# 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 $\frac{m}{n}$ 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 $\frac{m}{{n}_{0}}$ 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 $\frac{m}{n}$ 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

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

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.

Title | rings of rational numbers |
---|---|

Canonical name | RingsOfRationalNumbers |

Date of creation | 2014-03-18 15:35:06 |

Last modified on | 2014-03-18 15:35:06 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 17 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 13B30 |

Classification | msc 11A99 |

Synonym | subrings of rationals |

Synonym | subrings of $\mathbb{Q}$ |

Related topic | Localization^{} |

Related topic | ThereforeSign |

Defines | dyadic fraction |

Defines | p-integral rational numbers |

Defines | $p$-integral rational number |