Let $\mathfrak{a}\subseteq A$, and suppose that $\mathfrak{a}$ has a (minimal) primary decomposition. Then $Ass(A/\mathfrak{q})$ is 1;;non-empty/finite. Moreover, if $\mathfrak{a}={\mathfrak{q}}_1\bigcap \dots \bigcap {\mathfrak{q}}_n$, is a minimal primary decomposition, then:

PROOF:

1. There exists $x\notin \cup {\mathfrak{q}}_{\mathfrak{i}}$, for example $1\in A$. For such an $x$, we have $r(a:x)=\bigcap _{i=1}^{n}r({\mathfrak{q}}_i:x)=\bigcap _{i=1}^n{p}_i$, so by prime avoidance, there is some ${\mathfrak{p}}_i$ such that ${\mathfrak{p}}_{\mathfrak{i}}=r(\mathfrak{a}:x)$. This shows that 1;;$\{r({\mathfrak{q}}_{\mathfrak{1}}),\dots ,r({\mathfrak{q}}_{\mathfrak{n}})\}\supseteq Ass(A/\mathfrak{a})$
2. For the other direction, by minimality, we have for each $i$, an $x_i\in \bigcap _{j\ne i}{\mathfrak{p}}_j-{\mathfrak{p}}_{\mathfrak{i}}$, and then $r(\mathfrak{a}:x_{{}_i})={\mathfrak{p}}_i$.