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Let $I$ be a decomposable ideal of a commutative ring $R$ with minimal primary decomposition $I=\bigcap_{i=1}^n\mathfrak q_i$.

The first uniqueness theorem shows that $\{\sqrt {\mathfrak q_i}:1\le i\le n\}=\{\sqrt{(I:x)}\in \operatorname{Spec} R:x\notin I\}$

Is there any example imply that we could take off the radical, i.e., for some primary ideal $\mathfrak q_i$ of $I$, $\sqrt {\mathfrak q_i}$ can not be written of the form $(I:x)$?

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In particular, you are asking the following:

If $\mathfrak q$ is a $\mathfrak p$-primary ideal do we have $\mathfrak p=(\mathfrak q:x)$, that is, $\mathfrak p\in \mathrm{Ass}(R/\mathfrak q)$?

This holds if $R$ is noetherian; see How to prove Ass$(R/Q)=\{P\}$ if and only if $Q$ is $P$-primary when $R$ is Noetherian?
Otherwise, let $R=K[X_1,\dots,X_n,\dots]$ and $\mathfrak q=(X_1^2,\dots,X_n^2,\dots)$. Now notice that $\mathrm{Ass}(R/\mathfrak q)=\emptyset$.

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  • $\begingroup$ When $R$ is not noetherian, simply set that $I$ is decompsible, can we have the set of "associated prime ideal of $I$" $=\mathrm{Ass}(R/I)$. I feel confused because book of Atiyah doesn't use the concept of "Ass". $\endgroup$
    – Liding Yao
    Commented May 6, 2016 at 7:57
  • $\begingroup$ @yaoliding There is no need to know the concept of "Ass". Just show that $\mathfrak p=(X_1,\dots,X_n,\dots)$ and $\mathfrak p\ne(\mathfrak q:x)$ for any polynomial $x$. $\endgroup$
    – user26857
    Commented May 6, 2016 at 8:00
  • $\begingroup$ I see, so does the concept "associated prime ideal of I" have a name, since it's not always equal to $Ass(R/I)$ $\endgroup$
    – Liding Yao
    Commented May 6, 2016 at 8:15

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