The field of order 8 can be presented as follows:
The polynomial
X3 + X + 1 
Z/2Z[X] is irreducible,
so K is a field of order 8 according to a previous theorem.
Put
x = X + (X3 + X + 1)
so that K = Z/2Z(x). Each element of
K can be written in the form
Z/2Z.
The polynomial X3 + X + 1 has
3 roots in K, viz., x, x2, and x +
x2.
According to the theorem, each of them
leads to an automorphism. For example, the root x2 corresponds to the map
: K -> K with
(x) = x2. That is,
(a + bx + cx2)
= a + bx2 + cx4
= a + cx + (b + c)x2. another way of writing the expression in the right hand side is (a + bx + cx2)2. (Verify!)
The
assignment for boils down to the same as
(y) = y2 (y
L).
The automorphism 2 satisfies
2(
) = (()) =
(2) = 4 = 2 + .
Apparently, this is the automorphism corresponding to the third
root. The group of automorphisms has order 3, and hence is
isomorphic to Z/3Z.
| id | |
2 | |
| 0 | 0 | 0 | 0 |
| 1 | 1 | 1 | 1 |
| | |
2 | + 2
|
| 1 + | 1 +
| 1 + 2 | 1 +
+ 2 |
| 2
| 2 |
+ 2 | |
| 1 + 2 | 1
+ 2 | 1 + + 2
| 1 + |
|
+ 2 | + 2
| | 2 |
| 1 + + 2 | 1 +
+ 2 | 1 + | 1
+ 2 |