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1 Solution of Some Homework Problems (3.1) Prove that a commutative ring R has a unique 1. Proof: Let 1 R and 1 R be two multiplicative identities of R. Then since 1 R is an identity, 1 R = 1 R 1 R. Since 1 R is an identity, 1 R 1 R = 1 R. Thus 1 R = 1 R 1 R = 1 R, and so the multiplicative identity is unique. (3.4) (i) If X is a set, prove that the Boolean group B(X) in Example.18 with elements the subsets of X and with addition given by U + V = (U V ) (V U), where U V = {x U : x V }, is a commutative ring if one defines multiplication UV = U V. We call B(X) a boolean ring. (ii) Prove that B(X) contains exactly one unit. (iii) If Y is a proper subset of X, show that the unit in B(Y ) is distinct from the unit in B(X). Conclude that B(Y ) is not a subring of B(X). Proof: (i) By the definition of U + V, we have U + V B(X) and U + V = (U V ) (V U) = (V U) (U V ) = V + U, and so + is a commutative binary operation in B(X). Note that U + V = (U V ) (V U) = (U V ) (U V ). Therefore, U, V, W B(X), the set (U +V )+W = [((U V ) (U V )) W ] [((U V ) (U V )) W ] consists of the elements in X that occurs in an odd number of sets in U, V and W (in either one of them or in all of them). This is the same as the set of U + (V + W ), and so (U + V ) + W = U + (V + W ). Thus + is associative. Since the empty set B(X), and since U B(X), U + = (U ) ( U) = U, the empty set is the additive identity in B(X). For any U B(X), U + U = and so every element in B(X) has an additive inverse. This indicates that B(X) with addition is an abelian group. 1

2 By the definition of multiplication, U, V B(X), UV = U U B(X) is a commutative binary operation of B(X). For U, V, W B(X), (UV )W = (U V ) W = U (V W ) = U(V W ), and so multiplication is associative in B(X). As for any U B(X), UX = U X = U, X is the multiplicative identity of B(X). We also need to show that distributive law holds. U, V, W B(X), U(V + W ) = U [(V W ) (V W )] = (U (V W )) (U (V W )) = [(U V ) (U W )] [(U V ) (U W )] = UV + UW. Therefore, B(X) is a commutative ring. (ii). By (i), we know that X is a unit. Suppose that U is a unit of B(X). Then for some V inb(x), UV = U V = X. Since U B(X), U X. Since X = UV = U V U, we have U = X and so X is the only unit of B(X). (iii) Let Y X be a proper subset of X. Then Y X. By (i), B(Y ) is a commutative ring. By (ii), Y is the only unit in B(Y ), and Y is not a unit in B(X). Therefore, B(Y ) is not a subring of B(X). (3.5) Show that U(Z m ) = {[k] Z m : (k, m) = 1}. Proof: If (k, m) = 1 (as integers), then there exist integers s and t such that sk + tm = 1 (as integers). Therefore, sk sk + tm 1 (mod m). It follows that [s][k] = [1] in Z m, and so {[k] I m : (k, m) = 1} U(Z m ). Conversely, let [k] U(Z m ). Then [s] Z m such that [sk] = [s][k] = [1], and so sk 1 (mod m). Hence there exists an integer t such that sk = 1 + tm, or sk + ( t)m = 1 (as integers). This implies that (m, k) = 1 and so {[k] Z m : (k, m) = 1} U(Z m ). (3.10) (i) Prove that R = {a + b : a, b Z} is a domain. (ii) Prove that R = { 1 (a + b ) : a, b Z} is not a domain. (iii) Using the fact that α = 1 (1 + 19) is a root of x x + 5, prove that R = {a + bα : a, b Z} is a domain.

3 Proof: The binary operations of these sets are assumed to be the addition and multiplication of the real or the complex numbers. (i) Let R denote the field of real numbers with real number addition and multiplication. Then R is a subset of R. To see that R is a subring of R, it suffices to show that (1) 1 R. This is true as 1 = R. () For any a + b, a + b R, we have (a + b ) (a + b ) = (a a ) + (b b ) R. (3) For any a + b, a + b R, we have (a + b )(a + b ) = (aa + bb ) + (ab + a b) R. As R is a subring of a domain R, R itself is also a domain. (ii) Suppose that this R is a ring. Then multiplication must be a binary operation of R. Note that 1 R but 1 1 = 1 4 R. Thus multiplication is not a binary operation of R, and so R is not even a ring, not to mention a domain. (iii) Let C denote the set of all complex numbers with number addition and multiplication. Then C is a field, and this R is a subset of C. To see that R is a subring of C, it suffices to show that (1) 1 R. This is true as 1 = ( )/ R. () For any a + bα, a + b α R, we have a + bα (a + b α) = (a a ) + (b b )α R. (3) For any a + bα, a + b α R, we have (a + bα)(a + b α) = aa + (ab + a b)α + bb α. Since α = α 5, we can rewrite the product as (a + bα)(a + b α) = aa + (ab + a b)α + bb (α 5) = (aa 5bb ) + (ab + a b + bb )α R. Therefore, R is a subring of the field C. Since a field is a domain, R itself is also a domain. (3.1) (i) If R is a commutative ring, define the circle operation a b by a b = a + b ab. Prove that the circle operation is associative and that 0 a = a for all a R. 3

