Suppose $p$ is a prime and $p^k-1$ divides $p^n-1$. How can I show that $\mathbb F_{p^k}\subset \mathbb F_{p^n}$? The only thing I can deduce from the given condition is that $\mathbb F_{p^n}^\times$ contains an element of order $p^k-1$ since $\mathbb F_{p^n}^\times$ is cyclic of order $p^n-1$, and any cyclic group $C_r$ contains an element of order $s$ provided $s$ divides $r$ (take $x^{r/s}$ where $x$ is a generator). How can I proceed?
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1Maybe you find your answer in this post. – Amin235 Jul 17 '18 at 22:13
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Hint: $\mathbb F_{p^n}$ is the set of roots of $x^{p^n}-x$.
lhf
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Am I supposed to show that the powers of $\alpha$ (an element of the multiplicative group of order $p^k-1$) satisfy $x^{p^n-1}-1=0$? If I believe in the fact that $x^{p^k-1}-1$ divides $x^{p^n-1}-1$, then I can see that the $p^k-1$ powers of $\alpha$ satisfy it because they satisfy $x^{p^k-1}-1$. Also I know that the roots of $x^{p^r}-x$ in a field of characteristic $p $ form a subfield. Is that what you meant? – user557 Jul 17 '18 at 23:03
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You can show by induction on $\max(k,n)$ that $\gcd(p^k-1, p^n-1) = p^{\gcd(k,n)}-1$. In particular, $p^k-1\mid p^n-1$ iff $k\mid n$. It is well-known that $\mathbb{F}_{p^k}\leq \mathbb{F}_{p^n}$ iff $k\mid n$.
A. Pongrácz
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