ˜ Now lT (I˜) = . Since T is right T is strong right Kasch lT (K) is a minimal left lM (I) 0 ideal of T ; that is lB(K) ∩ lB(M) is minimal l (I˜) 6= 0, l (I) ⊂ Soc(M ) Kasch T but we have M A left ideal of B. We have lB(K) ⊆ Soc(BB) so we get lM (I) = 0 and so lA(I) 6= 0 for every since K is maximal right ideal of B. In case A. A maximal right ideal of Thus is right Kasch. Soc(BB) ⊆ lB(M) we get lB(K) ∩ lB(M) = I 0 (iii) Let I˜ = be any maximal right lB(K). On the other hand if lB(M) is essential MB in B we get l (K) ⊆ Soc( B) ⊆ l (M) and ideal of T. Now B B B B so l (M) ∩ l (K) = l (K). Thus l (K) is a ˜ lA(I) 0 B B B B lT (I) = . Since A is right Kasch B B lM (I) 0 minimal left ideal of showing that is strong l (I) 6= 0 and so right Kasch. A A 0 I 0 l (I˜) 6= 0. Also if K˜ = is a maximal Now let I˜ = be a maximal right T MK MB right ideal of T then ideal of Tthen 0 0 ˜ lA(I) 0 ˜ lT (I) = . Since T is strong right lT (K) = , which implies lM (I) 0 0 lB(K) ∩ lB(M) Kasch lT (I˜) is minimal left ideal of T. Also since that lT (K˜ ) 6= 0 and so T is right Kasch. A is right Kasch lA(I) 6= 0 and lT (I˜) is minimal Theorem III.4. (i) Assume that A and B are left ideal of T implies from Prop II.2 that lA(I) is minimal right ideal of A and strong right Kasch and that lB(M) is an essential left ideal of B. Then T is strong right Kasch lM (I) = MlA(I) = 0 for all maximal right ideals I of A. provided lM (I) = MlA(I) = 0 for all maximal right ideals I ofIJSERA. (ii) Let A be right Kasch. Assume that either 4. Right mininjective and strong right mininjective rings Soc(BB) ⊆ lB(M) or that lB(M) is essential in BB and lB(K) ⊆ Soc(BB) for every maximal right ideal K of B. Then T is strong right Kasch A ring R is said to be right mininjective if implies that A and B are strong right Kasch and and only if any right R-homomorphism f : I → R of simple right ideal I of R into R extends further lM (I) = MlA(I) = 0 for all maximal right ideals I of A. to a homomorphism g : R → R in Mod-R. Equivalently any such f is of the form f(x) = rx Proof. (i) Assume A, B strong right Kasch and for all x ∈ I for some ﬁxed r ∈ R. Clearly any lB(M) is an essential right ideal of B and lM (I) = right injective ring is right mininjective.[7] MlA(I) = 0 for all maximal right ideals I of A. If R is a ring, a right module MR is called I 0 Let I˜ = be a maximal right ideal of mininjective if for each right ideal K of R, MB every R-morphism γ : K → M extends to R. lA(I) 0 Equivalently if γ = m. is left multiplication by T then lT (I˜) = . Since A is strong lM (I) 0 some element of M. Mininjective left modules are right Kasch lA(I) is minimal left ideal of A and deﬁned similarly. Clearly every injective module given that lM (I) = MlA(I) = 0, so by using Prop is mininjective. [Let Q be an R-module. (Baer’s ˜ II.2 we get that lT (I) is minimal left ideal of T. Criterion) The module Q is injective if and A 0 Also if K˜ = is a maximal right ideal only if for every left ideal I of R any R-module MK homomorphism g : I → Q can be extended to 0 0 of T then l (K˜ ) = . an R-module homomorphism G : R → Q.] Our T 0 l (K) ∩ l (M) B B interest is in the right mininjective rings that is Since B is strong right Kasch lB(K) is a minimal the rings R for which RR is mininjective. left ideal of B, also since lB(M) is essential left ideal of B we have lB(M) ∩ lB(K) = lB(K) and

