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 $p$?

**Remarks**

**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?)

reduced. I think that at least in the finitely generated case this implies "YES" on Question 1. – Sándor Kovács Jul 4 '11 at 2:50