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(Ideal). a Subring a of a Ring R Is Called a (Two-Sided) Ideal of R If for Every R R and Every a A, Ra a and Ar A

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(Ideal). a Subring a of a Ring R Is Called a (Two-Sided) Ideal of R If for Every R R and Every a A, Ra a and Ar A CHAPTER 14 Ideals and Factor Rings Ideals Definition (Ideal). A subring A of a ring R is called a (two-sided) ideal of R if for every r R and every a A, ra A and ar A. 2 2 2 2 Note. (1) A “absorbs” elements of R by multiplication. (2) Ideals are to rings as normal subgroups are to groups. Definition. An ideal A of R is a proper ideal if A is a proper subset of R. Theorem (Ideal Test). A nonempty subset A of a ring R is an ideal of R if (1) a, b A = a b A. 2 ) − 2 (2) a A and r R = ar A and ra A. 2 2 ) 2 2 Proof. Follows directly from the definition of ideal and Theorem 12.3 (Subring Test). ⇤ Example. (1) 0 (the trivial ideal) and R itself are ideals of R. { } (2) For any positive integer n, nZ = 0, n, 2n, 3n, . is an ideal of Z. { ± ± ± } 155 156 14. IDEALS AND FACTOR RINGS (3) Let R be a commutative ring with identity and let a R. The set 2 a = ra r R h i { | 2 } is an ideal of R called the principal ideal generated by a. Note that the commutative assumption is necessary here. Also, context will distinguish between this use of a and its use in cyclic groups. h i (4) Let R[x] denote the set of all poynomials with real coefficients and let A be the subset of all polynomials with constant term 0. Then A is an ideal of R[x] and A = x . h i (5) Let R be a commutative ring with unity and let a , a , . , a R. Then 1 2 n 2 I = a , a , . , a = r a + r a + + r a r R h 1 2 ni { 1 1 2 2 · · · n n| i 2 } is an ideal of R called the ideal generated by a1, a2, . , an. Proof. If r a + r a + + r a , r0 a + r0 a + + r0 a I, 1 1 2 2 · · · n n 1 1 2 2 · · · n n 2 (r a + r a + + r a ) (r0 a + r0 a + + r0 a ) = 1 1 2 2 · · · n n − 1 1 2 2 · · · n n (r a r0 a ) + (r a r0 a ) + + (r a r0 a ) = 1 1 − 1 1 2 2 − 2 2 · · · n n − n n (r r0 )a + (r r0 )a + + (r r0 )a I. 1 − 1 1 2 − 2 2 · · · n − n n 2 If r a + r a + + r a I and r R, 1 1 2 2 · · · n n 2 2 r(r a + r a + + r a ) = (r a + r a + + r a )r = 1 1 2 2 · · · n n 1 1 2 2 · · · n n (rr )a + (rr )a + ı + (rr )a I. 1 1 2 2 · · · n n 2 Therefore I is an ideal by Theorem 14.1. ⇤ 14. IDEALS AND FACTOR RINGS 157 (6) Let Z[x] be the ring of all polynomials with integer coefficients and let I be the subset of Z[x] of all polynomials with even constant term. Then I = x, 2 h i is an ideal of Z[x]. Proof. [To show I = x, 2 .] h i If f(x) x, 2 , f(x) = xg(x) + 2h(x) where g(x), h(x) R. Then 2 h i 2 f(0) = 0 g(0) + 2 h(0) = 2h(0), · · so f(x) I. Also, if f(x) I, 2 2 n n 1 f(x) = anx + an 1x − + a1x + 2k = − · · · n 1 n 2 x(anx − + an 1x − + a1) + 2k x, 2 . − · · · 2 h i Thus, by mutual inclusion, I = x, 2 . h i [Show I is an ideal.] Suppose k(x) Z[x] and f(x) = xg(x) + 2h(x) I. Then 2 2 p(x) = f(x)k(x) = k(x)f(x) = xk(x)g(x) + 2k(x)h(x) I. 2 Also, if f(x) = xf1(x) + 2f2(x) and g(x) = f(x) = xg1(x) + 2g2(x), f(x) g(x) = x (f (x) g (x) + 2 (f (x) g (x) I, − 1 − 1 2 − 2 2 so I is an ideal by Theorem 14.1. ⇤ (7) Let R be the ring of all real-valued functions of a real variable. Let S be the subset of all di↵erentiable functions (this means for f S, f 0(x) is defined for all real x. S is not an ideal of R. 2 1, x > 0 Let f(x) = 1 S and let g(x) = sgn x = 0, x = 0. 