# regular elements of finite ring

###### Theorem.

If the finite ring $R$ has regular elements^{}, then it has a unity. All regular elements of $R$ form a group under the ring multiplication and with identity element^{} the unity of $R$. Thus the regular elements are exactly the units of the ring; the rest of the elements are the zero and the zero divisors.

Proof. Obviously, the set of the regular elements is non-empty and closed under the multiplication. Let’s think the multiplication table of this set. It is a finite distinct elements (any equation $ax=ay$ reduces to $x=y$). Hence, for every regular element $a$, the square ${a}^{2}$ determines another ${a}^{\prime}$ such that ${a}^{2}{a}^{\prime}=a$. This implies ${a}^{\prime}({a}^{2}{a}^{\prime})(a{a}^{\prime})={a}^{\prime}a(a{a}^{\prime})$, i.e. $({a}^{\prime}a){(a{a}^{\prime})}^{2}=({a}^{\prime}a)(a{a}^{\prime})$, and since ${a}^{\prime}a$ is regular (http://planetmath.org/ZeroDivisor), we obtain that ${(a{a}^{\prime})}^{2}=a{a}^{\prime}$. So $a{a}^{\prime}$ is idempotent^{}, and because it also is , it must be the unity of the ring (http://planetmath.org/Unity): $a{a}^{\prime}=1$. Thus we see that $R$ has a unity which is a regular element and that $a$ has a multiplicative inverse ${a}^{\prime}$, also regular. Consequently the regular elements form a group.

Title | regular elements of finite ring |
---|---|

Canonical name | RegularElementsOfFiniteRing |

Date of creation | 2013-03-22 15:11:09 |

Last modified on | 2013-03-22 15:11:09 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 19 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 13G05 |

Classification | msc 16U60 |

Related topic | GroupOfUnits |

Related topic | PrimeResidueClass |

Related topic | WedderburnsTheorem |

Related topic | Unity |