# characteristic of finite ring

The characteristic^{} (http://planetmath.org/Characteristic) of the residue class ring $\mathbb{Z}/m\mathbb{Z}$, which contains $m$ elements, is $m$, too. More generally, one has the

Theorem. The characteristic of a finite ring divides the number of the elements of the ring.

*Proof.* Let $n$ be the characteristic of the ring $R$ with $m$ elements. Since $m$ is the order (http://planetmath.org/OrderGroup) of the group
$(R,+)$, the Lagrange’s theorem implies that

$$ma=\mathrm{\hspace{0.33em}0}\mathit{\hspace{1em}}\forall a\in R.$$ |

Let $m=qn+r$ where $$. Because

$$ra=(m-qn)a=ma-q(na)=\mathrm{\hspace{0.33em}0}-0=\mathrm{\hspace{0.33em}0}\mathit{\hspace{1em}}\forall a\in R$$ |

and $n$ is the least positive integer $\nu $ making all $\nu a=0$, the number $r$ must vanish. Therefore,
$m=qn$, i.e. $n\mid m$.

Remark. A ring $R$, the polynomial ring $R[X]$ and the ring $R[[X]]$ of formal power series have always the same characteristic.

Title | characteristic of finite ring |
---|---|

Canonical name | CharacteristicOfFiniteRing |

Date of creation | 2013-03-22 19:10:19 |

Last modified on | 2013-03-22 19:10:19 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 7 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 16B99 |

Related topic | Multiple^{} |

Related topic | IdealOfElementsWithFiniteOrder |