We define the ideal quotient of two ideals $I, J \unlhd C$ (commutative domain), $(J:I) = \{x \in C \mid xI \subseteq J\}$. I am seeking conditions on the ring $C$ and the ideal $I$ such that $(J:I) = J$. For instance, if $J$ is prime then $(J:I)I \subseteq J \implies (J:I) \subseteq J \ \mathrm{or} \ I \subseteq J$, so if we demand that $ I \nsubseteq P$ for all prime ideals $P$ in $C$ we should obtain $(J:I) = J$ when $J$ is prime. EDIT this is an impossible condition here

I am now trying to generalise this to all ideals $J$. The idea I'm trying is as follows:

I take a primary decomposition of $J = K_1 \cap \dots \cap K_n$ where $K_i$ is $P_i$-primary (ie $\sqrt{K_i} = P_i$). Then $(J:I)I \subseteq J \subseteq K_i$, as $K_i$ is $P_i$-primary, gives $(J:I) \subseteq K_i$ or $I \subseteq \sqrt{K_i} = P_i$.

So if we demand $I \nsubseteq P$ for any prime ideal $P$ in $C$, can we conclude that $(J:I) \subseteq J$?

  • $\begingroup$ oops, I've edited the question - sorry $\endgroup$ – Help please Aug 9 '17 at 11:08
  • $\begingroup$ Also, your demand: "$I \nsubseteq P$ for any prime ideal $P$ in $C$" is unsatisfiable for a proper ideal $I$, $\endgroup$ – quasi Aug 9 '17 at 11:11
  • $\begingroup$ Is the reasoning that any ideal in a commutative noetherian ring is contained in a maximal (hence prime) ideal? If so, do you have any suggestions as to a condition that I could apply instead of this $\endgroup$ – Help please Aug 9 '17 at 11:16
  • $\begingroup$ Yes, every proper ideal can be extended to a maximal ideal, and all maximal ideals are prime. $\endgroup$ – quasi Aug 9 '17 at 11:17
  • 1
    $\begingroup$ If $J$ is a radical ideal, and $S$ is a set of prime ideals whose intersection is $J$, then if $I$ is an ideal such that $I \nsubseteq P$, for all $P \in S$, then $J:I=J$. In particular, $S$ can be the set of those prime ideals which are minimal among the prime ideals containing $J$. $\endgroup$ – quasi Aug 9 '17 at 11:49

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