Adv Alegbra HW Solutions 71

Adv Alegbra HW Solutions 71 - h are p 1 , p 2 , . . . , p n...

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71 2.110 (i) Generalize Theorem 2.128 by proving that if the prime factoriza- tion of an integer m is m = p e 1 1 ··· p e n n , then I m = I p e 1 1 ×···× I p en n . Solution. The proof is by induction on n 2, the base case being Exercise 2.109. For the inductive step, de f ne m 0 = p e 1 1 ··· p e n 1 n 1 . Since ( m 0 , p e n n ) = 1, the result now follows from Theorem 2.128. (ii) Generalize Corollary 2.131 by proving that if the prime factoriza- tion of an integer m is m = p e 1 1 ··· p e n n , then U ( I m ) = U ( I p e 1 1 ) ×···× U ( I p en n ). Solution. As in part (i), this is a straightforward induction on n 2. 2.111 (i) If p is a prime, prove that φ( p k ) = p k ( 1 1 p ) . Solution. Of the numbers between 1 and p n ,every p th one is a multiple of p . Thus, there are p n 1 multiples of p , and p n p n 1 of these numbers are prime to p . Hence, φ( p n ) = p n p n 1 = p n ( 1 1 / p ) . (ii) If the distinct prime divisors of a positive integer
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Unformatted text preview: h are p 1 , p 2 , . . . , p n , prove that ( h ) = h ( 1 1 p 1 )( 1 1 p 2 ) ( 1 1 p n ). Solution. Let h = p e 1 1 p e t t be the prime factorization of h . By Corollary 2.131, ( h ) = ( p e 1 1 ) ( p e t t ) = p e 1 1 ( 1 1 / p 1 ) p e t t ( 1 1 / p t ) = h ( 1 1 / p 1 )( 1 1 / p 2 ) ( 1 1 / p t ). 2.112 Let p be an odd prime, and assume that a i i mod p for 1 i p 1. Prove that there exist a pair of distinct integers i and j with ia i ja j mod p . Solution. Note f rst that a i i mod p implies Y a i Y i mod p ( p 1 ) ! mod p 1 mod p ,...
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