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, November 7. Main idea: Factorization into irreducibles is often obvious. Factorization uniquely is not common. Key words: associate, irreducible, factorization into irreducibles Goal: Learn that factorization into irreducibles need not always be unique. Previous Assignment 1. The Even Integers are 2Z (a) What are the irreducibles in 2Z ans: The irreducibles are numbers which contain exactly one factor of 2. Examples are 10, 30, 70 (b) Find an element of 2Z which has two different factorizations into irreducibles. ans: 36 = 6x6 = 2x18 100 = 10x10 = 2x50 2. In Z[Sqrt[2]] = {a+bSqrt[2] | a,b are integers} (a) Find the greatest common divisor w = GCD(266 + 175 Sqrt[2], 307 + 179 Sqrt[2]) 2 - Sqrt[2] ---------------------- 266+175 Sqrt[2] | 307 + 179 Sqrt[2] 182 + 84 Sqrt[2] Sqrt[2] ------------------- ------------------- 125 + 95 Sqrt[2] | 266 + 175 Sqrt[2] 190 + 125 Sqrt[2] 1 ------------------ ------------------ 76 + 50 Sqrt[2] | 125 + 95 Sqrt[2] 76 + 50 Sqrt[2] ---------------- 49 + 45 Sqrt[2] Sqrt[2] ------------------ 49 + 45 Sqrt[2] | 76 + 50 Sqrt[2] 90 + 49 Sqrt[2] -4 - 3 Sqrt[2] --------------- ----------------- -14 + Sqrt[2] | 49 + 45 Sqrt[2] 50 + 38 Sqrt[2] - Sqrt[2] --------------- ------------------ -1 + 7 Sqrt[2] | -14 + Sqrt[2] -14 + Sqrt[2] --------------- 0 The greatest common divisor is -1+7 Sqrt[2]. 2-Sqrt[2] --------------------- 266+175 Sqrt[2] | 307 + 179 Sqrt[2] 182 + 84 Sqrt[2] Sqrt[2] ------------------- --------------- 125 + 95 Sqrt[2] | 266+175 Sqrt[2] 190+125 Sqrt[2] Sqrt[2] --------------- ------------------- 76+ 50 Sqrt[2] | 125 + 95 Sqrt[2] 100 + 76 Sqrt[2] 2 Sqrt[2] ------------------ --------------- 25 + 19 Sqrt[2] | 76+50 Sqrt[2] 76+50 Sqrt[2] ------------- 0 The Greatest common divisor is 25 + 19 Sqrt[2]. Now 25+19 Sqrt[2] and -1+7 Sqrt[2] are both associates and so are both the greatest common divisor. (-1+7 Sqrt[2])(3+2 Sqrt[2]) = 25 + 19 Sqrt[2] and 3+2 Sqrt[2] is invertible since (3+2 Sqrt[2])(3-2 Sqrt[2]) = 1 The greatest common divisor is also 25+19 Sqrt[2] (3+2 Sqrt[2])(-1+7 Sqrt[2]) = 25 + 19 Sqrt[2] So 25 + 19 Sqrt[2] and -1 + 7 Sqrt[2] are "associates" Two elements are called associates if they differ by a unit. (b) Find x and y such that x(266 + 175 Sqrt[2])+y(307 + 179 Sqrt[2]) = w From the first continued fraction: 2-Sqrt[2] Sqrt[2] 1 Sqrt[2] -4 -3Sqrt[2] -Sqrt[2] 1 2-Sqrt[2] -1+2Sqrt[2] 1+Sqrt[2] 1+3Sqrt[2] -21-14 Sqrt[2] 29+24 Sqrt[2] 0 1 Sqrt[2] 1+Sqrt[2] 2+2 Sqrt[2] -19-13 Sqrt[2] 28+21 Sqrt[2] (-1+7Sqrt[2]){29+24 Sqrt[2],28+21 Sqrt[2]} = {307 + 179 Sqrt[2], 266 + 175 Sqrt[2]} (-21-14 Sqrt[2])(28+21 Sqrt[2])-(-19-13 Sqrt[2])(29+24 