Stelle die folgenden Polynome als Produkte von Primpolynomen über ihren jeweiligen Grundkörpern dar.

  1. ´x^6 + 1 in CC[x]´
  2. ´x^6 + 1 in RR[x]´
  3. ´x^6 + 1 in ZZ_2[x]´
Approach

Teil 1

Nullstellen: ´x^6 = -1 = e^(i pi)´

´x^6 = e^(i pi) = e^(i (pi + k * 2pi))´ ´k in ZZ´

´=> x_k = e^(i((pi + k * 2pi)/6))´

´k: 0,1,…,6´ ´=> x^6+1 = (x - e^(i pi/6)) * ´´(x - e^(i (3pi)/6)) * ´´(x - e^(i (5pi)/6)) * ´´(x - e^(i (7pi)/6)) * ´´(x - e^(i (9pi)/6)) * ´´(x - e^(i (11pi)/6))´

<mark>Umwandlung in arithmetische Form fehlt!</mark>

Teil 2

´(x - e^(i pi/6))(x - e^(i (11pi)/6))´ ´= x^2 - (e^(i pi/6) + e^(i (11pi)/6)) + e^(i pi (1/6 + 11/6))´ ´= x^2 - Re(e^(i pi/6))x + 1´

´x^6 + 1 = (x^2 - cos(20˚)x + 1) * ´´(x^2 - cos(40˚)x + 1) * ´´(x^2 - cos(60˚)x + 1) * ´´(x^2 - cos(80˚)x + 1) * ´´(x^2 - cos(100˚)x + 1) * ´´(x^2 - cos(120˚)x + 1)´

Teil 3

´-1 = 1´ in ´ZZ_2´

´x^6 + 1 = x^6 - 1´ ´= (x^3)^2 - 1^2´ ´= (x^3 + 1)(x^3 - 1)´ ´= (x^3 - 1)^2´ ´= (x^3 - 1)^2´

<mark>Ende fehlt</mark>

Für ´x^8 + 1´:

´x^8 + 1 = x^8 - 1´ ´= (x^4)^2 - 1^2´ ´= (x^4 + 1)(x^4 - 1)´ ´= (x^4 - 1)^2´ ´= ((x^2)^2 - 1^2)^2´ ´= ((x^2 + 1)(x^2 - 1))^2´ ´= ((x^2 - 1)^2)^2´ ´= (x^2 - 1)^4´ ´= ((x - 1)(x + 1))^4´ ´= (x + 1)^8´


Solution
    1. ´x^6+1 = (x - e^(i pi/6)) * ´´(x - e^(i (3pi)/6)) * ´´(x - e^(i (5pi)/6)) * ´´(x - e^(i (7pi)/6)) * ´´(x - e^(i (9pi)/6)) * ´´(x - e^(i (11pi)/6))´ <mark>Umwandlung in arithmetische Form fehlt!</mark>

    2. ´x^6 + 1 = (x^2 - cos(20˚)x + 1) * ´´(x^2 - cos(40˚)x + 1) * ´´(x^2 - cos(60˚)x + 1) * ´´(x^2 - cos(80˚)x + 1) * ´´(x^2 - cos(100˚)x + 1) * ´´(x^2 - cos(120˚)x + 1)´

    3. <mark>Fehlt</mark>

  • URL:
  • Language:
  • Subjects: math
  • Type: Transform
  • Duration: 25min
  • Credits: 6
  • Difficulty: 0.7
  • Tags: hpi polynomial irreducible polynomial field
  • Note:
    HPI, 2014-07-07, Mathe 2, Aufgabe 55
  • Created By: adius
  • Created At:
    2014-07-26 21:34:49 UTC
  • Last Modified:
    2014-07-27 23:54:33 UTC