y**t
发帖数: 50 | 1 This is really obvious.
Let F={0, 1, beta, beta^2, ..., beta^(2^n-2)}
First,for multiplication,F-{0}is a cyclic group
for addition,it has a unit 0,every element has a
inverse which is itself.For the sum of 2 elements
just notice that let s=a+b,east to see that s^{2^n}=s(i.e.,s^{2^n-1}=1)
while s is in GF,the set F contains all the elements
of order dividing 2^n-1,so F IS A FIELD.
In fact,a finite extension of dimension n over Z/pZ
is the solutions of X^{2^n}-X in the algebraic closure
of Z/pZ.Th |
|
