Let $A$ be a ring that is not a domain. Let $\mathfrak{p}$ be a prime ideal in $A$ such that $\cap_{n=0}^{\infty}\mathfrak{p}^n=(0)$. If $\mathfrak{p}$ is principal, then it is a minimal prime ideal.
How can i prove this result? some ideas will help.
Best Answer
Here is another, much more elementary proof. You do not even need to assume that $R$ is not a domain (if you are willing to accept the weaker conclusion that $\mathfrak p$ is minimal among the nonzero prime ideals).
Choose a generator $p$ of $\mathfrak p$. Take an arbitrary nonzero prime ideal $\mathfrak q\subseteq \mathfrak p$ and a nonzero element $a\in \mathfrak q$. Let $m$ be maximal such that $p^m$ divides $a$ (this exists since $a$ is nonzero, so $a\notin \bigcap_n(p^n)$), so that $a=p^mb$ for some $b\notin (p)$. Then also $b\notin \mathfrak q$ (because $\mathfrak p\supseteq \mathfrak q$), so by primality of $\mathfrak q$, we conclude that $p\in \mathfrak q$, and hence $\mathfrak p\subseteq \mathfrak q$.