• Size and characteristic of a finite field
Concerning the field F2, there is something striking and contrary to ordinary mathematics with the formula 1 + 1 = 0. Indeed the same type of formula holds in each finite field. There is a number p such that 1 + 1 + ..... + 1= 0 where 1 is repeated p times. This number p is a prime integer and is called the characteristic of the field. For instance we say that F2 is a field of characteristic 2 . It can be shown that, if a field has characteristic p then its number of elements is equal to q = pk for some integer k > 0 . Moreover, for all prime numbers p and all integers k > 0 there exists a field of size q = pk. Thus the first finite fields according to their size are
F2, F3, F4, F5, F7, F8, F9, F11, ......
There is no finite field of size 6 or 10 for instance. Fields of size p are called prime fields and they are easily represented by the integers 0, 1, 2,...., p-1 in a similar way as we did above for the field F2. A field of size q = pk with k > 1 is more sophisticated, but it contains the prime field Fp. We give below tables for operations in F3 and F4.
|
Operations in F3
= { 0, 1, 2 }
|
|
Addition
|
|
Multiplication
|
| + |
0 |
1 |
2 |
* |
0 |
1 |
2 |
| 0 |
0 |
1 |
2 |
0 |
0 |
0 |
0 |
| 1 |
1 |
2 |
0 |
1 |
0 |
1 |
2 |
| 2 |
2 |
0 |
1 |
2 |
0 |
2 |
1 |
|
Operations in F4 = { 0, 1, u, v }
|
|
Addition
|
|
Multiplication
|
| + |
0 |
1 |
u |
v |
* |
0 |
1 |
u |
v |
| 0 |
0 |
1 |
u |
v |
0 |
0 |
0 |
0 |
0 |
| 1 |
1 |
0 |
v |
0 |
1 |
0 |
1 |
u |
v |
| u |
u |
v |
0 |
1 |
u |
0 |
u |
v |
1 |
| v |
v |
u |
1 |
0 |
v |
0 |
v |
1 |
u |
• Characteristic of F(q)
Observe that if a field Fq has characteristic p then for every element x of this field we have p*x = 0. Consequently according to the way addition has been defined in the field F(q), we also have p*X = 0 for every element X of F(q). We say that F(q) is an infinite field of characteristic p. For instance F(2) has characteristic 2 and we have 2*X = 0 for all formal number X. This property has an important consequence : for all
X and Y in F(2) we have
( X + Y )2 = X2 + 2*X*Y + Y2 = X2 + Y2
Similarly for all
X and Y in F(q) we will have
This formula will be of great help for computations in F(q). As an illustration in F(2), if X = 1011 then it implies easily X2 = 1000101 .