Torsion module Finite composition length 
Let $R$ be a Dedekind domain. $A$ be a finitely generated R-module. Then $A = A_1 ⊕ A_2$,for some torsion module A 1 and torsion-free module $A_2.$ Prove that $A_1$ has finite composition length.

I can see why $A_1$ has ACC, but I can not prove $A_1$ has DCC.
This is from Module Theory, Extending Modules, and Generalizations. Theorem 4.12.
 A: Let $R$ be a Dedekind domain and $M$ a torsion module over $R$. Then $M$ is a finitely generated module over the ring $S = R/ \operatorname{Ann}_R(M)$, and any composition series of $M$ over $S$ is a composition series over $R$. Note that $\operatorname{Ann}_R(M)$ is non-zero because $M$ is torsion, and according to Wikipedia it follows that $S$ is a (principal) Artinian ring. As described on the same page, it follows from Hopkins' theorem that $M$ is of finite length over $S$, and therefore over $R$.

Here is an alternative approach.
Proposition. Let $R$ be a noetherian ring, and $M$ a finite $R$ module.
Then $M$ is of finite length if and only if $\operatorname{Ass}_R(M) \subseteq \operatorname{Spm} R$.
[Recall that the set of associated primes of $M$ is $\operatorname{Ass}_S(M) = \lbrace \mathfrak{p} \in \operatorname{Spec} R : \mathfrak{p} = \operatorname{Ann}_S(m) \text{ for some } m \in M \rbrace$.]
Proof.
By 00L0, there exists a filtration by submodules $$ 0 = M_0 \subset M_1 \subset \cdots \subset M_n = M $$ such that each quotient $M_i /M_{i-1} \cong R/ \mathfrak{p}_i$ for some $\mathfrak{p}_i \in \operatorname{Spec} R$. Now, there are two cases:

*

*Each $\mathfrak{p}_i$ is maximal. In this case, each quotient $M_i / M_{i-1}$ is simple, so the chain $M_\bullet$ is maximal. By 00J3, it follows that the length of $M$ is $n < \infty$.

*There is some $\mathfrak{p}_i$ that is not maximal. Then $R/ \mathfrak{p}_i$ is an integral domain that is not a field. Take any non-zero non-unit $x\in R/ \mathfrak{p}_i$. Then $R/ \mathfrak{p}_i \neq (x) \nsupseteq (x^2) \nsupseteq (x^3) \cdots \neq 0$ is an infinite chain of submodules of $R/\mathfrak{p}$, which corresponds to an infinite chain of submodules between $M_{i-1}$ and $M_i$. So the chain $M_\bullet$ can be refined to a chain of arbitrary length. In particular, $M$ does not have finite length.

Now we need to make a connection between the above and $\operatorname{Ass}_R(M)$.
By 02CE, $\operatorname{Ass}_R(M)$ is precisely the set of minimal primes among $\lbrace \mathfrak{p}_1,\dots,\mathfrak{p}_n \rbrace$. So $\operatorname{Ass}_R(M) \subseteq \operatorname{Spm}R$ if and only if each $\mathfrak{p}_i$ is maximal. $\Box$
Corollary. Let $R$ be a noetherian domain of Krull dimension 1, and $M$ an $R$-module that is finite and torsion. Then $M$ has finite length.
Proof. Since $R$ has dimension 1, any associated prime of $M$ is either 0 or maximal. $M$ being torsion, $0 \notin \operatorname{Ann}_R(M)$, so $\operatorname{Ann}_R(M) \subseteq \operatorname{Spm}R$. $\Box$
