Homework 3

 


p(x) = x6 + x5 + 1

is primitive (hence, irreducible) over GF(2) . Use p(x) to construct a log/antilog table for GF(26) .


p(x) = x2 + x + 2 

is primitive (hence, irreducible) over GF(3) . Use p(x) to construct a log/antilog table for GF(32) .


ab = 0

Prove that  is not a field.


 
 DEGREE 4 ... 3 37D ...

 

o       Since p(x) is irreducible and of degree 3, it follows that

GF(24) = GF(2)[x] mod p(x)

List all the elements of GF(24) in the above representation, i.e., in terms of

ksi = x mod p(x).

o       Let ksi = x mod p(x). Why is { ksik} not a complete list of all the non-zero elements of GF(24)?


Last Modified: March 1, 2001