This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Homework 10: Math 3360 Applicable Algebra Chapter 17: Congruences and the Chinese Remainder Theorem 9. Show that there is no complete set of representatives consisting of 0 and powers of x modulo x 4 + x 3 + x 2 + x + 1 in F 2 [ x ]. Solution: First we will examine how many elements would be in a complete set of represen tatives modulo x 4 + x 3 + x 2 + x + 1. Since every polynomial of degree 4 or higher can be reduced to a smaller polynomial via the identity x 4 = x 3 + x 2 + x + 1, then it is clear that every polynomial in F 2 [ x ] is equivalent to exactly one polynomial of degree 3 or less. Thus a complete set of representatives would consist of all polynomials of degree 3 or less, of which there are 2 4 = 16. Now let us examine the powers of x . We have x 1 = 0 x 2 = x 2 x 3 = x 3 x 4 = x 3 + x 2 + x + 1 x 5 = x ( x 3 + x 2 + x + 1) = ( x 3 + x 2 + x + 1) + x 3 + x 2 + x = 1 Thus there are only five distinct powers of x . So the set of powers of x and 0 has six elements and thus cannot be a complete set of representatives. 19ii. Solve for f ( x ) in F 2 [ x ] where f ( x ) x 2 ( modx 4 + x 2 + x ) f ( x ) x ( modx 4 + x 3 + 1) . Solution: We know that f ( x ) = x + g ( x ) ( x 4 + x 3 +1) so we get that x + g ( x ) ( x 4 + x 3 +1) x 2 ( modx 4 + x 2 + x ). Thus g ( x ) ( x 2 + x + x 3 +1) x + x 2 ( modx 4 + x 2 + x ). So in order to find g ( x ) we need to find the inverse of x 3 + x 2 + x + 1 modulo x 4 + x 2 + x . Doing the Euclidean Algorithm we get: x 4 + x 2 + x = x ( x 3 + x 2 + x + 1) + ( x 3 ) x 3 + x 2 + x + 1 = 1( x 3 ) + ( x 2 + x + 1) x 3 = x ( x 2 + x + 1) + ( x 2 + x ) x 2 + x + 1 = 1( x 2 + x ) + 1 So going backwards we get 1 = ( x 2 + x 1 ) + 1( x 2 + x ) = ( x 2 + x + 1) + 1(( x 3 ) + x ( x 2 + x + 1)) = (1 + x )( x 2 + x + 1) + 1( x 3 ) = (1 + x )(( x 3 + x 2 + x + 1) + ( x 3 )) + ( x 3 ) = (1 + x )( x 3 + x 2 + x + 1) + ( x )( x 3 ) = (1 + x )( x 3 + x 2 + x + 1) + x (( x 4 + x 2 + x ) + x ( x 3 + x 2 + x + 1)) = x ( x 4 + x 2 + x ) + (1 + x + x 2 )( x 3 + x 2 + x + 1) Thus the inverse is (1 + x + x 2 ), so we can multiply both sides of our equation by that to get that g ( x ) (1+ x + x 2 )( x 2 + x )( modx 4 + x 2 + x ) x 4 + x ( modx 4 + x 2 + x ) x 2 ( modx 4 + x 2 + x ). Thus we get that f ( x ) = x + x 2 ( x 4 + x 3 + 1) = x 6 + x 5 + x 2 + x and we are done....
View
Full
Document
 Spring '08
 BILLERA
 Algebra, Remainder Theorem, Remainder, Congruence

Click to edit the document details