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 14. ############################################### # # # Test, Wednesday, November 28 # # # ############################################### The main part of this lesson is on pages 368-381 starting with "Euclidean Domains." Main idea: Unique factorization of the integers. Key words: Irreducible, Prime, Euclidean Algorithm Goal: Give the proof that the integers have unique factorization. Unique Factorization using Euclidean property. 1. Show factorization into irreducibles. 2. Show Greatest Common Divisors Exist. 3. Show that the Greatest Common Divisor of A and B is expressible as GCD(A,B) = XA+YB 4. Show that any Irreducible is a Prime. 5. Show that elements can be expressed as a product of primes uniquely. Start of the proof that the integers are a unique factorization domain. 1. Any integer different {-1,0,1} can be factored into irreducibles. Proof: Given an integer a which is not -1,0,1, repeatedly factor a and its factors as long as they have nontrivial factors. a row 0 / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ ao a1 row 1 / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ aoo ao1 a1o a11 row 2 / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ aooo aoo1 ao1o ao11 a1oo a1o1 a11o a111 row 3 IMPORTANT: GIVE A REASON WHY THIS FACTORING WILL NOT SIMPLY GO ON FOREVER. If x = yz then |x| = |y||z|. If this factorization is non trivial, then |y| < |x| and |z| < |x|. Therefore in row i of the table, those elements that are not irreducible have absolute value less than |a|-i. After |a|-1 rows, the elements in the row must all be irreducible. Therefore a can be factored into a finite number of irreducibles. 2. Given a and b not both zero, there exists a number GCD(a,b) such that: (a) GCD(a,b) divides a and GCD(a,b) divides b. (b) If d divides a and d divides b, then d divides GCD(a,b). Proof: Given integers a and b not both zero, let S = {ax+by | x,y are integers}. Since a and b are not both zero, there are positive elements in S. Let axo+byo be a smallest positive element in S. We claim that axo+byo is a GCD(a,b). (a) We claim that axo+byo divides a. The proof that axo+byo divides b is similar. We can use the Euclidean algorithm to write a = (axo+byo)q + r where 0 <= r < axo+byo Thus a(1-xoq)+b(-yoq) = r. Since r is in S, by minimality of axo+byo we have r = 0. Thus axo+byo divides a. (b) If d divides a and d divides b, then d | ax+by for all x and y. Therefore d | axo+byo. 3. Every irreducible is a prime. Proof: Let p be an irreducible and suppose that p divides ab. GCD(p,a) divides p so GCD(p,a) is either a unit or an associate of a. (a) If GCD(a,b) is a unit, then px+ay = 1 multiply this by b gives pxb + aby = b. Since p divides the left hand side, then p divides b. (b) If GCD(a,b) is an associate of a, then p divides GCD(p,a) and GCD(p,a) divides b So p divides b. 4. Factorization into irreducibles is unique. Proof: Suppose p1 p2 ... pm = q1 q2 ... qn where the pi and qj are irreducibles. Since p1 is a prime, p1 = qj uj for some unit uj. Cancel p1 form both sides leaving p2 ... pm = q1 q2 ...uj... qn Continue canceling the pi until we have 1 = product of perhaps some q's and some u's. Since any remaining q would be a divisor of 1 and therefore a unit, all the q's have to be canceled off. In the process of this canceling we have found that m = n and, we have matched each pi with an associate pj. Therefore the factorization is unique up to associates. Assignment: Page 380-381 work all problems except Problem 15. If you do not know a terminology like PID which means "Principle ideal domain", omit that problem. You should know what a domain is. The meaning of domain is essentially what we have been studying. That is, Subsets of the complex numbers which are closed under addition, subtraction, and multiplication. Examples are the integers, the Gaussian Integers, Z[Sqrt[-6]]. Another example is polynomials. The criterion for a domain are addition, subtraction, commutative multiplication, and ab = 0 implies a = 0 or b = 0.