Lemma III.2. The following conditions are equiv- for all maximal right ideals T of R. alent for a ring R: (3) ⇒ (1): Given l(T ) is simple or zero for all (1) R is right mininjective. maximal right ideals T of R. To show that R is (2) If kR is simple, k ∈ R, then lr(k) = Rk. right mininjective: Let γ : kR → R be R-linear, (3) If kR is simple and r(k) ⊆ r(a); k, a ∈ R, where kR is minimal right ideal of R and let then Ra ⊆ Rk. i : kR → R be the inclusion map. If T = r(k) (4) If kR is simple and γ : kR → R is R-linear, then r(k) is a maximal right ideal of R so l(T ) is then γ(k) ∈ Rk. simple or zero. By Lemma III.3 l(T ) = Kd where Where r(k) = {b ∈ R| kb = 0} and lr(k) = {a ∈ K = kR, so Kd is simple. Thus γ = ci in Kd R| ab = 0; ∀ b ∈ r(k)} for some c ∈ R. So γ = c. proving that R is right mininjective. Proof. Given (1), Let 0 6= a ∈ lr(k). Then γ : kR → R is well defined by kr 7→ ar. Since R is right mininjective we get γ = c. where c ∈ R In this section we study triangular matrix rings 0 A 0 and γ extends to homomorphism γ : R → R then T = which are right mininjective. a = ck, thus lr(k) = Rk. Other implications are MB 0 0 routine verifications. rA(a) = {a ∈ A| aa = 0}, lM rA(a) = {n ∈ 0 0 M| na = 0; ∀ a ∈ rA(a)}, Duality considerations: lB(M) = {b ∈ B| bM = 0}, rB(K) = {b ∈ If MR is a right R-module, define the dual of M as B| Kb = 0}. d M = HomR(M,RR). This is a left R-module, where, if r ∈ R and λ ∈ M d, the map rλ ∈ M d Theorem III.5. Consider the following conditions is defined by (rλ)(m) = rλ(m) for all m ∈ M. (1),(2),(3)and (4). (1) A and B are right mininjective and for a ∈ Lemma III.3. If M = mR is a principal module A, m ∈ M with rA(a) and T = r(m), then and rA(m) maximal right ideals of A we have M d ∼ l(T ) = lr(m). = lM (rA(a)) ⊆ Ma, l r (m) = 0 and l r (m) ⊆ Bm. Proof. Let b ∈ l(T ), then the map λb : M → R is A A M A well defined by (2) T is right mininjective. (3) l (M) is an essential left ideal of B and λb(mr) = br. Then b 7→ λb is an isomorphism B l(T ) → M d ofIJSER left R-modules. lB(K) ⊆ Soc(BB) for every maximal right ideal K of B.

[If λb1 = λb2 means that λb1 (m) = λb2 (m); (4) lB(M) is an essential right ideal of B and that is b1 = b2 and so we get injectivity. Also if lB(J(B)) = Soc(BB). d λ ∈ M , λ : MR → R such that λ(m) = a, then Then we have following implications :(1) ⇒ (2), λ(mr) = ar. Now λ(m.r(m)) = 0 = a.r(m) so (2) and (3) ⇒ (1), a ∈ lr(m) and we get surjectivity.] (2) and (4) ⇒ (1). Proposition III.1. The following are equivalent for Proof. (1) ⇒ (2): In view of Lemma III.2[9], it a ring R suffices to show that for any minimal right ideal (1) R is right mininjective. uT of T the equality lr(u) = T u holds. Since (2) M d is simple or zero for all simple right R- u.r(u) = 0 we always have u ∈ lr(u), hence T u ⊆ modules M. lr(u). Therefore it suffices to show that lr(u) ⊆ (3) l(T ) is simple or zero for all maximal right T u. ideals T of R. Since uT is minimal right ideal of T , from d Corollary II.7 we see that one of the following is Proof. (1) ⇒ (2): Let γ, δ ∈ M , where MR is simple and assume that γ 6= 0 then we have valid: a 0 δoγ−1 : γ(M) → R, so since M is simple; (i) u = with (a, m)A a simple submod- R m 0 γ(M) is simple right ideal of R. Since R is right −1 ule of (A ⊕ M)A or mininjective we have δoγ = a. for some a ∈ R. 0 0 d d (ii) u = with bM = 0 and bB is minimal So δ = aγ in M and so M is simple. 0 b (2) ⇒ (3): If T is maximal right ideal of R right ideal of B. then (R/T )R is simple. Let M = (R/T )R = Suppose (i) holds, to find r(u): Let (m) = mR then T = r(m), by previous Lemma a1 0 d ∼ ∈ r(u), so M = l(T ) and so l(T ) is either simple or zero m1 b1 IJSER © 2014 http://www.ijser.org International Journal of Scientific & Engineering Research, Volume 5, Issue 12, December-2014 ISSN 2229-5518 28

