For polynomial $x^8-1$ I'm tasked to do $3$ things:
1) factor polynomial into irreducible polynomials over integer field with polynomials.
2) find all rupture and splitting fields over complex numbers.
3) find a splitting field over $\mathbb Z_3$, how many elements does it have?
For the 1st task there is a hint: Use substitution and Eisenstein's criteria to proof some polynomials of the form $x^n+1$ are irreducible over integer field with polynomials.
$1)$ I'm kinda puzzled by the hint since I went with distribution: $$x^8-1=(x^4-1)(x^4+1)=(x^2-1)(x^2+1)(x^4+1)=(x-1)(x+1)(x^2+1)(x^4+1).$$ It's easy to see (for example from derivative) what minimum of polynomials $(x^2+1)$ and $(x^4+1)$ making positive functions over real(but that requires continuity). Which would make these irreducible as the other two are already of degree $1$.
$2)$ Rupture fields are $\mathbb Q(1)= \mathbb Q,\mathbb Q(-1)=\mathbb Q, \mathbb Q(e^{πi})$ and $\mathbb Q(e^{πi/2})$ for polynomials $(x-1)$, $(x+1)$, $(x^2+1)$ and $(x^4+1)$ respectively. Making the splitting field same as the last rupture field $\mathbb Q(e^{πi/2})$.
$3)$ I believe the splitting field to be $\mathbb Z_3(2^{\frac{1}{4}})$ for that sufficises all the polynomials $(x+2)(x+1)(x^2+1)(x^4+1)$, where $2^\frac{1}{2}$ you get as $2^\frac{1}{4}*2^\frac{1}{4}$. The number of elements is $3+3*2+3*2+3*2=21.$
Can you check whether my assumptions are correct/give me an explanation on the hunt for $1)$?
Thanks for your time
For 1), indeed the factorisation into irreducible polynomials over $\mathbb{Z}$ is $$ x^8-1=(x^4+1)(x^2+1)(x-1)(x+1). $$ Now $x^{m}+1$ is irreducible if and only if $m$ is a power of $2$, via Eisenstein, see this duplicate:
Prove $x^n+1$ is irreducible over $\mathbb{Q}[X]$ iff $n=2^k$ for $k \in \mathbb{N}$
This explains the hint.