Proposition

Let $a$ be an irreducible ideal in a ring $A$. Then the following are equivalent:

$a$ is primary;
for every multiplicatively closed subset $S$ of $A$ we have $(S^{-1}a)^{c} = (a:x)$ for some $x \in S$;
the sequence $(a:x^n)$ is stationary, for every $x \in A$.

Solution

Let $p = r(a)$.

Suppose $S \cap p = \emptyset$. Then $(S^{-1}a)^c = a$ by Proposition 4.8 [Atiyah]. Since $S \cap p = \emptyset$, $1 \notin p$. $(a:1) = a$ by Lemma 4.4 [Atiyah]. Therefore, $(S^{-1}a)^c = (a:1)$ where $1 \in S$.
Suppose $S \cap p \ne \emptyset$. Proposition 4.8 [Atiyah] implies $(S^{-1}a)^c = (S^{-1}A)^c = A$. Let $x \in S \cap p$. Then $x \in p$, so $x^n \in q$ for some $n \in \mathbb{N}$. Therefore, $A = (a:x^n)$ by Lemma 4.4. Since $S$ is multiplicatively closed, $x^n \in S$. Thus we have $(S^{-1}a)^c = A = (a:x^n)$ with $x^n \in S$.

TODO

TODO