I am trying to show that $(0)$ and $p^n \mathbb{Z}$ are the precisely primary ideals in $\mathbb{Z}.$

Clearly $(0)$ is a prime ideal hence primary and radical of $p^n \mathbb{Z}$ being the maximal ideal $p\mathbb{Z}$ is primary.

How do I prove the converse ?, i.e., Any nonzero primary ideal in $\mathbb{Z}$ is of the form $p^n \mathbb{Z}.$

MY try: Let $I$ be any nonzero primary ideal in $\mathbb{Z}.$ Then $\text {rad}(I)=p \mathbb{Z}$ for some prime $p.$ Now since $\mathbb{Z}$ is Noetherian ring some power of $p \mathbb{Z}$ is contained in $I.$ Let $n$ be the smallest such that $(p\mathbb{Z})^n \subset I.$ The I should show that $I \subset (p\mathbb{Z})^n$ as well, which I couldn't prove. Note that $(p\mathbb{Z})^n=p^n \mathbb{Z}$.

Any help will be appreciated. Thanks.

  • $\begingroup$ An ideal $I \subseteq R$ is primary iff all zero divisors in $R/I$ are nilpotent. Let $n \in \mathbb Z$. If there are two distinct primes dividing $n$, can you show that $\mathbb Z/n$ has a non nilpotent zero divisor? $\endgroup$ Jun 10 '19 at 6:05
  • $\begingroup$ Yes you are absolutely correct but I want to prove it just through the way I am trying to. $\endgroup$
    – user371231
    Jun 10 '19 at 6:11

If $I$ is any ideal in $\mathbb{Z}$, then $ I = (m)$ for some $ m \in \mathbb{Z}$. Write $m = p_{1}^{k_{1}}\cdots p_{n}^{k_{n}}$ where $k_{i} \geq 1$. Then $\sqrt I$ = $(p_{1}\cdots p_{n})$ but $I$ is $p$-primary, hence involves only one prime divisor.


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