# characterization of primary ideals

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 $x+I\neq 0+I$ and there is $y\in R$, $y+I\neq 0+I$ 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 $y\not\in I$, because $y+I\neq 0+I$ 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 $x,y\not\in I$. 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. $\square$

Title characterization of primary ideals CharacterizationOfPrimaryIdeals 2013-03-22 19:04:29 2013-03-22 19:04:29 joking (16130) joking (16130) 5 joking (16130) Derivation msc 13C99