{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

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

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 ) .

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}