301 Abstract Algebra 2:00 - 2:50 MWF Hentzel 432 Carver hentzel@iastate.edu Web Page http://www.math.iastate.edu/hentzel/class.301 Text: A First Course in Abstract Algebra, sixth Edition by John B. Fraleigh Course: Chapter 1,2,3,4,5 Wednesday, December 5 Main Idea: Key Words: Goal: Assignment: A symmetric function f(x1,x2,...,xn) is a function such that f(pi(x1),pi(x2), ..., pi(xn)) = f(x1,x2,...,xn) for all permutations pi. (a) Show that f(a,b,c) = a/b+b/c+c/a takes on two values as one permutes the elements a,b,c. f(a,b,c) = f(b,c,a) = f(c,a,b) = a/b+b/c+c/a f(b,a,c) = b(a,c,b) = f(c,b,a) = b/a+c/b+a/c (b) Show that f(a,b,c,d) = ab+cd takes on 3 values as one permutes a,b,c,d. The possible values are ab+cd ac+bd ad+bc (c) p(x) = (x-a)(x-b)(x-c)(x-d). Compute the coefficients of this polynomial and show that they are symmetric functions in a,b,c,d. (x-a)(x-b)(x-c)(x-d) = x^4 -(a+b+c+d) x^3 + (ab+ac+ad+bc+bd+cd) x^2 - (abc+abd+acd+bcd) x + (abcd) (d) Express x^3+y^3+z^3 in terms of the elementary symmetric functions x+y+z, xy+xz+yz, xyz. 2 2 2 2 2 2 x^3+y^3+z^3 - (x+y+z)^3 = -3 x y-3xy -3 x z-6xyz-3y z-3xz -3yz x^3+y^3+z^3-(x+y+z)^3+3(x+y+z)(x y+x z+y z) = 3 x y z x^3+y^3+z^3 - (x+y+z)^3 + 3(x+y+z)(x y+x z+y z) - 3 x y z = 0 Suppose we wish to solve x^4 + px^3 + qx^2 + rx + x = 0. Let us assume that the roots are a,b,c,d Then x^4 - px^3 + qx^2 - rx + x = (x-a)(x-b)(x-c)(x-d) p = (a+b+c+d) q = (ab+ac+ad+bc+bd+cd) r = (abc+abd+acd+bcd) s = abcd If we look at (x-(ab+cd))(x-(ac+bd))(x-(ad+bc)) This will be a cubic polynomial and its coefficients will be symmetric functions in a,b,c,d. constant term = -((bc+ad)(ac+bd)(ab+cd)) = -(a+b+c+d)^2 abcd + 2(ab+ac+ad+bc+bd+cd)(abcd) -(abc+abd+acd+bcd)^2 2 (ab+ac+ad+bc+bd+cd)(abcd) = -p^2 s + 2 qs - r^2 + 2 qs Coefficient of x = 2 2 2 2 2 2 2 a b c + a b c + a b c + a b d + a b d + a c d + b c d + 2 2 2 2 2 a c d + b c d + a b d + a c d + b c d = (a+b+c+d)(abc+abd+acd+bcd)-4abcd = pr - 4s Coefficient of x^2 = -ab - ac - bc - ad - bd - cd = -q So we certainly can write down the coefficients and solve the degree three polynomial x^3+(-q)x^2+(pr-4s)x+(-p^2 s+2qs-r^2+2qs) = 0 and this will give us the roots a b + c d a c + b d a d + b c. Now we can solve the quadratics x^2-(ab+cd)x+abcd x^2-(ac+bd)x+abcd x^2-(ad+bc)x+abcd and get the roots ab, cd, ac, bd, ad, bc (ab)(ac) (bc)(bd) (ac)(bc) (ad)(bd) Now -------- = a^2 ---------- = b^2 -------- = c^2 --------- = d^2 bc cd ab ab So we can get a. b, c, d. Actually, we consider all possible products, 2*2*2 = 8 of them. Check each of the answers. 4 will be correct and 4 will be extraneous. We check by putting them into the original polynomial The theory works like this: The function f(a,b,c,d) = ab+cd has four arguments and took on less than 4 values. If we wish to find a solution to p(x) = x^n + a1 x^(n-1) + ... + an, We look for a function f(x1,x2, ..., xn) which takes on fewer than n values. Suppose the values were y1, y2, ... yn-1. Then the polynomial (x-y1)(x-y2) ... (x-yn-1) can be solved since its coefficients are symmetric functions in the coefficients of p(x) and are known. This gives us the values y1, y2, ... , yn-1 to work with. We can use those values to eventually get the roots of the given polynomial. This method will not work above degree 4 because there are no functions with n variables which take on fewer than n values except the trivials ones. One value or two values. What we have is a homomorphism from Sn to Sn-1. The kernel of this homomorphism is a normal subgroup of Sn. The only normal subgroup of Sn for n>= 5 is An. So we can get started. The function to use is y = (a-b)(a-c)(a-d)(a-e)(b-c)(b-d)(b-e)(c-d)(c-e)(d-e). y^2 is a symmetric function so we can find y. This is all we can find. The other possibility is only one value. This is of no use. No assignment for Friday.