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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?

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Where did you hear this? –  The Chaz 2.0 Jun 7 '11 at 1:33
    
Axioms 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. –  Arjang Jun 7 '11 at 1:41

4 Answers 4

up vote 13 down vote accepted

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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Your conterexamples in the first paragraph fail to satisfy the condition of being integral domains, though. –  Mariano Suárez-Alvarez Jun 7 '11 at 2:00
    
@Mariano: That condition was added after I posted my answer. I've now edited to answer the OP's new question. –  Zev Chonoles Jun 7 '11 at 2:02
    
and can you exmplain how does multiplication operator work for n = p^k? –  Ali.S Jun 7 '11 at 3:16
    
@Gajet: I've added an explanation, but if there's anything you're having trouble with I can expand on it some more. –  Zev Chonoles Jun 7 '11 at 3:40
    
@zev: correct me if i'm mistaken but with your * operator in a field of 4 elemnts we have (1+1)*2 = 3 but 1*2+1*2 = 0 and this means 4 elemnt set with (+ mod 4) and your table as * operator is not a ring, so it's not an integeral. –  Ali.S Jun 7 '11 at 3:59

A ring can be constructed with any finite number of elements, namely the integers modulo $n$.

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excuse me I meant a ring with out zero devisor namely an integral; –  Ali.S Jun 7 '11 at 1:57

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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why should 2a = 5b? –  Ali.S Jun 7 '11 at 3:10
    
@Gajet: well, one can pick $a$ and $b$ so that both $2a$ and $5b$ are zero, so in particular they will be equal to each other :) –  Mariano Suárez-Alvarez Jun 7 '11 at 3:36
    
I mean there has to be proof for any operator * there is such a<>0 and b<>0 that 2a = 0 and 5b = 0. –  Ali.S Jun 7 '11 at 3:53
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@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. –  Mariano Suárez-Alvarez Jun 7 '11 at 4:53

Another way to define field F4 is by defining addition with:

$$ 0 + a = a $$

$$ a + a = 0 $$

$$ 1 + 2 = 3 $$

and multiplication having:

$$ 0 \times a = 0 $$

$$ 1 \times a = a $$

$$ 2 \times 2 = 3 $$

$$ \begin{array}{c|cccc} {\Large +} & \overline{0} & \overline{1} & \overline{2} & \overline{3} \\ \hline \overline{0} & \overline{0} & \overline{1} & \overline{2} & \overline{3} \\ \overline{1} & \overline{1} & \overline{0} & \overline{3} & \overline{2} \\ \overline{2} & \overline{2} & \overline{3} & \overline{0} & \overline{1} \\ \overline{3} & \overline{3} & \overline{2} & \overline{1} & \overline{0} \end{array} \hskip0.5in \begin{array}{c|cccc} {\Large \times} & \overline{0} & \overline{1} & \overline{2} & \overline{3} \\ \hline \overline{0} & \overline{0} & \overline{0} & \overline{0} & \overline{0} \\ \overline{1} & \overline{0} & \overline{1} & \overline{2} & \overline{3} \\ \overline{2} & \overline{0} & \overline{2} & \overline{3} & \overline{1} \\ \overline{3} & \overline{0} & \overline{3} & \overline{1} & \overline{2} \end{array}$$

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