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, October 31. Main idea: Finishing up the proofs of antiquity. Key words: duplicating the cube, squaring the circle, trisecting the angle. Goal: Learn the solution to the three problems of antiquity Previous Assignment: 1. x = 2^(1/2) + 2^(1/3). (a) Why must x satisfy a polynomial over the rationals? We can capture x by making two extensions to the rationals. x is in K = Q[2^(1/2)][2^1/3] The degree of K over Q is 2*3 = 6. Thus x satisfies a polynomial of degree at most 6. (b) Why must x satisfy a polynomial of degree no more than 6? Because the degree of K over Q is 6, 1,x,x^2,x^3,x^4,x^5,x^6 have do be linearly dependent. This linear combination of powers of x is a polynomial over Q that x satisfies. (c) What is that polynomial? Hint: {1,x,x^2,x^3,x^4,x^5} are given below. x^0 = 1 x^1 = 2^(1/3) + Sqrt[2] x^2 = 2 + 2^(2/3) + 2*2^(5/6) x^3 = 2 + 6*2^(1/6) + 6*2^(1/3) + 2*Sqrt[2] x^4 = 4 + 2*2^(1/3) + 8*Sqrt[2] + 12*2^(2/3) + 8*2^(5/6) x^5 = 40 + 40*2^(1/6) + 20*2^(1/3) + 4*Sqrt[2] + 2*2^(2/3) + 10*2^(5/6) x^6 = 12 + 24*2^(1/6) + 60*2^(1/3) + 80*Sqrt[2] + 60*2^(2/3) + 24*2^(5/6) We want a dependence relation between 1,x,x^2,x^3,x^4,x^5,x^6. 1 0 2 2 4 40 12 0 0 0 6 0 40 24 0 1 0 6 2 20 60 0 1 0 2 8 4 80 0 0 1 0 12 2 60 0 0 2 0 8 10 24 The Row-Canonical Form is 1 0 0 0 0 0 4 0 1 0 0 0 0 24 0 0 1 0 0 0 -12 0 0 0 1 0 0 4 0 0 0 0 1 0 6 0 0 0 0 0 1 0 4 + 24 x -12 x^2 + 4 x^3 + 6 x^4 -x^6 = 0 is the polynomial satisfied by x. 2. Find the polynomial satisfied by Cos[20] degrees. Hint: Cos[A+B] = Cos[A] Cos[B] - Sin[A] Sin[B]. Sin[A+B] = Sin[A] Cos[B] + Cos[A] Sin[B]. Cos[60] = 1/2. Cos(3a) = Cos(2a+a) = Cos(2a)Cos(a)-Sin(2a)Sin(a) = (Cos(a)^2 - Sin(a)^2)Cos(a) -2Sin(a)Cos(a)Sin(a) = Cos(a)^3 -3 Cos(a)Sin(a)^2 = Cos(a)^3 -3 Cos(a)(1-Cos(a)^2) = 4 Cos(a)^3 -3 Cos(a) If a = 20 degrees, then 3a = 60 degrees and Cos(3a) = 1/2 So 1/2 = 4 Cos(20)^3 - 3 Cos(20) The polynomial satisfied by Cos(20) is 8 x^3 - 6 x - 1. 3. You have received a letter from a young friend in High School. She has just read in her geometry book that it is impossible to trisect an angle. She says, "How do they know that I cannot trisect the angle. They do not know all of the clever things I might try. Maybe the author of the book can't find a construction. But that does not mean that I can't. Explain to her the technique employed to show that 60 degrees cannot be trisected. Keep it simple. She only wants the basic idea. She is too young for the details. Given a construction, we can compute the coordinates of each point using the equations of lines and circles using analytic geometry. There is a property that the coordinates of each of these points have. A number has the "constructible property" if the smallest polynomial that it satisfies has degree 2^n. One can show that if all the points in a partial construction have the "constructible property", then points located by using the intersection of two known lines, or one known lines and one known circle will also have the "constructible property". If one ever constructs a point that does not have the constructible property, then there is a flaw in the proof. To trisect 60 degrees will allow one to construct the point (Cos[20], Sin[20). The degree of the polynomial for Cos[20] is 3. This is impossible and so it must be impossible to trisect 60 degrees. Remark: The intersection of two circles can be obtained by the intersection of a line and a circle. x^2 + y^2 + ax + by + c = 0 x^2 + y^2 + Ax + By + C = 0 are two circles. Subtract them (A-a)x+(B-b)y+(C-c) = 0. This is a line, of course, it passes through the points of intersection of the two circles. Now you can find the points of intersection by solving this line with either circle. The line is called the "radical axis". It is obvious where it is when the circles intersect. It still exists even when the circles do not intersect and has some special properties with reference to the circles. But I cannot remember what they are now. Theorem: (Page 416 8.4.9, 8.4.10 8.4.11) (a) You cannot duplicate the cube. (b) You cannot trisect angles. (c) You cannot square the circle. Proof: Duplicating the cube requires locating a line segment of length x = CubeRoot[2]. The minimal polynomial satisfied by x is x^3 -2. Of course, some angles can be trisected. The angle of 90 degrees is one of them. If there were a construction to trisect any angle, one could apply it to 60 degrees and construct the point (Cos[20],Sin[20]). But Cos[20] satisfies a minimal polynomial 8 x^3 - 6 x - 1. Since the degree of this polynomial is 3, this is impossible. We have to leave a big hole in this proof. A number is rational if it can be written as p/q where p and q are integers. A number is irrational if it is real and not rational. A number is algebraic if it satisfies a polynomial over the rationals. A number is transcendental if it is irrational, but not algebraic. Once we know that Pi is transcendental, then it satisfies no polynomial over the integers. Thus it cannot satisfy a polynomial of degree a power of 2. Squaring the circle requires constructing Sqrt[Pi]. If you could construct Sqrt[Pi] you could construct Pi. But that would be impossible.