Unformatted text preview: p ² 1 . Proof. Let K ± denote the multiplicative group of nonzero elements of K . Then K ± is abelian of order n ² 1 . Let m denote the period of K ± . On the one hand, m ³ n ² 1 D j K ± j . On the other hand, x m D 1 for all elements of K ± , so the polynomial equation x m ² 1 D has n ² 1 distinct solutions in the ﬁeld K . But the number of distinct roots of a polynomial in a ﬁeld is never more than the degree of the polynomial (Corollary 1.8.24 ), so n ² 1 ³ m . Thus the period of K ± equals the order n ² 1 of K ± . But the period and order of a ﬁnite abelian group are equal if, and only if, the group is cyclic. This follows from the fundamental theorem of ﬁnite abelian groups, Theorem 3.6.2 . n...
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- Fall '08
- Algebra, Prime number, Abelian group, Cyclic group, z2