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I have the extension $\mathbb F_3[A]/\mathbb F_3$ where the $A$ are the roots of $x^{80}-1$.

I need to prove this extension is Galois, find the Galois group, and describe the automorphisms. but I'm having no success at all. the only thing I can think of is to look at $x^{81}-x$ and use the fact $x^{p^m}-x\mid x^{p^n}-x\iff m\mid n$ and $x^3-x\equiv 0$ somehow, but again, I don't know how to proceed.. :(

Arrow
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user355165
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You're nearly there. You should know that the roots of $x^{81} - x$ form a finite field $E$ of order $81 = 3^{4}$, that any finite field is Galois over any of its subfields, and that the Galois group of $E$ over $\mathbb{F}_{3}$ is cyclic of order $4$, generated by the Frobenius automorphism $t \mapsto t^{3}$.

Andreas Caranti
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    Just as an addition, it is worth noting that the Galois group of any finite field is cyclic over its prime field, so as long as $ A $ is a finite set comprised of elements algebraic over $ \mathbf F_3 $, the extension $ \mathbf F_3(A) $ is always cyclic over $ \mathbf F_3 $. – Ege Erdil Jul 20 '16 at 11:24
  • @Starfall I'm still not sure what the actual answer is or how to describe the automorphisms.. could you help me out? – user355165 Jul 20 '16 at 12:11
  • What does it mean to you when one says "the Galois group is cyclic of order 4"? – Ege Erdil Jul 20 '16 at 12:51