4. Let 1<a∈Z. Prove for positive n,d∈Z that d | n⇔ad−1 | an−1. Conclude in particular that Fpd⊆Fpn⇔d | n.
Proof: (3) (⇒) Evidently (xd−1)(xn−d+xn−2d+...+xd+1)=xn−1. (⇐) Write f(x)(xd−1)=xn−1 and assume d ∤ n so n=qd+r for some 0<r<d. We thus have every solution of xd−1 is a solution of xn−1 in any field containing Z. Let ζ be a primitive dth root of unity. Then ζn−1=ζqd+r−1=ζr−1≠0, a contradiction.
(4a) First we note that an−1≡ar−1 mod ad−1 if n≡r mod d. This is because, after writing n=qd+r for 0≤r<d, we have 1≡(ad)q≡aqd≡an−r and now an≡ar. Since 0≤ar−1<ad−1 we have an−1≡0 if and only if n≡0 mod d.
(4b) Assume d | n; then pd−1 | pn−1 so xpd−1−1 | xpn−1−1 and the splitting field of the latter contains the former, i.e. Fpd⊆Fpn. Assume Fpd⊆Fpn; then we see these are exactly the splitting fields of xpd−1−1 and xpn−1−1 respectively, andxpd−1−1=∏α∈Fpd(x−α) | ∏α∈Fpn(x−α)=xpn−1−1implying pd−1 | pn−1 implying d | n. ◻
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