On Weakly (1, n)-Submodules and Weakly n-Submodules of Modules over Commutative Rings

1. Introduction

Throughout this paper, all rings are commutative with a nonzero identity and all modules are nonzero and unital. Let R denote such a ring and M such an R-module. We write $U\left(R\right)$ and $N\left(R\right)$ for the units and nilpotent elements of R, respectively. For a submodule N of M we set $\left(N{:}_{R}M\right)=\{r\in R:rM\subseteq N\}$. An element $a\in R$ is nilpotent on M if $a^{n}M=0$ for some $n\ge 1$; the set of all such elements is $Nil\left(M\right)=\sqrt{{ann}_R\left(M\right)}$, satisfying $N\left(R\right)\subseteq Nil\left(M\right)$ with equality when M is faithful.

We write $M-\mathrm{rad}\left(N\right)=\sqrt{\left(N{:}_{R}M\right)}·M$ for the M-radical of N, where $\sqrt{\left(N{:}_{R}M\right)}=\{r\in R:r^k\in \left(N{:}_{R}M\right)\mathrm{for}\mathrm{some}k\ge 1\}$.

The study of prime submodules and their generalisations is a central and active theme in module theory [2,3,4]. A proper submodule N of M is prime if $am\in N$ implies $a\in \left(N{:}_{R}M\right)$ or $m\in N$. Many generalisations have been introduced; we summarise the ones most relevant to this paper in Table 1.

The concepts of weakly n-ideal and weakly $(1,n)$-ideal were introduced at the ring level by Ersoy, Koç, Tekir, Yeşilot and Yıldız [1], where it was shown that every strongly 1-absorbing primary ideal is a weakly $(1,n)$-ideal and every weakly $(1,n)$-ideal is weakly 1-absorbing primary, with both inclusions strict, and their behaviour in trivial extensions $A⋉B$ and amalgamated algebras $A{\bowtie }^{f}J$ was investigated. Üregen and Koç [9] subsequently introduced a module-level analogue of weakly n-submodules with a different form of the defining condition (see Remark 3).

The present paper introduces weakly n-submodules and weakly $(1,n)$-submodules as module-level generalisations of the ring-theoretic notions of [1]. Our choice of conditions is governed by the principle that the module-level definition should specialise to the ring-level definition when $M=R$: a proper submodule N is a weakly n-submodule if $0\ne am\in N\Rightarrow a\in \left(N{:}_{R}M\right)$ or $m\in Nil\left(M\right)·M$, and a weakly $(1,n)$-submodule if $0\ne abm\in N$ for nonunits $a,b$ implies $ab\in \left(N{:}_{R}M\right)$ or $m\in Nil\left(M\right)·M$, where $Nil\left(M\right)·M$ denotes the submodule generated by elements $cm$ with $c\in Nil\left(M\right)$ and $m\in M$. These conditions differ from the module-level analogue of [9]; see Remark 3.

The new contributions of this paper are as follows.

• We establish the hierarchy of the new classes (Proposition 2.8): weakly $n\Rightarrow$ weakly $(1,n)\Rightarrow$ weakly 1-absorbing primary, with both inclusions strict (Examples 2.5 and 2.10).
• We prove a six-fold characterisation (Theorem 2.16).
• Via the Reduction Lemma (Lemma 2.18), which establishes a precise dictionary between the weakly $(1,n)$-submodule property of a submodule N of a faithful finitely generated multiplication module and the weakly $(1,n)$-ideal property of the ideal $\left(N{:}_{R}M\right)$, we transfer the ring-theoretic structure theory of [1] to the module setting (Theorems 2.19, 2.21, 2.22).
• We study the transfer under localisations, quotient modules, and homomorphisms (Section 3).

We emphasise that the substantive new content of the third item above is the correspondence established by the Reduction Lemma itself — a translation between two a priori different categories (modules over R, and ideals of R) that is not addressed in [1], which is purely ring-theoretic — rather than a re-derivation of the underlying ring-theoretic classification, which we cite directly from [1]. We note further that this reduction, and the corresponding reliance on [1], is confined to the faithful finitely generated multiplication setting: of the twenty-seven numbered results of the paper, only five (Lemmas 2.17–2.18 and Theorems 2.19, 2.21, 2.22) require this hypothesis. Every other result — the definitions themselves, the hierarchy of classes, the six illustrative examples, the structural containment $N\subseteq Nil\left(M\right)·M$ (Lemma 2.11), Propositions 2.13–2.15, the six-fold characterisation, and the full behaviour under localisation, quotients, and homomorphisms treated in Section 3—is established for a general R-module M (at most finitely generated, where relevant), with no reduction to [1] available and no counterpart there.

