This question is motivated by this one. The main point of the question (was) to try to weaken the notion of rank. After the answers and comments, it seems this is not a good way to do it, but perhaps it is still useful for anyone trying to do the same.

Remark: Conditions added after Hailong's comment and Tom's answer: $R$ reduced and $M$ indecomposable.

Let $R$ be a reduced noetherian ring and $M$ an indecomposable $R$-module. For any prime $p\in\mathrm{Spec} R$ with residue field $\kappa(p)$ define $$\delta_M(p)=\dim (M\otimes_R \kappa(p))$$ the local rank of $M$ at $p$.

Question 1: If $\delta_M(p)=1$ for all $p$, does it follow that $M_p\simeq R_p$? Or more generally, if $\delta_M$ is constant, does it imply that $M_p$ is a free $R_p$-module for all $p$?


1 If $M$ is finitely generated, then by Nakayama lemma these questions are easy. (See Exercise II.5.8 on page 125 in [Hartshorne]).

2 Tom Goodwillie points out that if $R=\mathbb Z$ and $M\subset \mathbb Q$ consists of all rational numbers $a/b$ such that $b$ is square-free, then $M_p\simeq R_p$, but it is not invertible, so that would be too much to ask.

3 Yves Cornulier shows here that if $M$ is projective and $M_p\simeq R_p$, then $M$ is finitely generated. In other words, if the answer to (the first part of) Question 1 is "YES" then $M$ cannot be projective. So this suggests a subquestion...

Question 1a: Does there exist an example of an $R$ and a non-finitely generated projective $M$ for which $\delta_M$ is constant?

And let me include also a somewhat vague, but related question:

Question 2: Is the class of modules $M$ for which $\delta_M$ is finite for all $p$ interesting? Is there some kind of a finiteness condition they satisfy? (Other than the one that this means directly). Maybe with some additional hypetheses? (projective?)

  • 1
    $\begingroup$ Unless I misunderstand something, but Question 1 is not true even if $M$ is finitely generated. How about $M=R/(x)$ with $x$ inside the nilradical? $\endgroup$ Commented Jul 4, 2011 at 1:24
  • $\begingroup$ Hailong, you're right. I vaguely thought about that possibility, but by the time I wrote the question I forgot. I will have to revise it. Thanks! $\endgroup$ Commented Jul 4, 2011 at 1:34
  • $\begingroup$ Hailong, I added that $R$ be reduced. I think that at least in the finitely generated case this implies "YES" on Question 1. $\endgroup$ Commented Jul 4, 2011 at 2:50
  • $\begingroup$ You don't want irreducible. $\endgroup$ Commented Jul 4, 2011 at 9:42
  • $\begingroup$ @Tom: I agree. Thx. $\endgroup$ Commented Jul 4, 2011 at 17:33

2 Answers 2


Here are a few comments, too long to fit in the comments box.

1) One can modify Tom Goodwillie's example as follows: for simplicity lets pick $R$ to be a local domain of dimension $1$ (so there are only $2$ prime ideals, $0$ and $\mathfrak m$, with residues $K$, the quotient field and $k$, the residue field of $R$ respectively). Then any module $M$ that fits into a short exact sequence:

$$0 \to k \to M \to K \to 0 $$ would satisfy: $\delta_M(0)= \delta_M(\mathfrak m) =1$. But $M_{\mathfrak m} =k$ is not a free $R_{\mathfrak m}=R$-module.

Note that one can not try the exact sequence with $k, K$ swapped, since $Ext^1_R(k,K)=0$. On the other hand $Ext^1(K,k)$ is complicated, since projective resolutions of $K$ depend on the continuum hypothesis!

2) To weed out examples like above, perhaps more relevant than indecomposability or projectivity is to require $M$ to be torsion-free, so $M$ injects into $M\otimes Q(R)$ ($Q(R)$ is the total ring of quotients).

Say we assume this and the "rank" is $1$. Then immediately we know that $M$ is a submodule of $Q(R)$, and this more or less describes $M$: at every prime $p$, once we tensor with $\kappa (p)$, basically only one denominator of $M$ (that is not in $R$) is left. When $R=\mathbb Z$, one can see the example in your 2) very clear from this point of view.

3) Finally, I am not sure we should call $\delta_M(p)$ the local "rank". When $M$ is finitely generated, $\delta_M(p)$ is the minimal number of generators of $M$ locally at $p$. Of course, if $M$ is locally free, it will be the rank, but we definitely do not want to assume that.

This perhaps explains why the easy counter-examples by me and Tom Goodwillie exist: $M$ can locally have a constant number of generators, but it generally does not mean $M$ is locally free.

  • $\begingroup$ Hailong: thanks for the nice analysis! The reason I called it "local rank" was that it was motivated by the question linked above and by trying to understand what one might possibly mean by "rank" if not the rank of the (assumed to be) free $M_p$. Your answer shows the one tried here is not a viable notion. $\endgroup$ Commented Jul 4, 2011 at 23:57

Question 1: No. Let $M=\oplus_p\kappa(p)$.

  • $\begingroup$ Tom, nice. Also, too easy. :) I have to think of a reasonable condition that rules that out.... $\endgroup$ Commented Jul 4, 2011 at 2:37
  • $\begingroup$ Well, I added the obvious "indecomposable". $\endgroup$ Commented Jul 4, 2011 at 2:48

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