# characterization of primary ideals

Proposition^{}. Let $R$ be a commutative ring and $I\subseteq R$ an ideal. Then $I$ is primary if and only if every zero divisor^{} in $R/I$ is nilpotent^{}.

Proof. ,,$\Rightarrow $” Assume, that we have $x\in R$ such that $x+I$ is a zero divisor in $R/I$. In particular $x+I\ne 0+I$ and there is $y\in R$, $y+I\ne 0+I$ such that

$$0+I=(x+I)(y+I)=xy+I.$$ |

This is if and only if $xy\in I$. Thus either $y\in I$ or ${x}^{n}\in I$ for some $n\in \mathbb{N}$. Of course $y\notin I$, because $y+I\ne 0+I$ and thus ${x}^{n}\in I$. Therefore ${x}^{n}+I=0+I$, which means that $x+I$ is nilpotent in $R/I$.

,,$\Leftarrow $” Assume that for some $x,y\in R$ we have $xy\in I$ and $x,y\notin I$. Then

$$(x+I)(y+I)=xy+I=0+I,$$ |

so both $x+I$ and $y+I$ are zero divisors in $R/I$. By our assumption^{} both are nilpotent, and therefore there is $n,m\in \mathbb{N}$ such that ${x}^{n}+I={y}^{m}+I=0+I$. This shows, that ${x}^{n}\in I$ and ${y}^{m}\in I$, which completes^{} the proof. $\mathrm{\square}$

Title | characterization of primary ideals |
---|---|

Canonical name | CharacterizationOfPrimaryIdeals |

Date of creation | 2013-03-22 19:04:29 |

Last modified on | 2013-03-22 19:04:29 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 5 |

Author | joking (16130) |

Entry type | Derivation |

Classification | msc 13C99 |