2. Definitions, Basic Properties and Structure Theorems

We begin by introducing the two central notions of this paper. The definitions are formulated so that they specialise to the ring-level concepts of [1] when $M=R$ (see Remark 1).

Definition 1.

Let M be an R-module and let N be a proper submodule of M.

(i) 

N is an n-submodule if $am\in N$ for $a\in R$, $m\in M$ implies $a\in \left(N{:}_{R}M\right)$ or $m\in Nil\left(M\right)·M$.

(ii) 

N is a weakly n-submodule if $0\ne am\in N$ for $a\in R$, $m\in M$ implies $a\in \left(N{:}_{R}M\right)$ or $m\in Nil\left(M\right)·M$.

(iii) 

N is a $(1,n)$-submodule if $abm\in N$ for nonunits $a,b\in R$ and $m\in M$ implies $ab\in \left(N{:}_{R}M\right)$ or $m\in Nil\left(M\right)·M$.

(iv) 

N is a weakly $(1,n)$-submodule if $0\ne abm\in N$ for nonunits $a,b\in R$ and $m\in M$ implies $ab\in \left(N{:}_{R}M\right)$ or $m\in Nil\left(M\right)·M$.

Thus the weakly versions are obtained from their non-weak counterparts by adding the hypothesis $\ne 0$.

Remark 1.

When $M=R$, the identities $\left(N{:}_{R}R\right)=N$, ${ann}_R\left(R\right)=0$, and $Nil\left(R\right)=N\left(R\right)$ hold, so our definitions specialise to the weakly n-ideals and weakly $(1,n)$-ideals of [1].

Remark 2.

Every n-submodule is a $(1,n)$-submodule. Indeed, suppose $abm\in N$ for nonunits $a,b\in R$ and $m\notin Nil\left(M\right)·M$. Setting $c=ab$, the product $c·m=abm$ belongs to N, and the n-submodule condition applied to $c·m\in N$ yields $c\in \left(N{:}_{R}M\right)$, i.e. $ab\in \left(N{:}_{R}M\right)$.

Example 1.

Let $R=\mathbb{Z}$, $M={\mathbb{Z}}_4$, and $N=2{\mathbb{Z}}_4=\{\overline{0},\overline{2}\}$. One computes directly that $\left(N{:}_{\mathbb{Z}}M\right)=2\mathbb{Z}$ and $Nil\left(M\right)·M=2{\mathbb{Z}}_4=N$. Now suppose $0\ne a\overline{m}\in N$ with $\overline{m}\notin Nil\left(M\right)·M$. Then $\overline{m}\in \{\overline{1},\overline{3}\}$, and in both cases the condition $a\overline{m}\in N=2{\mathbb{Z}}_4$ forces $a\in 2\mathbb{Z}=\left(N{:}_{\mathbb{Z}}M\right)$. Hence N is a weakly n-submodule, and therefore also weakly $(1,n)$.

Example 2.

Let $R=k\left[\right[X\left]\right]$ where k is a field, $M=R$, and $N=\left(X^2\right)$. Since R is an integral domain with ${ann}_R\left(R\right)=0$, we have $Nil\left(M\right)·M=\left\{0\right\}$. To see that N is not weakly n, take $a=m=X$. Then $0\ne X^2\in N$, yet $X\notin \left(X^2\right)$ and $X\notin \left\{0\right\}=Nil\left(M\right)·M$, so neither alternative of the weakly n-condition holds. On the other hand, N is weakly $(1,n)$. To see this, observe that since $R=k\left[\right[X\left]\right]$ is quasi-local with unique maximal ideal $\left(X\right)$, every nonunit of R lies in $\left(X\right)$, so for any nonunits $a,b\in R$ we have $ab\in {\left(X\right)}^2=\left(X^2\right)=\left(N{:}_{R}R\right)$, and the weakly $(1,n)$-condition is satisfied vacuously. Hence N is weakly $(1,n)$ but not weakly n.

Example 3.

