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I know the Galois Group is cyclic of order 2 and I got the splitting field to be $\mathbb{F}_9$ but I don't understand how to write the frobenius automorphism that describes the action

Nicholas Roberts
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qypevj123
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    If $q = p^m$ and $F_q$ is an finite extension of $F_p$, the frobenius automorphism of $F_q$ is always just $x \mapsto x^p$. What exactly are you confused about? – Dionel Jaime Apr 21 '20 at 18:47
  • how would i write the roots as a permutation? – qypevj123 Apr 21 '20 at 18:48
  • Maybe what you mean to ask is how you would describe the automorphism as a permutation of the roots? But I still don't understand where your confusion lies. As the answer below states, if $\alpha$ is a root $\alpha^3$ must be a root and $\alpha^9 = \alpha$. – Dionel Jaime Apr 21 '20 at 18:55
  • The polynomial is separable so there are four roots in the splitting field, say $\alpha_1$, $\alpha_2$, $\alpha_3$, $\alpha_4$. Because the Galois group is cyclic of order two, and none of the roots are in the prime field, the Frobenius must permute the roots in $(12)(34)$ fashion (a renumbering of the roots may be needed). It isn't transitive because the polynomial is reducible. – Jyrki Lahtonen Apr 22 '20 at 04:23

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The Froebenius automorphism $\sigma : x \mapsto x^3$ swaps the roots since it transforms a root into a root and $\sigma(x) \neq x$ for all $x \notin \mathbb{F}_3$.

And of course $\sigma^2$ is the identity on $\mathbb{F}_9$.

And that's it, the galaois group is $\{\sigma, Id\}$.

Olivier Roche
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