torsion free modules $M$ over Noetherian domain of dimension $1$ for which $l(M/aM) \le (\dim_K K \otimes_R M) \cdot l(R/aR), \forall 0 \ne a \in R$

Question

Let $R$ be a Noetherian domain of Krull-dimension $1$ (i.e. every non-zero prime ideal maximal). Let $M$ be a torsion free $R$-module . Let $K$ be the fraction-field of $R$ and let $r=\dim_K S^{-1}M=\dim_K K \otimes_R M $ (where $S=R \setminus \{0\}$ ). Suppose $r$ is finite. Under these conditions, if $M$ is finitely generated, then I can prove that $l(M/aM) \le r \cdot l(R/aR), \forall 0 \ne a \in R$ . My question is :

Is $l(M/aM) \le r \cdot l(R/aR), \forall 0 \ne a \in R$ even for such non-finitely generated modules $M$ with the other conditions remaining same ? If this is not true in general, is it true at least when $M$ is countably generated ?

Here $l(\cdot)$ denotes the "length" of the module.

Also asked https://math.stackexchange.com/questions/2650702/torsion-free-modules-m-over-noetherian-domain-of-dimension-1-for-which-lm

Answer 1

$l(M/aM) \le r \cdot l(R/aR), \forall 0 \ne a \in R$ should be true for all torsion free modules $M$. Every module is the direct limit of its finitely generated submodules, and the functor $ M \mapsto M/aM$ commutes with direct limits, so we get $M/aM \cong \lim_{\rightarrow} N/aN$ where $N$ runs over all finitely generated submodules of $M$. This limit is over a direct system, so now your claim follows from the following fact:

If $N^i, i\in I$ is a direct system of $R$ modules with $l(N^i) < c$, then $l(\lim_{\rightarrow} N^i) < c$.

Proof: if $0 = N_0 \subset N_1 \subset \dots \subset N_c = \lim_{\rightarrow} N^i$ is a proper chain of submodules of the direct limit, we can pick elements $n_j \in N_j \setminus N_{j-1}$ for $1 \leq j \leq c$. Since we are taking a direct limit and c is finite, there is an $N^i$ such that all the $n_j$ are in the image of $N^i \rightarrow \lim_{\rightarrow} N^i$. By pulling the chain back to $N^i$, we would get that the length of $N^i$ is at least $c$.