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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.

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  • $\begingroup$ 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. $\endgroup$
    – Youngsu
    Commented Feb 17, 2018 at 22:16
  • $\begingroup$ @Youngsu : In general, there does exist ... $R=\mathbb Z$, $M=\mathbb Q$ ... $\endgroup$
    – user
    Commented Feb 18, 2018 at 13:40
  • $\begingroup$ That makes sense. Thank you. $\endgroup$
    – Youngsu
    Commented Feb 18, 2018 at 17:59
  • $\begingroup$ 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. $\endgroup$
    – Ganapathy K
    Commented Nov 7, 2021 at 18:45

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