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If $JA_P \subseteq IA_P$ for all $P \in \mathrm{Ass}(R/I)$ then must $\mathrm{Ass}(R/J) \subseteq \mathrm{Ass}(R/I)$?

For some background, I was working on this problem.

Let $I$ and $J$ be ideals of a Noetherian ring $A$. If $J_P = JA_P \subseteq IA_P = I_P$ for all $P \in \mathrm{Ass}(A/I)$, then $J \subseteq I$.

In other words, if we have containment locally at every primary ideal containing $I$, then we have global containment.

Here was my attempt at a solution: let $I = \mathfrak{p_1} \cap \dots \cap \mathfrak{p_n}$ be the minimal primary decomposition of $I$ with $\sqrt{\mathfrak{p_i}} = P_i$ the associated prime. Since $J_P \subseteq I_P$, the $P$-primary component of $J$ is contained in the $P$-primary component of $I$ by a theorem in Matsumura for example. If I knew the associated primes of $A/J$ were contained in $Ass(A/I)$ we'd be done, but I don't see why this should be true. Any hints?

user26857
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    Related: http://math.stackexchange.com/questions/353250/associated-prime-ideals-in-a-noetherian-ring-exercise-6-4-in-matsumura/1038377#1038377 and http://math.stackexchange.com/questions/1586262/if-ja-p-subset-ia-p-for-every-associated-prime-then-j-subset-i?noredirect=1&lq=1 – user26857 Sep 04 '16 at 08:40
  • For $J=0$ you get $\mathrm{Ass}(R)⊆\mathrm{Ass}(R/I)$ for any ideal $I$. What's going on if $R$ is an integral domain? – user26857 Sep 04 '16 at 08:41

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