4. Let $1 < a ∈ \mathbb{Z}$. Prove for positive $n,d∈\mathbb{Z}$ that $d~|~n⇔a^d-1~|~a^n-1$. Conclude in particular that $\mathbb{F}_{p^d}⊆\mathbb{F}_{p^n}⇔d~|~n$.
Proof: (3) ($⇒$) Evidently $(x^d-1)(x^{n-d}+x^{n-2d}+...+x^d+1)=x^n-1$. ($⇐$) Write $f(x)(x^d-1)=x^n-1$ and assume $d~\not \mid~n$ so $n=qd+r$ for some $0 < r < d$. We thus have every solution of $x^d-1$ is a solution of $x^n-1$ in any field containing $\mathbb{Z}$. Let $\zeta$ be a primitive $d^{th}$ root of unity. Then $\zeta^n-1=\zeta^{qd+r}-1=\zeta^r-1 \neq 0$, a contradiction.
(4a) First we note that $a^n-1 ≡ a^r-1~\text{mod }a^d-1$ if $n≡r~\text{mod }d$. This is because, after writing $n=qd+r$ for $0 ≤ r < d$, we have $1≡(a^d)^q ≡ a^{qd} ≡ a^{n-r}$ and now $a^n ≡ a^r$. Since $0 ≤ a^r - 1 < a^d - 1$ we have $a^n - 1 ≡ 0$ if and only if $n ≡ 0~\text{mod }d$.
(4b) Assume $d~|~n$; then $p^d-1~|~p^n-1$ so $x^{p^d-1}-1~|~x^{p^n-1}-1$ and the splitting field of the latter contains the former, i.e. $\mathbb{F}_{p^d}⊆\mathbb{F}_{p^n}$. Assume $\mathbb{F}_{p^d}⊆\mathbb{F}_{p^n}$; then we see these are exactly the splitting fields of $x^{p^d-1}-1$ and $x^{p^n-1}-1$ respectively, and$$x^{p^d-1}-1=\prod_{α∈\mathbb{F}_{p^d}} (x-α)~|~\prod_{α∈\mathbb{F}_{p^n}} (x-α) = x^{p^n-1}-1$$implying $p^d-1~|~p^n-1$ implying $d~|~n$.$~\square$
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