a 0 a 0 A 0 a 0 1 = 0 which . Now let 1 ∈ lr(u) m 0 m1 b1 M rB(b) m1 b1 aa 0 0 0 a 0 A 0 implies 1 = this so 1 = ma1 0 0 0 m1 b1 M rB(b) gives us aa1 = 0, ma1 = 0; that is a A 0 0 0 1 = gives us a1 ∈ rA(a) ∩ rA(m) and m1, b1 can be m1a + b1M b1rB(b) 0 0 anything, writing I = rA(a) ∩ rA(m) we have a1 = 0 , b1 ∈ lB(rB(b)) and m1A + b1M = 0 I 0 r(u) = . gives m1 = 0, b1M = 0; that is b1 ∈ lB(M) and MB 0 0 so lr(u) = and a 0 0 l (M) ∩ l (r (b)) Now to find lr(u):Let 1 ∈ lr(u). So B B B m b 0 0 1 1 T u = . Since B is right mininjective a 0 I 0 0 Bb 1 = m1 b1 MB lB(rB(b)) ⊆ Bb hence a I 0 0 0 lr(u) ⊆ T u. 1 = , this m1I + b1M b1B 0 0 (2) and (3) ⇒ (1) Let a ∈ A satisfy the condition gives a 0 that aA is minimal ideal of A. Then u = a1 ∈ lA(I), b1 = 0, m1I + b1M = 0; that is 0 0 m1 ∈ lM (I) . Therefore generates a minimal right ideal of T ; because l (I) 0 (a, 0)A is simple submodule of (A ⊕ M) . Since lr(u) = A and T u = A lM (I) 0 T is right mininjective we get lr(u) ⊆ T u. Now λa 0 a1 0 |λ ∈ A, n ∈ M, µ ∈ B . to find lr(u): Let ∈ r(u), then na + µm 0 m1 b1 Since (a, m)A ∼ A/r (a, m) ∼ R/I is simple so a 0 a1 0 aa1 0 = A = = = I must be maximal ideal of A. So rA(a) ∩ rA(m) 0 0 m1 b1 0 0 is maximal and hence exactly one of the following 0 0 gives us a ∈ r (a), m ∈ M, b ∈ B, is valid. 0 0 1 A 1 1 (α) I = r (a) and m = 0 rA(a) 0 A therefore r(u) = . Now let (β) I = r (m) and a = 0 MB A (γ) I = r (a) = r (m). a1 0 A A ∈ lr(u) so lA(rA(a)) 0 m1 b1 If (α) holds then lr(u) = IJSERlM (rA(a)) 0 a1 0 rA(a) 0 = Aa 0 m1 b1 MB and T u = . Since A is right Ma 0 a1rA(a) 0 mininjective, we get lA(rA(a)) ⊆ Aa and given m1rA(a) + b1M b1B lM (rA(a)) ⊆ Ma so we get lr(u) ⊆ T u. 0 0 l r (m) 0 = gives us a1 ∈ lArA(a) , b1 = If (β) holds then lr(u) = A A and 0 0 lM rA(m) 0 0 and m ∈ l r (a). Therefore 1 M A 0 0 lArA(a) 0 T u = , now given that lArA(m) = 0 lr(u) = , u = Bm 0 lM rA(a) 0 and lM rA(m) ⊆ Bm so lr(u) ⊆ T u. a 0 Aa 0 l r (a) 0 and T u = . Since If (γ) holds then lr(u) = A A 0 0 Ma 0 lM rA(m) 0 lr(u) ⊆ T u we get lArA(a) ⊆ Aa and Aa 0 and T u = , since A is lM rA(a) ⊆ Ma. Since aA is minimal ⇔ rA(a) Ma + Bm 0 is maximal right ideal of A, we see that right mininjective lArA(a) ⊆ Aa and given that lM rA(a) ⊆ Ma whenever rA(a) is a maximal lM rA(m) ⊆ Bm we get lr(u) ⊆ T u. right ideal of A. From lArA(a) ⊆ Aa we conclude 0 0 In case if (ii) is valid u = with that A is right mininjective. 0 b bM = 0 , bB is minimal right ideal of B. To find Now let m ∈ M satisfy the condition that a 0 lr(u): Let 1 ∈ r(u), now rA(m) is maximal right ideal of A. Then m1 b1 (0, m)A =∼ A/r (m) is simple submodule of A 0 0 a1 0 0 0 0 0 = = 0 (A ⊕ M)A. Hence if v = , then vT is 0 b m1 b1 bm1 bb1 m 0 which gives bm1 = 0, bb1 = 0; a minimal right ideal of T hence lr(v) ⊆ T v. that is b ∈ r (b). Therefore r(u) = a 0 1 B Now to find lr(v): Let 1 ∈ r(v) then m1 b1 IJSER © 2014 http://www.ijser.org International Journal of Scientific & Engineering Research, Volume 5, Issue 12, December-2014 ISSN 2229-5518 29

