Exercise $7.3$ from Atiyah-Macdonald:
Let $\mathfrak a$ be an irreducible ideal in a ring $A$. Then the following are equivalent:
$(1)$ $\mathfrak a$ is primary;
$(2)$ for every multiplicatively closed subset $S$ of $A$ we have$(S^{−1}\mathfrak a)^c= (\mathfrak a:x)$ for some $x\in S$;
$(3)$ the sequence $(\mathfrak a:x^n)$ is stationary, for every $x\in A$.
I have already shown that $(1)\implies (2)$ and $(2)\implies (3)$.
$(3)\implies (1):$
Suppose $xy\in \mathfrak a$ and assume $x\notin \mathfrak a$. By assumption, we have $(\mathfrak a:y^m)=(\mathfrak a:y^{m+1})$ for some $m\in \mathbb N$. Now I have to use the fact that $\mathfrak a$ is irreducible. But I cannot find any natural candidates $\mathfrak b$ and $\mathfrak c$ to write $\mathfrak a=\mathfrak b\cap\mathfrak c$. One choice is to try to prove $\mathfrak a=(\mathfrak a:y^m)\cap (\mathfrak a:x)$. But the claim seems wrong. Are there any obvious choices I'm missing?
Hints are appreciated.
P.S. I have seen some online solutions to the problem, but every one of them goes to the quotient space $A/\mathfrak a$. I understand those solutions, but I would like to prove without taking quotient.
Claim:$\newcommand{\a}{\mathfrak a}$ $\a=(\a+(x))\cap(\a+(y^m))$.
Clearly, $\a\subseteq(\a+(x))\cap(\a+(y^m))$. Now, assume $z\in (\a+(x))\cap(\a+(y^m))$. Then $z=a_1+r_1x=a_2+r_2y^m$ for some $a_1,a_2\in\a$ and $r_1,r_2\in A$. So $a_1+r_1x-a_2=r_2y^m$. Multiplying by $y$, we have $$r_2y^{m+1}=a_1y+r_1xy-a_2y\in\a\:(\because a_1,a_2,xy\in\a).$$ So $r_2\in(\a:y^{m+1})=(\a:y^{m})$. Therefore $r_2y^m\in\a$ and hence $z=a_2+r_2y^m\in\mathfrak a$. The claim follows.
Since $\a$ is irreducible, either $\a=\a+(x)$ or $\a=\a+(y^m)$. As $x\notin\a$, we have $\a=\a+(y^m)$. So $y^m\in\a$. Hence $\a$ is primary.