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?
Can we make an integral domain with any number of members?
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0Axioms do not specify conditions such as one described above, such conditions can be derived as result of aximos. But if someone decides to make it an axiom that within their system somethings are (not) possible then it is true by default. – 2011-06-07
4 Answers
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 $\begin{array}{c|cccc} {\Large +} & \overline{0} & \overline{1} & \overline{x} & \overline{x+1} \\ \hline \overline{0} & \overline{0} & \overline{1} & \overline{x} & \overline{x+1}\\ \overline{1} & \overline{1} & \overline{0} & \overline{x+1} & \overline{x} \\ \overline{x} & \overline{x} & \overline{x+1} & \overline{0} & \overline{1} \\ \overline{x+1} & \overline{x+1} & \overline{x} & \overline{1} & \overline{0}\end{array}\hskip0.5in \begin{array}{c|cccc} {\Large \times} & \overline{0} & \overline{1} & \overline{x} & \overline{x+1} \\ \hline \overline{0} & \overline{0} & \overline{0} & \overline{0} & \overline{0}\\ \overline{1} & \overline{0} & \overline{1} & \overline{x} & \overline{x+1} \\ \overline{x} & \overline{0} & \overline{x} & \overline{x+1} & \overline{1} \\ \overline{x+1} & \overline{0} & \overline{x+1} & \overline{1} & \overline{x}\end{array}$
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0@Gajet There are tons of material available for efficient software (and even hardware) representation of finite fields. If the size of the field is a power of two (a case of particular interest for applications in coding theory, and also cryptography), then you are in luck, because using a polynomial basis like Zev did here, your addition becomes bitwise XOR, which is particularly efficient :-) – 2011-06-07
A ring can be constructed with any finite number of elements, namely the integers modulo $n$.
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0excuse me I meant a ring with out zero devisor namely an integral; – 2011-06-07
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?
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4@fajet: Cauchy's theorem says that whenever$a$prime p divides the order of a finite group, there is an element in that group of order exactly p. If R is a ring of order 10, then 2 and 5 divide the order of the additive group of R, so there exist elements a and$b$whose orders are 2 and 5, respectively. This means that a and b are both non zero, that a+a=0 and b+b+b+b+b=0. – 2011-06-07
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}$