
Irreducible Factors
Consider the polynomial $\displaystyle p(x)=x^61$.
(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) & (x1) 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.

$\displaystyle x^6  1$
As you noted it is the difference of two squares:
$\displaystyle (x^31)(x^3+1)$
However these are the difference and sum respectively of two cubes:
$\displaystyle [(x1)(x^2+x+1)][(x+1)(x^2x+1)]$
$\displaystyle x^2 \pm x+1$ only factorises over the complex numbers.

I'm still a little confused...
(a) So all monic irreducible factors of p(x) over Q are:
$\displaystyle
(x+1), (x1), (x^2+x+1), (x^2x+1)
$
(b) All monic irreducible factors of p(x) over R are:
$\displaystyle
(x+1), (x1), (x^2+x+1), (x^2x+1)
$
(c) Now to find monic irreducible factors of p(x) over C are we use the quadratic formula for $\displaystyle x^2 \pm x+1$:
$\displaystyle (x^2+x+1), (x^2x+1), \frac{1 \pm \sqrt{3}}{2}$, $\displaystyle \frac{1 \pm \sqrt{3}}{2}$, (x+1), (x1)
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?

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