INTRODUCTION TO MODERN ALGEBRA MATH 3613 SECTION ...

INTRODUCTION TO MODERN ALGEBRAMATH 3613 SECTION 001, SPRING 2009INSTRUCTOR: WEIPING LIANSWER TO HOMEWORK 9• 4.1.3 (a) List all polynomials of degree 3 in Z 2 [x]. So the polynomial f(x) = x 3 +a 2 x 2 + a 1 x + a 0 , for a 0 , a 1 , a 2 ∈ Z 2 . So there are total 2 3 different polynomials with(a 2 , a 1 , a 0 ) = (0, 0, 0), (1, 0, 0), (0, 1, 0), (1, 1, 0), (0, 0, 1), (1, 0, 1), (1, 1, 1), (0, 1, 1).(b) Similarly list all polynomials of degree less than 3 in Z 3 [x], these are polynomialsf(x) = a 2 x 2 +a 1 x+a 0 with a 2 , a 1 , a 0 ∈ Z 3 . There are 3 3 different polynomials.You may list as in (a).• 4.1.7 Let f(x), g(x) ∈ R[x] be two arbitrary elements. We need to show that f(x) ·g(x) = g(x) · f(x) to make that R[x] is commutative. We can expressf(x) =n∑m∑a i x i , g(x) = b j x j ,i=0j=0for a i (0 ≤ i ≤ n), b j (0 ≤ i ≤ n) in R. Thus f(x) · g(x) = a 0 b 0 + (a 0 b 1 + a 1 b 0 )x +(a 0 b 2 + a 1 b 1 + a 2 b 0 )x 2 + · · · + a n b m x n+m . Since R is commutative, so a i b j = b j a i forall 0 ≤ i ≤ n, 0 ≤ j ≤ m. Hence we havef(x) · g(x) = a 0 b 0 + (a 0 b 1 + a 1 b 0 )x + (a 0 b 2 + a 1 b 1 + a 2 b 0 )x 2 + · · · + a n b m x n+m= b 0 a 0 + (b 1 a 0 + b 0 a 1 )x + (b 2 a 0 + b 1 a 1 + b 0 a 2 )x 2 + · · · + b m a n x m+n= b 0 a 0 + (b 0 a 1 + b 1 a 0 )x + (b 0 a 2 + b 1 a 1 + b 2 a 0 )x 2 + · · · + b m a n x m+n= g(x) · f(x).• 4.1.8 We need to prove that 1 R = 1 R[x] . Let f(x) = ∑ ni=0 a ix i ∈ R[x]. Then weneed to show that 1 R · f(x) = f(x) · 1 R = f(x).1 R · f(x) = 1 R · (==n∑a i x i )i=0n∑1 R · a i x ii=0n∑a i x i ,i=01

2 ANSWER TO HOMEWORK 9since 1 R · a i = a i for a i ∈ R.f(x) · 1 R = (===n∑a i x i ) · 1 Ri=0n∑(a i x i · 1 R )i=0n∑a i · 1 R x ii=0n∑a i x i ,i=0where the second identity follows from the distribution law, the third from Theorem4.1 (ii) (So 1 R x i = x i 1 R since 1 R ∈ R), and the last from a i · 1 R = a i since a i ∈ R.• 4.1.9 If c ∈ R is a zero divisor, then c ≠ 0 R and there exists a d ≠ 0 R and d ∈ Rwith c · d = 0 R (since R is commutative, so c · d = d · c). Thus we can have bothc, d ∈ R[x]. Note that 0 R = 0 R[x] . So both c, d ≠ 0 R[x] and c · d = 0 R[x] . Orconstruct a polynomial d + dx ∈ R[x] which is ≠ 0 R[x] by Theorem 4.1 (v), andc · (d + dx) = cd + cdx = 0 R + 0 R x = 0 R = 0 R[x] . Hence it is also a zero divisor inR[x].• 4.1.11 By 4.1.7, F [x] is a commutative ring since F is a field. 1 F ≠ 0 F . By 4.1.8,we have 1 F [x] = 1 F ≠ 0 F = 0 F [x] . By Corollary 4.3, F [x] is an integral domain if Fis a field. So the only thing we need to verify that F [x] is not a field, is to find anonzero element in F [x] which has no inverse. The obvious nonzero element not inF is x by Theorem 4.1 for the construction of F [x]. We are going to show that x isnot a unit. Suppose the contrary. If x is a unit in F [x], then there exists an elementf(x) ∈ R[x] such that x · f(x) = f(x) · x = 1 F [x] = 1 F . By Theorem 4.2, we havedeg(x · f(x)) = deg(x) + deg f(x) = 1 + deg f(x), deg f(x) ≥ 0.deg(x · f(x)) = deg 1 F = 0.Thus 1 + deg f(x) = 0 which is impossible for a non-negative integer deg f(x). Thisleads to the claim that x is not a unit. Therefore F [x] is not a field.• 4.1.12. (i) If f(x) is a unit in F [x], then there exists g(x) ∈ F [x] such that f(x) ·g(x) = 1 F = g(x) · f(x). By Theorem 4.2, we havedeg f(x) + deg g(x) = deg(f(x) · g(x)) = deg 1 F = 0.For non-negative degree, we have deg f(x) = deg g(x) = 0.(ii) If deg f(x) = 0, then f(x)is a nonzero constant in F . Thus f(x) is a unit inF . We have a g ∈ F such that f(x) · g = g · f(x) = 1 F . Since g ∈ F ⊂ F [x] and1 F = 1 F [x] . So f(x) is also a unit in F [x]

