I'm working on the following problem:
Let $K$ be the splitting field of $f(x)=x^{6}+x^{3}+1$ over $\mathbb{Z}_{2}$, and let $\alpha$ be a root of $f(x)$.
(1) Show that $\alpha^{9}=1$ but $\alpha^{3}\neq1$.
(2) Show that $f(x)$ is irreducible over $\mathbb{Z}_{2}$.
(3) Find the order of the Galois group $G(K/\mathbb{Z}_{2})$ of $K$ over $\mathbb{Z}_{2}$, and describe all intermediate fields between $\mathbb{Z}_{2}$ and $K$ in the form $\mathbb{Z}_{2}(\beta)$, where $\beta\notin\mathbb{Z}_{2}$.
I have tried so far:
(1) Since $x^{9}-1=(x^{3}-1)(x^{6}+x^{3}+1)$ and $\alpha^{6}+\alpha^{3}+1=0$, we have $\alpha^{9}=1$, and $\alpha^{3}\neq1$ because if not, we would have $\alpha^{6}+\alpha^{3}+1=1$ in $\mathbb{Z}_{2}$, which yields a contradiction.
(2) Put $p(x)=\text{irr}(\alpha,\mathbb{Z}_{2})$. Then, we have $p(x)\mid f(x)$.
I was stuck here.
First, is it solving problem 2 using the result of problem 1? or independent?
Second, since $K$ is an extension of $\mathbb{Z}_{2}$, $K$ is of order $2^{m}$. Thus, $9\mid 2^{m}-1$ since $\alpha^{9}=1$.
Then, the only possible $k$ is 6? If so, does that imply $K=\mathbb{Z}_{2}(\alpha)$?
In (2), it seems to be $p(x)=f(x)$, but i'm not sure about that.
How to show that (2) and find the order of $G(K/\mathbb{Z}_{2})$.
Give some advice! Thank you!
$f(x) = x^6+x^3+1$ is indeed the $9$th cyclotomic polynomial. This means that the roots $r$ of $f(x)$ are primitive $9$th roots of unity. This means that for any integer $k < 9$, $r^k\neq1$, and $9$ is the smallest integer $k$ such that $r^k=1$, answering your first question. $2$ is a primitive root $\pmod 9$, so $f(x)$ over $\mathbb{Z}_{2}$ is therefore irreducible, reference here. For (3), I am not sure what you mean, if you are asking about the multiplicative order of $2 \pmod 9$, well it is $6$.