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Subrings and Ideals Chapter 2 Subrings and Ideals 2.1 INTRODUCTION In this chapter, we discuss, subrings, sub fields. Ideals and quotient ring. We begin our study by defining a subring. If (R, +, ×) is a ring and S is a non-empty subset of R, then ‘+’ and ‘×’, may induce binary operations ‘+’ and ‘×’ respectively on S. If S is a ring with respect to these induced operations, then we call S, a subring of R. 2.2 SUBRING Definition 2.1 Let (R, +, ×) be a ring and let S be a non-empty subset of R. If (S, +, ×) is a ring, then S is called a subring of R. Every non-zero ring R has two trivial subrings, viz. the ring itself and the zero ring consisting of the zero element of the ring R. {0} and R are called the improper subrings of R. If S is a subring of R, then (i) S is a subgroup of additive group R. i.e., (S, +) is a subgroup of (R, +). (ii) S is closed with respect to multiplication. éùa 0 Example 1: The set S of all 2 ´ 2 matrices of the type êú where a, b, c are ëûbc ´ integers is subring of the ring M2 of all 2 matrices over Z. Example 2: The set of integers Z is a subring of the ring of real numbers. Theorem 2.1: A non-empty subset S of a ring R is a subring of R if and only if a - b Î S and ab Î S for all a, b Î S Proof: Let S be a subring of R and let a, b Î S. Then S is a subgroup of R under addition. Hence, b Î S Þ - b Î S and, a Î S, b Î S Þ a Î S, - b Î S Þ a + (- b) Î S Þ a - b + S Chapter-02 2nd Proof Anvi T12.p6539 10/5/10, 11:53 AM Subrings and Ideals 41 Hence, S is a subring of R. Example 2: Show that S = {0, 3} is a subring of (z6, +6, x6) under the operations +6 and x6. Solution: We construct the composition tables as follows: + ´ 6 03 6 03 003 000 330 303 From the above composition tables Î Î Þ - - Î a S, b S a +6 ( b) = a b S ´ Î " Î a 6 b S a, b S Hence, S is a subring of R. Example 3: Let m be any fixed integer and let S be any subset of Z, the set of integers, such that S = {..., - 3m, - 2m, - m, 0, m, 2m, 3m, ...} show that S is a subring of (z, +, ×) Solution: Let rm, s m Î S, then r, s Î Z now rm - sm = (r - s)m Î Z (Q r - s Î Z " r, s Î Z) and, (rm) (sm)= (rsm) m Î Z (Q r s m Î Z " r, s, m Î Z) \ S is a subring of (Z, +, ×) Theorem 2.2: The necessary and sufficient conditions for a non-empty subset S of a ring R to be a subring of R are: (i) S + (- S) = S and (ii) SS Í S Solution: Let S be a subring of R Then (S, +) is a subgroup of (R, +) let a + (- b) Î S + (- S), then a + (- b) Î S + (- S) Þ a Î S, - b Î - S Þ a Î S, b Î S Þ a - b Î S Þ a + (- b) Î S Thus, S + (- S) Í S (1) 0 Î S Þ 0 Î - S Now a Î S, 0 Î - S Þ a + 0 Î ST (- S) Hence, S Í S + (- S) (2) Chapter-02 2nd Proof Anvi T12.p6541 10/5/10, 11:53 AM 42 Linear Algebra From (1) and (2), we have S + (- S)= S Again, S is a subring of R Þ S is closed under multiplication. Thus, a Î S, b Î S Þ a b Î S but ab Î SS Now ab Î SS Þ a Î S, b Î S Þ ab Î S Þ SS Í S Hence, proved. Conversely, let S + (- S) = S and SS Í S " a, b Î S, we have ab Î SS Þ ab Î S (Q SS Í S) Again S + (- S)= S Þ S + (- S) Í S we have a + (- b) Î S + (–S) Í S Thus, Þ a + (- b) Î S " a, b Î S Þ a - b Î S " a, b Î S Thus, a - b Î S, ab Î S " a, b Î S Hence, S is a subring of R. Theorem 2.3: The intersection