# ideal of elements with finite order

Theorem. The set of all elements of a ring, which have a finite order in the additive group^{} of the ring, is a (two-sided) ideal of the ring.

Proof. Let $S$ be the set of the elements with finite order in the ring $R$. Denote by $o(x)$ the order of $x$. Take arbitrary elements $a,b$ of the set $S$.

If $\mathrm{lcm}(o(a),o(b))=n=ko(a)=lo(b)$, then

$$n(a-b)=na-nb=ko(a)a-lo(b)b=k\cdot 0-l\cdot 0=0-0=0.$$ |

Thus $$ and so $a-b\in S$.

For any element $r$ of $R$ we have

$$o(a)(ra)=\underset{o(a)}{\underset{\u23df}{ra+ra+\mathrm{\dots}+ra}}=r(\underset{o(a)}{\underset{\u23df}{a+a+\mathrm{\dots}+a}})=r(o(a)a)=r\cdot 0=0.$$ |

Therefore, $$ and $ra\in S$. Similarly, $ar\in S$.

Since $S$ satisfies the conditions for an ideal, the theorem has been proven.

Title | ideal of elements with finite order |

Canonical name | IdealOfElementsWithFiniteOrder |

Date of creation | 2013-03-22 17:52:30 |

Last modified on | 2013-03-22 17:52:30 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 8 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 20A05 |

Classification | msc 16D25 |

Related topic | OrderGroup |

Related topic | Lcm |

Related topic | Multiple |

Related topic | OrdersOfElementsInIntegralDomain |

Related topic | CharacteristicOfFiniteRing |