New PDF release: Algebra II - Noncommunicative Rings, Identities

By A. I. Kostrikin, I. R. Shafarevich

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Sample text

Show that if a = xn xn−1 · · · x2 x1 in base 10 notation, then a ≡ x1 − x2 + x3 − · · · + (−1)n−1 xn mod 11. 7. Show that 43, 171, 234 ≡ 7 mod 11. 8. Let m be a positive integer. Show that 3 divides 22m+1 + 1 and 22m − 1. m m+1 9. Show that if 22 ≡ −1 mod p, then 22 m (22 10. ) n + 1, 22 ≡ 1 mod p. + 1) = 1 for all m, n ∈ N with m = n. (Hint: 11. Construct addition and multiplication tables for Z5 and Z6 . 12. For which nonzero elements x ∈ Z12 does there exist a nonzero element y with xy = 0? What about in Z5 and Z7 ?

Let x ∈ Z be such that ax ≡ 1 mod n and let k ∈ N be such that x + kn > 0. Then, a(x + kn) = ax + akn ≡ 1 + 0 mod n ≡ 1 mod n. (iii) implies (iv). By the definition of multiplication in Zn it is clear that (iii) implies the existence of an [x] ∈ Zn with [a][x] = [1]. Equivalence of (iv) and (v). Clearly (v) implies (iv). Now suppose that we have elements [x], [y] ∈ Zn such that [a][x] = [1] = [a][y]. Then, [x] = [x] · [1] = [x][a][y] = [a][x][y] = [1] [y] = [y]. Thus the uniqueness in part (v) comes as a bonus.

First suppose that p satisfies (C). Let x ∈ N and x | p. Then there exists y ∈ N with p = xy. Clearly we must have 1 ≤ x, y ≤ p. But, p = xy implies that p | xy and therefore, by (C), p | x or p | y. If p | x then necessarily p = x and y = 1, whereas if p | y then necessarily p = y and x = 1. Therefore, either x = 1 or x = p. This means that p must be prime. Now let us assume that p is a prime number and that p | ab where a, b ∈ Z. Let d = (a, p). Then d | p so that d = 1 or p. If d = p, then p = d | a, as required.

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Algebra II - Noncommunicative Rings, Identities by A. I. Kostrikin, I. R. Shafarevich

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