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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....
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 Spring '08
 BILLERA
 Algebra, Remainder Theorem, Remainder, Congruence, Hamming Code, Error detection and correction

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