Let $M$ be a finitely generated, non-zero , torsion-free module over a complete local $1$-dimensional Noetherian domain $(R,\mathfrak m)$. Let $n=\mu(M)=l_R(M/\mathfrak mM)$ Then $n\ge r:=\dim_K M\otimes_R K$ , where $K$ is the fraction field of $K$ . Then $M$ embeds into $ R^r$. So the canonical map $M\to M^{**}$ is injective, so we can identify $M$ with a submodule of $M^{**}$.
My question is: Does $M^{**}/M$ necessarily have finite length ?
Here $(-)^*:=\operatorname{Hom}_R(-,R)$.
I can show my claim is true when $M$ is an ideal of $R$ because in that case, I know that $I^{**}$ can be embedded inside $R$. Now if $I$ is a non-zero ideal in a Noetherian local domain, then $\mathfrak m^n\subseteq I$ for some $n\ge 1$ hence $R/I$ has finite length as $R$-module and so $l_R(I^{**}/I)\le l_R(R/I)< \infty$ . I'm not sure though how to tackle the case for general modules.
Best Answer
Yes. Note that the canonical map $M\to M^{**}$ becomes an isomorphism after tensoring with $K$ (it is just the canonical isomorphism from a finite-dimensional vector space to its double dual). Thus $M^{**}/M\otimes K=0$, so $M^{**}/M$ is annihilated by some nonzero $r\in R$. But then $R/(r)$ has finite length and $M^{**}/M$ is a finitely generated $R/(r)$-module, so $M^{**}/M$ also has finite length.