Let $A$ be a ring, $x$ a nonzero element of $A$ and consider the annihilator of $x$, i.e $Ann(x)$.
Now let $S$ denote the collection of all prime ideals of $A$ containing $Ann(x)$. It can be shown using Zorn's lemma that $S$ contains minimal elements (with respect the inclusion $\subseteq$).
My question is about the following argument:
Let $P$ be a minimal prime ideal in the set of all prime ideals that contain $Ann(x)$ and suppose $z \in P$. My question is: why does it follows that $z \in \sqrt{Ann(x)}$ ?
We know that $\sqrt{Ann(x)}$ consists of all prime ideals of $A$ that contain $Ann(x)$.
So let $P$ be a minimal prime ideal in the set of all prime ideals that contain $Ann(x)$ and assume $z \in P$.
Now let $J$ be any prime ideal of $A$ that contains $Ann(x)$. We need to show that $z \in J$.
Since $P$ is minimal with respect the inclusion we can't have that $J$ is properly contained in $P$. Why does it follows then that $P$ must contain $J$ ?
Can you please explain?