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 Friday, October 26. This material can be found in Section 8.2 page 399 starting with the words "An Application to Field Theory" through page 403. Main idea: Satisfying a polynomial is the same as being dependent. Key words: Minimal Polynomial, Simple Extension Goal: Show that extending by roots of polynomials are finite extensions. Previous Assignment: Page 399 Exercises 8.2 Problems 1,2,3,4,6,8,10 1. | 1| | 0| { | 0|, | 1| } | 1| | 1| { | 1|, |-1| } | 1| | 1| { | 2|, | 3| } |1| |1| |0| 2. { |1|,|0|,|1| } |0| |1| |1| 1 1 0 a 1 0 1 b 0 1 1 c 1 1 0 a 0 -1 1 -a+b 0 1 1 c 1 1 0 a 0 1 -1 +a-b 0 1 1 c 1 0 1 b 0 1 -1 +a-b 0 0 2 -a+b+c 1 0 1 b 0 1 -1 +a-b 0 0 1 -a/2+b/2+c/2 1 0 0 +a/2+b/2-c/2 0 1 0 +a/2-b/2+c/2 0 0 1 -a/2+b/2+c/2 |1| |1| |0| |a| 1/2(a+b-c) |1| +1/2(a-b+c) |0| +1/2(-a+b+c) |1| = |b| |0| |1| |1| |c| This shows that the vectors span R^3. We now want to show they are linearly independent. Suppose |1| |1| |0| |0| a|1|+b|0|+c|1| = |0| |0| |1| |1| |0| Then a+b = 0 a +c = 0 b+c = 0 Eq(1)-Eq(2)-Eq(3) = -2 c = 0. Therefore c = 0 and then a = 0 and b = 0. This shows they are linearly independent. |-1| | 2| | 10| 3. { | 1|,|-3|,|-14| } | 2| | 1| | 0| |-1| | 2| | 10| | 0| 2| 1|-4|-3|+|-14| = | 0| | 2| | 1| | 0| | 0| Since a linear combination gives zero, the vectors are linearly dependent and are not a basis. 4. Q[ Sqrt[2] ] over Q { 1, Sqrt[2] } 6. Q[ CubeRoot[2] ] over Q { 1, CubeRoot[2], CubeRoot[4] } 8. Q[ i ] over Q { 1, i} 10. According to Theorem 8.2.23, the element 1+a of Z2[a] of Example 8.1.19 is algebraic over Z2. Find the irreducible polynomial for 1+a in Z2[x]. (1+a)(1+a) = 1 + a + a + a^2 = 1 + a^2 = 1 + a + 1 = a (1+a)(1+a) + (1+a) + 1 = 1 + 1+a + 1 = 0 The polynomial satisfied by (1+a) is x^2 + x + 1. Theorem 8.2.23 Page 399: Suppose F c K and the dimension of K over F is finite. Let k be an element of K. (a) Then k satisfies a polynomial with coefficients in F. (b) F c F[k] c K and the dimension of F[k] over F is the degree of the minimal polynomial satisfied by k over F. Proof: The number of independent vectors can never be more than the dimension. Therefore, for any k in K, 1, k, k^2, k^3, ..., k^n are n+1 vectors in a vector space of dimension n. Therefore they must be linearly dependent. There exists coefficients fi not all zero such that n n SUM fi k^i = 0. Then k satisfies the polynomial SUM fi x^i i=0 i=0 This finishes the proof of part (a). We now will show part (b). By definition F[k] is the smallest field containing F and k. Certainly the smallest field containing k will also contain 1, k, k^2, k^3, ..., k^n, ... . If we can show that <1, k, k^2, k^3, ..., k^d> is a field for some d, then F[k] = <1, k, k^2, k^3, ..., k^d>. d Suppose that SUM ci x^i is the minimal polynomial satisfied by k. i=0 Then 1, k, k^2, k^3, ..., k^(d-1) have to be linearly independent because if not, k would satisfy a polynomial of smaller degree then d. We know that the dimension of G = <1,k, k^2, k^3, ..., k^(d-1)> is d. We will now show that G is a field. The properties of a field which are not obvious for G are closure under multiplication and the existence of multiplicative inverses. Closure under multiplication ---------------------------- d-1 G is closed under multiplication because k^d = SUM -ci k^i i=0 allows us to reduce all powers of k to linear combinations of powers less than d and such things are in G. Existence of inverses --------------------- We know that given w =/=0 in G, then 1, w, w^2, ... are all in G. Let q(x) = cr x^r + cr-1 x^(r-1) ... + co be some polynomial of minimal degree satisfied by w. If co were zero then we could factor an x out of each term of q(x) giving q(x) = x (cr x^(r-1) + ... + c1). Then q(w) = w (cr w^(r-1) + ... + c1) = 0. Since w =/= 0, then cr w^(r-1) + ... + c1 = 0 and this contradicts that r was minimal among the degrees of polynomials satisfied by w. Since co =/= 0 we can rewrite q(w) = 0 to give: w (cr w^(r-1) + ... + c1) = -co The inverse of w is (-1/co)(cr w^(r-1) + ... + c1) which is in G. Therefore G = F[k] so the dimension of F[k] over F is d. Assignment: 1. Show that the intersection of 2 F-lines is an F-point 2. Show that the intersection of an F-line and an F-circle is an F[ Sqrt[gamma] ] point where gamma is in F. 3. Show that two F-points determine 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. 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. 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. 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. 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)