{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

so6 - Math 55 Homework 6 Solutions Friday 1 Polynomials mod...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 0 and g ( x ) = b k x k + . . . + b 1 x + b 0 , 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 0 , 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 0 and g ( x ) = b k x k + . . . + b 1 x + b 0 . By assumption for every i , a i b i ( mod p ) . Now f ( m ) = a k m k + . . . + a 1 m + a 0 b k m k + . . . + b 1 m + b 0 = 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 0 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 0 . (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 ( x ) . So deg p e f ( x ) · e g ( x ) = deg p ( f ) + deg p ( g ) . And e f ( x ) · e g ( x ) f ( x ) · g ( x )( mod p ) . So also deg p ( f ( x ) · g ( x )) = deg p ( f ) + deg p ( g ) .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}