4 (ii) Prove that a commutative ring R is a field if and only if {r R : r 1} is an abelian group under the circle operation. Proof: (i) a, b, c R, (a b) c = (a + b ab) c = (a + b ab) + c c(a + b ab) = a + b + c ab ac bc + abc = a + (b + c bc) a(b + c bc) = a (b c). Thus the operation is associative. Moreover, a R, 0 a = (0 + a) 0 a = a. (ii) Let A = {r R : r 1}. Suppose that R is a field. Then R {0}, and so 0 A. For a 1, a A, if 1 = a 1 a = a 1 + a a 1 a, then 1 a = a 1 a 1 a = a 1 (1 a ), or (1 a 1 )(1 a ) = 0. Since R has no zero divisor, either a 1 = 1 or a = 1, contrary to the assumption that a 1, a A. Therefore, a 1 a A also. Moreover, a 1 a = a 1 +a a 1 a = a + a 1 = a a 1. Thus (also by (i)), is an associative and commutative binary operation of A, with 0 serves as the identity (see (i)). Let a A. To see that a has an inverse in A with respect to the operation, we solve the equation a + b ab = 0 for the unknown b in R. Since R is a field, the inverse of a in A is b = a(1 a) 1 (it exists as a 1). Thus when R is a field, A with is an abelian group. Conversely, we assume that A with is an abelian group. Since R is a commutative ring, it suffices to show that any r R {0} has a multiplicative inverse. Let r R {0, 1}. Then r A. Since r 0, r + 1 A. Since (A, ) is a group, there exists a b A such that 0 = (r + 1) b = r b (r + 1)b = r b rb b = r(1 b) + 1. Thus r 1 = b 1 R, and so R is a field. (3.16) Show that F = {a + b : a, b Q} is a field, where Q denotes the field of rational numbers. Proof: Note that F R is a subset of the real number field. We first verify that F is a subring of R by checking each of the following: (1) 1 F, as 1 = +0 F. () If a+b, a +b F, then a + b a + b = a a + b b F. 4

5 (3) If a+b, a +b F, then a + b Therefore, F is a subring of R. a + b = aa +bb + ab +a b F. It remains to show that every non zero element in F has an inverse. Let r = a+b F {0}. Since is not a rational number, for any rational numbers a and b with a = b 0, a b 0. Let a = 4a, b a b = 4b 1 (a + b ) F. a + b a b. a + b and so r has a multiplicative inverse in F. Then a, b are rational numbers and so = 4(a b ) 4(a b ) = 1. (3.1) (i) If R is a domain, show that if a polynomial in R[x] is a unit, then it is a nonzero constant (the converse is true if R is a field). (ii) Show that (x + 1) = 1 in Z 4 [x]. Conclude that the hypothesis in part (i) that R be a domain is necessary. Proof: (i) Suppose that u(x) R[x] is a unit. Then v(x) R[x] such that u(x)v(x) = 1. Since R is a domain, deg(u(x)) + deg(v(x)) = deg(1) = 0, and so deg(u(x)) = 0. It follows that u(x) = u R {0}. If R is a field, then u R {0}, there exists r 1 R, and so r 1 R[x]. Thus as a constant polynomial, u R {0} is a unit in R[x]. (ii) Note that Z 4 is not a domain, as [] [] = [0]. Since (x + 1) = 4x + 4x + 1 = 1 in Z 4 [x], thus when R is not a domain, a unit of R[x] does not have to be a non zero constant in R. (3.3) If R is a commutative ring and f(x) = n s i x i R[x] has degree n 1, define its derivative f (x) R[x] by f (x) = s 1 + s x + 3s 3 x + + ns n x n 1 ; if f(x) is a constant polynomial, define its derivative to be the zero polynomial. Prove that the usual rules of calculus hold: (i) (f + g) = f + g. (ii) (rf) = r(f). 5

