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Let $K$ be a finite field, $f \in K[X]$ irreducible with degree $m$.

Show that $Gal(f)$ is cyclic of order $m$.

I have shown that $f$ is separable over $K$ by using that $K$ is finite and thus perfect. Thus $Gal(f)=Gal(K(a_1,...,a_m)/K)$, where the $a_i$ are pairwise distinct roots of $f$.

Furthermore since $K(a_1,...,a_m)/K$ is finite and separable, there exists a primitive element $a \in K(a_1,...,a_m)$, such that $K(a_1,...,a_m)=K(a)$.

By that $Gal(f)=Gal(K(a_1,...,a_m)/K)=Gal(K(a)/K)$.

Also since $K$ is finite, $K(a)$ is finite and thus $K(a)^\times$ is cyclic, but I am not sure if this helps.

I am stuck here. Am I on the right track? I welcome any hints.

B.Swan
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  • This follows immediately from the fact that any extension of finite fields is cyclic Galois. Have you seen that result in the book? Anyway, it has also been explained on our site many times. – Jyrki Lahtonen Mar 03 '19 at 21:20
  • This exercise comes before Galois groups are examined in detail, I guess this is part of what I am asked to prove. Thank you. – B.Swan Mar 03 '19 at 21:32
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    You may benefit from a study of this ecent thread and others linked to it. I am undecided whether this is question is a duplicate of that or some earlier. Anyway, as you get the roots of an irreducible polynomial by raising one of them to $q$th power, we soon see that adjoining a single root gives the rest of them free of charge. Hence normality. I think Darij Grinberg did a very good job in the linked thread. What do you think? – Jyrki Lahtonen Mar 03 '19 at 21:36
  • This is great, thank you. Does that mean that my primitive element can be any of the roots of $f$? – B.Swan Mar 03 '19 at 21:59
  • Depends what you mean by a primitive element. In the context of finite fields the meaning is different from that used elsewhere in field theory. I tried to explain this here. It is really a bit unfortunate overloading of terminology, but poses no real problems when you are aware of it. – Jyrki Lahtonen Mar 04 '19 at 19:55

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