Can we make an integral domain with any number of members? 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?
 A: A ring can be constructed with any finite number of elements, namely the integers modulo $n$.
A: Suppose $R$ is a ring of order $10$. Then there are elements $a\neq0$ and $b\neq0$ such that $2a=5b=0$, because of Cauchy's theorem applied to the additive group of $R$. Let $c=ab$. Can $c$ be non-zero?
A: 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}$$
A: Let $R$ be a finite integral domain, with $n=|R|$. Then $R$ is a finite field, and therefore we must have $n=p^k$ for some prime number $p$ and $k\geq 1$. Conversely, for any prime power $p^k$, there is an integral domain with that number of members, namely $\mathbb{F}_{p^k}$. Thus, there is an integral domain with $n$ elements if and only if $n$ is a power of a prime number.
Thus $\mathbb{F}_4$ is an integral domain with 4 elements, but there is no integral domain with 6 elements because 6 is not a prime power.
The proof that any finite integral domain $R$ is in fact a finite field is quite simple. Given any $a\in R$, $a\neq 0$, let $f:R\rightarrow R$ be the map defined by $f(x)=ax$. Because $R$ is an integral domain, this map must be injective. But because $R$ is finite, an injective map from $R$ to $R$ must be a bijection. Thus, there is some $x\in R$ such that $f(x)=ax=1$, and this $x$ is a multiplicative inverse of $a$. 
We can define $\mathbb{F}_{p^k}$ to be the ring $\mathbb{F}_p[x]/(f)$ for any irreducible $f\in \mathbb{F}_p[x]$ of degree $k$ - no matter what such $f$ we choose, the result is the same up to isomorphism. Note that $\mathbb{F}_p$ is just an alternate notation for $\mathbb{Z}/p\mathbb{Z}$, the integers modulo $p$. Thus, the multiplication in $\mathbb{F}_{p^k}$ is just multiplication of polynomials, taken modulo the polynomial $f$. For example, in $\mathbb{F}_4$, we take $\mathbb{F}_4=\mathbb{F}_2[x]/(x^2+x+1)$, and letting $\overline{g}$ denote $g\in\mathbb{F}_2[x]$ taken modulo $x^2+x+1$, addition and multiplication look like
$$\begin{array}{c|cccc} {\Large +} & \overline{0} & \overline{1} & \overline{x} & \overline{x+1} \\ \hline \overline{0} & \overline{0} & \overline{1} & \overline{x} & \overline{x+1}\\ \overline{1} & \overline{1} & \overline{0} & \overline{x+1} & \overline{x} \\ \overline{x} & \overline{x} & \overline{x+1} & \overline{0} & \overline{1} \\ \overline{x+1} & \overline{x+1} & \overline{x} & \overline{1} & \overline{0}\end{array}\hskip0.5in \begin{array}{c|cccc} {\Large \times} & \overline{0} & \overline{1} & \overline{x} & \overline{x+1} \\ \hline \overline{0} & \overline{0} & \overline{0} & \overline{0} & \overline{0}\\ \overline{1} & \overline{0} & \overline{1} & \overline{x} & \overline{x+1} \\ \overline{x} & \overline{0} & \overline{x} & \overline{x+1} & \overline{1} \\ \overline{x+1} & \overline{0} & \overline{x+1} & \overline{1} & \overline{x}\end{array}$$
