1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $k$?

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.