Proving the Reducibility of Polynomial f_n(x) for n>=2 and n not prime

[burritoloco]

Homework Statement

Prove that for n>=2, n not prime, the following polynomial is reducible
$$f_n(x) = x^{n-1} + x^{n-2} + ... + x + 1$$

Homework Equations

The Attempt at a Solution

If n is even, the polynomial has -1 as a root, so it is reducible. But when n is odd (and not prime) I'm not sure how to show it? Thanks for the help.

 

[I like Serena]

Suppose you first multiply by (x-1).
Can you then reduce the polynome (other than by (x-1))?

 

[Dick]

-1 works when 2 is a factor of n because it's a primitive 2nd root of 1. Can you show if p is a prime factor of n and u is a primitive pth root of n, then u is a root of your polynomial?

 

[burritoloco]

@Serena: Thank you. If n is not prime, then in particular, the p-th cyclotomic polynomial, p a prime and p|n, divides fn(x). Would there be an easier way to show this without knowing about the factorization of x^n - 1?

@Dick: Oh, I see that I forgot to mention in the question the most important part! I need to factor the polynomial over the rationals! Sry :) Otherwise, by what Serena implies, the answer would be obvious as the n roots of unity except 1 are the roots of fn(x).

 

[burritoloco]

To clarify: I meant reduce fn(x) over the rationals!

 

[Dick]

To clarify: I meant reduce fn(x) over the rationals!

If you can deduce that the roots are the pth roots of unity, except for 1, when you multiply all of those factors together, you do get a rational polynomial.

 

[burritoloco]

Yes I know :), I've just read it off from my book. Each irreducible factor will be over the rationals (over Z actually). Follows from the factorization of x^n - 1 with cyclotomic factors. Using Mobius inversion formula, it follows that each cyclotomic polynomial has integer coefficients. Thanks!