0 0 a 0 0 0 0 0 a 0 1 = = u = , to find r(u); let 1 ∈ m 0 m1 b1 ma1 0 0 b m1 b1 0 0 r(u) then gives a1 ∈ rA(m). Thus 0 0 a 0 0 0 0 0 1 = implies that r (m) 0 0 b m1 b1 0 0 r(v) = A . MB bm1 = 0 and bb1 = 0. Now bm1 = 0 gives a 0 m1 ∈ M since bM = 0 and bb1 = 0 gives To find lr(v), let 1 ∈ lr(v) then A 0 m1 b1 b1 ∈ rB(b), therefore r(u) = . M rB(b) a1 0 rA(m) 0 = a1 0 m1 b1 MB Now to find lr(u): Let ∈ lr(u) m1 b1 a1rA(m) 0 0 0 = then m1rA(m) + b1M b1B 0 0 a1 0 A 0 gives = m1 b1 M rB(b) a1 ∈ lArA(m), b1 = 0, m1 ∈ lM rA(m). a1A 0 0 0 lArA(m) 0 = Therefore lr(v) = , m1A + b1M b1rB(b) 0 0 lM rA(m) 0 gives us a1A = 0; that is a1 = 0 , b1rB(b) = 0; 0 0 0 0 v = , T v = . Since that is b1 ∈ lBrB(b) and m 0 Bm 0 m A + b M = 0 imply that m A = T is right mininjective lr(v) ⊆ T v; that is 1 1 1 0 giving m1 = 0 and b1M = lArA(m) = 0 , lM rA(m) ⊆ Bm. 0 giving b1 ∈ lB(M). Therefore Now it remains to show that B is right 0 0 mininjective. From Proposition III.1 [9] it lr(u) = and 0 lB(M) ∩ lBrB(b) suffices to show that lB(K) is simple or zero for 0 0 T u = . Since lr(u) = T u we all maximal right ideals K of B. Let K be any 0 Bb A 0 maximal right ideal of B then K˜ = have lB(M) ∩ lBrB(b) = Bb. Also since MK r (b) is maximal ideal of B, we have is a maximal right ideal of T . Since T is right B J(B) ⊆ rB(b) hence lBrB(b) ⊆ lB(J(B)), mininjective lT (K˜ ) is simple or zero. To find by assumption lB(J(B)) = Soc(BB) and ˜ a1 0 ˜ lT (K):Let ∈ lT (K), then so lBrB(b) ⊆ Soc(BB) ⊆ lB(M). Hence m1 b1 lB(M) ∩ lBrB(b) = lBrB(b) = Bb proving that a1 0 IJSERA 0 = B is right mininjective. m1 b1 MK a A 0 0 0 1 = gives Definition III.3. A ring R is called strong right m1A + b1M b1K 0 0 mininjective if lR(I) = 0 for every maximal right a1 = 0, b1K = 0 gives b1 ∈ lB(K) , m1A = 0 ideal I of R. gives m1 = 0 , b1M = 0 gives b1 ∈ lB(M). Since every simple right ideal of R is of the 0 0 Therefore lT (K˜ ) = , form aR with rR(a) maximal right ideal of R. 0 lB(M) ∩ lB(K) lRrR(a) = Ra if R is right mininjective. It follows so if lT (K˜ ) is simple or zero implies that that R is strong right mininjective if lRrR(a) = 0 lB(M) ∩ lB(K) is minimal right ideal of implies that Ra = 0; that is a = 0. This means B. Since lB(M) is essential left ideal and that R has no simple right ideals so Socle(RR)=0. lB(K) ⊆ Soc(BB) so lB(K) ∩ lB(M) = lB(K), That is a ring R is strong right mininjective ⇔ hence lB(K) is either simple or zero for all Socle(RR) = 0. R is strong right mininjective ⇔ maximal ideals K of B showing that B is right the dual of every simple right R-module is zero. mininjective. A 0 (2) and (4) ⇒ (1) We need only to prove that Corollary III.1. Let T = . Then T is (2) and (4) imply right mininjectivity of B. The MB strong right mininjective if and only if A is strong other statements in (1) follow as in proof of (2) right mininjective, Soc(M ) = 0 and Soc(L ) = 0 and (3) ⇒ (1). Let b ∈ B with bB simple. Since A B where L = lB(M). lB(M) is an essential right ideal of B we get bB ⊆ Soc(BB) ⊆ lB(M). Proof. By Corollary II.1 soc(TT ) = 0 0 0 0 soc(AA) 0 u = then uT = is min- ,L = lB(M) then 0 b bM bB soc(MA) soc(LB) imal right ideal of T . By the right mininjectivity T is strong right mininjective if and only if of T we get lr(u) = T u. Now to find lr(u): soc(TT )=0.