2 8 <> 1, x < 0 − h(x) = g(x)f(x) = sgn x S. Thus S is not an ideal, but is a subring of R. 62 :> 158 14. IDEALS AND FACTOR RINGS Factor Rings Theorem (14.2 — Existence of Factor Rings). Let R be a ring and A a subring of R. The set of cosets r+A r R is a ring under the operations { | 2 } (s + A) + (t + A) = s + t + A and (s + A)(t + A) = st + A () A is an ideal of R. [In this case , we say R/A is a factor ring of R.] Proof. We know the set of cosets form a group under addition. If our multiplication is well-defined, i.e., multiplication is a binary operation, it is clear that the multiplication is associative and distributive over addition. [To show multiplication is well-defined A is an ideal of R.] () ( =) Suppose A is an ideal of R and let s + A = s0 + A and t + A = t0 + A. ( Now s = s0 + a and t = t” + b where a, b A. Then 2 st = (s0 + a)(t0 + b) = s0t0 + s0b + at0 + ab = ) st + A = s0t0 + s0b + at0 + ab + A = s0t0 + A since s0b + at0 + ab A. Thus multiplication is well-defined. 2 (= ) (using contrapositive) Suppose A is a subring of R that is not an ideal. then) a A and r R ar A or ra A. WLOG, assume ar A. Consider9 a2+ A = 0 +2A and−3−r + A62. 62 62 (a + A)(r + A) = ar + A, but (0 + A)(r + A) = 0 r + A = A = ar + A, · 6 so multiplication is not well-defined and R/A is not a ring. ⇤ 14. IDEALS AND FACTOR RINGS 159 Example. (1) Z/5Z = 0 + 5Z, 1 + 5Z, 2 + 5Z, 3 + 5Z, 4 + 5Z is a factor ring since 5Z { } is an ideal of Z. (3 + 5Z) + (4 + 5Z) = 7 + 5Z = 2 + 5 + 5Z = 2 + 5Z and (3 + 5Z)(4 + 5Z) = 12 + 5Z = 2 + 10 + 5Z = 2 + 5Z. We have essentially modular 5 arithmetic. (2) 3Z/9Z = 0 + 9Z, 3 + 9Z, 6 + 9Z is a factor ring since 9Z is an ideal of { } 3Z, the arithmetic essentially modulo 9. (6 + 9Z) + (6 + 9Z) = 12 + 9Z = 3 + 9 + 9Z = 3 + 9Z and (6 + 9Z)(6 + 9Z) = 36 + 9Z = 9Z. a1 a2 (3) Let R = ai Z and I be the subring of R consisting of a3 a4 2 ⇢ matrices with even entries. I is an ideal of R. Proof. a a Clearly, subtraction is closed in I. So let 1 2 R and a3 a4 2 2b 2b b b 1 2 = 2 1 2 I. Then 2b3 a4 b3 b4 2 a a b b a a b b 1 2 (2) 1 2 = 2 1 2 1 2 I a3 a4 b3 b4 a3 a4 b3 b4 2 ✓ ◆ and b b a a b b a a 2 1 2 1 2 = 2 1 2 1 2 I, b3 b4 a3 a4 b3 b4 a3 a4 2 ✓ ◆ ✓ ◆ so I is an ideal. ⇤ 160 14. IDEALS AND FACTOR RINGS What is R/I ? | | Solution. 2a + r 2a + r Every member of R can be written in the form 1 1 2 2 where 2a3 + r3 2a4 + r4 ai Z and ri 0, 1 . But 2 2 { } 2a + r 2a + r 2a 2a r r r r 1 1 2 2 + I = 1 2 + 1 2 + I = 1 2 + I. 2a3 + r3 2a4 + r4 2a3 2a4 r3 r4 r3 r4 4 r1 r2 Thus there are 2 choices for or 16 choices for the lements of R/I. ⇤ r3 r4 Example. Consider R = Z[i]/ 2 i , the factor ring of the Gaussian integers over 2 i . What are its elemenh − ts?i h − i For r R, r = a + bi + 2 i as a start. Now 2 i + 2 i = 0 + 2 i , so we 2can consider 2 i h= 0−moi d 2 i or i = 2.−Then,h as−ani example,h − i − h − i 3 + 4i + 2 i = 3 + 8 + 2 i = 11 + 2 i . h − i h − i h − i Similarly, for all r R, r = a + 2 i where a Z. 2 h − i 2 Next, we also have, for i = 2, i2 = 4 or 1 = 4 or 0 = 5. Thus − 7+6i+ 2 i = 7+12+ 2 i = 19+ 2 i = 4+5 3+ 2 i = 4+ 2 i . h − i h − i h − i · h − i h − i It follows that 0 + 2 i , 1 + 2 i , 2 + 2 i , 3 + 2 i , 4 + 2 i . { h − i h − i h − i h − i h − i} Are any of these the same? 5(1 + 2 i ) = 5 + 2 i = 0 + 2 i h − i h − i h − i so 1 + 2 i is 1 or 5. If 1 + 2 i = 1, then | h − i| | h − i| 1 + 2 i = 0 + 2 i = 1 2 i or 1 = (2 i)(a + bi) h − i h − i ) 2 h − i − where a+bi Z[i].
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