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.
Best Answer
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.