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, October 29. Main idea: The DNA identifier of constructible numbers. Key words: Minimal polynomial of degree 2^n. Goal: The algebraic way to show something is impossible is to find a property that must hold for objects that are legally created. If somebody creates an object without that property, then it must have been created illegally. Previous Assignment: 1. Show that the intersection of 2 F-lines is an F-point Ans: Suppose a,b,c,d,e,f are elements of F. If we have two F-lines: ax+by=c dx+ey=f The intersection is (x,y) where |c b | |f e | ce-bf x = ------------ = ------- |a b | ae-bd |d e | |a c | |d f | af-cd y = ------------ = ----------- |a b | ae-bd |d e | both x and y are in F because F is closed under multiplication, subtraction, and division. 2. Show that the intersection of an F-line and an F-circle is an F[ Sqrt[gamma] ] point where gamma is in F. Solution: Suppose A,B,C,m,b are elements of F. x^2 + y^2 + Ax + By + C = 0 is an F-circle. y = m x + b is an F-line Substituting y = mx+b into the equation of the circle gives x^2 + (m x+b)^2 + A x + B(m x+b) + C = 0 (1+m^2) x^2 + (A + 2 b m + B m) x + b^2 + b B + C = 0 This is a quadratic and the solution is given by the quadratic formula: 2 2 2 -A - 2 b m - B m (+/-)Sqrt[(A + 2 b m + B m) - 4 (b + b B + C) (1 + m )] x = ----------------------------------------------------------------------- 2 2 (1 + m ) x = r + s Sqrt[ gamma ] where 2 2 2 2 2 gamma = A - 4 b - 4 b B - 4 C + 2 A (2 b + B) m + B m - 4 C m -A - 2 b m - B m r = --------------------- 2 2 (1 + m ) -1 s = ----------- 2 2 (1 + m ) This means that if the coefficients for the line and the circle are all in a field F, the point of intersection will not necessarily have its coordinates in F. The point of intersection will have its coefficients in F[ Sqrt[gamma] ] where gamma is in F. 3. Show that two F-points determine an F-line. Solution: (p,q) and (r,s) are points with p,q,r,s in F the equation of the line through these two points is: y-q= (q-s)/(p-r) (x-p) y = mx + b where q-s m = -------- p-r ps-qr b = ------------ p-r Since the coefficients of the line are in F, it is an F-line. 4. Show that a circle with a F-point for a center and two F-Points to determine the radius, is an F-circle. (a,b) is the center and the distance from (c,d) to (e,f) is the radius. r = Sqrt[(c-e)^2 + (d-f)^2] The equation of the circle is (x-a)^2 + (y-b)^2 = r^2 or (x-a)^2 + (y-b)^2 = (c-e)^2 + (d-f)^2 Expanding this out will give the equation of a circle x^2 + b^2 + Ax+By+C = 0 and A,B,C are all in F. 5. Explain why a construction to locate a point (x,y) implies the existence of a sequence of fields Q = Fo c F1 c F2 c F3 .... c Fn where the degree of Fi+1 over Fi is 2 and x and y are in Fn. Solution: We start with the rationals Q = Fo. The construction will give us instructions to locate points. The construction will eventually terminate after locating a finite number of points. At each stage of the construction we have a field Fi which contains the coordinates of the points and the coefficients of the equations of any lines and circles which have been used. If the next statement of the construction requires us to find the intersection of two lines, that point will be an Fi-point. If the next statement of the construction requires the intersection of a line and a circle, that point will be an Fi[ Sqrt[gamma] ] point where gamma is in F. If we keep track of each of these extensions we have a chain of fields Q = Fo c F1 c F2 c F3 c ... c Fn where Fi+1 = Fi[ Sqrt[ gamma ] ] where gamma is in Fi. Theorem 8.3.4 Page 403: If E c F c K are fields and the dimension of F over E is r and the dimension of K over F is s then the dimension of K over E is rs. E c F c K <-- r --><-- s --> <------ rs ------> Proof: Suppose that f1, f2, ..., fr is a basis of F over E. Suppose that k1, k2, ..., ks is a basis of K over F. We will show that the rs elements B = {fi kj | i=1..r, j=1..s} are a basis of K over E. ___________________________________________________________________ | | |____________________________ | | | K | |_______ | | | | F | | | E | | | | | f1 f2 ... fr | k1 k2 ... ks | | | | | |______| | | | | | |___________________________| | | | |_________________________________________________________________| We will first show that B is a spanning set. If x is an element of K, since the k's are a basis of K over F we can write x as a linear combination of the k's with coefficients in F. s x = SUM cj kj where the cj are in F. j=1 Since the f's are a basis of F over E, each of the cj's is a linear combination of the f's with coefficients in E. r So cj = SUM eij fi with eij in E. i=1 s r s r This means that x = SUM (SUM eij fi) kj = SUM SUM eij (fi kj) j=1 i=1 j=1 i=1 and we have shown that x is a linear combination of elements in B with coefficients in E. We will now show that the rs elements in B are linearly independent. s r Suppose that 0 = SUM SUM eij (fi kj) j=1 i=1 s r Then 0 = SUM (SUM eij fi) kj j=1 i=1 Since the k's are linearly independent over F we must have all of the coefficients are zero. Thus for each j, r 0 = SUM eij fi i=1 Now, since the f's are linearly independent over E, all of the coefficients eij are zero. We have just proved that B is a set of independent vectors. Since B is an independent spanning set, B is a basis. The dimension of a vector space is the number of elements in a basis, so the dimension of K over E is rs. End of Proof. Now going back to a Geometric Construction. The existence of a geometric construction of a point (a,b) means that there is a sequence of fields Q = F0 c F1 c F2 c ... c Fi c Fi+1 c ... c Fn and both coordinates of the point (a,b) are elements of Fn. Since Fi+1 = Fi[ Sqrt[gi] ] for some gi in Fi, we have two possibilities. If Sqrt[gi] is already in Fi, then Fi = Fi+1. If Sqrt[gi] is not in Fi, then the dimension of Fi+1 over Fi is 2 and it has a basis of 1 Sqrt[gi]. Q = F0 c F1 c F2 c ... c Fi c Fi+1 c ... c Fn <-?--><-?--> <--?-> <-?--> <---------------------------------------> 2^n The total dimension of Fn over Q will be the product of all of the successive dimensions. Since these successive dimensions are either 1 or 2, the product of all of them will be a power of 2. Now a and b are both in Fn. Let us consider the element b. The explanation for a will be identical. Q c Q[b] c Fn The field generated by Q and b will lie somewhere between Q and Fn. Q c Q[b] c Fn <---r--><---s---> We let r be the dimension of Q[b] over Q and We let s be the dimension of Fn over Q[b]. Then r*s = 2^n because the degree of Fn over Q is a power of 2. This means that r itself is a power of 2. Since the degree of the field Q[b] over Q will be the degree of the minimal polynomial satisfied by b, we know that if (a,b) is constructible, then b satisfies a polynomial over the rationals and the degree of the minimal such polynomial is a power of 2. Assignment: 1. x = 2^(1/2) + 2^(1/3). (a) Why must x satisfy a polynomial over the rationals? (b) Why must x satisfy a polynomial of degree no more than 6? (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) 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. 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.