# orders of elements in integral domain

###### Theorem.

Let $(D,+,\cdot )$ be an integral domain^{}, i.e. a commutative ring with non-zero unity 1 and no zero divisors^{}. All non-zero elements of $D$ have the same order (http://planetmath.org/OrderGroup) in the additive group^{} $(D,+)$.

Proof. Let $a$ be arbitrary non-zero element. Any multiple (http://planetmath.org/GeneralAssociativity) $na$ may be written as

$$na=n(1a)=\underset{n}{\underset{\u23df}{1a+1a+\mathrm{\cdots}+1a}}=(\underset{n}{\underset{\u23df}{1+1+\mathrm{\cdots}+1}})a=(n1)a.$$ |

Thus, because $a\ne 0$ and there are no zero divisors, an equation $na=0$ is equivalent (http://planetmath.org/Equivalent3) with the equation $n1=0$. So $a$ must have the same as the unity of $D$.

Note. The of the unity element is the characteristic (http://planetmath.org/Characteristic) of the integral domain, which is 0 or a positive prime number.

Title | orders of elements in integral domain |
---|---|

Canonical name | OrdersOfElementsInIntegralDomain |

Date of creation | 2013-03-22 15:40:28 |

Last modified on | 2013-03-22 15:40:28 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 9 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 13G05 |

Related topic | OrderGroup |

Related topic | IdealOfElementsWithFiniteOrder |