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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, $\mathfrak p_1\subset\mathfrak p_2$.]

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    $\begingroup$ There seem to be some mix-ups in the indices; not sure if they are typo's or the source of your confusion. First, you write "If $\mathfrak b\subset S_{\mathfrak p}(0)$ let $\mathfrak{p}$ be a minimal prime ideal of $A$ contained in $\mathfrak{p}_n$." The first inclusion should be $\mathfrak b\subset S_{\mathfrak p_n}(0)$ instead. Second, in your particular case you suppose $\mathfrak{q}_1\cap\mathfrak{q}_3\subset\mathfrak{q}_2$. Instead you should consider $\mathfrak{q}_1\cap\mathfrak{q}_2\subset\mathfrak{q}$ where $\mathfrak{q}\subset\mathfrak{q}_3$ is minimal. $\endgroup$
    – Servaes
    Commented Aug 29, 2019 at 12:19
  • $\begingroup$ @Servaes - Thanks! I've corrected the typo. I don't understand you second comment. Do you agree that the inclusion $\mathfrak{q}_1\cap\mathfrak{q}_3\subset\mathfrak{q}_2$ would imply that the decomposition is not reduced? $\endgroup$ Commented Aug 29, 2019 at 12:46
  • $\begingroup$ Ah yes I agree, and I now understand your intention there. Never mind my second comment. Does $\mathfrak{p_1}\subset\mathfrak{p_2}$ not imply that $\mathfrak{b}=\mathfrak{q}_1\cap\mathfrak{q}_2$ is not reduced to begin with? (Sincere question as I haven't thought about primary ideals in about a decade.) $\endgroup$
    – Servaes
    Commented Aug 29, 2019 at 13:00
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    $\begingroup$ @Servaes - I'm unable to answer your question precisely, but in general there are non-trivial inclusions among the prime ideals of a decomposable ideal (whence the notions of isolated vs embedded prime ideal). $\endgroup$ Commented Aug 29, 2019 at 13:09
  • $\begingroup$ [For the interested reader: the parenthesis "Noetherian rings satisfy this condition: see Chapter 10" in the hint refers to Corollary 10.21.] $\endgroup$ Commented Aug 29, 2019 at 14:39

1 Answer 1

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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.

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