Exercise 4.19 in Atiyah-MacDonald I'm unable to prove the last sentence in the hint to Exercise 4.19 in the book by Atiyah and MacDonald.
Here is the statement of the exercise (with the notation $\subset$ instead of $\subseteq$ for inclusion):

Let $A$ be a ring and $\mathfrak p$ a prime ideal of $A$. Show that every $\mathfrak p$-primary ideal contains $S_{\mathfrak p}(0)$, the kernel of the canonical homomorphism $A\to A_{\mathfrak p}$.
Suppose that $A$ satisfies the following condition: for every prime ideal $\mathfrak p$, the intersection of all $\mathfrak p$-primary ideals of $A$ is equal to $S_{\mathfrak p}(0)$. (Noetherian rings satisfy this condition: see Chapter 10.) Let $\mathfrak p_1,\dots,\mathfrak p_n$ be distinct prime ideals, none of which is a minimal prime ideal of $A$. Then there exists an ideal $\mathfrak a$ in $A$ whose associated prime ideals are $\mathfrak p_1,\dots,\mathfrak p_n$.
[Proof by induction on $n$. The case $n=1$ is trivial (take $\mathfrak a=\mathfrak p_1$). Suppose $n>1$ and let $\mathfrak p_n$ be maximal in the set $\{\mathfrak p_1,\dots,\mathfrak p_n\}$. By the inductive hypothesis there exists an ideal $\mathfrak b$ and a minimal primary decomposition $\mathfrak b=\mathfrak q_1\cap\dots\cap\mathfrak q_{n-1}$, where each $\mathfrak q_i$ is $\mathfrak p_i$-primary. If $\mathfrak b\subset S_{\mathfrak p_n}(0)$ let $\mathfrak p$ be a minimal prime ideal of $A$ contained in $\mathfrak p_n$. Then $S_{\mathfrak p_n}(0)\subset S_{\mathfrak p}(0)$, hence $\mathfrak b\subset S_{\mathfrak p}(0)$. Taking radicals and using Exercise 10, we have $\mathfrak p_1\cap\dots\cap\mathfrak p_{n-1}\subset\mathfrak p$, hence some $\mathfrak p_i\subset \mathfrak p$, hence $\mathfrak p_i=\mathfrak p$ since $\mathfrak p$ is minimal. This is a contradiction since no $\mathfrak p_i$ is minimal. Hence $\mathfrak b\not\subset S_{\mathfrak p_n}(0)$ and therefore there exists a $\mathfrak p_n$-primary ideal $\mathfrak q_n$ such that $\mathfrak b\not\subset\mathfrak q_n$. Show that $\mathfrak a=\mathfrak q_1\cap\dots\cap\mathfrak q_n$ has the required properties.]

To explain my problem, consider the particular case $n=3$. We get $\mathfrak b=\mathfrak q_1\cap\mathfrak q_2$ and $\mathfrak a=\mathfrak q_1\cap\mathfrak q_2\cap\mathfrak q_3$. Suppose $\mathfrak q_1\cap\mathfrak q_3\subset\mathfrak q_2$. I should be able to derive a contradiction from this, but I'm not. [All I can say is that, setting $\mathfrak p_i:=r(\mathfrak q_i)$, we get $\mathfrak p_1\cap\mathfrak p_3\subset\mathfrak p_2$, hence $\mathfrak p_1\subset\mathfrak p_2$ or $\mathfrak p_3\subset\mathfrak p_2$, and thus, $\mathfrak p_3$ being maximal, $\mathfrak p_1\subset\mathfrak p_2$.]
 A: Assume that $\mathfrak{a}=\mathfrak{q}_1\cap \cdots \cap \mathfrak{q}_n$ is not a minimal primary decomposition. Since $\mathfrak{q}_n\not \supset \mathfrak{q}_1\cap \cdots \cap \mathfrak{q}_{n-1}$, there is an index $i, 1\leq i\leq n-1$ such that $\mathfrak{q}_i$ contains the intersection of the other primary ideals. Assume that $i=1$, i.e. $\mathfrak{q}_1\supset \mathfrak{q}_2\cap \cdots \cap\mathfrak{q}_n$. By induction hypothesis, $\mathfrak{q}_1\not \supset \mathfrak{q}_2\cap \cdots \cap\mathfrak{q}_{n-1}$, so there exists $x\in \mathfrak{q}_2\cap \cdots \cap\mathfrak{q}_{n-1}- \mathfrak{q}_1$. On the other hand, $\mathfrak{p}_1=r(\mathfrak{q}_1)\not \supset \mathfrak{q}_n$. Indeed, if $\mathfrak{p}_1\supset \mathfrak{q}_n$ then $\mathfrak{p_1}\supset \mathfrak{p}_n$, since $\mathfrak{p}_n$ is maximal in $\{\mathfrak{p}_1,\dots,\mathfrak{p}_n\}$ we must have $\mathfrak{p_1}=\mathfrak{p}_n$, a contradiction. Hence there exists $y\in \mathfrak{q}_n$ such that $y\notin \mathfrak{p}_1$.
Consider $xy\in \mathfrak{q}_2\cap \cdots \cap \mathfrak{q}_n\subset \mathfrak{q}_1$. Since $\mathfrak{q}_1$ is $\mathfrak{p}_1$-primary, so either $x\in \mathfrak{q}_1$ or $y\in \mathfrak{p}_1$. But by construction of $x$ and $y$, this is not the case.
