In a ring $A$, let $D(A)$ denote the set of prime ideals $\mathfrak p$ which satisfy the following condition:
there exists $a \in A$ such that $\mathfrak p$ is minimal in the set of prime ideals containing $(0:a)$.
If the zero ideal has a primary decomposition, show that $D(A)$ is the set of associated prime ideals of $(0)$.
By 1st Uniqueness Theorem, I proved that any associated prime ideal of $(0)$ belongs to $D(A)$.
For the converse, let $\mathfrak p$ be a prime in $D(A)$, I know that the union of associated prime ideals of $(0)$ is the set of zero-divisor of $A$, and the union of $D(A)$ is also.
But I have no idea to continue. Can anyone give me some hint or suggestion? I appreciated any help.
Let $\mathfrak{p}$ be a prime in $D(A)$. Then there exists $a\in A$ such that $\mathfrak{p}$ is minimal in the set of prime ideals containing $(0:a)$. Let $(0)=\bigcap_{i=1}^{n}\mathfrak{q}_{i}$ be a minimal primary decomposition of $(0)$ and $\mathfrak{p}_{i}=\sqrt{\mathfrak{q}_{i}}$ for $i=1,\dotsc,n$. Then we have $\sqrt{(0:a)}=\bigcap_{i=1}^{n}\sqrt{(\mathfrak{q}_{i}:a)}=\bigcap_{a\not\in\mathfrak{q}_{j}}\mathfrak{p}_{j}\subset\mathfrak{p}$ and hence $(0:a)\subset\mathfrak{p}_{j}\subset\mathfrak{p}$ for some $j$. Since $\mathfrak{p}$ is minimal, $\mathfrak{p}=\mathfrak{p}_{j}$.