Math 202 -algebra Iv -hw 6 Solutions

(4) Let a > 1 be an integer. Prove that for any positive integers n, d that d divides n if and only if a − 1 divides a − 1 (cf. the previous exercise). Conclude in particular that Fpd ⊆ Fpn if and only if p divides n. Proof. The proof from the previous problem shows that d divides n if and only if a − 1 divides a − 1; the only change is that in the last step we argue by size rather than degree (if 0 < r < d, then a − 1 cannot divide a − 1 since a − 1 is nonzero, positive, and less than a − 1). Applying this result with a = p, we see that p − 1 divides p − 1 if and only if d divides n. We can think of Fpn as the union of {0} and Fpn = μpn−1 ⊂ Fp, the group of (p − 1)st roots of unity in Fp. Therefore we have d | n⇔ p − 1 | p − 1⇔ μpd−1 ⊂ μpn−1 ⇔ Fpd ⊂ Fpn .