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 Monday, December 3 Main Idea: The Quartic is not that hard to solve. Key Words: Symmetric Function, Elementary Symmetric Functions. Goal: Show how the process of reduction of degree is used for cubic and quartic. Hand-in-homework 1. Solve 4 x^3 - 3x = 1/Sqrt[2] x = z+1/(4z) z^3 -Sqrt[2]/8 + 1/(64 z^3) = 0 z^6 - Sqrt[2]/8 z^3 + 1/64 = 0 1+I Sqrt[2] 1-I Sqrt[2] z^3 = ------------ z^3 = ------------ 16 16 z^3 = 1/8 Cis[45] z^3 = 1/8 Cis[315] z = 1/2 Cis[15] z = 1/2 Cis[105] x = z+1/(4z) x = z+1/(4z) = 1/2 Cis[15]+ 1/(2 Cis[15]) x = 1/2 Cis[105]+1/(2 Cis[105]) = 1/2 Cis[15]+1/2 Cis[-15] x = 1/2 Cis[105]+1/2 Cis[-105] = Cos[15]. x = Cos[105] Actually, the three cube roots of 1/8 Cis[45] and 1/8 Cis[315] are 1/2 Cis[15] 1/2 Cis[105] 1/2 Cis[135] 1/2 Cis[225] 1/2 Cis[255] 1/2 Cis[345] The corresponding values of x are Cos[15] Cos[105] = Cos[-105] = Cos[255] Cos[135] Cos[225] = Cos[-225] = Cos[135] Cos[255] Cos[345] = Cos[-345] = Cos[15] So you actually get 6 roots, but each one is obtained twice. 2. Solve 4 x^3 - 3x = Sqrt[3]/2 = 0 z = z + 1/(4z) z^3 -Sqrt[3]/8 + 1/(64z^3) = 0 z^6 -Sqrt[3]/8 z^3 + 1/64 = 0 Sqrt[3] - I Sqrt[3] + I z^3 = ------------ z^3 = ------------ 16 16 z^3 = 1/8 Cis[-30] z^3 = 1/8 Cis[30] z = 1/2 Cis[-10] z = 1/2 Cis[10] z = 1/2 Cis[110] z = 1/2 Cis[130] z = 1/2 Cis[130] z = 1/2 Cis[250] x = Cos[-10] x = Cos[10] x = Cos[110] x = Cos[130] = Cos[-130] = Cos[230] x = Cos[230] x = Cos[250] = Cos[-250] = Cos[110] 3. Find a substitution x = y+r so that the equation x^3 + ax^2 + bx + c = 0 loses the x^2 term to become y^3 + b' y + c' = 0 so that then the previous technique can be used. f[x_] := x^3 + a x^2 + bx + c 2 3 bx + c + a (r + y) + (r + y) = 0 2 3 2 2 3 bx + c + a r + r + (2 a r + 3 r ) y + (a + 3 r) y + y = 0 So let r = -a/3. The quartic Formula: We solve x^4 + ax^2 + bx + c = 0. (* Notice the x^3 is missing *) We wish to split up the x^4 + ax^2 + bx + c so that we get the sum of two squares: (x^2 + r)^2 + (s x + t)^2 = x^4 + a x^2 + b x + c x^4 + 2 r x^2 + r^2 s^2 x^2 + 2 s t x + t^2 ------------------------------------------------- x^4 + a x^2 + bx + c c = r^2 + t^2 b = 2 s t a = 2 r + s^2 (-4 a +8 r) t^2 = (-4 a + 8 r)(c - r^2) - b^2 = - 4 s^2 t^2 4 a t^2 = 8 r t^2 + 4 s^2 t^2 -b^2 = (-4a+8r)(c-r^2) (r^2-c)(-4a+8r) - b^2 = 0. 8 r^3 - 4 a r^2 - 8 c r + (4 a c -b^2) = 0 Once we have solved this cubic for r, we can (x^2 + r)^2 + (s x + t)^2 = x^4 + a x^2 + b x + c = 0 Solve the new equation for x. (x^2+r) = + i (s x + t). (x^2+r) = - i (s x + t). x^2 - i s x + r -i t = 0 x^2 + i s x + r + i t = 0 i s +/- Sqrt[-s^2 - 4 r + 4 i t] x = ---------------------------------- 2 -i s +/- Sqrt[-s^2 - 4 r - 4 i t] x = ---------------------------------- 2 We have shown the existence of a quadratic, a cubic, and a quartic formula. There is no quintic formula. There is a polynomial of degree 5 whose roots cannot be expressed by by a complicated expression involving +,-,*,/ and extraction of n th roots. How do we know that? One uses the same approach as the proof that CubeRoot[2] and Cos[20] cannot be constructed. If this polynomial had a root expressible using radicals, then one could capture