4023f11ps5a

# 4023f11ps5a - Homework#5 Solutions Due October 5 2011 1...

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Homework #5 Solutions Due: October 5, 2011 1. Compute the product (5 x 3 + 4 x 2 + 3)(6 x 2 + 3 x + 5) in Z 7 [ x ]. I Solution. (5 x 3 + 4 x 2 + 3)(6 x 2 + 3 x + 5) = 6 ¢ 5 x 5 + (5 ¢ 3 + 6 ¢ 4) x 4 + (5 ¢ 5 + 3 ¢ 4) x 3 + (4 ¢ 5 + 6 ¢ 3) x 2 + 3 ¢ 3 x + 3 ¢ 5 = 2 x 5 + 4 x 4 + 2 x 3 + 3 x 2 + 2 x + 1 : J 2. Let p ( x ) = x 3 + x 2 + 1 2 Z 2 [ x ], and let K = Z 2 [ x ] = ( p ( x )). If t = x 2 K , then K can be represented by K = ' a 0 + a 1 t + a 2 t 2 : a 0 ; a 1 ; a 2 2 Z 2 ; where t satisﬂes the relation t 3 + t 2 + 1. (See Theorem 0.8.4.) Find the multiplicative inverse of each of the following elements of K . Express your answers in the above form: a 0 + a 1 t + a 2 t 2 . (a) t I Solution. Dividing t 3 + t 2 + 1 by t gives t 3 + t 2 + 1 = t ( t 2 + t ) + 1 so in K we get t ( t 2 + t ) + 1 = 0 since t 3 + t 2 + 1 = 0 in K . Hence, t ( t 2 + t ) = ¡ 1 = 1 so t ¡ 1 = t 2 + t in K . J (b) t + 1 I Solution. Dividing t 3 + t 2 + 1 by t + 1 gives t 3 + t 2 + 1 = t 2 ( t + 1) + 1 so that t 2 ( t + 1) = ¡ 1 = 1 and hence ( t + 1) ¡ 1 = t 2 . J (c) t 2 + t + 1 I Solution. Applying the Euclidean Algorithm gives t 3 + t 2 + 1 = t ( t 2 + t + 1) + ( t + 1) t 2 + t + 1 = t ( t + 1) + 1 ; and working backwards we get 1 = ( t 2 + t + 1) + t ( t + 1) = ( t 2 + t + 1) + t (( t 3 + t 2 + 1) + t ( t 2 + t + 1)) = ( t 2 + t + 1)(1 + t 2 ) + t ( t 3 + t 2 + 1) : Since t 3 + t 2 +1 = 0 in K , this gives 1 = ( t 2 + t +1)(1+ t 2 ) so ( t 2 + t +1) ¡ 1 = 1+ t 2 . J 3. List all of the polynomials of degree 2 over Z 2 . Which of these are irreducible? Math 4023 1

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Homework #5 Solutions Due: October 5, 2011 I Solution. There are 4 polynomials of degree 2 over Z 2 : x 2 , x 2 +1, x 2 + x , x 2 + x +1.
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4023f11ps5a - Homework#5 Solutions Due October 5 2011 1...

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