Let $R=\mathbb{Z}$, $M={\mathbb{Z}}_6$, and $N=\left\{\overline{0}\right\}$. One computes $\left(N{:}_{\mathbb{Z}}M\right)=6\mathbb{Z}$. Moreover, $Nil\left(M\right)=\sqrt{{ann}_{\mathbb{Z}}\left({\mathbb{Z}}_6\right)}=\sqrt{6\mathbb{Z}}=6\mathbb{Z}$, and therefore $Nil\left(M\right)·M=6\mathbb{Z}·{\mathbb{Z}}_6=\left\{\overline{0}\right\}$. Since the only element of N is $\overline{0}$, every product $am\in N$ satisfies $am=\overline{0}$, hence the hypothesis $0\ne am\in N$ is never fulfilled. Therefore N is both weakly n and weakly $(1,n)$ vacuously.

However, N is not an n-submodule, that is, the condition $\ne 0$ cannot be dropped. Indeed, taking $a=2$ and $m=\overline{3}$ gives $2·\overline{3}=\overline{0}\in N$, but $2\notin \left(N{:}_{\mathbb{Z}}M\right)=6\mathbb{Z}$ and $\overline{3}\notin Nil\left(M\right)·M=\left\{\overline{0}\right\}$.

Remark 3.

Üregen–Koç [9] define a weakly n-submodule by the condition $0\ne am\in N\Rightarrow a\in Nil\left(M\right)$ or $m\in N$, which is genuinely different from ours. To see this, consider $R=\mathbb{Z}$, $M={\mathbb{Z}}_4\times {\mathbb{Z}}_2$, and $N=\{(\overline{0},\overline{0}),(\overline{0},\overline{1})\}$. Taking $a=1$ and $m=(\overline{0},\overline{1})$, one checks that N is a weakly n-submodule in the sense of [9] but fails to be one in our sense. A key distinction between the two definitions is that every weakly n-submodule in our sense satisfies $N\subseteq Nil\left(M\right)·M$ (Lemma 2.11), a property which fails under the definition of [9].

The next proposition establishes the containment relations among all classes introduced above and connects them to the existing notion of weakly 1-absorbing primary submodules.

Proposition 1.

Let N be a proper submodule of M. Then the following implications hold.

(i) 

If N is a weakly n-submodule, then N is a weakly $(1,n)$-submodule.

(ii) 

If N is a $(1,n)$-submodule, then N is a weakly $(1,n)$-submodule.

(iii) 

If N is a weakly $(1,n)$-submodule, then N is a weakly 1-absorbing primary submodule.

Proof. (i) Suppose $0\ne abm\in N$ for nonunits $a,b\in R$ and $m\notin Nil\left(M\right)·M$. Setting $c=ab$, we have $0\ne cm\in N$ with $m\notin Nil\left(M\right)·M$. Applying the weakly n-submodule condition to $0\ne c·m\in N$ yields $c\in \left(N{:}_{R}M\right)$ or $m\in Nil\left(M\right)·M$. Since the latter is excluded, we obtain $ab=c\in \left(N{:}_{R}M\right)$.

(ii) Since every $(1,n)$-submodule satisfies the defining condition without the restriction $abm\ne 0$, the weakly $(1,n)$-condition (which only requires $0\ne abm\in N$) is automatically satisfied.

(iii) Suppose $0\ne abm\in N$ with nonunits $a,b$ and $m\notin M-\mathrm{rad}\left(N\right)$. Since ${ann}_R\left(M\right)\subseteq \left(N{:}_{R}M\right)$, taking radicals gives $Nil\left(M\right)=\sqrt{{ann}_R\left(M\right)}\subseteq \sqrt{\left(N{:}_{R}M\right)}$, and therefore $Nil\left(M\right)·M\subseteq \sqrt{\left(N{:}_{R}M\right)}·M=M-\mathrm{rad}\left(N\right)$. Hence $m\notin Nil\left(M\right)·M$, and the weakly $(1,n)$-condition gives $ab\in \left(N{:}_{R}M\right)\subseteq \sqrt{\left(N{:}_{R}M\right)}$. □

Remark 4.

Prime submodules and the classes introduced above are mutually independent. As a first example, take $R=\mathbb{Z}$ and $M={\mathbb{Z}}_6$. The zero submodule $N=\left\{\overline{0}\right\}$ is weakly n (by Example 2.6) but is not prime, since $2·\overline{3}=\overline{0}\in N$ while $2\notin \left(N{:}_{\mathbb{Z}}M\right)=6\mathbb{Z}$ and $\overline{3}\notin N$. Conversely, take $R=\mathbb{Z}$ and $M=\mathbb{Z}$. The submodule $N=2\mathbb{Z}$ is prime but is not an n-submodule, since $1·2=2\in N$ while $1\notin \left(N{:}_{\mathbb{Z}}M\right)=2\mathbb{Z}$ and $2\notin Nil\left(M\right)·M=\left\{0\right\}$.

