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I've to find the Galois correspondence between the subgroups of $Gal(E,K)$ and the intermediate fields of $E$, where $K=\mathbb{F}_7$ and $E=K[\alpha]$ and $\alpha$ is a root of the polynomial $x^9-1$.

So far, what I've done is:

We can factorice the polynomial as

$x^9-1=(x-1)(x^2+x+1)(x^3-2)(x^3-4)$

over $\mathbb{F}_7$.

We observe that $1$ is root of $x-1$, $2$ and $4$ are the roots of $x^2+x+1$. So, $\alpha$ is a root of $(x^3-2)(x^3-4)$.

If $\alpha$ is a root $(x^3−2)$ then $[E:K]=3$ since $(x^3-2)$ is the minimal polynomial of $\alpha$ over $\mathbb{F}_7$.

I know that if $\sigma \in Gal(E/K)$, then $\sigma$ is completely determined by its action on $\alpha$ and that $\sigma(\alpha)$ is a root of $(x^3-2)$.

So, I would like to know how the roots of this polynomial look. I also know that

$E=\lbrace a_2 \alpha ^2+ a_1 \alpha +a_0 | a_2,a_1,a_0 \in \mathbb{F}_7 \rbrace $

since $[E:K]=3$. I suppose that is analogous for int the case that $\alpha$ is a root of $(x^3-4)$.

Thanks in advance for any help.

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  • $\begingroup$ So $x^2+x+1 = (x-2)(x-4)$ $\endgroup$ – Kenny Lau Oct 9 '17 at 23:12
  • $\begingroup$ Yes, I mentioned that I know that. But, I would like to know that happens when $\alpha$ is a root of $x^3-2$ or $x^3-4$. $\endgroup$ – A. Smith Oct 9 '17 at 23:16
  • $\begingroup$ I don't understand the question. What is $\alpha$? Which root? Is the question asking for 9 cases, one for each root? $\endgroup$ – Kenny Lau Oct 9 '17 at 23:20
  • $\begingroup$ Are the other roots of the polynomial in E? That is what I'm looking for. I need to describe explicit the elements of $Gal(E/K)$ and then describe the intermediate fields $K \subset F \subset E$. $\endgroup$ – A. Smith Oct 9 '17 at 23:20
  • $\begingroup$ It would make more sense if the problem had $E$ being the splitting field of $x^9 - 1$. Otherwise, as Kenny Lau said, the problem seems to be underspecified. $\endgroup$ – Daniel Schepler Oct 9 '17 at 23:28
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In this case the best thing is to note that if you have a cubic root of $2$ and a cubic root of $1$, their product is again a cubic root of $2$ (in a similar way of affine and linear associated system or in linear differential equations). So, once you ``created'' $\alpha$, it's sufficient to note that $X_3=\{1,2,4\}$ is the set of cubic root of unity. Then the roots of $x^3-2$ is precisely $\alpha X=\{\alpha,2\alpha,4\alpha\}$. Another way (perhaps in other situations) could be to view $\alpha$ more as $\mu_\alpha:\mathbb{F}_7[\alpha]\longrightarrow\mathbb{F}_7[\alpha]$ defining $\mu_\alpha(a):=\alpha a$ (this gives you an immersion $\mathbb{F}_7[\alpha]\hookrightarrow M_{3\times 3}(\mathbb{F}_7)$ Perhaps having the more concrete matrices could be more pratical sometimes. A third, intermediate, solution coul be to use simply the relation $\alpha^3=2$ and solve the equation $x^3=2$ on elements of form (as you noticed) $a+b\alpha+c\alpha^2$. Hope this was useful.

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  • $\begingroup$ A doddering old mathematician with failing eyesight begs you never to put an alpha next to an ‘a’. $\endgroup$ – Lubin Oct 13 '17 at 5:12
  • $\begingroup$ I'm sorry :D actually I don't like too... $\endgroup$ – Tancredi Oct 13 '17 at 12:01
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I suppose you meant to write “Galois correspondence between the subgroups of the Galois group and the intermediate fields”.

But as you seem to have realized, the field extension $E\supset K=\Bbb F_7$ is of degree three and normal, so there are no fields strictly between $E$ and $K$. I guess it should be pointed out that the Galois group is cyclic of order three, generated by the Frobenius automorphism $\sigma$, for which $\forall x\in E$, $\sigma(x)=x^7$.

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  • $\begingroup$ Yes, it's my mistake. I've edited the post. Thank you for answer the question. I got the same result with another reasoning, but I like this answer because it's more direct. $\endgroup$ – A. Smith Oct 14 '17 at 2:02

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