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Homerings of rational numbers

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The criterion for a non-empty subset $R$ of a given ring $Q$ for being a subring of $Q$, is that $R$ contains always along with its two elements also their difference and product. Since the field $\mathbb{Q}$ of the rational numbers is (isomorphic^{} to) the total ring of quotients of the ring $\mathbb{Z}$ of the integers, any rational number is a quotient $\displaystyle\frac{m}{n}$ of two integers $m$ and $n$. If now $R$ is an arbitrary subring of $\mathbb{Q}$ and

$\frac{m_{1}}{n_{1}},\,\frac{m_{2}}{n_{2}}\in R$ |

with $m_{1},\,n_{1},\,m_{2},\,n_{2}\in\mathbb{Z}$ (and $n_{1}n_{2}\neq 0$), then one must have

$\frac{m_{1}n_{2}-m_{2}n_{1}}{n_{1}n_{2}}\in R,\qquad\frac{m_{1}m_{2}}{n_{1}n_{% 2}}\in R.$ |

Therefore, the set of possible denominators of the elements of $R$ is closed under multiplication, i.e. it forms a multiplicative set. We can of course confine us to subsets $S$ containing only positive integers. But along with any positive integer $n_{0}$, the set $S$ has to contain also all positive divisors, inclusive 1 and the prime divisors of the number $n_{0}$, since the factorisation $n_{0}=uv$ of the denominator of an element $\displaystyle\frac{m}{n_{0}}$ of $R$ implies that the multiple $\displaystyle u\cdot\frac{m}{uv}=\frac{m}{v}$ belongs to $R$. Accordingly, $S$ consists of 1, a certain set of positive prime numbers and all finite products of these, thus being a free monoid on the set of those prime numbers.

Since $R$ contains all multiples^{} of each of its elements, it is apparent that the set of possible numerators form an ideal of $\mathbb{Z}$.

$\therefore$ Theorem. If $R$ is a subring of $\mathbb{Q}$, then there are a principal ideal^{} $(k)$ of $\mathbb{Z}$ and a multiplicative subset $S$ of $\mathbb{Z}$ such that $S$ is a free monoid on certain set of prime numbers and any element $\displaystyle\frac{m}{n}$ of $R$ is characterised by

$\displaystyle\begin{cases}m\in(k),\\ n\in S.\end{cases}$ |

The positive generator^{} $k$ of $(k)$ does not belong to $S$ except when it is 1.

Note. Since $k$ may be greater than 1, the ring $R$ is not necessarily the ring of quotients $S^{{-1}}\mathbb{Z}$, e.g. in the case

$R=\left\{\frac{2a}{3^{s}}\,\vdots\;\;a\in\mathbb{Z},\;\,s\in\mathbb{Z}_{+}% \right\}.\\$ |

Examples.

1. The ring $R:=S^{{-1}}\mathbb{Z}$ of the p-integral rational numbers where

$S=\{\mathrm{the\;power\;products\;of\;all\;positive\;primes\;except\;}p\}$. E.g. the 2-integral rational numbers consist of fractions with arbitrary integer numerators and odd denominators, for example $\frac{1000}{1001}$.

2. The ring $R:=S^{{-1}}\mathbb{Z}$ of the decimal fractions where
$S=\{\mathrm{the\;power\;products\;of\;2\;and\;5}\}$.

3. The ring of the terminating binary or dyadic fractions with any integer numerators but denominators from the set $S=\{1,\,2,\,4,\,8,\,\ldots\}$.

4. If $S=\{1\}$, the subring of $\mathbb{Q}$ is simply some ideal $(k)$ of the ring $\mathbb{Z}$.

All the subrings of $\mathbb{Q}$ (except the trivial ring $\{0\}$) have $\mathbb{Q}$ as their total ring of quotients.

## Mathematics Subject Classification

13B30*no label found*11A99

*no label found*

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