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I read this proposition from Lahtonen's comments here and I wish to make a proof for that.

Let $p$ be a prime and $f$ be a irreducible polynomial of degree $m$ over $\Bbb{F}_p$. Then $f$ remains irreducible over $\Bbb{F}_{p^n}$ iff $\gcd(m,n) = 1$.

I think I am able to prove $\Leftarrow,$ how do I prove $\Rightarrow$?

$\Leftarrow$: First, observe that $f$ divides $x^{p^m} - x$ and $x^{p^m} -x$ splits into linear factors in $\Bbb{F}_{p^m}$. So $f$ has a root $\beta \in \Bbb{F}_{p^m}$. We have $[\Bbb{F}_p(\beta):\Bbb{F}_p] = m$ and $[\Bbb{F}_{p^n}: \Bbb{F}_p] = n$. Since $\gcd(m,n)=1$, $[\Bbb{F}_{p^n}(\beta):\Bbb{F}_p] =mn$. By multiplicative property of degrees, $[\Bbb{F}_{p^n}(\beta):\Bbb{F}_{p^n}] =m$ and so $f$ is the minimal polynomial of $\beta$ over $\Bbb{F}_{p^n}$. In particular, $f$ is irreducible.

For $\Rightarrow$, I know that $[\Bbb{F}_{p^n}(\beta): \Bbb{F}_p] = mn$ so $\Bbb{F}_{p^n}(\beta) \cong \Bbb{F}_{p^{mn}}$....

Li Chun Min
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Observe that $\mathbb{F}_{p^n}$ and $\mathbb{F}_{p^m}$ are both contained in $\mathbb{F}_{p^k}$ where $k=\frac{mn}{\gcd(m,n)}$ is the least common multiple of $m$ and $n$. Since $\beta\in\mathbb{F}_{p^m}$, it follows that $\mathbb{F}_{p^n}(\beta)\subseteq \mathbb{F}_{p^k}$, and so $mn\leq k$. But $k=\frac{mn}{\gcd(m,n)}$, so this can only happen if $\gcd(m,n)=1$.

Eric Wofsey
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