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 2 R and every a 2 A, ra 2 A and ar 2 A. 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 2 A =) a b 2 A. (2) a 2 A and r 2 R =) ar 2 A and ra 2 A. 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 2 R. The set hai = {ra r 2 R} 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 hai and its use in cyclic groups. (4) Let R[x] denote the set of all poynomials with real coe cients and let A be the subset of all polynomials with constant term 0. Then A is an ideal of R[x] and A = hxi. (5) Let R be a commutative ring with unity and let a 1, a 2,..., a n 2 R. Then I = ha 1, a 2,..., a n i = {r 1 a 1 + r 2 a 2 + + r n a n r i 2 R} is an ideal of R called the ideal generated by a 1, a 2,..., a n. If r 1 a 1 + r 2 a 2 + + r n a n, r 0 1a 1 + r 0 2a 2 + + r 0 na n 2 I, (r 1 a 1 + r 2 a 2 + + r n a n ) (r 0 1a 1 + r 0 2a 2 + + r 0 na n ) = (r 1 a 1 r 0 1a 1 ) + (r 2 a 2 r 0 2a 2 ) + + (r n a n r 0 na n ) = If r 1 a 1 + r 2 a 2 + + r n a n 2 I and r 2 R, (r 1 r 0 1)a 1 + (r 2 r 0 2)a 2 + + (r n r 0 n)a n 2 I. r(r 1 a 1 + r 2 a 2 + + r n a n ) = (r 1 a 1 + r 2 a 2 + + r n a n )r = Therefore I is an ideal by Theorem 14.1. (rr 1 )a 1 + (rr 2 )a 2 + ı + (rr n )a n 2 I.
14. IDEALS AND FACTOR RINGS 157 (6) Let Z[x] be the ring of all polynomials with integer coe cients and let I be the subset of Z[x] of all polynomials with even constant term. Then is an ideal of Z[x]. [To show I = hx, 2i.] I = hx, 2i If f(x) 2 hx, 2i, f(x) = xg(x) + 2h(x) where g(x), h(x) 2 R. Then so f(x) 2 I. Also, if f(x) 2 I, f(0) = 0 g(0) + 2 h(0) = 2h(0), f(x) = a n x n + a n 1 x n 1 + a 1 x + 2k = Thus, by mutual inclusion, I = hx, 2i. x(a n x n 1 + a n 1 x n 2 + a 1 ) + 2k 2 hx, 2i. [Show I is an ideal.] Suppose k(x) 2 Z[x] and f(x) = xg(x) + 2h(x) 2 I. Then p(x) = f(x)k(x) = k(x)f(x) = xk(x)g(x) + 2k(x)h(x) 2 I. Also, if f(x) = xf 1 (x) + 2f 2 (x) and g(x) = f(x) = xg 1 (x) + 2g 2 (x), f(x) g(x) = x (f 1 (x) g 1 (x) + 2 (f 2 (x) g 2 (x) 2 I, 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 2 S, f 0 (x) is defined for all real x. S is not an ideal of R. 8 >< 1, x > 0 Let f(x) = 1 2 S and let g(x) = sgn x = 0, x = 0. >: 1, x < 0 h(x) = g(x)f(x) = sgn x 62 S. Thus S is not an ideal, but is a subring of R.
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 2 R} is a ring under the operations (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.] 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 = s 0 + A and t + A = t 0 + A. Now s = s 0 + a and t = t + b where a, b 2 A. Then st = (s 0 + a)(t 0 + b) = s 0 t 0 + s 0 b + at 0 + ab =) st + A = s 0 t 0 + s 0 b + at 0 + ab + A = s 0 t 0 + A since s 0 b + at 0 + ab 2 A. Thus multiplication is well-defined. (=)) (using contrapositive) Suppose A is a subring of R that is not an ideal. then 9 a 2 A and r 2 R 3 ar 62 A or ra 62 A. WLOG, assume ar 62 A. Consider a + A = 0 + A and r + A. (a + A)(r + A) = ar + A, but (0 + A)(r + A) = 0 r + A = A 6= ar + A, so multiplication is not well-defined and R/A is not a ring.
