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  $n_1{n}_2\ne 0$), then one must have

$$\frac{m_1{n}_2-m_2{n}_1}{n_1{n}_2}\in R,\frac{m_1{m}_2}{n_1{n}_2}\in R.$$

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)  $u\cdot {\displaystyle \frac{m}{uv}}={\displaystyle \frac{m}{v}}$  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

$\{\begin{array}{cc}m\in (k),\hfill & \\ n\in S.\hfill & \end{array}$

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

$$R=\{\frac{2a}{3^s}\mathrm{⋮}a\in \mathbb{Z},s\in {\mathbb{Z}}_{+}\}.$$

Examples.

1.  The ring  $R:=S^{-1}\mathbb{Z}$  of the p-integral rational numbers (http://planetmath.org/PAdicValuation) where
$S=\{\mathrm{the}\mathrm{power}\mathrm{products}\mathrm{of}\mathrm{all}\mathrm{positive}\mathrm{primes}\mathrm{except}p\}$.  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 $S=\{\mathrm{the}\mathrm{power}\mathrm{products}\mathrm{of}\mathrm{\hspace{0.33em}2}\mathrm{and}\mathrm{\hspace{0.33em}5}\}$.

3.  The ring of the or dyadic fractions with any integer numerators but denominators from the set  $S=\{1,\mathrm{\hspace{0.17em}2},\mathrm{\hspace{0.17em}4},\mathrm{\hspace{0.17em}8},\mathrm{\dots }\}$.

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.