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

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.

• Just out of curiosity; does there exist a non-finitely generated torsion free $R$-module such that $r$ is finite? When $R$ is regular, I do not believe there exists any. Commented Feb 17, 2018 at 22:16
• @Youngsu : In general, there does exist ... $R=\mathbb Z$, $M=\mathbb Q$ ...
– user
Commented Feb 18, 2018 at 13:40
• That makes sense. Thank you. Commented Feb 18, 2018 at 17:59
• It is true even if $M$ is not finitely generated. Please refer to the proof of Krull - Akizuki theorem (Theorem 11.7) in Matsumura's Commutative Ring Theory. Commented Nov 7, 2021 at 18:45