MATH 241 Homework 11, Fall 2011

Math 241 Problem Set 11 solution manual Exercise. A11.1 Lemma 1. � n i + 1 � � n i � = � n + 1 i + 1 � . Proof. � n i + 1 � � n i � = n! (i+1)!(n−(i+1))!+ n i!(n−i)! = n!(n−i) (i+1)!(n−i)!+ n!(i+1) (i+1)!(n−i)! = (n+1)! (i+1)!((n+1)−(i+1))! = � n + 1 i + 1 � . a- We prove the binomial formula by induction: Base step : for n = 1, (a + b)1 = a + b = � 1 0 � a1−0b0 + � 1 1 � a1−1b1 So it is true for n = 1 Inductive step: Suppose it is true up to n, and let us prove it for n + 1: (a + b)n+1 = (a + b)(a + b)n = (a + b) Σ i=1...n � n i � an−ibi = Σ i=0...n � n i � an−i+1bi + Σ i=0...n � n i � an−ibi+1 =an+1 + [ Σ i=1...n � n i � an−i+1bi] + [ Σ i=0...(n−1) � n i � an−ibi+1] + bn+1 =an+1 + [ Σ i=1...n � n i � an−i+1bi] + [ Σ i=1...(n−1) � n i − 1 � an−i+1bi] + bn+1 =an+1 + [ Σ i=1...n( � n i � + � n i − 1 � )an−i+1bi] + bn+1 =an+1 + [ Σ i=1...n � n + 1 i � an−i+1bi] + bn+1 = Σ i=0...n+1 � n + 1 i � a(n+1)−ibi b- For non-commutative rings, the binomial formula fails. (a + b)2 = a2 + ab + ba + b2 (a+b)3 = (a+b)2(a+b) = (a2 +ab+ba+b2)(a+b) = a3 +a2b+aba+ab2 +ba2 +bab+b2a+b3 Section. 19 1 Exercise. 9 We first notice that for any element x ∈ Z3 × Z4 x.12 = 0, but since we know that the order of (1, 1) is 12, hence 12 is the smallest such number, and hence the char(Z3 × Z4)=12. Exercise. 11 Using the binomial formula we get the following: (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4, but since the ring is of char 4, and since a3b, a2b2, ab3 ∈ R, we get the following : (a + b)4 = a4 + (2 + 4)a2b2 + b4 = a4 + 2a2b2 + b4 = (a2 + b2)2 Exercise. 14 Consider the element � 2 −1 2 −1 � ∈ M2(Z), we can easily see that: � 2 −1 2 −1 � � 1 2 2 4 � = � 0 0 0 0 � , but matrices are non-zero, hence � 1 2 2 4 � is a zero divisor. Exercise. 23 It is easier to note that a2 = a =⇒ a(a − 1) = 0 =⇒ a = 0 or a − 1 = 0 (this only needs an integral domain) =⇒ a ∈ {0, 1}. Exercise. 29 Suppose that char(D)=k, with k neither 0 nor a prime number. Then we can find m, n ∈ N − {0, 1} such that k = m · n. So we get (m.1)(n.1) = 0, and hence since D is an integral domain we have either n.1 = 0 or m.1 = 0. Suppose (WLOG) we have m.1 = 0, hence m has the property m.α = 0 ∀α ∈ D, but since m < m.n = k we get a contradiction to the fact that char(D) = k. Hence char(D) must be either 0 or a prime number. Section. 20 In the exercises from 11 till 14 we are using theorem 20.12 form the book. Exercise. 11 2x ≡ 6mod(4), GCD(2, 4) = 2, and 2 divides 6, hence the we have 2 solutions for this equation in Z4, they are x = 1 + 4Z, and x = 3 + 4Z. Exercise. 12 22x ≡ 5mod(15), GCD(22, 11) = 1, and hence we have only one solution for this equation in Z15, and the solution is 5 + 15Z. Exercise. 13 36x ≡ 15mod(24) but GCD(36, 24) = 12 and 12 doesn’t divide 15, hence we have no solution. Exercise. 14 45x ≡ 15mod(24), GCD(45, 24) = 3, and 3 divides 15, hence we can divide the congruence by 3 to get the equation : 15x ≡ 5mod(8) which is equivalent to 7x ≡ 5mod(8) which is the same as solving 7x = 5 in Z8, but 7 is invertible with inverse 7 in Z8, hence we get x = 3 in Z8, so the solutions for our equation are 3 + 24Z, 11 + 24Z, and 19 + 24Z. 2 Exercise. 27 For an element a to be its own inverse it must satisfy a2 = 1. SO in Zp the only elements that are their own ,multiplicative inverse are the solutions for the equation x2 − 1 = 0 =⇒ (x − 1)(x + 1) = 0, but since Zp is a field whenever p is prime, this can only happen if x − 1 = 0 or x + 1 = 0, i.e x = 1 or x = −1 ≡ p − 1 mod(p). So we deduce that the only elements that are their own multiplicative inverse in Zp are 1, p − 1. Section. 21 Exercise. 