# 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:

1. The set of minimal prime ideals over $$J$$ is finite and non-empty.

2. There are primes $$I \subset P_i$$ s.t $$P_1P_2\cdots P_n \subset I$$.

3. 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.

## 1 Answer

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, $$d(R/I), and $$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.

• thanks for the partial answer - I agree this is weird. Unfortunately I don't have an example on hand, I'll look further into my notes. Indeed the direct sum property you mentioned is strange, I came upon it too.. – Mariah Feb 16 '19 at 1:23