Irreducible Factors

**Roam** · September 26th 2009, 03:49 PM

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.

**e^(i*pi)** · September 27th 2009, 02:37 AM

$x^6 - 1$

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.

**Roam** · September 29th 2009, 07:19 PM

I'm still a little confused...

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

$<br /> (x+1), (x-1), (x^2+x+1), (x^2-x+1)<br />$

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

$<br /> (x+1), (x-1), (x^2+x+1), (x^2-x+1)<br />$

(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?

**Roam** · September 30th 2009, 11:32 AM

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

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