Math 254a: Cyclotomic Fields and Fermat’s Last Theorem
نویسنده
چکیده
Proof. (i) We have (1− ζ pn)/(1− ζpn) = 1+ ζpn + · · ·+ ζ i−1 p ∈ Z[ζpn ]. On the other hand, if ii ≡ 1 (mod p), we have (1 − ζpn)/(1 − ζ i pn) = (1 − ζ ii′ pn)/(1 − ζ i pn) = 1 + ζ p + · · ·+ ζ i(i′−1) p ∈ Z[ζpn ], so we find that (1 − ζ i pn) and (1− ζpn) divide one another in Z[ζpn] and hence in OQ(ζpn). (ii) By (i), 1 + ζpn = (1− ζ 2 pn)/(1− ζpn) is a unit in Z[ζpn ], hence in OQ(ζpn). (iii) We have Φpn(x) = (x p − 1)/(x n−1 − 1) = 1+x n−1 + · · ·+x n−1 = ∏ (i,p)=1(x− ζ i p ), so Φpn(1) = p = ∏ (i,p)=1(1− ζ i p ). By (i), the right side generates the same ideal as (1 − ζpn) (p−1)p .
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