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Y r ) where each xi and each yi is an element of M i . Then x ⊗ y = ( x1 y1 , x2 y2 , . . , x r y r ) where each product xi yi is performed according to the operation that makes the corresponding M i a monoid. 63. Recall that N × M is a Cartesian product; if we consider the monoids (N, ×) and (M, ×), we can show that the direct product is a monoid, much like N and M! To see why, we check each of the properties. (closure) Let t , u ∈ N × M. By definition, we can write t = (a, x α ) and u = b , x β for appropriate a, α, b , β ∈ N.

67 is spectacularly easy. 67. Substitution and the lemma give us α n − 1 = cos 2πn + i sin n = (1 + i · 0) − 1 = 0, 2πn n −1 n 2π n 2π n 5. Elliptic Curves 47 so α is indeed a root of x n − 1. 69. The nth roots of unity are Ω n = 1, α, α2 , . . 67. They form a cyclic group of order n under multiplication. Proof. For m ∈ N+ , we use the fact that the complex numbers are commutative under multiplication: (α m )n − 1 = α mn − 1 = α nm − 1 = (α n ) m − 1 = 1 m − 1 = 0. Hence α m is a root of unity for any m ∈ N+ ; they are distinct for m = 0, .

Hence α m is a root of unity for any m ∈ N+ ; they are distinct for m = 0, . . , n − 1 because the real and imaginary parts do not agree (think of a circle, and see Figure ). Since there can be only n distinct roots, Ω n is a complete list of nth roots of unity. We only sketch the proof that Ω n is a cyclic group. 71 that xy ∈ Ω n . The complex numbers are associative under multiplication; since Ω n C, the elements of Ω n are also associative under multiplication. The multiplicative identity 1 ∈ Ω n since 1n = 1 for all n ∈ N+ .

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