Question: 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?

Answer: 
Let $R$ be a finite integral domain, with $n=|R|$. Then $R$ is a [finite field](http://en.wikipedia.org/wiki/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
$$\begin{array}{c|cccc} {\Large +} & \overline{0} & \overline{1} & \overline{x} & \overline{x+1} \\ \hline \overline{0} & \overline{0} & \overline{1} & \overline{x} & \overline{x+1}\\ \overline{1} & \overline{1} & \overline{0} & \overline{x+1} & \overline{x} \\ \overline{x} & \overline{x} & \overline{x+1} & \overline{0} & \overline{1} \\ \overline{x+1} & \overline{x+1} & \overline{x} & \overline{1} & \overline{0}\end{array}\hskip0.5in \begin{array}{c|cccc} {\Large \times} & \overline{0} & \overline{1} & \overline{x} & \overline{x+1} \\ \hline \overline{0} & \overline{0} & \overline{0} & \overline{0} & \overline{0}\\ \overline{1} & \overline{0} & \overline{1} & \overline{x} & \overline{x+1} \\ \overline{x} & \overline{0} & \overline{x} & \overline{x+1} & \overline{1} \\ \overline{x+1} & \overline{0} & \overline{x+1} & \overline{1} & \overline{x}\end{array}$$
A ring can be constructed with any finite number of elements, namely the integers modulo $n$.