MODERN ALGEBRA 3• 4.1.15 Since R is an integral domain, it is commutative with 1 R ≠ 0 R . We only needto prove that every nonzero element in R is a unit. Let b ≠ 0 R . Thus we can applythe Division Algorithm in R[x] for g(x) = b ≠ 0 R and f(x) = 1 R , there exist uniquepolynomials (q(x), r(x)) such that f(x) = 1 R = b · q(x) + r(x), and either r(x) = 0or deg r(x) < deg g(x). Since deg g(x) = deg b = 0, so r(x) must be zero. We have1 R = b · q(x).0 = deg 1 R = deg(b · q(x) = deg b + deg q(x) = deg q(x),by Theorem 4.2 for integral domain R. Hence deg q(x) = 0 if and only if q(x) = q ∈R is a nonzero constant in R. Therefore we have q ∈ R such that b · q = q · b = 1 R .So b is a unit, and R is a field.• 4.1.18 The derivative rule for product, shows that D(x · x 2 ) = D(x 3 ) = 3x 2 andD(x)·D(x 2 ) = 1·(2x) = 2x. So D(x·x 2 ) ≠ D(x)·D(x 2 ). D is a not homomorphismof rings, and is not an isomorphism.

4 ANSWER TO HOMEWORK 9• 4.2.2 Notice that the definition of gcd of f(x) and g(x) is a monic polynomial d(x)satisfies the Definition on page 91. So gcd(f(x), 0 F ) = c −1n · f(x).• 4.2.4 (a) If f(x)|g(x), then g(x) = f(x)p(x) and deg g(x) = deg f(x) + deg p(x) ≥deg f(x). Thus g(x)|f(x), we have deg f(x) ≥ deg g(x). This implies deg f(x) =deg g(x) and deg p(x) = 0. So p(x) is a nonzero constant in F . We denote it byc ∈ F .(b) From f(x) = cg(x), deg f(x) = deg g(x). Then for f(x) = x n + a 1 x n−1 + · · · +a n , g(x) = x n + b 1 x n−1 + · · · + b n , we havex n + a 1 x n−1 + · · · + a n = c(x n + b 1 x n−1 + · · · + b n ) = cx n + cb 1 x n−1 + · · · cb n .By Theorem 4.1 (iv), we have 1 F = c by comparing the coefficient of x n . Thereforef(x) = cg(x) = 1 F · g(x) = g(x) by 4.1.8.• 4.2.5(b) Applying the Euclidean Algorithem (Theorem 1.6) for a(x) = x 5 + x 4 +2x 3 − x 2 − x − 2 and b(x) = x 4 + 2x 3 + 5x 2 + 4x + 4, we havea(x) = b(x)q 0 (x) + r 0 (x),where q 0 (x) = x − 1 and r 0 (x) = −x 3 − x + 2.b(x) = r 0 (x)q 1 (x) + r 1 (x),where q 1 (x) = −x − 2 and r 1 (x) = 4x 2 + 4x + 8.r 0 (x) = r 1 (x)q 2 (x),where q 2 (x) = − 1 4 x+ 1 4 . Thus r 1(x) is the greatest common divisor. So gcd(a(x), b(x)) =14 (4x2 +4x+8) = x 2 +x+2 (Remember: gcd for polynomial must be monic. Answerr 1 (x) = 4x 2 + 4x + 8 is not monic so that it is incorrect!).• 4.2.9. If gcd(f(x), 0 F ) = 1 F , then by Theorem 4.5, there exist polynomials u(x), v(x)such thatf(x)u(x) + 0 F v(x) = 1 F .Hence f(x)u(x) = 1 F , and deg f(x) + deg u(x) = 0. This implies deg f(x) = 0. Sof(x) must be a nonzero constant in F .• 4.2.10. Do longdivision, or factor to see x + a + b is a factor of x 3 − 3abx + a 3 + b 3 .Thusgcd(x + a + b, x 3 − 3abx + a 3 + b 3 ) = x + a + b.• 4.2.14. If f(x)|h(x), then h(x) = f(x)p(x). If g(x)|h(x), then h(x) = g(x)q(x).Hencef(x)p(x) = h(x) = g(x)q(x).So g(x)|(f(x)p(x)). Since gcd(f(x), g(x)) = 1 F , by Theorem 4.7, g(x)|p(x). p(x) =g(x)l(x). Now we have h(x) = f(x)p(x) = f(x)g(x)l(x). So f(x)g(x)|h(x)..

MODERN ALGEBRA 5• 4.2.15. There are many ways to do this problem.Method 1. Let c(x) be a common factor of h(x) and g(x). Then h(x) = c(x)p(x)and f(x) = h(x)q(x) = c(x)p(x)q(x). So c(x) is a common factor of f(x) and g(x).So c(x) = 1 F from gcd(f(x), g(x)) = 1 F .Mehotd 2. By Theorem 4.5, there exist u(x), v(x) such thatf(x)u(x) + g(x)v(x) = 1 F .Since h(x)|f(x), there is a p(x) ∈ F [x] such that f(x) = h(x)p(x), andh(x)p(x)u(x) + g(x)v(x) = 1 F .Note that Theorem 4.5 is only for one direction to express gcd, but not necessaryf(x)u(x) + g(x)v(x) = d(x) = gcd(f(x), g(x)). 1 F is an exceptional case. Let c(x)be a common factor of h(x) and g(x), therefore h(x) = c(x)h 1 (x), g(x) = c(x)g 1 (x).andc(x)h 1 (x)p(x)u(x) + c(x)g 1 (x)v(x) = c(x) [h 1 (x)p(x)u(x) + g 1 (x)v(x)] = 1 F .Therefore c(x)|1 F and deg c(x) = 0. So c(x) must be a nonzero constant. Hencegcd(h(x), g(x)) = 1 F .