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 This material can be found on pages 21-26 starting with Complex and Matrix Algebra. Monday, November 5. Main idea: Irreducible is not the same as prime. Key words: Factor, Irreducible, Prime. Unit. Goal: Understand better irreducible and prime as preparatory towards unique factorization. Previous Assignment (a) Find the |z| and the argument of z when z = 5 + 12 i . |z| = 13 Arg[z] = Arctan[12/5] = 1.17601 radians = 67.3801 degrees (b) Plot the five fifth roots of 1. f[z_] := {Re[z],Im[z]}; A = Table[f[Cos[2 Pi/5 i]+I Sin[2 Pi/5 i]],{i,1,6}]; ans = ListPlot[A,PlotJoined->True,AspectRatio->Automatic]; Display["five.ps",ans]; (c) Plot the six sixth roots of 1. f[z_] := {Re[z],Im[z]}; A = Table[f[Cos[2 Pi/6 i]+I Sin[2 Pi/6 i]],{i,1,7}]; ans = ListPlot[A,PlotJoined->True,AspectRatio->Automatic]; Display["six.ps",ans]; (d) Find the Sqrt[-24 + 70 i] x^2 = 74 Cis[1.90109] x = Sqrt[74](Cos[1.90109/2]+I Sin[1.90109/2]) x = 5.00001 + 6.99999 I x = 5 + 7 I (e) Find the Cube Root of -9 + 46 i. x^3 = 13 Sqrt[13] Cis[1.76401] x = (13 Sqrt[13])^(1/3) (Cos[1.76401/3] + I Sin[1.76401/3]) x = 3 + 2 I (f) Find Log[e i] 1+Pi/2 I (g) Express Cos[4 t] in terms of Cos[t]. (Cos[t]+I Sin[t])^4 4 3 2 2 3 4 = Cos[t] +4 I Cos[t] Sin[t]-6 Cos[t] Sin[t] - 4 I Cos[t] Sin[t] +Sin[t] 2 4 3 3 4 = -6 Cos[t] + 7 Cos[t] + 4 I Cos[t] Sin[t] - 4 I Cos[t] Sin[t] + Sin[t] Cos[4 t] = -6 Cos[t]^2 + 7 Cos[t]^4 + Sin[t]^4 -Cos[4 t] -6 Cos[t]^2 + 7 Cos[t]^4 + Sin[t]^4 (h) Find Gaussian integers x and y such that x(3+5i) + y(2+7i) = 1 1 ------ 2+7I | 3+5I 2+7I -2+2I ------ ---------- 1-2I | 2+7I 2+6I -2-I ----- ------ I | 1-2I 1-2I ---- 1 -2+2i -2-I 1 1 -1+2I 5-3I 0 1 -2+2I 7-2I (-1+2I)(7-2 I) - (-2+2I)(5-3I) = 1 Now we have to adjust the product to have 3+5I and 2+7I (-1+2I)(-I)*I(7-2 I) - (-2+2I)(-I)*I(5-3I) = 1 (2+I)(2+7I)-(2+2I)(3+5I) = 1 y x y = 2+I x = -(2+2I) A number is called a unit if it has a multiplicative inverse. In the integers, only -1, and 1 are units. In the Gaussian Integers, only 1,i,-i,-1 are units. Why? In Z[Sqrt[2]] there are a lot of units. What are they? Define N[a+bSqrt[2]] = | a^2 -2 b^2 |. Show that this norm is multiplicative. N[(a+bSqrt[2])(c+dSqrt[2])| = N(ac+2bd +(ad+bc)Sqrt[d]) = |(a c+2b d)^2 -2(a d+b c)^2| 2 2 2 2 2 2 2 2 = | a c - 2 b c - 2 a d + 4 b d | 2 2 2 2 = |(a - 2 b ) (c - 2 d ) | = N[ a+b Sqrt[2]] N[c+d Sqrt[2] ] So if a+b Sqrt[2] is invertible, then N[a+b Sqrt[2] ] = [+/-] 1 a^2 - 2 b^2 = [+/-] 1 The units are just the integral solution of a^2 - 2 b^2 = +/- 1 a=3 and b = 2 is one possible solution. and if one has one solution then (a+Sqrt[2] b)(a-Sqrt[2] b) = [+/-]1 Take this to any power gives (a+Sqrt[2] b)^n (a-Sqrt[2] b)^n = ([+/-] 1)^n so if we expand (a+bSqrt[2])^n and collect the integer part and collect the radical part, we will have another solution. In[63]:= Do[ Print[Expand[(3+2 Sqrt[2])^n]],{n,1,20}] 3 + 2 Sqrt[2] 17 + 12 Sqrt[2] 99 + 70 Sqrt[2] 577 + 408 Sqrt[2] 3363 + 2378 Sqrt[2] 19601 + 13860 Sqrt[2] 114243 + 80782 Sqrt[2] 665857 + 470832 Sqrt[2] 3880899 + 2744210 Sqrt[2] 22619537 + 15994428 Sqrt[2] 131836323 + 93222358 Sqrt[2] 768398401 + 543339720 Sqrt[2] 4478554083 + 3166815962 Sqrt[2] 26102926097 + 18457556052 Sqrt[2] 152139002499 + 107578520350 Sqrt[2] 886731088897 + 627013566048 Sqrt[2] 5168247530883 + 3654502875938 Sqrt[2] 30122754096401 + 21300003689580 Sqrt[2] 175568277047523 + 124145519261542 Sqrt[2] 1023286908188737 + 723573111879672 Sqrt[2] In Z[Sqrt[-6]] = {a+bSqrt[-6] such that a,b in Z], show that 10 can be factored into irreducibles in two ways. 10 = 2 5 = (2+I Sqrt[6])(2-I Sqrt[6]) Now we have to show that 1+I Sqrt[6] and 1-I Sqrt[6] are irreducible. Proofs like this are usually done with a norm. Fortunately, there is one. N(a+bSqrt[-6]) = a^2 + 6 b^2 N(2+ Sqrt[-6]) = 10 If 2+ Sqrt[-6] factored into (a+b Sqrt[-6])(c+d Sqrt[-6]) then N(2+I Sqrt[6] ) = N(a+b Sqrt[-6]) N(c+d Sqrt[-6]) 10 = (a^2 + 6 b^2)(c^2 + 6 d^2) The only choices are 1 10 2 5 5 2 10 1 If a^2 + 6 b^2 = 1, then b = 0 and a = [+/-] 1. Thus the factorization is trivial since a is a+bSqrt[-6] is a unit. If a^2 + 6 b^2 = 2, then b=0 and a^2 = 2. There is no such a in the integers. If a^2 +6 b^2 = 5 then b=0 and a^2 = 5. There is no such a in the integers. If a^2 + b^2 = 10, then c^2 + 6 d^2 = 1 and d = 0 and c = [+/-] 1 and the factorization is trivial since c + d Sqrt[-6] is a unit. The proof that 2- Sqrt[-6] is irreducible is similar. Notice that (a+bSqrt[-6]) = a^2 + 6 b^2 is a norm. Assignment: 1. The Even Integers are 2Z (a) What are the irreducibles in 2Z (b) Find an element of 2Z which has two different factorizations into irreducibles. 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]) (b) Find x and y such that x(266 + 175 Sqrt[2])+y(307 + 179 Sqrt[2]) = w 3. In Z[Sqrt[3]] = {a+bSqrt[3] | a,b are integers} (a) Find a norm. (b) Prove it is a norm