Let $R$ be a commutative, Noetherian ring.
Let $d$ be a dimension function: for each finitely generated $R$-module $M$, we assign a natural number, or zero, such that for every $N \leq M$ a submodule we have $d(M) = \max\{d(N), d(M/N)\}$.
Let $J \triangleleft R$ be an ideal and let $m \in \mathbb{N}$. How does it follow that $d(A/J) = d(A/{J^{m}})$?
So far, I think that if we assume that $d$ is invariant under isomorphisms (if $M_1 \cong M_2$ then $d(M_1) = d(M_2)$, which isn't technically given), we get that $d(A/J^m) = \max\{d(A/J),d(A/J^m/A/J)\} \geq d(A/J)$, since for any $m \geq 1$ we have $J^m \leq J$ and so $A/J$ is isomorphic to a submodule of $A/J^m$. How do you get the other direction?
Some useful facts that may be relevant:
The set of minimal prime ideals over $J$ is finite and non-empty.
There are primes $I \subset P_i$ s.t $P_1P_2\cdots P_n \subset I$.
If $M$ is a finitely generated $R$-module then setting $I = \mathrm{Ann}_R(M)$ we get that $R/I$ is isomorphic to an $R$-submodule of $M^n$ for $n$ the number of generators of $M$.
The first two facts follow from $R$ being Noetherian. The third from commutativity as well.