What is the Galois group $Gal(E/F)$ when $F=GF(5^3)$ and $E=GF(5^{24})$. I have attempted to describe the Galois group, but I've become stuck, and it's entirely possible that I've made mistakes as well. Here's what I've done.

I know that for any finite field, there exists irreducible polynomials of any degree $n$ over that field. So then there exists an irreducible polynomial $p$ over $F$ of degree $8$. Since $E$ is a finite extension of $F$, it is a simple extension, so let's say $E = F(\alpha)$ where $\alpha$ is some root of $p$. The basis of $E$ over $F$ is then $\{1, \alpha, \alpha^2, \alpha^3, \alpha^4, \alpha^5, \alpha^6, \alpha^7\}$, and $[E:F] = 8 = |Gal(E/F)|$. So, the Galois group should be isomorphic to either $\mathbb{Z}_8, \mathbb{Z}_4\times\mathbb{Z}_2$ or $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$, unless I'm mistaken.

Assuming that I've not made mistakes so far (which is far from certain), I don't know where to go from here in order to explicitly describe the automorphisms of the Galois group. How do I know what those automorphisms are? Some help would be nice.

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    $\begingroup$ Look at the Frobenius map $GF(5^{24}) \rightarrow GF(5^{24})$, $x\mapsto x^{5^3}$ and show that it (lies in ${\rm Gal}(E/F)$ and) generates ${\rm Gal}(E/F)$. $\endgroup$ – Claudius May 25 '16 at 11:07
  • $\begingroup$ @user218931 Thank you, I'll try that. Is that the only angle of approach as far as you know? $\endgroup$ – Auclair May 25 '16 at 11:35
  • $\begingroup$ Yes, I think so. At least, this is how one proves in general that ${\rm Gal}(E/F)$ is cyclic whenever $E$ (and hence $F$) is a finite field. $\endgroup$ – Claudius May 25 '16 at 11:53
  • $\begingroup$ @user218931 I see. I'm not too familiar with the Frobenius endomorphism, so if you don't mind, would you post an answer just outlining the ideas of solution? I don't need a complete solution. $\endgroup$ – Auclair May 25 '16 at 11:56
  • $\begingroup$ You've got a good answer, so I won't bother typing another. I do want to remark that your list of possible Galois groups is missing the non-abelian possibilities. Any extension involving only finite fields has a cyclic Galois group, but for example over $\Bbb{Q}$ you should not forget about the non-abelian possibilities. $\endgroup$ – Jyrki Lahtonen May 25 '16 at 12:21

Notice, that we have $x^{p^n} = x$ for all $x\in GF(p^n)$ ($p$ a prime and $n\ge1$ a natural number): this is how $GF(p^n)$ is constructed, namely by adjoining all roots of $X^{p^n}-X$ to $GF(p) = \mathbb Z/p\mathbb Z$.

One checks that the Frobenius $\varphi\colon GF(p^n) \rightarrow GF(p^n)$, $x\mapsto x^p$ is an injective ring homomorphism, and hence an automorphism of fields. Since $\varphi(x) = x$ for all $x\in GF(p)$ (by the little Fermat), $\varphi$ is an element of ${\rm Gal}(GF(p^n)/GF(p))$, of which you know that this is a group of order $n$.

It therefore suffices to show that the powers $\varphi^k:= \varphi\circ\dotsm \circ\varphi$ ($k$ times) are all distinct for for $k$ running through $0,1,\dotsc,n-1$. Assume that $\varphi^j = \varphi^k$ for some $0\le j\le k\le n-1$. It follows that $\varphi^{k-j}= {\rm id}_{GF(p^n)}$, i. e. $\varphi^{k-j}\colon x\mapsto x^{p^{k-j}}$ fixes all points in $GF(p^n)$. In other words, $X^{p^{k-j}}-X$ is a polynomial of degree $p^{k-j}<p^n$ (since $k-j<n$) with $p^n$ many roots. Over a field, this is only possible if $X^{p^{k-j}}-X$ is the zero polynomial, i. e. $k-j=0$. This shows that ${\rm id}_{GF(p^n)} = \varphi^0, \varphi, \varphi^2,\dotsc,\varphi^{n-1}$ are $n$ pairwise distinct elements of ${\rm Gal}(GF(p^n)/GF(p))$. Hence this group is cyclic.

For $k\mid n$, ${\rm Gal}(GF(p^n)/GF(p^k))$ is a subgroup of the cyclic group ${\rm Gal}(GF(p^n)/GF(p))$ and hence itself cyclic (generated by $\varphi^k\colon x\mapsto x^{p^k}$).

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    $\begingroup$ Thank you! So in my case the order of Gal$(GF(p^n)/GF(p)$ is 24, and $k=3$. Would this imply that the order of Gal$(GF(p^n)/GF(p^k)$ is 8, and is therefore isomorphic to $\mathbb{Z}_8$? $\endgroup$ – Auclair May 25 '16 at 12:48
  • $\begingroup$ Yes, this is the right conclusion. $\endgroup$ – Claudius May 25 '16 at 12:50
  • $\begingroup$ Excellent. Your proof that the group is cyclic is great, but I feel slightly complicated for an answer on an exam. How much of your proof is "common" knowledge? What I mean is, do you think it's reasonable to just state that the Galois groups in your last paragraph are known to be cyclic, and argue from there that the one I'm looking for is isomorphic to $\mathbb{Z}_8$? $\endgroup$ – Auclair May 25 '16 at 12:54
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    $\begingroup$ Yes, if you study (finite) galois extensions of finite groups, then one of the first things you prove is that the galois groups are cyclic, generated by the Frobenius map; so yes, this should be common knowledge. So for an exam, it suffices to note this general fact (unless you are explicitly asked to prove it in this generality, of course). $\endgroup$ – Claudius May 25 '16 at 12:58

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