The field with 9 elements starts with the integers mod 3, forms polynomials with coefficients in the integers mod 3, and then looks at only the remainders of these polynomials when divided by an irreducible (prime) polynomial of degree two in GF(3).
Exercise:
Verify that the polynomial x^2+1 is irreducible by showing that
it has no roots in GF(3).
Hint: Plug in 0, 1 and 2 for x and
show that these are not roots.
The elements of GF(9) are therefore:
0, 1, 2, x, x+1, x+2, 2x, 2x+1, 2x+2
Here are some examples of addition:
1+2=0 (x) + (2x+1) = 1 (2x+2) + (2x+2) = 2(2x+2) = x+1Here are some examples of multiplication:
2 * 2 = 1 x * 2 = 2x x * x = x^2 = x^2 + 2(x^2+1) = 3x^2 + 2 = 2 (x+1) * (2x) = 2x^2 + 2x = 2 * 2 + 2x = 2x + 4 = 2x + 1
Here is the complete multiplication table:
Because of invertibility, every row and every column of this table is a permutation of the elements 2, x, x+1, .... It is as if of the 7! different permutations, 7 are being selected so that when placed as rows, reading down columns, the result is also a permutation. Because multiplication is commutative, the table is symetric. Other observations are possible and intriguing.
Hints:
The modular operation is setting x^2+1 to zero. Solving, x^2=2. A simple way of manipulating these polynomials is to replace every x^2 by 2, every x^3 by 2x and so forth. It is really the same process are long division, but for some reason it results in faster and more reliable results.