In a commutative, Noetherian ring, $d(A/J) = d(A/{J^{m}})$ 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.
 A: Partial answer:
I'll assume that more precisely the condition is that if 
$$0\to M'\to M\to M''\to 0$$ 
is a short exact sequence of modules over $R$ (commutative, Noetherian) with $M$ finitely generated, then $d(M)=\max\{d(M'),d(M'')\}$. This addresses the isomorphism vs set theoretic equality problem that the original statement had.
I suspect the claim in the question is false, but I can't come up with a counterexample, so I'm leaving this partial answer here. Mostly the problem is that I can't think of a function $d$ which has the desired property, so if the OP has one, adding that context to the question would be really helpful.
This is a weird property. First consider a direct sum. $M\oplus N$. $d(M\oplus N)=\max\{d(M),d(N)\}$, which implies that $d(M^n)=d(M)$, so in particular, $d(R^n)=d(R)$. However, if $M$ is finitely generated, then there is a surjection $R^n\to M$ for some $n$, which implies that $d(M)\le d(R^n)=d(R)$. Thus $d(R)$ is an upper bound on all the "dimensions" of modules over $R$. (I put dimensions in quotation marks, because this behaves very differently from the notions of dimension that I am familiar with.)
Now if $I$ is an ideal of $R$, then $d(R)=\max\{d(I),d(R/I)\}$, which means that one of $d(I)$ or $d(R/I)$ equals $d(R)$. If $d(R/I)=d(R)$, then since $R/I$ is a quotient of $R/I^m$ for all $m\ge 1$, we have $d(R/I)=d(R)\le d(R/I^m)\le d(R)$, so $d(R/I)=d(R/I^m)=d(R)$ in this case.
In the other case, $d(R/I)<d(R)$, and $d(I)=d(R)$. This is the case in which we could falsify the claim in the question, because if $d(I^m) < d(R)$, $d(R/I^m)=d(R)$, then there is no clear contradiction.
However, I can't think of any examples of functions $d$ which have the desired property, so I don't have a counterexample at hand.