Lemma III.4. SocTT is an essential right ideal of range, being a potent ring or being a clean ring T if and only if the following are valid: (i) SocAA all pass over to triangular matrix rings. Throughout is essential in AA. this section T will denote the triangular matrix ring A 0 (ii) SocMA is essential in MA. . Let R be a ring and n an integer (iii) If lB(M) 6= 0, Soc(lB(M)B) is essential in MB lB(M)B. ≥ 1. Recall that a row (a1, a2...an) of elements of R is called a right unimodular n row if there Proof. By Corollary II.1 we have SocT = T exists elements λi ∈ R for 1 ≤ i ≤ n satisfying Soc(AA) 0 n where LB = lB(M). X Soc(MA) soc(LB) aiλi = 1. Let Ur(n, R) denote the set of Assume that Soc(TT ) is essential in TT . For any i=1 α 6= 0 in A and m 6= 0 in M we have right unimodular n rows over R. Recall that n a 0 aA 0 0 0 is said to be in the stable range of R if for T = and T = 0 0 0 0 m 0 any (a1, ...., an+1) ∈ Ur(n + 1,R) we can find 0 0 elements ci ∈ R for 1 ≤ i ≤ n satisfying the . From mA 0 condition that a 0 0 0 (a1 + an+1c1, ..., an + an+1cn) ∈ Ur(n, R). T ∩ Soc(T ) 6= 0 and T ∩ 0 0 T m 0 ai O Soc(TT ) 6= 0 we immediately see that aA ∩ Lemma III.5. Let ti = ∈ T for 1 ≤ mi bi Soc(AA) 6= 0 and mA ∩ Soc(MA) 6= 0. Also if i ≤ n. Then 0 0 0 6= b ∈ LB; we have T ∩ Soc(TT ) 6= 0 (t1, ..., tn) ∈ Ur(n, T ) if and only if (a1, ..., an) ∈ 0 b U (n, A) and and so we get bB ∩ Soc(L ) 6= 0. These yields r B (b , ..., b ) ∈ U (n, B). (i),(ii) and (iii). 1 n r Conversely, assume (i),(ii)and (iii). Since any Proof. Assume that (t1, ..., tn) ∈ Ur(n, T ). Then a 0 principal right ideal T can be written there exists m b λi O si = ∈ T for 1 ≤ i ≤ n satisfy- a 0 0 ωi µi as A + . Because of Lemma n m bM bB X ing tisi = 1 in T . This in particular yields II.1, to show that SocTT is essential in TT it i=1 a n n suffices to showIJSER that A ∩ SocTT 6= 0 and X X m aiλi = 1 in A and biµi = 1 in B. Hence 0 0 i=1 i=1 ∩ SocTT 6= 0 whenever (a, m) 6= bM bB (a1, ..., an) ∈ Ur(n, A) and (b1, ..., bn) ∈ (0, 0) ∈ A ⊕ M or b 6= 0 in B. Conditions (i) and U (n, B). r a conversely, assume that (a1, ..., an) ∈ Ur(n, A) (ii) clearly imply A∩SocTT 6= 0 whenever m and (b1, ..., bn) ∈ Ur(n, B). To show that (a, m) 6= (0, 0) A ⊕ M bM 6= 0 in . If we have (t1, ..., tn) ∈ Ur(n, T ). let λi ∈ A and µi ∈ B 0 0 0 0 ∩ SocT ⊇ ∩ satisfy bM bB T bM 0 n n X X 0 0 aiλi = 1 and biµi = 1. Now to find Soc(M ) 0 i=1 i=1 A n 0 0 λi 0 X = 6= 0. If on the other si = such that tisi = 1 in T . ωi µi bM ∩ Soc(MA) 0 i=1 0 0 n hand bM = 0 then 0 6= b ∈ L and ∩ X B 0 bB Let v = miλi define ωi = −µiv, then i=1 0 0 n n n Soc(TT ) ⊇ 6= 0. X X X 0 bB ∩ Soc(LB) biωi = − bi(µiv) = − biµiv = −v Corollary III.2. The ordinary triangular matrix i=1 i=1 i=1 λi 0 A 0 and define si = then simple checking ring is strong right mininjective ⇔ A ωi µi AA n is so. X shows that tisi = 1 in T. Hence (t1, ..., tn) ∈ In this section we will see some properties pass- i=1 U (n, T ). ing over to triangular matrix ring. We will prove r that certain properties like having n in the stable

