How can I efficiently verify distributivity holds for a 4-element field?
I already have the additive groups for my fields ($Z_4$ and $Z_2 \times Z_2$), but now I need to come up with a multiplication table and make sure distributivity works. I see no other way than checking every combination.
If $F$ is a four-element field, then its multiplicative group has order $3$, hence it is cyclic.
Therefore $F=\{0,1,a,a^2\}$. Let's try reconstructing the Cayley table for the addition. We know that $x+x=0$, because the characteristic must be $2$, as subrings of fields have no zero-divisor.
Note that $1+a\ne a$ and that $a+a^2$ has to be different from $a$ and $a^2$. However it cannot be $0$, otherwise $a(1+a)=0$, which is a contradiction. Hence $a^2+a=1$. Similarly, $1+a=a^2$ and the Cayley table can be filled in: $$ \begin{array}{c|cccc} + & 0 & 1 & a & a^2 \\ \hline 0 & 0 & 1 & a & a^2 \\ 1 & 1 & 0 & a^2 & a \\ a & a & a^2 & 0 & 1 \\ a^2 & a^2 & a & 1 & 0 \end{array} $$ This proves that, if a four-element field exists, it is unique up to isomorphism.
A four-element field exists, namely $F_2[X]/(X^2+1)$, where $F_2=\mathbb{Z}/2\mathbb{Z}$. Thus you need not verify distributivity from the Cayley tables.