Galois Field GF(4)

This is the only way to define operations on a four-element set making it a field (up to a permutation of elements). I will use some general properties of fields in the following.

1. First, multiplication and addition are commutative, which saves us some guesswork (we only need to determine half the tables).
2. Furthermore, there's got to be 0 and 1. Multiplication by 1 and 0 works the same in any field, so that takes cares of two rows in the multiplication table.
3. In any field, elements distinct from zero with multiplication form an abelian group. In case of $\mathbf F_4$, it is a three-element group, and there is only one such group, so it must be the cyclic group of order three, hence $B^2=D$, $BD=1$, $D^2=B$, so we're done with multiplication.
4. The characteristic of a field is always a prime number, so it must be $2$ in case of $\mathbf F_4$, so there must be zeroes on diagonal of additive table.
5. Additive identity and inverse are unique, so $B+1\neq B,1,0$, so it must be $D$, similarly $D+1=B$
6. Knowing the above, we can easily see that $B+D=B+B+1=0+1=1$, so we're done.

How do we construct $F_4?$ We can interpret it as a quadratic extension of $F_2$ by the roots of the polynomial $X^2 + X + 1$. If $\alpha$ denotes one root of this, then a second root is $1 + \alpha$, and from the knowledge that $1 + 1 = 0$ and $\alpha^2 = \alpha + 1$ we can work out the addition and multiplication tables of $F_4$.