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!

  • $\begingroup$ Lord Shark's answer is, of course, ok (and the way I would have solved this). For a different approach you can also argue as follows. By Freshman's dream $\alpha^2$ is also a zero of $p(x)$. Consequently so are $\alpha^4,\alpha^8$, $\alpha^{16}=\alpha^7$ (using the fact that $\alpha^9=1$). Keeping going, $\alpha^{14}=\alpha^5$ must then also be a root. As $\alpha$ has order nine, all those roots are distinct. Hence $p(x)$ has at least six distinct roots. Hmm.... $\endgroup$ – Jyrki Lahtonen Apr 27 '18 at 9:49
  • $\begingroup$ (+1) for sharing your (correct) solution to part (1). $\endgroup$ – Jyrki Lahtonen Apr 27 '18 at 9:51

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 $k$?

  • $\begingroup$ The only possible $k$ is 6? $\endgroup$ – Primavera Apr 21 '18 at 16:04

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