1. ## Irreducible Factors

Consider the polynomial $\displaystyle p(x)=x^6-1$.

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

I'm really not sure what to do, I know that $\displaystyle x^6 - 1 = (x^3)^2 - 1 = (x^3 - 1)(x^3 + 1)$, 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 $\displaystyle a(x), b(x) \in Q$) 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.

2. $\displaystyle x^6 - 1$

As you noted it is the difference of two squares:

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

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

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

$\displaystyle x^2 \pm x+1$ only factorises over the complex numbers.

3. I'm still a little confused...

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

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

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

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

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

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

4. I don't know if I listed the factors correctly for each part. I'm confused...