so6 - Math 55 - Homework 6 Solutions Friday, July 22, 2011...

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Unformatted text preview: Math 55 - Homework 6 Solutions Friday, July 22, 2011 1. Polynomials mod p . In this problem: p is a prime. all polynomials have integer coefficients. Definitions: For polynomials f ( x ) = a k x k + ... + a 1 x + a and g ( x ) = b k x k + ... + b 1 x + b , we say that f ( x ) g ( x )(mod p ) if for every i , a i b i (mod p ). For a polynomial f ( x ) = a k x k + a k- 1 x k- 1 + ... + a x + a , define deg p ( f ) to be the largest integer (at most k ) so that p 6 | a deg p ( f ) (and deg p ( f ) =- if no such integer exists.) (a) If f ( x ) g ( x )(mod p ), and m is an integer, prove that f ( m ) g ( m )(mod p ). Let f ( x ) = a k x k + ... + a 1 x + a and g ( x ) = b k x k + ... + b 1 x + b . By assumption for every i , a i b i ( mod p ) . Now f ( m ) = a k m k + ... + a 1 m + a b k m k + ... + b 1 m + b = f ( m )( mod p ) as desired. (b) Prove the for any polynomial f ( x ), there is a polynomial g ( x ) of degree deg p ( f ( x )) so that g ( x ) f ( x )(mod p ). Let f ( x ) = a k x k + ... + a 1 x + a and let deg p ( f ( x )) = j k . So a i 0( mod p ) for i > j . And a j 6 0( mod p ) so a j 6 = 0 . Put g ( x ) = a j x j + ... + a 1 x + a . (If j =- then g ( x ) = 0 .) So the degree of g ( x ) is deg p ( f ( x )) by how it is defined. For i > j , a i 0( mod p ) . and for i j , a i a i ( mod p ) . So f ( x ) g ( x )( mod p ) . (c) Prove that if f ( x ) and g ( x ) are polynomials, then deg p ( f ( x ) g ( x )) = deg p ( f ( x )) + deg p ( g ( x )). Using part (b), replace f and g with e f and e g whose degrees are deg p ( f ) and deg p ( g ) respectively. Then e f ( x ) e g ( x ) has degree deg p ( f ) + deg p ( g ) . Also, the leading coefficients of e f ( x ) and e g ( x ) are not divisible by p , so neither is the lead- ing coefficent of e f ( x ) e g...
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This note was uploaded on 03/14/2012 for the course MATH 55 taught by Professor Strain during the Summer '08 term at University of California, Berkeley.

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so6 - Math 55 - Homework 6 Solutions Friday, July 22, 2011...

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