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Let $m,n$ be positive integers and $p$ be a prime. Consider the finite field $\Bbb{F}_{p^n}$ and let $f(x)\in \Bbb{F}_p[x]$ be an irreducible polynomial of degree $m$. Let the factorisation of $f(x)$ in $\Bbb{F}_{p^n}[x]$ into distinct irreducible polynomials be $f(x)=f_1(x) \cdots f_r(x)$.

I have to find $r$ in terms of $m$ and $n$.

I know that $G=Gal(\Bbb{F}_{p^n}/\Bbb{F}_p)$ acts transitively on the factors $\{f_1(x),\ldots,f_r(x)\}$, i.e., for $f_i(x),f_j(x)$ there exists $\tau_{ij}\in G$ such that $f_i(x)^{\tau_{ij}}=f_j(x)$. We have by the orbit-stabiliser theorem of a transitive action $$r=\frac{n}{\mid Stab_{G}(f_1)\mid}.$$

Since $G$ is cyclic, in fact $G$ is generated by the Frobenius $\sigma:\Bbb{F}_{p^n} \to \Bbb{F}_{p^n}$, given by $a \mapsto a^p$ and since ${\mid Stab_{G}(f_1)\mid}=\cdots={\mid Stab_{G}(f_r)\mid}$, it must be the case $Stab_{G}(f_1)=\cdots=Stab_{G}(f_r)=H$(say), let $H=\langle\sigma^i\rangle$ for some $i\in \{0,1,\ldots,n-1\}$. Then it is clear that $\forall \; i, \;f_i(x)\in \Bbb{F}_{p^n}^{H}=\Bbb{F}_{p^{(i,n)}}$(by Artin's theorem), where $(i,n)$ denotes the gcd of $i,n$. One can also see that $f_i(x)\in \Bbb{F}_{p^m}[x] \; \forall\; i $, since $f(x)$ splits in $\Bbb{F}_{p^m}$. Thus $f_i(x)\in \Bbb{F}_{p^m}[x]\cap \Bbb{F}_{p^n}[x] \; \forall\; i $, implying that $f_i(x) \in \Bbb{F}_{p^{(m,n)}}[x]\;\forall\; i$. From here can I conclude that $(i,n)=(m,n)$?

I need some help with some detail. I will appreciate that. Thanks.

user371231
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  • Basically everything you say is correct. One way of filling in the gaps is to observe that the zeros of $f(x)$ are in $\Bbb{F}{p^m}$. If $\alpha$ is one of those zeros, then the other zeros of the minimal polynomial of $\alpha$ are $F^j(\alpha)$, where $F=\sigma^n:a\mapsto a^{p^n}$ is the lowest power of $\sigma$ that keeps all of $\Bbb{F}{p^n}$ fixed pointwise. The orbits of $\langle F\rangle$ are in 1-1 correspondence with the factors $f_i(x)$. – Jyrki Lahtonen Sep 07 '21 at 05:10
  • I'm fairly sure we have done this on the site already, but apparently without the finite-fields tag. Related. – Jyrki Lahtonen Sep 07 '21 at 05:12
  • The mother thread. A version specific to finite fields. The finite field case is special because the degree of an extension specifies it uniquely. Therefore we can be more specific about the Galois theory. – Jyrki Lahtonen Sep 07 '21 at 05:13
  • Thanks for all the links. Now I am trying to understand those. – user371231 Sep 07 '21 at 05:22

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