Factorization into irreducible

ijk · New Member Offline Joined: 17 Mar 2008 Posts: 17

How do you factor $x^n -1 , n>=3$ into irreducibles in Z[x] ? Why do we need to know this? What are we looking for?

**Posted:** Wed Mar 19, 2008 2:09 am

ZetaX

Cyclotomic polynomials.

**Posted:** Wed Mar 19, 2008 2:44 am

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LydianRain · Yang-Mills Theory Offline Joined: 02 Mar 2008 Posts: 785

In response to the rather philosophical questions "why?" and "what are we looking for?":

Say $n = 4$ . Then the solutions to $x^4 = 1$ are the 4 complex 4th roots of unity: $1, - 1, i, - i$ . But you could argue that two of these are "fake" 4th roots of unity. One of them is already 1, and one is really a square root of 1. So it makes sense to factor $x^4 - 1 = (x - 1)(x + 1)(x^2 + 1)$ . The first factor has the "1th" root of unity, the second factor has the square root of unity, and the third factor has the authentic primitive 4th roots of unity.

**Posted:** Wed Mar 19, 2008 10:46 pm