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Let $p\in\mathbb{N}$ be a prime. Let $F=\mathbb{F}_{p^n}$, $n\geq1$, be a finite field of $p^n$ elements containing $M=\mathbb{Z}/(p)$. I would like to show that $\mathrm{Gal}(F/M)\cong\mathbb{Z}/(n)$.

Here's my work so far. It is easy to show that $F/M$ is a simple field extension (i.e. there exists an element $\alpha\in F$ such that $F=M(\alpha)$) because of the existence of a generator of the multiplicative group of the field. Then the Galois group of $F/M$ contains the automorphisms of $F$ sending such generator to roots contained in $F$ of the minimal polynomial of the generator over $M$. I don't know how to proceed from here.

Ray Bern
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  • I suggest you to take a look at what happens with the Frobenius automorphism – Tommaso Scognamiglio Nov 11 '18 at 18:11
  • I suppose you could write $;\Bbb Z/(p)=:\Bbb F_p=;$ the prime field with $;p;$ elements...This is a rather standard notation, though perhaps not a universal one. You wouldn't also need to point out $;F;$ contains this field: it must to and we should know it. Finally, read the other comment: it may be important. – DonAntonio Nov 11 '18 at 18:22
  • Since $x^p -x$ has at most $p$ roots in $F$, $M$ is the fixed field of the Frobenius thus $Gal(F/M) $ is the cyclic group generated by the Frobenius. – reuns Nov 11 '18 at 19:16

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