a 0 Theorem III.6. n is in the stable range of T if ∈ T . and only if n is in the stable range of A and n is m b 2 2 in the stable range of B. Then a = e+α, b = f+β where, e = e ∈ A, f = f ∈ B, α is a unit in A and β is a unit in B. Then Proof. Assume n is in the stable range of T . Let e 0 α 0 e 0 u = + with an (a1, ..., an+1) ∈ Ur(n + 1,A) and (b1, ..., bn+1) ∈ 0 f m β 0 f ai 0 α 0 Ur(n + 1,B). Let ti = ∈ T , since idempotent and a unit in T . 0 bi m β n is in the stable range of T there exists q = i ci 0 ∈ T for 1 ≤ i ≤ n such that (t1 + ui di Recall that a ring R is said to be potent if idem- tn+1q1, t2 + tn+1q2, ..., tn + tn+1qn) ∈ Ur(n, T ). potents mod J(R) can be lifted and every right By Lemma III.5 (a + a c , ..., a + a c ) ∈ 1 n+1 1 n n+1 n (equivalently left) ideal L of R with L ( J(R) Ur(n, A) and contains a non zero idempotent. (b1 + bn+1d1, ..., bn + bn+1dn) ∈ Ur(n, B), hence n is in the stable range of both A and B. Theorem III.7. T is a potent ring if and only if Conversely, assume that n is in the stable range of A and B are potent rings. both A and B. Let Proof. ai 0 From Corollaries II.1, II.2 we have J(T ) = ti = ∈ T for 1 ≤ i ≤ n + 1, satisfy J(A) 0 mi bi and idempotents mod J(T ) can MJ(B) (t1, ..., tn+1) ∈ Ur(n + 1,T ); that is there exists be lifted in T if and only if idempotents mod λi 0 si = for 1 ≤ i ≤ n + 1 in T such that J(A) can be lifted in A and idempotents mod ωi µi n+1 J(B) can be lifted in B. Assume T is potent. X t s = 1 in T . Let I be a right ideal of A not contained in i i i=1 ˜ I 0 n+1 J(A). Then I = is a right ideal of X 0 0 From Lemma III.5. we see that aiλi = 1 in T not contained in J(T ). Hence there exists an i=1 e 0 e 0 n+1 element 0 6= e ∈ I with = X 0 0 0 0 A and biµi = 1 in B. That is (a1, ..., an+1) ∈ e 0 2 i=1 in T , equivalently e = e in A. Let U (n + 1,A) andIJSER(b , ..., b ) ∈ U (n + 1,B). 0 0 r 1 n+1 r Since n is in the stable range of both A, B we get K be a right ideal of B with K * J(B). 0 0 elements ci ∈ A, di ∈ B for Then K˜ = is a right ideal of T KMK 1 ≤ i ≤ n satisfying the condition that (a1 + with K˜ * J(T ). Hence there exists an element an+1c1, ..., an + an+1cn) is in Ur(n, A) and (b1 + b d , ..., b + b d ) is in U (n, B). Let q = 0 0 ˜ 0 0 0 0 n+1 1 n n+1 n r i ∈ K with 6= and ci 0 m f m f 0 0 ∈ T for 1 ≤ i ≤ n, then ti + 0 di 0 0 0 0 0 0 = . But since ai + an+1ci 0 m f m f m f tn+1qi = . From 0 0 0 0 mi + mn+1ci bi + bn+1di = Lemma III.5 we see that (t1 + tn+1q1, ..., tn + m f m f 0 0 tn+1qn) ∈ Ur(n, T ). Hence n is in the stable range , we get f 2 = f and fm = m. From of T . fm f 2 0 0 0 0 6= , we see that either f 6= 0 m f 0 0 A ring R is said to be Clean [10] if every element or m 6= 0. From fm = m we conclude that f 6= 0. in R is the sum of a unit and an idempotent. If Thus 0 6= f ∈ K and f 2 = f in B. This proves θ : R → S is a surjective ring homomorphism and implication T potent ⇒ A and B are potent. R is clean, then so is S. Now assume that A and B are potent rings. Let I˜ be a right ideal of T with I˜ J(T ). Then there Proposition III.2. T is clean if and only if A and * a 0 B are clean. exists an element ∈ I˜ with m b Proof. Since A and B are factor rings of T , to a 0 ∈/ J(T ). This means that either a∈ / prove the proposition we have only to show that m b if A and B are clean then so is T . Let u = J(A) or b∈ / J(B). To show that I˜ contains a

