# finite ring has no proper overrings

The regular elements^{} of a finite commutative ring $R$ are the units of the ring (see the parent (http://planetmath.org/NonZeroDivisorsOfFiniteRing) of this entry). Generally, the largest overring of $R$, the total ring of fractions^{} $T$, is obtained by forming ${S}^{-1}R$, the extension by localization, using as the multiplicative set $S$ the set of all regular elements, which in this case is the unit group of $R$. The ring $R$ may be considered as a subring of $T$, which consists formally of the fractions $\frac{a}{s}=a{s}^{-1}$ with $a\in R$ and $s\in S$. Since every $s$ has its own group inverse ${s}^{-1}$ in $S$ and so in $R$, it’s evident that $T$ no other elements than the elements of $R$. Consequently, $T=R$, and therefore also any overring of $R$ coincides with $R$.

Accordingly, one can not extend a finite commutative ring by using a localization^{}. Possible extensions^{} must be made via some kind of adjunction (http://planetmath.org/RingAdjunction). A more known special case is a finite integral domain^{} (http://planetmath.org/AFiniteIntegralDomainIsAField) — it is always a field and thus closed under the divisions.

Title | finite ring has no proper overrings |
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Canonical name | FiniteRingHasNoProperOverrings |

Date of creation | 2013-03-22 15:11:12 |

Last modified on | 2013-03-22 15:11:12 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 10 |

Author | pahio (2872) |

Entry type | Result |

Classification | msc 13G05 |

Related topic | ExtensionByLocalization |

Related topic | ClassicalRingOfQuotients |

Related topic | AFiniteIntegralDomainIsAField |

Related topic | RingAdjunction |

Related topic | FormalPowerSeries |