Example. 14. IDEALS AND FACTOR RINGS 159 (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. and (6 + 9Z) + (6 + 9Z) = 12 + 9Z = 3 + 9 + 9Z = 3 + 9Z (6 + 9Z)(6 + 9Z) = 36 + 9Z = 9Z. apple a1 a (3) Let R = 2 a a 3 a i 2 Z and I be the subring of R consisting of 4 matrices with even entries. I is an ideal of R. Clearly, subtraction is closed in I. So let apple a1 a 2 a 3 a 4 2 R and apple apple 2b1 2b 2 b1 b = 2 2 2 I. Then 2b 3 a 4 b 3 b 4 apple apple a1 a 2 b1 b (2) 2 = 2 a 3 a 4 b 3 b 4 apple a1 a 2 apple b1 b 2 a 3 a 4 b 3 b 4 and apple apple apple apple b1 b 2 2 a1 a 2 b1 b = 2 2 a1 a 2 b 3 b 4 a 3 a 4 b 3 b 4 a 3 a 4 so I is an ideal. 2 I 2 I,
160 14. IDEALS AND FACTOR RINGS What is R/I? Solution. Every member of R can be written in the form a i 2 Z and r i 2 {0, 1}. But apple 2a1 + r 1 2a 2 + r 2 + I = 2a 3 + r 3 2a 4 + r 4 apple apple 2a1 2a 2 r1 r + 2 + I = 2a 3 2a 4 r 3 r 4 apple 2a1 + r 1 2a 2 + r 2 2a 3 + r 3 2a 4 + r 4 where apple r1 r 2 r 3 r 4 + I. Thus there are 2 4 choices for apple r1 r 2 r 3 r 4 or 16 choices for the lements of R/I. Example. Consider R = Z[i]/h2 ii, the factor ring of the Gaussian integers over h2 ii. What are its elements? For r 2 R, r = a + bi + h2 ii as a start. Now 2 i + h2 ii = 0 + h2 ii, so we can consider 2 i = 0 mod h2 ii or i = 2. Then, as an example, 3 + 4i + h2 ii = 3 + 8 + h2 ii = 11 + h2 ii. Similarly, for all r 2 R, r = a + h2 ii where a 2 Z. Next, we also have, for i = 2, i 2 = 4 or 1 = 4 or 0 = 5. Thus 7+6i+h2 ii = 7+12+h2 ii = 19+h2 ii = 4+5 3+h2 ii = 4+h2 ii. It follows that {0 + h2 ii, 1 + h2 ii, 2 + h2 ii, 3 + h2 ii, 4 + h2 ii}. Are any of these the same? 5(1 + h2 ii) = 5 + h2 ii = 0 + h2 ii so 1 + h2 ii is 1 or 5. If 1 + h2 ii = 1, then 1 + h2 ii = 0 + h2 ii =) 1 2 h2 ii or 1 = (2 i)(a + bi) where a+bi 2 Z[i]. Then 2a+b+( a+2b)i = 1 or 2a+b = 1 or a+2b = 0. Then 5b = 1 =) b = 1, a contradiction. Thus 1 + h2 ii = 5 and R = 5. 5
14. IDEALS AND FACTOR RINGS 161 Example. Let R[x] be the ring of polynomials with real coe cients and hx 2 + 1i the principal ideal generated by x 2 + 1. Then hx 2 + 1i = {f(x)(x 2 + 1) f(x) 2 R[x]}. Now R[x]/hx 2 + 1i = {g(x) + hx 2 + 1i g(x) 2 R[x]}, and (using the division algorithm for real polynomials), g(x) = q(x)(x 2 + 1) + r(x) where r(x) = 0 or the degree of r(x) is less than 2, the degree of x 2 + 1. Thus r(x) = ax + b where a, b 2 R. Thus g(x) = q(x)(x 2 + 1) + ax + b and g(x) + hx 2 + 1i = ax + b + q(x)(x 2 + 1) + hx 2 + 1i = ax + b + hx 2 + 1i, so R[x]/hx 2 + 1i = {ax + b + hx 2 + 1i a, b 2 R}. How to multiply in R[x]/hx 2 + 1i? Since x 2 + 1 + hx 2 + 1i = 0 + hx 2 + 1i, x 2 + 1 = 0 mod x 2 + 1 or x 2 = 1. Thus (2x + 3 + hx 2 + 1i)(5x 2 + hx 2 + 1i) = 10x 2 + 11x 6 + hx 2 + 1i = 11x 16 + hx 2 + 1i. 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 2 R and ab 2 A implies a 2 A or b 2 A. A maximal ideal A of R is a proper ideal of R if, whenever B is an ideal of 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 2 A, then ab = 0 =) a = 0 or b = 0 since Z is an integral domain =) a 2 {0} or b 2 {0}. For n > 1, nz is a prime ideal () n is prime. ((=) Suppose n is prime. Recalling hni = nz, suppose a, b 2 Z with ab 2 n. Then n ab =) (Euclid s Lemma) n a or n b =) a 2 hni or b 2 hni. Thus hni is prime. (=)) (by contrapositive) Suppose n is not prime. Then n = st where s < n or t < n. We have st 2 hni, but s 62 hni and t 62 hni, so hni is not prime. Example. Consider the lattice of ideals of Z 100. The diagram shows that h2i and h5i are the maximal ideals.