this root by a series of extensions Q = Fo c F1 c F2 c ... c Fi c Fi+1 ... Fn 1/ni where Fi+1 = Fi[ (gi) ]. for some gi in Fi. This is, we keep taking some root of an element which we already have. ---------------------------------------------------------- | | |----------------------------------------------- | | | | | | | |-------------------------- | | | | | | | | | | |------------------ | | | | | | | | |--------- | | | | | | 1/n1 | 1/n2 | 1/ni | | | Q | g1 | g2 | ... gi ... | x | |--------- | | | | | | | | | |------------------ | | | | | | | | | | | |------------------------- | | | | | |---------------------------------------------- | | | ---------------------------------------------------------- Finally, there is property which holds for extensions like this. For each step one gets a normal subgroup of the permutation group S5, I = No c N1 c N2 ... c Sn and Ni+1/Ni is a abelian. Since such a chain of subgroups does not exist for Sn, one must not be able to capture a root of the polynomial by radicals. The chain of subgroups for S5 are only I c A5 c S5. And A5/I is not abelian. One can build such a chain for S4 and S3. One would have to because quartic and cubic formulas exist. I c K4 c A4 c S4 I c A3 c S3. The particular polynomial used to show that one cannot solve a quintic by radicals can be chosen to be any irreducible polynomial of degree 5 with exactly 2 complex roots. Such a polynomial is f(x) = 2x^5 - 5x^4 + 5 See Page 494 Problem 8. We will not give the proof that we cannot solve the general degree five polynomials by radicals. Instead we will show how the process of solving a degree 4 was essentially accomplished by solving a degree 3. This is called reduction of degree. Reduction of degree stops at five and we will explain why this happens. Notice that we do not prove that you cannot solve a degree five. We show that you cannot solve a degree five by reduction of degree. 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. (b) Show that f(a,b,c,d) = ab+cd takes on 3 values as one permutes a,b,c,d. (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. (d) Express x^3+y^3+z^3 in terms of the elementary symmetric functions x+y+z, xy+xz+yz, xyz. x = a/b + b/c + c/a y = b/a + c/b + a/c 2 2 2 2 2 2 a b + a b + a c + b c + a c + b c x+y = --------------------------------------- a b c x+y = -3 a b c + (a + b + c) (a b + a c + b c) --------------------------------------- a b c t = -(a b c) s = (a b+a c+b c) r = -(a+b+c) (x-a)(x-b)(x-c) = x^3 +r x^2 + s x + t 3 t - r s x+y = ---------------------------- -t 3 3 4 4 2 2 2 3 3 3 3 4 a b + a b c + a b c + 3 a b c + a c + b c + a b c x y = ------------------------------------------------------------- 2 2 2 a b c +(a+b+c)^3 a b c+(a b+b c+c a)^3-6(a+b+c)(a b+b c+c a)(a b c)+9( a^2 b^2 c^2) x y = ------------------------------------------------------------------------------ a^2 b^2 c^2 t = -(a b c) s = (a b+a c+b c) r = -(a+b+c) r^3 t + s^3 - 6 r s t + 9 t^2 x y = -------------------------------------- t^2 So we can solve the polynomial with roots x and y. Now we know a/b+b/c+c/a and b/a+c/b+a/c 2 2 2 a b + a c + b c x = ------------------ a b c 2 2 2 a b + b c + a c y = ------------------ a b c