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Study of Formal Triangular Matrix Rings

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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 mininjec- tive, clean, potent or a ring of stable rank≤ n. The f(I ⊕ M) ⊕ Kj 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 Bg 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, fW ⊕ 0j W a simple submodule of (A ⊕ All the rings considered will be associative rings M) g and with identity, all the modules considered will be A f0 ⊕ Kj 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 = 0g: 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 2 A, b 2 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 2 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 <A A; I2 ≤B (M ⊕ B) and MI1 ⊆ is a minimal right ideal of BB satisfying KM = 0. I2; that is MI1 ⊕ 0 ⊆ I2. Conversely, it is clear that W ⊕ 0 for any simple (2) Right ideals of T are of the form J1 ⊕ J2 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 2 = 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 2 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 2 R such that R. Now maximal right ideals of T are x2 − x 2 I. Such an element is said to be lifted fI ⊕ M ⊕ Bj I maximal right ideal of Ag [ provided that there exists an idempotent y 2 R fA ⊕ M ⊕ Kj K maximal right ideal of Bg so such that y − x 2 I and we say that x has been J(T ) = J(A) ⊕ M ⊕ J(B) lifted to y.] J(A) 0 a 0 = . Let 2 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 fW ⊕0j W a simple submodule of (A⊕M) g[ A J(T ) a2 − a 2 f0⊕Kj K is a minimal right ideal of B satisfying which belongs to ; therefore J(A); b2 − b 2 J(B) a KM = 0g. 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 2 T such that − that KM = 0 which implies that K ⊆ lB(M). m0 b0 m b Therefore a0 0 a − a0 0 = 2 J(T ) ; that soc(TT ) = soc(AA)⊕ soc(MA)⊕ soc(lB(M)) m0 b0 m − m0 b − b0 IJSER is a − a0 2 J(A) and b − b0 2 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 2 direct summand of T=J(T ) hence I¯ is a direct J(A); b 2 J(B) summand of A=J(A) and J¯ is a direct summand of IJSER © 2014 http://www.ijser.org International Journal of Scientific & Engineering Research, Volume 5, Issue 12, December-2014 ISSN 2229-5518 23 aλ 0 a B=J(B).
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