HW12_solns - a prmitive element in F 16 , we have F 16 = {...

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MATH 3360 Sketch of solutions for Assignment 12 Ch.3 4.1 Work on the ring F 11 [ x ] / ( x 11 - 1), and for the RS code to be 3-error correcting, you need a chain of 3 × 2 = 6 consecutive powers of a primitive element a F 11 , i.e. the generator polynomial is of the form g ( x ) = ( x - a )( x - a 2 ) ··· ( x - a 6 ) Pick your favorite primitive element a , e.g. a = 2, and remember to check whether it is primitive or not. Ch.3 4.5 Work on the ring F 16 [ x ] / ( x 16 - 1). From the given condition, the generator polynomial has to be of degree 15 - 11 = 4. Again, you just need to pick a primitive element a F 16 . For instance, take F 16 = F 2 [ b ], where b is a root of the irreducible quartic polynomial x 4 + x + 1. Then check yourself that b is order 15, and the generator polynomial can be chosen as g ( x ) = ( x - b )( x - b 2 )( x - b 3 )( x - b 4 ) Additional Problem 1 Again we use F 16 = F 2 [ b ] as in the last question. Since we have checked b is
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Unformatted text preview: a prmitive element in F 16 , we have F 16 = { , 1 ,b,b 2 ,b 3 , ,b 14 } (Note there are exactly 16 elements). To nd the conjugate class, we can start with b , i.e. the conjugate class containing b is: { b,b 2 ,b 2 2 ,b 2 3 , } Yet we know b 2 4 = b 16 = b , so the conjugate class contains exactly 4 elements. We then take away the elements in this conjugate class from F 16 , and carry out the same calculation for the remaining elements, we will get the 6 con-jugate classes: { } , { 1 } , { b,b 2 ,b 4 ,b 8 } , { b 3 ,b 6 ,b 12 ,b 9 } , { b 5 ,b 10 } , { b 7 ,b 11 ,b 13 ,b 14 } And each conjugate class gives you an irreducible polynomial of degree equal to the cardinality of the class, which is a factor of x 16-x 1...
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