There are a number of ways to represent elements of the field; we'll start by representing them as polynomials with degree at most 1, and with integer coefficients modulo 2. There are four such polynomials: {0, 1, x, x + 1}.
Here are the addition and multiplication tables:
+ 0 1 x x + 1
0 0 1 x x + 1
1 1 0 x + 1 x
x x x + 1 0 1
x + 1 x + 1 x 1 0
⋅ 0 1 x x + 1
0 0 0 0 0
1 0 1 x x + 1
x 0 x x2 mod m x2 + x mod m
x + 1 0 x + 1 x2 + x mod m x2 + 1 mod m
Hold on. What's that funny-looking m?
It's a "reduction polynomial" which brings the product back down to degree 1 or less. It has to be a polynomial of degree 2. There are four such polynomials: let's try each and see what we get.
x2
0 x
x 1
x2 + 1
1 x + 1
x + 1 0
x2 + x
x 0
0 x + 1
x2 + x + 1
x + 1 1
1 x
Note that the first three polynomials all factor into products of lower-degree polynomials: x2 = x(x), x2 + 1 = (x + 1)(x + 1), x2 + x = x(x + 1). Only x2 + x + 1 is prime; and this prime reduction polynomial generates a complete multiplication table with no 0s. This is a necessary condition to be a field. Our final tables are:
+ 0 1 x x + 1
0 0 1 x x + 1
1 1 0 x + 1 x
x x x + 1 0 1
x + 1 x + 1 x 1 0
⋅ 0 1 x x + 1
0 0 0 0 0
1 0 1 x x + 1
x 0 x x + 1 1
x + 1 0 x + 1 1 x
We can also write our elements in binary form: 0 => 00, 1 => 01, x => 10, and x + 1 => 11. In this notation our tables become:
+ 00 01 10 11
00 00 01 10 11
01 01 00 11 10
10 10 11 00 01
11 11 10 01 00
⋅ 00 01 10 11
00 00 00 00 00
01 00 01 10 11
10 00 10 11 01
11 00 11 01 10