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?