Here is the 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 $p(x)\mid f(x)$.
It seems to be $p(x)=f(x)$. But, i'm not sure about that.
If so, is it true that $K=\mathbb{Z}_{2}(\alpha)$?
I don't know what I missed given conditions.
Give some advice! Thank you!
If $K=\Bbb F_2(\alpha)$ has $2^k$ elements, then $9\mid(2^k-1)$ as $K^*$ is a group of order $2^k-1$ with a subgroup of order $9$ (generated by $\alpha$). What does that tell you about $k$?