14. IDEALS AND FACTOR RINGS 163 Problem (Page 275 # 15). If A is an ideal of a ring R and 1 2 A, then A = R. Let r 2 R. Then r = r 1 2 A. Example. hx 2 + 1i is a maximal ideal of R[x]. Suppose A is an ideal of R[x] and hx 2 + 1i & A, [To show 9 c 2 R, c 6= 0, with c 2 A.] Let f(x) 2 A, f(x) 62 hx 2 + 1i. Then f(x) = q(x)(x 2 + 1) + r(x) where r(x) 6= 0 and deg r(x) < 2. Then r(x) = ax + b where a and b are not both 0, and ax + b = r(x) = f(x) q(x)(x 2 + 1) 2 A. Thus a 2 x 2 b 2 = (ax + b)(ax b) 2 A, and a 2 (x 2 + 1) 2 A since hx 2 + 1i A. Then 0 6= a 2 + b 2 = (a 2 x 2 + a 2 ) (a 2 x 2 b 2 ) 2 A. Let c = a 2 + b 2. Since c 2 A, c 2 R[x] =) 1 c 2 R[x], so 1 = 1 c 2 A. By c Page 275 # 15, A = R[x], and so hx 2 + 1i is a maximal ideal of R[x]. Example. hx 2 + 1i is not prime in Z 2 [x], since it contains but does not contain x + 1. (x + 1) 2 = x 2 + 2x + 1 = x 2 + 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. 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. (=)) Suppose R/A is an integral domain and ab 2 A. Then (a+a)(b+a) = ab+a = A =) a+a = A or b+a = A =) a 2 A or b 2 A. 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 2 A =) a 2 A or b 2 A since A is prime. Thus 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 2 R andb = {br + a r 2 R, a 2 A}, then B is an ideal of R. For br 1 + a 1, br 2 + a 2 2 B and and r, r 0 2 R, (br 1 + a 1 ) (br 2 + a 2 ) = b(r 1 r 2 ) + (a 1 a 2 ) 2 B and r 0 (br + a) = b(r 0 r) + r 0 a 2 B. 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. (=)) Suppose R/A is a field and B is an ideal of R with A & B. Let b 2 B, b 62 A. Then b + A 6= A, so 9 c 2 A 3 (b + A)(c + A) = 1 + A, the multiplicative identity of R/A. Since b 2 B, bc 2 B. Because 1 + A = (b + A)(c + A) = bc + A, 1 bc 2 A & B. Thus 1 = (1 bc) + bc) 2 B. By Page 275 # 15, B = R, so A is maximal. ((=) Suppose A is maximal and let b 2 R, b 62 A. [To show b + A has a multiplicative inverse.] Consider B = {br + a r 2 R, a 2 A}. B is an ideal of R by Page 275 #25, and A & B Since A is maximal, B = R. Thus 1 2 B, say 1 = bc + a 0 where a 0 2 A. Then Thus R/A is a field. 1 + A = bc + a 0 + A = bc + A = (b + A)(c + A). Corollary. A maximal ideal is a prime ideal.