(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 coecients 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 coecients 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 and3r + 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= 0moi d 2 i or i = 2.Then,h asani 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]. Then 2a+b+( a+2b)i = 1 or 2a+b = 1 or a+2b = 0. 2 1 Then 5b = 1 = b = , a contradiction. Thus 1 + 2 i = 5 and R = 5. ) 5 | h i| | | 14. IDEALS AND FACTOR RINGS 161 Example. Let R[x] be the ring of polynomials with real coecients and x2 + 1 the principal ideal generated by x2 + 1. Then h i 2 2 x + 1 = f(x)(x + 1) f(x) R[x] . h i { | 2 } Now 2 2 R[x]/ x + 1 = g(x) + x + 1 g(x) R[x] , h i { h i| 2 } and (using the division algorithm for real polynomials), g(x) = q(x)(x2 + 1) + r(x) where r(x) = 0 or the degree of r(x) is less than 2, the degree of x2 + 1. Thus r(x) = ax + b where a, b R. Thus g(x) = q(x)(x2 + 1) + ax + b and 2 g(x) + x2 + 1 = ax + b + q(x)(x2 + 1) + x2 + 1 = ax + b + x2 + 1 , h i h i h i so 2 2 R[x]/ x + 1 = ax + b + x + 1 a, b R . h i { h i| 2 } How to multiply in R[x]/ x2 + 1 ? h i Since x2 + 1 + x2 + 1 = 0 + x2 + 1 , x2 + 1 = 0 mod x2 + 1 or x2 = 1. Thus h i h i (2x + 3 + x2 + 1 )(5x 2 + x2 + 1 ) = h i h i 10x2 + 11x 6 + x2 + 1 = 11x 16 + x2 + 1 . h i h i Note that, with x playing the role of i, this ring is isomorphic to the complex numbers.

Prime Ideals and Maximal Ideals Definition (Prime Ideal, Maximal Ideal). A prime ideal A of a commutative ring R is a proper ideal of R such that a, b R and ab A implies a A or b A. A maximal ideal A of R is a proper 2ideal of R if,2whenever B is2an ideal2of R and A B R, then B = A or B = R. ✓ ✓ 162 14. IDEALS AND FACTOR RINGS Example. Consider the ring Z. 0 is a prime ideal of Z. If ab A, then ab = 0 = a = 0 or b = 0 since Z is{ an} integral domain = a 02 or b 0 . ) ) 2 { } 2 { } For n > 1, nZ is a prime ideal n is prime. () Proof. ( =) Suppose n is prime. Recalling n = nZ, suppose a, b Z with ab n. Then( n ab = (Euclid’s Lemma) n ha ori n b = a n or2 b n . Th2 us n is prime.| ) | | ) 2 h i 2 h i h i (= ) (by contrapositive) Suppose n is not prime. Then n = st where s < n or)t < n. We have st n , but s n and t n , so n is not prime. 2 h i 62 h i 62 h i h i ⇤

Example. Consider the lattice of ideals of Z100.

The diagram shows that 2 and 5 are the maximal ideals. h i h i 14. IDEALS AND FACTOR RINGS 163 Problem (Page 275 # 15). If A is an ideal of a ring R and 1 A, then A = R. 2

Proof. Let r R. Then r = r 1 A. 2 · 2 ⇤ Example. x2 + 1 is a maximal ideal of R[x]. h i Proof. Suppose A is an ideal of R[x] and x2 + 1 & A, h i [To show c R, c = 0, with c A.] Let f(x) A, f(x) x2 + 1 . Then 9 2 6 2 2 62 h i f(x) = q(x)(x2 + 1) + r(x) where r(x) = 0 and deg r(x) < 2. Then r(x) = ax + b where a and b are not both 0, and6 ax + b = r(x) = f(x) q(x)(x2 + 1) A. 2 Thus a2x2 b2 = (ax + b)(ax b) A, and a2(x2 + 1) A 2 2 since x2 + 1 A. Then h i ✓ 0 = a2 + b2 = (a2x2 + a2) (a2x2 b2) A. 6 2 1 1 Let c = a2 + b2. Since c A, c R[x] = R[x], so 1 = c A. By 2 2 ) c 2 c · 2 Page 275 # 15, A = R[x], and so x2 + 1 is a maximal ideal of R[x]. h i ⇤ 2 Example. x + 1 is not prime in Z2[x], since it contains h i (x + 1)2 = x2 + 2x + 1 = x2 + 1, but does not contain x + 1. 164 14. IDEALS AND FACTOR RINGS Problem (Page 275 # 26). If R is a commutative ring with unity and A is a proper ideal of R, then R/A is a commutative ring with unity. Proof. Note that (b + A)(c + A) = bc + A = cb + A = (c + A)(b + A). Thus R/A is commutative. Also, if 1 is the unit of R, 1 + A is the unit of R/A. ⇤

Theorem (14.3 — R/A is an Integral Domain A is Prime). Let R be a commutative ring with identity and let A be an()ideal of R. Then R/A is an integral domain A is prime. () Proof. (= ) Suppose R/A is an integral domain and ab A. Then ) 2 (a+A)(b+A) = ab+A = A = a+A = A or b+A = A = a A or b A. ) ) 2 2 Thus A is prime. ( =) Note that R/A is a commutative ring with unity for any proper ideal A from( Pge 275 # 26. Suppose A is prime and (a + A)(b + A) = ab + A = 0 + A = A. Then ab A = a A or b A since A is prime. Thus 2 ) 2 2 a + A = A or b + A = A = R/A has no zero-divisors and is thus an integral ) domain. ⇤ 14. IDEALS AND FACTOR RINGS 165 Problem (Page 275 # 25). Let R be a commutative ring with unity and let A be an ideal of R. If b R andB = br + a r R, a A , then B is an ideal of R. 2 { | 2 2 } Proof.

For br + a , br + a B and and r, r0 R, 1 1 2 2 2 2 (br + a ) (br + a ) = b(r r ) + (a a ) B 1 1 2 2 1 2 1 2 2 and r0(br + a) = b(r0r) + r0a B. 2 Thus B is an ideal by the ideal test. ⇤ Theorem (14.4 — R/A is a Field A is maximal). Let R be a commutative ring with unity and let A (be)an ideal of R. Then R/A is a field A is maximal. () Proof. (= ) Suppose R/A is a field and B is an ideal of R with A & B. Let b B, b )A. Then b + A = A, so c A (b + A)(c + A) = 1 + A2, the multiplicativ62 e identity of6 R/A. Since9 2b B3,bc B. Because 2 2 1 + A = (b + A)(c + A) = bc + A, 1 bc A & B. Thus 1 = (1 bc) + bc) B. By Page 275 # 15, B = R, so Ais maximal.2 2 ( =) Suppose A is maximal and let b R, b A. ( 2 62 [To show b + A has a multiplicative inverse.] Consider B = br + a r R, a A . B is an ideal of R by Page 275 #25, { | 2 2 } and A & B Since A is maximal, B = R. Thus 1 B, say 1 = bc + a0 where 2 a0 A. Then 2 1 + A = bc + a0 + A = bc + A = (b + A)(c + A). Thus R/A is a field. ⇤ Corollary. A maximal ideal is a prime ideal.