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I have to compute Galois group of $\mathbb F_q/\mathbb F_p$ where $q=p^n$. I know already that $\mathbb F_q/\mathbb F_p$ is galois, so I don't need to prove it. Moreover, I know that \begin{align*}\sigma :\mathbb F_q&\longrightarrow \mathbb F_q\\ x&\longmapsto x^p\end{align*} is an automorphism that fix elements of $\mathbb F_p$ and thus $\sigma \in Gal(\mathbb F_q/\mathbb F_p)$.

Now, I know that if $k<p^n$ then $\sigma ^k(x)=(x^{p})^k=x^{pk}\neq 1$ and thus $\sigma $ is of order $p^n$. Now if I can prove that $|Gal(\mathbb F_q/\mathbb F_p)|=p^n$, then $Gal(\mathbb F_q/\mathbb F_p)=\left<\sigma \right>$ and it would be finish, but I can't prove it since even $X^q-X$ split over $\mathbb F_q$, theis polynomial is not irreducible, and thus, I can't conclude. So how can I find an irreducible polynomial of degree $q$ that split over $\mathbb F_q$ ?

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    $\begingroup$ You seem to mix $p^n$ with $n$ quite a bit. The degree of the field extension is $n$. So you want $n$ automorphisms not $p^n$. $\endgroup$
    – quid
    Dec 19 '15 at 12:36
  • $\begingroup$ so what is wrong in my argumentation ? Because, $(X^p)^k=X^{pk}$ and not $X^{p^k}$, no ? $\endgroup$
    – MSE
    Dec 19 '15 at 12:44
  • $\begingroup$ Yes $(X^p)^k $ is $X^{pk}$ but you do not get that to begin with as $\sigma^k (x)$ is not $\sigma(x)^k$ but rather the $k$-fold application of $\sigma$ to $x$. See the answer for details. $\endgroup$
    – quid
    Dec 19 '15 at 12:50
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There is a bit of confusion in your attempt but some ideas are alright. Let me try to set this straight.

The degree of the extension is $n$. Thus the Galois group has $n$ elements.

You know one, the Frobenius automorphism $\sigma$. Now, for $k < n$ (not $p^n$) one has that $\sigma^{k}(x) = x^{p^k}$ (not $x^{pk}$, note that $\sigma(\sigma(x))= \sigma(x^p)=(x^p)^p = x^{p^2}$ and so on), and then, since $x^{p^k} \neq x$ (not $1$) one concludes $\sigma^k$ is not the identity map (that is it is not the "one-element" of the Galois group).

Consequently the order of $\sigma$ is (at least) $n$, and the $n$ automorphisms $\sigma^{0}, \sigma^1, \dots, \sigma^{n-1}$ are all distinct. Thus this must be the full Galois group. That is, the Galois group is cyclic of order $n$; more precisely, it is generated by the Frobenius automorphism.

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  • $\begingroup$ I should have said: "since there exists an $x$ such that $x^{p^k} \neq x$". $\endgroup$
    – quid
    Dec 19 '15 at 13:04
  • $\begingroup$ The last thing is how to prove that $\sigma ^i$ and $\sigma ^j$ are distincts for all $i\neq j$ ? Let $i\neq j$ and suppose $\sigma^i (x)=\sigma ^j(x)$ for all $x$. Then $x^{p^i}=x^{p^j}$ and thus (if $i>j$) $x^{p^i}(1-x^{p^{i-j}})=0$ therefore $x^{p^{i-j}}=1$, in particular $\sigma ^{i-j+1}(x)=x$ and thus $\sigma ^{i-j+1}$ fix $\mathbb F_q$. How can I continue ? $\endgroup$
    – MSE
    Dec 19 '15 at 14:02
  • $\begingroup$ There is no real need for doing this. It is a general fact that if $g \in G$ and $g^j$ is not the neutral element for $1 \le j <n$ then $g^0, g^1, \dots, g^{n-1}$ are distinct. For if not $g^i = g^j$ for $i < j$ yields $g^{j-i}$ equals the neutral element, contradiction as $1 \le j-i < n $. If you want you can rewrite that argument using $\sigma$. $\endgroup$
    – quid
    Dec 19 '15 at 14:08
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The exponent $k$ in $\sigma^k$ accounts for the number of compositions of the function $\sigma$ with oneself : $$\sigma^k := \underbrace{\sigma \circ \dots \circ \sigma}_{k \ \textrm{times}} $$

Hence, the function $\sigma^k$ corresponds to the polynomial $X^{p^k}$, ie $\sigma^k(x) = x^{p^k}$.

The polynomial you're looking for is simply $P := X^q - X$.

Proof: $P$ is a polynomial of degree $q$ whose $q$ roots are precisely the elements of the fixed field of $\sigma^n$, namely $\mathbb{F}_q$. One hence gets the factorization : $$P = \prod_{a \in \mathbb{F}_q} (X-a) $$

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