I was trying to understand a construction in the exercise of Atiyah and MacDonald, Introduction to Commutative Algebra, in chapter $4$ (exercise $19$, pg. $57$) and I stuck at the last line of that. I state it below.

Let $A$ be a Noetherian ring. Given distinct prime ideals $\{p_1, \ldots, p_n\}$ where none of the $p_i$'s are minimal prime in $A,$ want to construct an ideal $I$ whose associated primes are $\{p_1, \ldots, p_n\}.$

For this we use the fact that $S_p(0)= \bigcap J $ where $J$ varies over all $p$-primary ideals in $A$ and $S_p(0)= \ker (A \to A_p).$ We will induct on $n.$ For $n=1$ take $I=p_1$ and we are done. Now assume that $n>1$ and $p_n$ be maximal in $\{p_1, \ldots,p_n\}.$ By the inductive hypothesis there exists an ideal $b$ and an irredundant primary decomposition $b=q_1 \cap\cdots \cap q_{n-1},$ where each $q_i$ is $p_i$ primary. Now if $b \subset S_{p_n}(0),$ let $p$ be a minimal prime ideal of $A$ contained in $p_n.$ Since $S_{p_n}(0) \subset S_p(0),$ $b \subset S_p(0).$ Taking radicals and using localization property we can see that $p_1 \cap \cdots \cap p_{n-1} \subset p,$ hence some $p_i \subset p$ and therefore $p_i=p,$ as $p$ is minimal. This is a contradiction since no $p_i$ is minimal. Hence $b$ is not contained in $S_{p_n}(0)$ and therefore there exists a $p_n$-primary ideal $q_n$ such that $b \not\subset q_n.$ Now we claim that $I=q_1 \cap \cdots \cap q_n$ has the required property, i.e., all we want to show that $I=q_1 \cap \cdots \cap q_n$ is an irredundant primary decomposition of $I.$

I failed to show that $q_2 \cap \cdots \cap q_n \subset q_1$ will not happen while we know that $q_2 \cap \cdots \cap q_{n-1} \not\subset q_1.$ I need some help.

Thanking you.

  • 2
    $\begingroup$ Set $J=q_2 \cap \cdots \cap q_{n-1}$. Then $J\nsubseteq q_1$, while $Jq_n\subseteq q_1$. Then $q_n\subseteq p_1$ and therefore $p_n\subseteq p_1$, a contradiction. (I think you missed this: if $IJ\subseteq Q$ with $Q$ $P$-primary, and $I\nsubseteq Q$, then $J\subseteq P$.) $\endgroup$ – user26857 Dec 6 '17 at 21:49
  • $\begingroup$ @user26857 yes I couldn’t see that ...I don’t think i ever thought that, anyway thank you very much . $\endgroup$ – user371231 Dec 7 '17 at 7:16
  • $\begingroup$ I want to ask one more thing if the above construction is possible if there are minimal primes in the collection? $\endgroup$ – user371231 Dec 7 '17 at 7:22
  • 1
    $\begingroup$ I guess not. Instead can find another argument based on symbolic powers of primes. $\endgroup$ – user26857 Dec 7 '17 at 8:10

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