# Algèbres et Modules by Ibrahim Assem

By Ibrahim Assem

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Extra resources for Algèbres et Modules

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But then p1 is divisible by q1 , and since both are prime, they must be equal. We then have p1 (1 − q1s1 −1 . . qlsl ) = 0, so that (1 − q1s1 −1 . . qlsl ) = 0, and hence q1s1 −1 . . qlsl = 1. Since the qi are prime and since no positive number divides 1 other than 1 itself, we must have l = 1 and s1 = 1. Suppose t > 1. 16, together with an induction on s1 + · · · + sl , shows that p1 must divide qi for some i. But since p1 and qi are prime, we must have p1 = qi . A similar argument shows that q1 = pj for some j, and our order assumption then shows that i = j = 1, so that p1 = q1 .

Proof (a + b) + c = a + b + c = a + b + c. Since a + (b + c) has the same expansion, addition is associative. The other assertions follow by similar arguments. 7 says, simply, that the operations of addition and multiplication give Zn the structure of a commutative ring. When we refer to Zn as a group, we, of course, mean with respect to addition. 4, Zn has order n. 8. Let n be a positive integer. Then there are groups of order n. 4, Zn = Z/ ≡, where ≡ denotes equivalence modulo n. CHAPTER 2. 9.

3 shows that Rθ ·Rφ = Rφ ·Rθ . Thus, unlike Gl2 (R), SO(2) is an abelian group. 6. There is a homomorphism, exp, from R onto SO(2), deﬁned by exp(θ) = Rθ . The kernel of exp is 2π = {2πk | k ∈ Z}. Thus, Rθ = Rφ if and only if θ − φ = 2πk for some k ∈ Z. 3. Suppose that θ ∈ ker exp. Since the identity element of SO(2) is the identity matrix, the deﬁnition of Rθ forces cos θ to be 1 and sin θ to be 0. 9, we have that exp θ = exp φ if and only if −φ + θ ∈ ker(exp), so the result follows. We conclude with a calculation of the order of the elements of SO(2).