# rings of rational numbers

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 (http://planetmath.org/Divisibility), inclusive 1 and the prime divisors (http://planetmath.org/FundamentalTheoremOfArithmetics) 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 (http://planetmath.org/GeneralAssociativity)  $\displaystyle u\cdot\frac{m}{uv}=\frac{m}{v}$  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 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 (http://planetmath.org/PAdicValuation) 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 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.

Title rings of rational numbers RingsOfRationalNumbers 2014-03-18 15:35:06 2014-03-18 15:35:06 pahio (2872) pahio (2872) 17 pahio (2872) Theorem msc 13B30 msc 11A99 subrings of rationals subrings of $\mathbb{Q}$ Localization ThereforeSign dyadic fraction p-integral rational numbers $p$-integral rational number