MATH 241 Homework 12, Fall 2011

Math 241 Problem Set 12 solution manual Exercise. A12.1 a- Notice that using Fermat’s theorem we can see that for any x ∈ Zp we have that xp − x = 0, and hence we can write xp − x = x(x − 1)(x − 2)...(x − (p − 1)). b- The roots of x3 − x in Z6 are 0,1,2,3,4,5. It is not surprising to have 6 roots for an equation of degree 3 since our ring is not integral domain. Section. 22 Exercise. 3 f(x) + g(x) = (2x3 + 3x + 4) + (3x2 + 2x + 3) = 5x2 + 5x + 7 = 5x2 + 5x + 1. f(x).g(x) = (2x3 + 3x + 4).(3x2 + 2x + 3) = 6x4 + 13x3 + 24x2 + 17x + 12 = x3 + 5x. Exercise. 4 f(x) + g(x) = 3x4 + 2x3 + 4x2 + 1. f(x).g(x) = x7 + 2x6 + 4x5 + x3 + 2x2 + x + 3 Exercise. 5 A polynomial of degree three is of the form a0 + a1x + a2x2 + a3x3, where a0, a1, a2, and a3 are in Z2, hence we have 16 such polynomials. Exercise. 9 Φ3[f(x)] = f(3) = (34 + 6)(33 + 3.32 + 3) = (81 + 6)(27 − 27 + 3) = (87)(3) = (3)(3) = 9 = 2. Exercise. 11 Using Fermat’s theorem and since 4 is not equal to 0 mod7, =⇒ (46) ≡ 1 mod7. (4106) = (4102).44 = (46)17.44 = 256 = 4 mod7 , also similarly you can find out that 499 = 1, and 453 = 2. Hence Φ4(3x99 + 5x11 + 2x53) = 3.4 + 5.1 + 2.2 = 0. Exercise. 12 f(1) = 12 + 1 = 0, hence 1 is a root. f(0) = 1, hence 0 is not a root. So the only root is 1. Exercise. 14 f(0) = 0 f(1) = 2 f(2) = 4 f(3) = 4 f(4) = 0 Hence the roots of f are 0, and 4. 1 Exercise. 15 By simple calculation you can see that 0 and 2 are the roots of f, and that 0 and 4 are the roots of g. Hence 0, 2, 4 are the roots of f.g. Exercise. 27 a- Let f(x) = a0+a1x+...+anxn, and g(x) = b0+b1x+...+bmxm, without loss of generality we can assume that m ≥ n. Hence f + g = (a0 + b0) + (a1 + b1)x + ... + (an + bn)xn + bn+1xn+1 + ... + bmxm. D(f + g) = a1 + b1 + 2(a2 + b2)x + ... + n(an + bn)xn−1 + (n + 1)bn + 1xn + ... + mbmxm−1 = (a1 + 2a2x + ... + nanxn−1) + (b1 + 2b2x + ... + mbmxm−1) = D(f) + D(g). So D is a well defined map satisfying D(f + g) = D(f) + D(g), hence D is a group homomorphism of the group < F[x], + >. This map can’t be a ring homomorphism since it doesn’t satisfy D(fg) = D(f) + D(g), in fact we are going to prove that D(fg) = fD(g) + gD(f) for all f, g ∈ F[x]. So let f, g ∈ F[x] (f, g as defined above). Then f.g = c0 + c1x + ... + cn+mxn+m where ci = Σ j=0...iajbi−j. D(f.g) = c1 +2c2x+...+(n+m)cn+mxn+m−1, and on the other hand fD(g)+gD(f) = (a0 + a1x+...+anxn)(b1+2b2x+...+mbmxm−1)+(b0+b1x+...+bmxm)((a1+2a2x+...+nanxn−1) = (a0b1 + b0a1) + (2a0b2 + 2a2b0 + a1b1 + a1b1)x + ... + ((n + m)anbm)xn+m−1 = D(f.g). b- Ker(f) = {f = a0 + a1 + ... + anxn | D(f) = 0} = {f = a0 + a1 + ... + anxn | a1 + 2a2x + ... + nanxn−1 = 0} = {f = a0 | a0 ∈ F}, and this is because our field F is of characteristic 0 hence the polynomial a1 + 2a2x + ... + nanxn−1 is only a 0 polynomial if all its coefficients are zero. c- The image of our map is F[x], since given g = b0 + b1x + ... + bmxm, we let f = c + b0x + 1 2b1x2 + ... + 1 m+1bmxm+1, and hence it is easy to see that D(f) = g, so f is surjective. Section. 23 Exercise. 2 5x4 + 5x2 + 6x x6 + 3x5 + 4x2 − 3x + 2 3x2 + 2x − 3 x6 + 3x5 + 6x4 x4 + 4x2 − 3x + 2 x4 + 3x3 + 6x2 4x3 + 5x2 + 4x + 2 4x3 + 5x2 + 3x x + 2 Exercise. 4 We do the same as above to get q(x) = 9x2 + 5x + 10, and r(x) = 2. 2 Exercise. 7 Check Solution manual for problem set 5, ex 3 section 20. Exercise. 10 f(x) = x3 +2x2 +2x+1, notice that f(2) = 0, and hence x−2 is a factor of f, so using the division algorithm as in ex 3 above, we get that f = (x − 2)(x2 + 4x + 3). Now let g(x) = x2 + 4x + 3, and we can see now that g(−1) = g(6) = 0, and hence again x − 6 is factor of g, so dividing again we get that f(x) = (x − 2)(x − 6)(x + 3). Exercise. 12 Following the same technique as in ex 10 , we can factorize f(x) into (x − 2)(x − 4)(x + 1) = (x − 2)(x + 1)2. Exercise. 13 f is a polynomial of degree 3, which has no roots in the field Z5, hence by theorem 23.10 we deduce that f is irreducible. 3