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B.A.,Sem-II,Mathematics(Algebra) Subring As in Group, We Have B.A.,Sem-II,Mathematics(Algebra) Subring As in Group, we have studied about a subgroup. In similar way, we shall study about a subring. Definition 1 Subring Let R be a ring. A non-empty subset S of the set R will be a subring of Ring R if 1. S is closed under the addition composition 2. S is closed under the multiplication composition and S itself is a ring with respect to these two operations. Note 1 We see that if R is a ring then R will also be an abelian group with respect to addition composition because it satisfies all the postulates to be an abelian group.For this go through the definitions of an abelian group and ring. It is also clear here that if S is a subring of a ring R,then S is a subgroup of the additive group of R. Note 2 If R is any ring , then f0g and R itself are always subring of R. These are known as improper subring of R.Other subrings, if any,of R are called proper subrings of R. Theorem The necessary and sufficient conditions for a non-empty subset S of a ring R to be a subring of R are 1. a 2 S; b 2 S ) a − b 2 S 2. a 2 S; b 2 S ) ab 2 S a b Example 1 Show that the set of matrices is a subring of the ring of 2×2 matrices with 0 c integral elements. SolutionNote that here ring is a set of all matrices of the type 2 × 2 with integral elements. Let R be the ring of 2 × 2 matrices with integral elements and let M be a subset of R. To show that M is a subring, we need to satisfy two conditions of the above theorem.For this,we take a b a b any two elements of M. Let A = 1 1 and B = 2 2 be any two members of M.Then 0 c1 0 c2 a b a b a − a b − b A−B = 1 1 - 2 2 = 1 2 1 2 , which is a member of M because each entry 0 c1 0 c2 0 c1 − c2 of the matrix is an integral element. a b a b a a a b + b c Also AB = 1 1 2 2 = 1 2 1 2 1 2 ,which is a member of M. Hence M is a 0 c1 0 c2 0 c1c2 subring of ring R. The above solution can be presented in exam (in short form) as follows: Solution Let R be the ring of 2 × 2 matrices with integral elements and let M be a subset of a b a b R. Let A = 1 1 and B = 2 2 be any two members of M.Then 0 c1 0 c2 a b a b a − a b − b A − B = 1 1 - 2 2 = 1 2 1 2 2 M , 0 c1 0 c2 0 c1 − c2 a b a b a a a b + b c Also AB = 1 1 2 2 = 1 2 1 2 1 2 2 M 0 c1 0 c2 0 c1c2 Hence M is a subring of ring R. The next theorem is an important result and analogous to the result "The intersection of two subgroups is a subgroup" which we have studied in the section of group.Before proving this theorem, once you should revise the group's result for intersection. Theorem The intersection of two subrings is a subring. ProofLet S1 and S2 be two subrings of a ring R.Then S1 \ S2 is not empty because at least 0 2 S1 \ S2.Let a; b be any two elements of S1 \ S2 ) a; b 2 S1 and a; b 2 S2.Since S1 is a subring of ring R, so a; b 2 S1 ) a − b 2 S1 and ab 2 S1. Similarly, since S2 is a subring of ring R,therefore, a; b 2 S2 ) a − b 2 S2 and ab 2 S2. This implies from above that a − b 2 S1 \ S2 and ab 2 S1 \ S2 Therefore S1 \ S2 is a subring of ring R. There are some problems of subrings.Students should themselves solve them. 1. The set of integers is a subring of the ring of rational numbers. 2. Let R be the ring of integers. Let m be any fixed integer and let S be any subset of R such that S = f:::; −3m; −2m; −m; 0; m; 2m; 3m; :::g: Then S is a subring of ring R..
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