# the multiplicative identity of a cyclic ring must be a generator

###### Theorem.

Let $R$ be a cyclic ring with multiplicative identity^{} $u$. Then $u$ generates (http://planetmath.org/Generator^{}) the additive group^{} of $R$.

###### Proof.

Let $k$ be the behavior of $R$. Then there exists a generator (http://planetmath.org/Generator) $r$ of the additive group of $R$ such that ${r}^{2}=kr$. Let $a\in \mathbb{Z}$ with $u=ar$. Then $r=ur=(ar)r=a{r}^{2}=a(kr)=(ak)r$. If $R$ is infinite, then $ak=1$, causing $a=k=1$ since $k$ is a nonnegative integer. If $R$ is finite, then $ak\equiv 1\mathrm{mod}|R|$. Thus, $\mathrm{gcd}(k,|R|)=1$. Since $k$ divides $|R|$, $k=1$. Therefore, $a\equiv 1\mathrm{mod}|R|$. In either case, $u=r$. ∎

Note that it was also proven that, if a cyclic ring has a multiplicative identity, then it has behavior one. Its converse is also true. See this theorem (http://planetmath.org/CyclicRingsOfBehaviorOne) for more details.

Title | the multiplicative identity of a cyclic ring must be a generator |
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Canonical name | TheMultiplicativeIdentityOfACyclicRingMustBeAGenerator |

Date of creation | 2013-03-22 15:56:59 |

Last modified on | 2013-03-22 15:56:59 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 16 |

Author | Wkbj79 (1863) |

Entry type | Theorem |

Classification | msc 16U99 |

Classification | msc 13F10 |

Classification | msc 13A99 |

Related topic | CyclicRing3 |

Related topic | CriterionForCyclicRingsToBePrincipalIdealRings |

Related topic | CyclicRingsOfBehaviorOne |