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[parent] rings of rational numbers (Theorem)

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

$\displaystyle \frac{m_1}{n_1},\, \frac{m_2}{n_2} \in R$
with $ m_1,\,n_1,\,m_2,\,n_2 \in \mathbb{Z}$ (and $ n_1n_2 \neq 0$), then one must have
$\displaystyle \frac{m_1n_2-m_2n_1}{n_1n_2} \in R, \quad \frac{m_1m_2}{n_1n_2} \in R.$
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}$.

% latex2html id marker 449 $ \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{align*}\begin{cases}m \in (k),\\ n \in S. \end{cases}\end{align*}    

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

$\displaystyle 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 $ \{0\}$) have $ \mathbb{Q}$ as their total ring of quotients.



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See Also: localization, therefore sign

Other names:  subrings of rationals, subrings of $\mathbb{Q}$
Also defines:  dyadic fraction, p-integral rational numbers, $p$-integral rational number

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Cross-references: decimal fractions, odd, fractions, ring of quotients, generator, multiplicative subset, principal ideal, ideal, numerators, free monoid, finite, prime numbers, implies, number, positive, multiplicative set, multiplication, closed under, denominators, quotient, rational number, integers, total ring of quotients, isomorphic, rational numbers, field, product, difference, contains, subring, ring, subset
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This is version 13 of rings of rational numbers, born on 2008-05-18, modified 2008-05-25.
Object id is 10600, canonical name is RingsOfRationalNumbers.
Accessed 926 times total.

Classification:
AMS MSC11A99 (Number theory :: Elementary number theory :: Miscellaneous)
 13B30 (Commutative rings and algebras :: Ring extensions and related topics :: Quotients and localization)

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