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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?

Zev Chonoles
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Ali1S232
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    Where did you hear this? – The Chaz 2.0 Jun 07 '11 at 01:33
  • 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. – jimjim Jun 07 '11 at 01:41

4 Answers4

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

Zev Chonoles
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  • Your conterexamples in the first paragraph fail to satisfy the condition of being integral domains, though. – Mariano Suárez-Álvarez Jun 07 '11 at 02:00
  • @Mariano: That condition was added after I posted my answer. I've now edited to answer the OP's new question. – Zev Chonoles Jun 07 '11 at 02:02
  • and can you exmplain how does multiplication operator work for n = p^k? – Ali1S232 Jun 07 '11 at 03:16
  • @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 07 '11 at 03:40
  • @zev: correct me if i'm mistaken but with your * operator in a field of 4 elemnts we have (1+1)2 = 3 but 12+12 = 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. – Ali1S232 Jun 07 '11 at 03:59
  • @Gajet: The field with 4 elements is not the integers modulo 4 - in the field with 4 elements, $2=0$. The elements of $\mathbb{F}_4$ are the polynomials $0$, $1$, $x$, and $x+1$ in $\mathbb{F}_2[x]$, reduced modulo $x^2+x+1$, so I'm not sure why you're referring to 2 and 3. The multiplication is precisely how I specified. I will add the addition table, perhaps this will help. – Zev Chonoles Jun 07 '11 at 04:02
  • (In $\mathbb{F}_4$, we have $2=0$ and $3=1$) – Zev Chonoles Jun 07 '11 at 04:21
  • The field with three elements is just $\mathbb{F}_3=\mathbb{Z}/3\mathbb{Z}$, so we have that $1+1=2$, and $2+1=0$. – Zev Chonoles Jun 07 '11 at 04:34
  • the point is I need addition and multiplication operators for a problem i'm trying to solve, and when you are developing a program you can't expect the computer to understand x(x+1) modulo x^2+x+1! so I have to map these terms to Zn. – Ali1S232 Jun 07 '11 at 05:23
  • You should be able to model a polynomial $a_0+a_1x+\cdots+a_nx^n$ over $\mathbb{Z}/p\mathbb{Z}$ (i.e., where the $a_i$ are in $\mathbb{Z}/p\mathbb{Z}$) by using an array of elements $(a_0,\ldots,a_n)$ in $\mathbb{Z}/p\mathbb{Z}$, being careful to define multiplication correctly - note that $(a+bx)(c+dx)\neq (ac)+(bd)x$, so don't define $(a,b)\cdot(c,d)=(ac,bd)$. The operation of taking a polynomial modulo another polynomial seems tougher to implement to me, but I do know that Mathematica, Maple, PARI, etc. can do it. Perhaps if you look at their documentation? – Zev Chonoles Jun 07 '11 at 06:06
  • @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 :-) – Jyrki Lahtonen Jun 07 '11 at 18:16
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A ring can be constructed with any finite number of elements, namely the integers modulo $n$.

Jim Belk
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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?

Mariano Suárez-Álvarez
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  • why should 2a = 5b? – Ali1S232 Jun 07 '11 at 03:10
  • @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-Álvarez Jun 07 '11 at 03:36
  • I mean there has to be proof for any operator * there is such a<>0 and b<>0 that 2a = 0 and 5b = 0. – Ali1S232 Jun 07 '11 at 03:53
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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-Álvarez Jun 07 '11 at 04:53
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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}$$

ypercubeᵀᴹ
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