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