nonzero idempotent it suffices to show that (2) If M = 0 or MA is simple then the right repeti- a 0 a 0 0 T = A + con- tiveness of A and B implies the right repetitiveness m b m bM bB of T . tains a non zero idempotent. If b∈ / J(B) there exists a non zero idempotent f ∈ bB and Proof. (1) If a ∈ A and I be finitely generated 0 0 right ideal of A then is a nonzero idempotent in uT where a 0 I 0 0 f u = ∈ T and I˜ = is finitely a 0 0 0 0 0 u = . Suppose a∈ / J(A). Then there generated right ideal of T . Since T is right repeti- m b k−1 exists a nonzero idempotent e in A of the form X tive there exists K ≥ 1 such that ukI˜ ⊆ unI˜. aλ for some λ ∈ A. Then aλe = e.e = e 6= 0. n=0 e 0 aλe 0 k The element = is a 0 k a I 0 mλe 0 mλe 0 We have u = u I˜ = and 0 0 0 0 a e 0 k−1 in A, hence ∈ uT . Also akI 0 X anI 0 m mλe 0 ⊆ which implies e 0 e 0 e 0 0 0 0 0 = . Thus n=0 mλe 0 mλe 0 mλe 0  k−1  k X n e 0 a I 0 a I 0 is a nonzero idempotent in uT . This that ⊆   . Therefore mλe 0 0 0  n=0  proves that T is potent. MB k−1 k X n Definition III.4. An element u of a ring R is said a I ⊆ a I; that is A is right repetitive. n=0 to be right repetitive if for each finitely generated Now let b ∈ B, J be any finitely right ideal of X right ideal I of R, the right ideal unI is finitely 0 0 0 0 B then v = ∈ T and J˜ = n≥0 0 b JMJ generated; equivalently there exists an integer k ≥ is finitely generated right ideal of T . Since T is k−1 k ˜ X n right repetitive there exists k ≥ 1 such that v J ⊆ 1 (depending on u and I) satisfying ukI ⊆ u I k−1 X 0 0 0 0 n=0 vnJ˜; that is ⊆ [6]. 0 bk JMJ n=0 R itself is said to be right repetitive if every k−1 IJSERX 0 0 0 0 element in R is right repetitive. The importance of which implies that 0 bn JMJ right repetitive rings arises from the fact that over n=0 k−1 such rings all cyclic right modules are hopfian. Re- 0 0 X 0 0 ⊆ ; that is call that a module N is hopfian if every surjective bkJM bkJ bnJM bnJ n=0 endomorphism f : N → N is automatically an k−1 X isomorphism. bkJ ⊆ bnJ. Therefore B is right repetitive. n=0 Lemma III.6. Let θ : R → S be a surjective ring (2) Assume that either M = 0 or MA is simple and homomorphism. If R is right repetitive so is S. then A and B are right repetitive. To show that T Proof. Let K be a finitely generated right ideal of is right repetitive we need to show that any element S and v ∈ S. By lifting a finite set of generators of T acts repetitively on any principal right ideal a 0 α 0 of S to R we get a finitely generated right ideal of T . Let I˜ = T and u = m b y β I R θ(I) = K u ∈ R of satisfying . Let satisfying be any element of T . We have θ(u) = v. Then since R is right repetitive and I a 0 0 I˜ = A + from Lemma II.1. is finitely generated there exists an integer k ≥ 1 m bM bB k−1 X α 0 α 0 α 0 with ukI ⊆ unI. From θ(unI) = vnK we see Also u = , u2 = y β y β y β n=0 k−1 α2 0 X = that vkK ⊆ vnK. Hence S is right repetitive. yα + βy β2 n=0 α3 0 u3 = yα2 + βyα + β2y β3 4 A 0 4 α 0 Theorem III.8. Let T = . u = 3 2 2 3 4 MB yα + βyα + β yα + β y β (1) If T is right repetitive so are A and B. so in general uk =