Example 2.5 shows that the containment in (i) is strict. The next example shows that the containment in (iii) is also strict.

Example 4.

Let $R=\mathbb{Z}$, $M={\mathbb{Z}}_{24}$, and $N=\left(\overline{3}\right)=\{\overline{0},\overline{3},\overline{6},\overline{9},\overline{12},\overline{15},\overline{18},\overline{21}\}$. One verifies that $\left(N{:}_{\mathbb{Z}}M\right)=3\mathbb{Z}$, $Nil\left(M\right)·M=6{\mathbb{Z}}_{24}=\{\overline{0},\overline{6},\overline{12},\overline{18}\}$, and $M-\mathrm{rad}\left(N\right)=N$. To see that N is not weakly $(1,n)$, take $a=b=2$ and $m=\overline{3}$. Then $0\ne \overline{12}\in N$, yet $ab=4\notin 3\mathbb{Z}=\left(N{:}_{\mathbb{Z}}M\right)$ and $\overline{3}\notin Nil\left(M\right)·M$, so neither alternative of the weakly $(1,n)$-condition holds.

To verify that N is weakly 1-absorbing primary, suppose $0\ne ab\overline{m}\in N$ with $\overline{m}\notin N$, so that $3\nmid m$. Since $3\mid abm$ and $3\nmid m$, it follows that $3\mid ab$, giving $ab\in 3\mathbb{Z}=\sqrt{3\mathbb{Z}}=\sqrt{\left(N{:}_{\mathbb{Z}}M\right)}$.

A distinctive structural feature of weakly n-submodules — one that sets them apart from their weakly $(1,n)$ counterparts — is that they are always contained in the nilpotent part of the module.

Lemma 1.

If N is a weakly n-submodule, then $N\subseteq Nil\left(M\right)·M$.

Proof. 

Suppose for contradiction that there exists $m\in N\setminus Nil\left(M\right)·M$. In particular $m\ne 0$ (since $0\in Nil\left(M\right)·M$), so $0\ne 1·m\in N$. Applying the weakly n-submodule condition with $a=1$ and noting that $m\notin Nil\left(M\right)·M$, we obtain $1\in \left(N{:}_{R}M\right)$. Hence $M=1·M\subseteq N$, contradicting the properness of N. □

Remark 5.

Lemma 2.11 does not extend to weakly $(1,n)$-submodules. Indeed, in Example 2.5 we have $\left(X^2\right)⊈\left\{0\right\}$, so the conclusion $N\subseteq Nil\left(M\right)·M$ fails. The reason is that the proof of Lemma 2.11 uses $a=1$, which is a unit and is therefore excluded by the $(1,n)$-hypothesis.

By Proposition 2.8(iii), every weakly $(1,n)$-submodule is weakly 1-absorbing primary. The next result provides a partial converse.

Proposition 2.

Let N be a proper submodule of M with $N\subseteq Nil\left(M\right)·M$. The following statements hold.

(i) 

If N is a weakly $(1,n)$-submodule, then N is a weakly 1-absorbing primary submodule.

(ii) 

If, in addition, $\left(N{:}_{R}M\right)$ is a radical ideal and N is a weakly 1-absorbing primary submodule, then N is a weakly $(1,n)$-submodule.

Proof. (i) This is a special case of Proposition 2.8(iii); the hypothesis $N\subseteq Nil\left(M\right)·M$ is not needed here.

(ii) Suppose $0\ne abm\in N$ for some nonunits $a,b\in R$ and $m\in M$ with $m\notin Nil\left(M\right)·M$. Since $\left(N{:}_{R}M\right)$ is radical, $\sqrt{\left(N{:}_{R}M\right)}=\left(N{:}_{R}M\right)$ and $M-\mathrm{rad}\left(N\right)=\left(N{:}_{R}M\right)·M$. Suppose for contradiction that $m\in M-\mathrm{rad}\left(N\right)$. Then $m={\sum }_i{r}_i{m}_i$ with $r_i\in \sqrt{\left(N{:}_{R}M\right)}=\left(N{:}_{R}M\right)$ and $m_i\in M$. Since $r_i\in \left(N{:}_{R}M\right)$, we have $r_{i}M\subseteq N\subseteq Nil\left(M\right)·M$, so $r_i{m}_i\in Nil\left(M\right)·M$ for each i. Therefore $m={\sum }_i{r}_i{m}_i\in Nil\left(M\right)·M$, contradicting our assumption. Hence $m\notin M-\mathrm{rad}\left(N\right)$, and the weakly 1-absorbing primary condition gives $ab\in \sqrt{\left(N{:}_{R}M\right)}=\left(N{:}_{R}M\right)$. □

Proposition 3.

