$(0)$ and $p^n\mathbb{Z}$ (where $p$ prime, $n$ positive integer) are the only primary ideals in $\mathbb{Z}$

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.

• 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? – paul blart math cop Jun 10 '19 at 6:05
• Yes you are absolutely correct but I want to prove it just through the way I am trying to. – 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.