Sqrt[2]) = -1 -(-21-14 Sqrt[2])(28+21 Sqrt[2])+(-19-13 Sqrt[2])(29+24 Sqrt[2]) = 1 -(-21-14 Sqrt[2])(28+21 Sqrt[2])(-1+7Sqrt[2])+(-19-13 Sqrt[2])(29+24 Sqrt[2])(-1+7Sqrt[2]) = (-1+7Sqrt[2]) (21+14 Sqrt[2])(266+175 Sqrt[2]) + (-19-13 Sqrt[2])(307 + 179 Sqrt[2]) = -1 + 7 Sqrt[2] x y Or, from the other continued fraction: 2-Sqrt[2] Sqrt[2] Sqrt[2] 2 Sqrt[2] 1 2-Sqrt[2] -1+2 Sqrt[2] 6-2 Sqrt[2] -9+14 Sqrt[2] 0 1 Sqrt[2] 3 7 Sqrt[2] (6-2 Sqrt[2])(7 Sqrt[2]) -3(-9+14 Sqrt[2]) = -1 (-6+2 Sqrt[2])(7 Sqrt[2]) +3(-9+14 Sqrt[2]) = +1 (-6+2 Sqrt[2])(7 Sqrt[2])(25+19 Sqrt[2]) +3(-9+14 Sqrt[2])(25+19 Sqrt[2]) = 25+19 Sqrt[2] (-6+2 Sqrt[2])(266 + 175 Sqrt[2]) +3(307 + 179 Sqrt[2]) = 25+19 Sqrt[2] x y 3. In Z[Sqrt[3]] = {a+bSqrt[3] | a,b are integers} (a) Find a norm. n(a+bSqrt[3]) = |a^2 - 3 b^2|. (b) Prove it is a norm n[(a+b Sqrt[3])(c+d Sqrt[3])] = n[ a c + 3 b d +(a d+b c)Sqrt[3] ] = |(a c+3 b d)^2 -3 (a d + b c)^2 | 2 2 2 2 = |(a - 3 b ) (c - 3 d )| = n[a+b Sqrt[3]] n[c + d Sqrt[3] ] The Euclidean algorithm will work if you can get a remainder which is less than the divisor. quotient --------- divisor | dividend . . . ---------- remainder Given two numbers a and b, if we can find a q and r such that a = bq + r and n(r) < n(b) then we have a Euclidean ring. The integers are a Euclidean ring. The Even integers are not a Euclidean ring. given 6 and 2, the best I can do is 6 = 2*4 -2 q = 4 r = -2 6 = 2*2 +2 q = 2 r = +2 In either case, the norm of the remainder is not strictly less than the divisor. We use the absolute value to be the norm. Theorem: Z[Sqrt[2]] = {a+b Sqrt[2] | a,b are integers} is a Euclidean ring. Proof: Given a+bSqrt[2] and c+d Sqrt[2]. a+b Sqrt[2] ----------- c+d Sqrt[2] a+b Sqrt[2] c-d Sqrt[2] = ----------- ----------- c+d Sqrt[2] c-d Sqrt[2] (a c - 2 b d) + (b c - a d)Sqrt[2] = ------------------------------------- c^2 - 2 d^2 a c - 2 b d b c - a d = -------------- + ------------ Sqrt[2] c^2 - 2 d^2 c^2 -2 d^2 a c - 2 b d b c - a d = ------------- -q + (------------ -r) Sqrt[2] + q-r Sqrt[2] c^2 - 2 d^2 c^2 -2 d^2 The absolute The absolute value of this value of this is at most 1/2 is at most 1/2. a + b Sqrt[2] = (q - r Sqrt[2]) (c+d Sqrt[2]) | a c - 2 b d b c - a d | = | ------------- -q + (------------ -r) Sqrt[2] | (c+d Sqrt[2]) | c^2 - 2 d^2 c^2 -2 d^2 | The norm of a + b Sqrt[2] - (q - r Sqrt[2]) (c+d Sqrt[2]) <= (1/4 + 2/4) n(c+d Sqrt[2]) < n(c+d Sqrt[2]) No assignment for Friday; 1. Show that the Gaussian Integers are a Euclidean Ring. 2. Find all the associates of 13+5 i in the Gaussian integers. 3. For any n which is not a perfect square, show that {a+bSqrt[2n] | a, b are integers} has a multiplicative norm.