6 (iii) (fg) = fg + f g. (iv) (f n ) = nf n 1 f, for all n 1. Proof: Let f(x) = n s i x i and g(x) = m t i x i be two polynomials in R[x]. Without loss of generality, we may assume that m n. (i). Let a j = 0 for j = n + 1,, m. We have f(x) + g(x) = m (s i + t i )x i. Thus m m m (f + g) = i((s i + t i )x i 1 = is i x i 1 + it i x i 1 = f + g. (ii). Fix an r R. Then rf(x) = n rs i x i. Thus n n (rf(x)) = ris i x i 1 = r is i x i 1 = r(f(x)). (iii). Let c k = k s i t k i. Then f(x)g(x) = nm c i x i. Thus (using k = i 1) (f(x)g(x)) = = = m+n m+n 1 k=0 m+n 1 k=0 ic i x i 1 = m+n 1 (k + 1)c k+1 x k k=0 k (k + 1)s j t k j+1 x k j=0 k m+n 1 js j t k j+1 x k + j=0 k=0 = f (x)g(x) + f(x)g (x). k (k j + 1)s j t k j+1 x k j=0 (iv). To show (iv), we argue by induction on n 1. When n = 1, this is the definition of f. Assume that n and assume that the formula holds for smaller values of n. Define f 0 (x) = 1 and f n (x) = (f(x)) n. Then by (iii) and by induction, (f n (x)) = (f(x)f n 1 (x)) = f(x) (n 1)f n (x)f (x) + f n 1 (x)f (x) = nf n 1 (x)f (x). (3.4) Let R be a commutative ring and let f(x) R[x]. (i) Prove that if (x a) f(x), then (x a) f (x) in R[x]. (ii) Prove that if (x a) f(x) and (x a) f (x), then (x a) f(x). Proof: (i) Let g(x) = (x a). Then by Exercise 3.3(iv), g (x) = (x a). Since g(x) f(x), h(x) R[x] such that f(x) = g(x)h(x). By Exercise 3.3(iii), f (x) = g (x)h(x) + g(x)h (x) = (x a)(h(x) + (x a)h (x)). 6

7 Therefore, (x a) f (x). (ii) Since (x a) f(x), we can write f(x) = (x a) n h(x), where n 1 in n integer, and where h(x) R[x] such that (x a) h(x). By Exercise 3.5(iii) and (iv), f (x) = n(x a) n 1 h(x)+(x a) n h (x). Since (x a) f (x), we have (x a) ((x a) n 1 h(x)). Since (x a) h(x), we must have n 1 1 and so n. (3.8) Find the gcd of x x and x 3 7x + 6 in Z 5 [x], and express it as a linear combination of them. Proof: Let f(x) = x x and g(x) = x 3 7x+6. Note that in Z 5 [x], g(x) = x 3 x+1. Apply division algorithm, x 3 x + 1 = (x x )(x + 1) + (x + 3) x x = (x + 3)(x 4) 10 = (x + 3)(x 4). Therefore, (f(x), g(x)) = x + 3. And x + 3 = 1 (x 3 x + 1) + ( x 1)(x x ) = g(x) + ( x 1)f(x). (3.31) Prove the converse of Euclid s lemma. Let k be a field and let f(x) k[x] be a polynomial of degree 1; if, whenever f(x) divides a product of two polynomials, it necessarily divides one of the factors, then f(x) is irreducible. Proof: Suppose, by contradiction, that f(x) is not irreducible. Then f(x) = a(x)b(x) such that both a(x) and b(x) has positive degree. It follows that f(x) a(x)b(x). By the assumption, f(x) a(x) or f(x) b(x). But neither is possible as the comparison of the degrees both sides would lead to a contradiction. (3.33) Let k be a field, and let f(x), g(x) k[x] be relatively prime. If h(x) k[x], prove that f(x) h(x) and g(x) h(x) imply f(x)g(x) h(x). Proof: Since f(x) and g(x) are relatively prime, s(x), t(x) k[x] such that s(x)f(x) + t(x)g(x) = 1. Since f(x) h(x), there exists a h 1 (x) k[x] such that h(x) = h 1 (x)f(x). Multiply both sides of s(x)f(x) + t(x)g(x) = 1 by h 1 (x) to get s(x)f(x)h 1 (x) + t(x)g(x)h 1 (x) = h 1 (x). Since g(x) h(x) and since h(x) = h 1 (x)f(x), g(x) divides both functions on the left hand side of the displayed equality above. Therefore, g(x) h 1 (x), and so g(x)f(x) h(x). 7