αk 0 and which is clearly true. yαk−1 + βyαk−2 + .. + βk−1y βk a 0 ukI˜ = uk T = m b αka 0 T yαk−1a + βyαk−2a + ... + βk−1ya + βkm βkb αka = A yαk−1a + βyαk−2a + ... + βk−1ya + βkm 0 0 + . Thus a typical element of βkbM βkbB ukI˜ is of the following form : αkaλ 0 , (yαk−1a + βyαk−2a + ... + βk−1ya + βkm)λ + βkbx βkbµ where λ ∈ A, x ∈ M and mu ∈ B. Thus u acts I˜ ⇔ k ≥ 2 repetitively on for some integer the If m 6= 0 or bM 6= 0 the right hand side of following three conditions are valid: (3)’ is whole of M when M is simple. Hence αk+1aA ⊆ aA + αaA + ... + αkaA A (1) (3) is valid in this case. Let m = 0 and bM = 0, βk+1bB ⊆ bB + βbB + ... + βkbB (2) If one of yαia + βyαi−1a + ... + βiya is not 0, (3) (yαka + βyαk−1a + ... + βkya + βk+1m)A + k+1 say for i = l then (3)’ is valid with k = l + 1. β bM ⊆ mA + bM + (ya + βm)A + βbM + i i−1 i k−1 If yα a + βyα a + ... + β ya = 0 for all X (yαia + βyαi−1a + ... + βiya + βi+1m)A + i ≥ 1 then the L.H.S. of (3)’ is zero, hence (3)’ is i=1 valid. This completes the proof of the theorem. k−1 X βi+1bM. i=1 REFERENCES k X {Since uk+1I˜ ⊆ uiI˜ = I˜+ uI˜+ ... + ukI˜ [1] K.R. Goodearl, Ring Theory, Nonsingular rings and modules, Marcel Dekker 1976. i=0 [2] C.Faith, Algebra II Ring Theory, Springer Verlag, Berlin/ a 0 0 I˜ = A + New York, 1976. m bM bB [3] David S. Dummit and Richard M. Foote, Abstract Alge- αa IJSER 0 0 bra, Wiley student edition 1999. uI˜ = A + [4] F. Kasch, Modules and Rings, A translation of modules ya + βm βbM βbB and rings, Translated by D.A.R Wallace, Academic Press, ukI˜ = 1982. αka [5] T.Y. Lam, A First Course in Noncommutative Rings, A+ yαk−1a + βyαk−2a + ... + βk−1ya + βkm Springer Verlag, New York, 1991. [6] K.R. Goodearl, Surjective endomorphisms of finitely 0 0 generated modules , Comm. Algebra 15 (1987),589-609. βkbM βkbB [7] M. Harada, Self Mininjective Rings, Osaka J. Math 19 uk+1I˜ = (1982), 587-597. αk+1a [8] M. Mullar, Rings of quotients of generalised matrix rings, k k−1 k k+1 A+ Comm. Algebra 15 (1987), 1991-2015. yα a + βyα a + ... + β ya + β m [9] W.K. Nicholson and M.F.Yousif, Mininjective Rings, J. 0 0 Algebra 187(1997), 548-578. } Now (1) is equivalent βk+1bM βk+1bB [10] W.K. Nicholson and K. Varadarajan, Countable linear transformations are clean, Proc. A.M.S. 126 (1998), 61- to saying that α ∈ A acts repetitively on aA, (2) 64. is equivalent to saying that β ∈ B acts repetitively [11] B. Sarath and K.Varadarajan, Dual Goldie Dimension II, on bB, (3) could be written as (3)’ Comm. Algebra 7 (1979)1885-1899. k k−1 k k+1 [12] K.Varadarajan, Dual Goldie Dimension, Comm. Algebra (yα a + βyα a + ... + β ya + β m)A + 7(1979) 565-610. k−1 [13] E.L.Green, On the representation theory of rings in k+1 X i β bM ⊆ mA + (ya + βm)A + (yα a + matrix form, Pacific J. Math, 100 (1982), 123-138. i=1 [14] W.K. Nicholson and J.F. Watters, Classes of simple βyαi−1a + ... + βiya + βi+1m)A + bM + βbM + modules and triangular rings, Comm. Algebra 20(1992), 141-153. [15] B. Stenstrom, Rings of Quotients, Springer Verlag, (1975). k−1 X [16] I.N. Herstein, A counter example in noitherian rings, βi+1bM Proc. Nat. Acad. Sci. U.S.A. , 54(1965), 1036-1037. i=1 [17] A. Haghany and K. Varadarajan, hopficity and co- hopficity over the ring of a Morita context. When M = 0 this is equivalent to show 0 ⊆ 0