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, November 9. Main idea: The integers are very special. Closely related structures behave very differently. Key words: Euclidean Norm, Greatest Common Divisor. Goal: Learn techniques useful for proving that the integers and later the polynomials over a field are unique factorization domains. Previous non-assignment . 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 multiplicative 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] ] In these quadratic extensions, you always get a multiplicative structure. ----------------------------------------------------------------------- The map f[a+b Sqrt[s]] = a^2 - s b^2 is multiplicative on F[Sqrt[s]]. The only requirement is that s be in F and that there be no square root of s already in F. ___________ Proof: Define a+b Sqrt[s] = a-b Sqrt[s] ___ _ _ ____ _ _ Clearly A+B = A+B and AB = A B _ Now f(A) = A A ____ _ _ _ _ f[ AB ] = AB AB = A B A B = (A A)(B B) = f[A] f[B]. The Euclidean algorithm will work if you can get a remainder which is less than the divisor. quotient --------- divisor | dividend . . . ---------- remainder If we want the Euclidean algorithm to work, we require the following: "Given two numbers a and b,(b=/=0), there must exist q and r such that a = bq + r AND either r = 0 or 0 < n(r) < n(b)" Z, Z[Sqrt[2], Z[Sqrt[3]], Z[i] all have this property. 2Z, Z[Sqrt[-6]] does not. There are only a finite number of extensions Z[Sqrt[s]] which have the property. I forget the exact number. Maybe 15 or so. All the rest do not have the property. Polynomials over a field with the n(p(x)) = degree(p(x)) also have this property. This is probably the most important example. For the even integers n(x) = |x| is not a Euclidean norm. 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 | r | = | d |. To be a Euclidean norm, The nonzero remainder must have norm strictly less than the the norm of the divisor. The Integers under n(x) = |x| is a Euclidean norm. Proof: Given a and b (b=/=0) the points nb (n an integer) are evenly spaced b apart on the x axis. -3 b -2 b -b 0 b 2b 3b ...........|.....|.....|.....|.....|.....|.....|........ a must exist somewhere on this number line. Either a = qb and r = 0, or else a is in some interval. Without loss of generality assume b > 0. qb < a < (q+1) b 0 < a-qb < b So | a-qb| < b. So a = qb + (a-qb) the remainder is a-qb and |a-qb| < |b|. As a refinement, we can actually choose q such that |r| <= |b|/2. Theorem: Z[Sqrt[2]] = {a+b Sqrt[2] | a,b are integers} has a Euclidean norm. Proof: Given A = a+bSqrt[2] and B = c+d Sqrt[2]. We have to find Q and R such that A = B Q + R where n(R) < n(B). 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 = ------------- -p + (------------ -q) Sqrt[2] + p+q Sqrt[2] c^2 - 2 d^2 c^2 -2 d^2 "integral part" The absolute The absolute value of this value of this is at most 1/2 is at most 1/2. |-----------------------------------| "rest" a + b Sqrt[2] -------------- = p+q Sqrt[2] + rest c + d Sqrt[2] a c - 2 b d b c - a d where rest = ------------- -p + (------------ -q) Sqrt[2] c^2 - 2 d^2 c^2 -2 d^2 2 2 | a c - 2 b d | | b c - a d | and n(rest) = | ------------- -p | + 2|(------------ -q)| < 3/4. | c^2 - 2 d^2 | | c^2 -2 d^2 | So a+b Sqrt[2] = (c+d Sqrt[2]) (p+q Sqrt[2]) + rest (c+d Sqrt[2]). So A = B Q + R n(R) = n(rest) n(c+d Sqrt[2]) <= 3/4 n(c+d Sqrt[2]). Definition: The greatest common divisor of a and b is a number c such that (1) c divides both a and b (2) if d also divides a and b, then d divides c. The Even integers do not have greatest common divisors. Give an example. Theorem: If there is a Euclidean norm, then the ring has greatest common divisors. Proof: Among all the numbers of the form AX+BY, pick the one with smallest positive norm. This element AXo+BYo with smallest positive norm will be the GCD(A,B). (1) We have to show that AXo+BYo divides both A and B. (2) We also have to show that if D divides A and B, then D divides AXo+BYo Obviously, we should use the Euclidean nature of the norm. A = Q(AXo+BYo) + R where n(R) < n(AXo+BYo). We can reorganize this to A(1-QXo)+B(-Yo) = R. Since n(AXo+BYo) was the number with smallest positive n value, we must have R = 0. So AXo+BYo divides A. Similarly, AXo+BYo divides B. The other part is easy. If D divides A and D divides B, then D divides anything of the form AX+BY. In particular, D divides AXo+BYo Assignment for Monday; 1. Find all the associates of 19 - 17 i in the Gaussian integers. 2. Show that the Gaussian Integers have a Euclidean norm. 3. In the Gaussian integers find Q and R such that 42 + 87 i = (13 + 15 i) Q + R and n(R) < n(13+15 i).