8 (3.37) (i) Let f(x) = (x a 1 ) (x a n ) K[x], where k is a field. Show that f(x) has no repeated roots if and only if the gcd (f, f ) = 1, where f (x) is the derivative of f. (ii) Prove that if p(x) Q[x] is an irreducible polynomial, then p(x) has no repeated roots in C. Proof: (i) Suppose that f(x) has a repeated root a. Then (x a) f(x). By Exercise 3.4(i), (f, f ) 1. Conversely, suppose that d = (f, f ) 1. Since K is a a field, deg(d) > 0. Let x a be a factor of d. Then (x a) f and (x a) f. By Exercise 3.4(ii), (x a) f, and so f has a repeated root. (ii) By contradiction, suppose that p(x) has a repeated root. Then by (i), (p, p ) 1. Since p(x) is an irreducible polynomial in Q[x], either (p, p ) = 1 or (p, p ) = p. Therefore as (p, p ) 1, we must have p p. But deg(p (x)) = deg(p(x)) 1, and so p p is not possible. This contradiction implies that p(x) cannot have a repeated root. (3.39) (i) let φ : A R be an isomorphism, and let ψ : R A be its inverse. Show that ψ is an isomorphism. (ii) Show that the composite of two homomorphisms (isomorphisms) is again a homomorphism(isomorphism). (iii) Show that A = R defines an equivalence relation on the class of all commutative rings. Proof: (i) For any a, b R, since φ is an isomorphism, a, b A such that φ(a ) = a, φ(b ) = b, φ(a + b ) = a + b and φ(a b ) = ab. Since ψ is the inverse map of φ, we have ψ(a) = a and ψ(b) = b, and ψ(a + b) = a + b = ψ(a) + ψ(b), ψ(ab) = a b = ψ(a)ψ(b). Therefore, ψ is a ring homomorphism. Since ψ is the inverse map of a bijection, ψ is also a bijection, and soψ is a ring isomorphism. (ii) Let f : R R and g : R R be ting homomorphisms. Then for any a, b R, (gf)(a + b) = g(f(a) + f(b)) = g(f(a)) + g(f(b)), (gf)(ab) = g(f(a)f(b)) = g(f(a))g(f(b)). Thus (gf) is also a ring homomorphism. If both f and g are isomorphisms, then as the composition of bijections is also a bijection, (gf) is also a ring isomorphism. (iii) Let Ω denote the set of all commutative rings. Define a relation = on Ω as follows: for R, R Ω, R = R if and only if R is isomorphic to R. Then since the identity map is 8

9 an isomorphism, R = R, R Ω, and so = is reflexive. If R = R, then by (i), R = R also, and so = is symmetric. Now let R, R, R Ω be such that R = R and R = R. Then by (ii), we also have R = R and so = is transitive. Thus = is an equivalence relation. (3.4) Let R be a commutative ring. Show that the function ɛ : R[x] R, defined by ɛ(a 0 + a 1 x + a x + + a n x n ) = a 0, is a homomorphism. Describe kerɛ in terms of roots of polynomials. Proof: Let f(x) = n s i x i and g(x) = m t i x i be two polynomials in R[x]. Then i+0 ɛ(f + g) = s 0 + t 0 = ɛ(f) + ɛ(g), ɛ(fg) = s 0 t 0 = ɛ(f)ɛ(g). Thus ɛ is a ring homomorphism. Let E 0 denote the kernel of ɛ. Then E 0 = {f(x) R[x] : f(x) = n s i x i } = {f(x) R[x] : f(0) = 0}. (3.45) Suppose that p is a prime integer. (i) Show that every element a Z p has a pth root (i.e., there is b Z p with a = b p ). (ii) Let K be a field that contains Z p as a subfield. For every positive integer n, show that the function φ n : K K, given by φ(a) = a pn, is a ring homomorphism. Proof: (i) Recall that Fermat s Theorem (Theorem 1.4) states that if p is a prime, then for any a Z p, and for any integer n 1, a pn = a in Z p. In particular, a = a p in Z p. (ii) Then ch(k) = p. That is, when 1 = 1 K, then p 1 = 0. This implies that for any a, b K, (a + b) p = a p = +b p. Arguing by induction on n, we have, for any positive integer n, and for any a, b K, (a + b) pn = a pn + b pn. Thus φ n (a + b) = φ n (a) + φ n (b). The law of exponents indicated that φ n (ab) = (ab) pn = a pn b pn = φ n (a)φ n (b), and so φ n is a ring homomorphism. 9

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