TITLE: Can we make an integral domain with any number of members?
QUESTION [5 upvotes]: Is it true that rings without zero divisors (integral domains) can have any number of members except for 4,6? and if this is true then what would the multiplication operator be?

REPLY [2 votes]: Another way to define field F4 is by defining addition with:
$$    0 + a = a $$
$$    a + a = 0 $$ 
$$    1 + 2 = 3 $$
and multiplication having:
$$    0 \times a = 0 $$ 
$$    1 \times a = a $$
$$    2 \times 2 = 3  $$
$$
\begin{array}{c|cccc}
 {\Large +} & \overline{0} & \overline{1} & \overline{2} & \overline{3} \\
 \hline
 \overline{0} & \overline{0} & \overline{1} & \overline{2} & \overline{3} \\
 \overline{1} & \overline{1} & \overline{0} & \overline{3} & \overline{2} \\
 \overline{2} & \overline{2} & \overline{3} & \overline{0} & \overline{1} \\
 \overline{3} & \overline{3} & \overline{2} & \overline{1} & \overline{0}
\end{array}

\hskip0.5in 

\begin{array}{c|cccc}
 {\Large \times} & \overline{0} & \overline{1} & \overline{2} & \overline{3} \\
 \hline
 \overline{0} & \overline{0} & \overline{0} & \overline{0} & \overline{0} \\
 \overline{1} & \overline{0} & \overline{1} & \overline{2} & \overline{3} \\
 \overline{2} & \overline{0} & \overline{2} & \overline{3} & \overline{1} \\
 \overline{3} & \overline{0} & \overline{3} & \overline{1} & \overline{2}
\end{array}$$