Journal of the
Korean Mathematical Society JKMS

Let $R$ be a commutative ring with identity and let $M$ be an $R$-module. Then $M$ is called a {\it multiplication module }if for every submodule $N$ of $M$ there exists an ideal $I$ of $R$ such that $N=IM$. Let $M$ be a non-zero multiplication $R$-module. Then we prove the following:
(1) there exists a bijection : $N(M) \cap V(\text{ann}_R(M)) \rightarrow \text{Spec}_R(M)$ and in particular, there exists a bijection : $$N(M) \cap \text{Max}(R) \rightarrow \text{Max}_R(M),$$
(2) $N(M)\cap V(\text{ann}_R(M)) = \text{Supp}(M) \cap V(\text{ann}_R(M))$, and
(3) for every ideal $I$ of $R$, $$ (((\root \of {I+\text{ann}_R(M)})M):_RM )= \cap_{P \in N(M) \cap V(\text{ann}_R(M)) } P. $$ The ideal $\theta(M)= \sum_{m \in M} (Rm:_RM)$ of $R$ has proved useful in studying multiplication modules. We generalize this ideal to prove the following result: Let $R$ be a commutative ring with identity, $P \in \text{Spec}(R)$, and $M$ a non-zero $R$-module satisfying
(1) $M$ is a finitely generated multiplication module,
(2) $PM$ is a multiplication module, and
(3) $P^nM \neq P^{n+1}M$ for every positive integer $n$,
then $ \cap_{n=1}^\infty (P^n + \text{ann}_R(M))\in V(\text{ann}_R(M))=\text{Supp}(M)\subseteq N(M). $

Keywords: prime submodules, maximal submodules, finitely generated modules, multiplication modules

MSC numbers: 13E15, 13A15, 16D10