1 We consider the function f : F −→ F ′ = {q1 + q2i | q1, q2 ∈ Q}, where F is the field of fractions of the integral subdomain D = {n + mi | n, m ∈ Z}. we define f(n+mi, n′+m′i) = n+mi n′+m′i, this is a well defined function since we can write n+mi n′+m′i = nn′+mm′ n′2+m′2 + mn′−m′n n′2+m′2 i, which is an element in F ′, and this is possible because (n + mi, n′ + m′i) ∈ F implies that n′ + m′i is non-zero. f is surjective since q1 + q2i = m n + a bi = mb+ani nb = f(mb + ani, nb + 0i) where n, m, a, b ∈ Z. f is injective since by definition of F we have that n+mi n′+m′i = a+bi a′+b′i =⇒ (n + mi, n′ + m′i) = (a + bi, a′ + b′i). Finally we still have to prove that f is a ring homomorphism, to do that let α1 = (n + mi, n′ + m′i), α2 = (a + bi, a′ + b′i) ∈ F we have to prove that f(α1 + α2) = f(α1) + f(α2), and that f(α1.α2) = f(α1.α2), which can be easily done through some calculations. Hence we can describe the elements of the field F to be all complex numbers with rational components. Section. 26 Exercise. 12 Let R = Z,then R is an integral domain, it is easy to see that 2Z is an ideal of R with R/2Z = Z2 is a field. Exercise. 13 Also Let R = Z, and consider the ideal 4Z, then it is easy to see that R/4Z has zero divisors. Exercise. 14 Consider R = Z × Z, and let I = Z × {0}, it is easy to see that I is an ideal of R, and R/I ∼= Z which is an integral domain, while R has zero divisors ( (0,1).(1,0)=(0,0) ). Exercise. 17 We proved in previous home work that R is ring, and since it is a subset of R, then it is a subring of R. Now let us prove that φ an injective homorphism whose image is R′, hence we get that R′ is a subring of M2(R). φ is well defined map from R into M2(R). Let a + b √ 2 ,and a′ + b′√ 2 ∈ R′ then φ(a + b √ 2 + a′ + b′√ 2) = φ(a + a′ + (b + b′) √ 2) = � a + a′ 2(b + b′) b + b′ a + a′ � = � a 2b b a � + � a′ 2b′ b′ a′ � = φ(a + b √ 2) + φ(a′ + b′√ 2). 3 φ((a + b √ 2)(a′ + b′√ 2)) = φ(aa′ + 2bb′ + (ab′ + ba′) √ 2) = � aa′ + 2bb′ 2(ab′ + ba′) ab′ + ba′ aa′ + 2bb′ � = � a 2b b a � � a′ 2b′ b′ a′ � = φ(a + b √ 2)φ(a′ + b′√ 2). Hence φ is a ring homomorphism. Next suppose φ(a + b √ 2) = 0, hence a = b = 0, hence φ is injective. Finally let g ∈ R′, then g = � a 2b b a � for some a, b ∈ Z so g = f(a + b √ 2) ∈ f(R), also f(α) ∈ R′ for all α ∈ R, hence f(R) = R′ so we get that R′ is a subring, and φ is am isomorphism between R and R′. Exercise. 22 a- We know that φ(N) is an adidtive subgroup of φ(R), so we only have to show that for all r′ ∈ φ(R), we have that r′φ(N) ⊆ φ(N), and φ(N)r′ ⊆ φ(N). let r′ ∈ φ(R) and y ∈ φ(N). Then there exists r ∈ R and x ∈ N such that r′ = φ(r) and y = φ(x). but since x ∈ N and N is an ideal =⇒ rx ∈ N =⇒ φ(rx) ∈ φ(N), =⇒ r′y ∈ φ(N) so r′φ(N) ⊆ φ(N). Similarly we can prove that φ(N)r′ ⊆ φ(N), and hence φ(N) is an ideal of φ(R). b- Let R = Z and let R′ = R, and consider N = 2Z an ideal of R , and let φ be such that φ(n) = n. Then we get φ(N) = N, but then 2Z is not an ideal of R. c- let N′ be an ideal of φ(R), then let us prove that φ−1(N′) is an ideal of R. We know that φ−1(N′) is an adidtive subgroup of R, so we only have to prove that rφ−1(N′) ⊆ φ−1(N′), and φ−1(N′)r ⊆ φ−1(N′) for all r ∈ R. let r ∈ R and x ∈ φ−1(N′). We want to show that rx and xr belong to φ−1(N′). Now x ∈ φ−1(N′) means that φ(x) ∈ N′. Since N′ is an ideal of R′, and φ(r) ∈ R′, we know that φ(r)φ(x) and φ(x)φ(r) both belong to N′. But then φ(rx) and φ(xr) belong to N′, which means that rx, xr ∈ phi−1(N′), which is what we wanted to show. Exercise. 37 First it is easy to see that φ is a well defined map between C and M2(R). Then let c1 = a1 + b1i, c2 = a2 + b2i ∈ C, then φ(c1 + c2) = φ((a1 + b1i) + (a2 + b2i)) = φ(a1 + a2 + (b1 + b2)i) = � a1 + a2 b1 + b2 −(b1 + b2) a1 + a2 � = � a1 b1 −b1 a1 � + � a2 b2 −b2 a2 � = φ(c1) + φ(c2), and φ(c1.c2) = φ(a1a2−b1b2+(a1b2+a2b1)i) = � a1a2 − b1b2 a1b2 + a2b1 −(a1b2 + a2b1) a1a2 − b1b2 � = � a1 b1 −b1 a1 � � a2 b2 −b2 a2 � = φ(c1)φ(c2). Hence we deduce that φ is a ring homomorphism. Moreover, φ is injective since, φ(a + bi) = 0 =⇒ a = b = 0. Then φ is an isomorphism. 4