# Irreducible Factors

• Sep 26th 2009, 03:49 PM
Roam
Irreducible Factors
Consider the polynomial $p(x)=x^6-1$.

(a) Find all monic irreducible factors of $p(x)$ over $Q$.
(b) Find all monic irreducible factors of $p(x)$ over $R$.
(c) Find all monic irreducible factors of $p(x)$ over $C$.

I'm really not sure what to do, I know that $x^6 - 1 = (x^3)^2 - 1 = (x^3 - 1)(x^3 + 1)$, so the the roots of p(x) are ±1 & (x+1) & (x-1) are two linear factors. Anyway the first question here asks for monic (degree 1) irreducible (so that it cannot be factorized as P(x)=a(x)b(x) where $a(x), b(x) \in Q$) factors of p(x) over Q (rationals). Can anyone show me the method for solving one of them, so that I'll be able to try solving the rest of it on my own. Thanks.
• Sep 27th 2009, 02:37 AM
e^(i*pi)
$x^6 - 1$

As you noted it is the difference of two squares:

$(x^3-1)(x^3+1)$

However these are the difference and sum respectively of two cubes:

$[(x-1)(x^2+x+1)][(x+1)(x^2-x+1)]$

$x^2 \pm x+1$ only factorises over the complex numbers.
• Sep 29th 2009, 07:19 PM
Roam
I'm still a little confused...

(a) So all monic irreducible factors of p(x) over Q are:

$
(x+1), (x-1), (x^2+x+1), (x^2-x+1)
$

(b) All monic irreducible factors of p(x) over R are:

$
(x+1), (x-1), (x^2+x+1), (x^2-x+1)
$

(c) Now to find monic irreducible factors of p(x) over C are we use the quadratic formula for $x^2 \pm x+1$:

$(x^2+x+1), (x^2-x+1), \frac{-1 \pm \sqrt{-3}}{2}$, $\frac{1 \pm \sqrt{-3}}{2}$, (x+1), (x-1)

are all the monic irreducible factors of p(x) over complex numbers.

Are my answers correct? Did I list all the factors correctly for each question?
• Sep 30th 2009, 11:32 AM
Roam
I don't know if I listed the factors correctly for each part. I'm confused... (Headbang)