# ideals with maximal radicals are primary

Assume that $R$ is a commutative ring and $I\subseteq R$ is an ideal, such that the radical $r(I)$ of $I$ is a maximal ideal. Then $I$ is a primary ideal.

Proof. We will show, that every zero divisor in $R/I$ is nilpotent (please, see parent object for details).

First of all, recall that $r(I)$ is an intersection of all prime ideals containing $I$ (please, see this entry (http://planetmath.org/ACharacterizationOfTheRadicalOfAnIdeal) for more details). Since $r(I)$ is maximal, it follows that there is exactly one prime ideal $P=r(I)$ such that $I\subseteq P$. In particular the ring $R/I$ has only one prime ideal (because there is one-to-one correspondence between prime ideals in $R/I$ and prime ideals in $R$ containing $I$). Thus, in $R/I$ an ideal $r(0)$ is prime.

Now assume that $\alpha\in R/I$ is a zero divisor. In particular $\alpha\neq 0+I$ and for some $\beta\neq 0+I\in R/I$ we have

 $\alpha\beta=0+I.$

But $0+I\in r(0)$ and $r(0)$ is prime. This shows, that either $\alpha\in r(0)$ or $\beta\in r(0)$.

Obviously $\alpha\in r(0)$ (and $\beta\in r(0)$), because $r(0)$ is the only maximal ideal in $R/I$ (the ring $R/I$ is local). Therefore elements not belonging to $r(0)$ are invertible, but $\alpha$ cannot be invertible, because it is a zero divisor.

On the other hand $r(0)=\{x+I\in R/I\ |\ (x+I)^{n}=0\mbox{ for some }n\in\mathbb{N}\}$. Therefore $\alpha$ is nilpotent and this completes the proof. $\square$

Corollary. Let $p\in\mathbb{N}$ be a prime number and $n\in\mathbb{N}$. Then the ideal $(p^{n})\subseteq\mathbb{Z}$ is primary.

Proof. Of course the ideal $(p)$ is maximal and we have

 $r\big{(}(p^{n})\big{)}=r\big{(}(p)^{n}\big{)}=(p),$

since for any prime ideal $P$ (in arbitrary ring $R$) we have $r(P^{n})=P$. The result follows from the proposition. $\square$

Title ideals with maximal radicals are primary IdealsWithMaximalRadicalsArePrimary 2013-03-22 19:04:31 2013-03-22 19:04:31 joking (16130) joking (16130) 4 joking (16130) Theorem msc 13C99