of two subrings of a ring R is a subring of R. Proof: Let S1 and S2 be two subrings of a ring R. Î Î Î Ç 0 S1, 0 S2, therefore, 0 S1 S2 Ç ¹ f Thus, S1 S2 Î Ç Now let a, b S1 S2, then Î Ç Þ Î Î a S1 S2 a S1, a S2 Î Ç Þ Î Î b S1 S2 b S1, b S2 But S1, S2 are subrings of R, therefore, Î Î Þ - Î Î a S, b S1 a b S1, ab S1 Î Î Þ - Î Î and, a S2, b S2 a b S2, ab S2 - Î - Î Þ - Î Ç a b S1, a b S2 a b S1 S2 Î Î Þ Î Ç ab S1, ab S2 ab S1 S2 - Î Ç Î Ç " Î Ç Consequently, a b S1 S2, ab S1 S2 a, b S1 S2 Ç Hence, S1 S2 is a subring of R. È Theorem 2.4: Let R be a ring and S1, S2 be two subrings. Then S1 S2 is a Í Í subring of R if and only if S1 S2 or S2 1. Chapter-02 2nd Proof Anvi T12.p6542 10/5/10, 11:53 AM 44 Linear Algebra 3. Prove that the z of integers is a subring of R, the set real numbers under addition and multiplication. 4. Show that the set of n ´ n matrices over the rational numbers is a subring of n ´ n matrices over the real numbers under addition and multiplication matrices. ´ ´ 5. Show that ({0, 2, 4}, +6, 6) is a subring of (Z6, +6, 6) where, Z6 = {0, 1, 2, 3, 4, 5}. 6. R is an integral domain. Show that the set S = {mx : x Î R, m is a fixed integer} is a subring of R. 7. If R is a ring, then show that the set N(a)= {x Î R; ax = xa} is a subring of R. 2.3 IDEALS 2.3.1 Left Ideal Definition 2.2: A non-empty subset U of a ring R is called a left ideal if (i) a Î U, b Î U Þ a - b Î U (ii) a Î U, r Î R Þ r a Î U Example: In the ring M of 2 ´ 2 matrices over integers consider the set U ìüéùa 0 = íýêú:,abÎ Z îþïïëûb 0 éù00 êú Î U Þ U ¹ f ëû00 Let éùa 0 éùc 0 A = êú, B = êú Î U, then ëûb 0 ëûd 0 éùac- 0 A - B = êú Î U ëûbd- 0 Also éùab éùa 0 P = êú Î M, A = êú Î U ëûgd ëûb 0 éùéùaba 0 Þ PA = êúêú ëûëûgdb 0 Chapter-02 2nd Proof Anvi T12.p6544 10/5/10, 11:53 AM Subrings and Ideals 45 éùabab+ 0 = êú Î U ëûgdab+ 0 This shows that U is a left ideal of M 2.3.2 Right Ideal Definition 2.3: A non-empty set U of a ring R is called a right ideal of R if (i) a Î U, b Î U Þ a - b Î U, and (ii) a Î U, r Î R Þ ar Î U. Example: Let M be the ring of 2 ´ 2 matrices one integers. Consider ìüéùab U = íýêú:,abÎ z îþïïëû00 éù00 êú Î U Þ U ¹ f ëû00 éùab éùcd éùacbd-- A = êú Î U B = êú Î U Þ A - B = êú Î 0 ëû00 ëû00 ëû00 éùab éùa 0 and, P = êú, A = êú Î U ëûgd ëûb 0 éùéùabab we have AP = êúêú ëûëû00gd éùababag++ bd = êú Î U ëû00 Hence, U is a right ideal of M. 2.3.3 Ideal Definition 2.4: A non-empty set U of a ring R is called an ideal (two-sided ideal) of R if (i) a Î U, b Î U Þ a - b Î U, and (ii) a Î U, r Î R Þ a r Î U and r a Î U Example: The set E of even integers is an ideal of the ring Z of integers. a, b Î E Þ a = 2m, b = 2n for some integers m and n we have a - b = 2m - 2n = 2(m - n) Î E also r Î Z, a Î E Þ ra = r(2m) = 2(rm) Î E and, ar = (2m) r = 2(mr) Î E Hence, E is an ideal of Z. Chapter-02 2nd Proof Anvi T12.p6545 10/5/10, 11:53 AM Subrings and Ideals 47 Consider a(xr) a(xr)= (ax) r = 0 × r = 0 Þ Î " Î Î x r S r R1 x S Thus, S is a right ideal of R. Example 3: If R is a commutative ring with unity, then the ideal Ra is the small- est ideal containing a. Proof: Let < a > = Ç {U: U is a ideal of R and a Î U} Clearly, < a > is the smallest ideal which contains a.
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