I have to use $Z_2 \times Z_2$ as my starting point of a four-element field. The addition Cayley table is simple enough if I just add components ($(0,0)$ is the additive identity), but I need help making the multiplication table. The layout of the table is forced (it is identical to the group $Z_3$ because it doesn't contain the additive identity), but how do I know which element is the multiplicative identity?
In fact, I can arrange any of the elements, so how can I figure out what the correct arrangement is? I could test distributivity, but I'm curious if there is a less tedious way than making 6 different multiplication tables and going through distributivity for them all.
The format $\Bbb Z_2\times \Bbb Z_2$ is helpful for the addition, but is irrelevant for a field's multiplication operation.
In lights of that, you can choose any nonzero to play the role of $1$, because denoting the nonzero elements by $a,b,c$ in any order, we always get $a+b=c$, so any permutation that fixes $(0,0)$ is an automorphism of the additive group.