Let ${\left\{N_i\right\}}_{i\in \Delta }$ be a family of weakly $(1,n)$-submodules of M. If ${\bigcap }_{i\in \Delta }{N}_i$ is a proper submodule of M, then ${\bigcap }_{i\in \Delta }{N}_i$ is a weakly $(1,n)$-submodule of M.

Proof. 

Let $0\ne abm\in {\bigcap }_{i\in \Delta }{N}_i$ for some nonunits $a,b\in R$ and $m\in M$ with $m\notin Nil\left(M\right)·M$. For each $i\in \Delta$, we have $0\ne abm\in N_i$, and since $N_i$ is a weakly $(1,n)$-submodule and $m\notin Nil\left(M\right)·M$, the weakly $(1,n)$-condition gives $ab\in \left(N_i{:}_{R}M\right)$. Therefore $ab\in {\bigcap }_{i\in \Delta }\left(N_i{:}_{R}M\right)=\left({\bigcap }_{i\in \Delta }{N}_i{:}_{R}M\right)$. □

For a submodule N of M and a nonempty subset $U\subseteq R$, the residual of N by U is the submodule $\left(N{:}_{M}U\right)=\{m\in M:um\in N\mathrm{for}\mathrm{all}u\in U\}$.

Proposition 4.

Let N be a weakly $(1,n)$-submodule of M and let U be a nonempty subset of R satisfying the following conditions:

(a) 

$U⊈\left(N{:}_{R}M\right)$;

(b) 

every $u\in U$ is a non-zero-divisor on M;

(c) 

for all $u\in U$ and $m\in M$, $um\in Nil\left(M\right)·M$ implies $m\in Nil\left(M\right)·M$.

Then $\left(N{:}_{M}U\right)$ is a weakly $(1,n)$-submodule of M.

Proof. 

We first verify that $\left(N{:}_{M}U\right)$ is a proper submodule of M. By condition (a), there exists $u_0\in U\setminus \left(N{:}_{R}M\right)$; then $u_{0}M⊈N$, so $\left(N{:}_{M}U\right)\ne M$.

Now let $0\ne abm\in \left(N{:}_{M}U\right)$ for some nonunits $a,b\in R$ and $m\in M$ with $m\notin Nil\left(M\right)·M$. For every $u\in U$, we have $ab\left(um\right)=u\left(abm\right)\in N$. Since u is a non-zero-divisor on M by (b) and $abm\ne 0$, we get $ab\left(um\right)\ne 0$. Since $m\notin Nil\left(M\right)·M$, the contrapositive of condition (c) gives $um\notin Nil\left(M\right)·M$. We may therefore apply the weakly $(1,n)$-condition to $0\ne ab\left(um\right)\in N$, obtaining $ab\in \left(N{:}_{R}M\right)$. To conclude that $ab\in \left(\left(N{:}_{M}U\right){:}_{R}M\right)$, take any $m^{\prime }\in M$ and any $u\in U$. Then $u(ab·m^{\prime })=ab\left(u{m}^{\prime }\right)\in ab·M\subseteq N$ (since $ab\in \left(N{:}_{R}M\right)$), so $ab·m^{\prime }\in \left(N{:}_{M}U\right)$. Therefore $ab\in \left(\left(N{:}_{M}U\right){:}_{R}M\right)$, and $\left(N{:}_{M}U\right)$ is a weakly $(1,n)$-submodule of M. □

We now establish a six-fold characterisation of weakly $(1,n)$-submodules.

Theorem 1.

Let N be a proper submodule of M. The following are equivalent.

(i) 

N is a weakly $(1,n)$-submodule of M.

(ii) 

For all nonunits $a,b\in R$ with $ab\notin \left(N{:}_{R}M\right)$, one has $\left(N{:}_{M}ab\right)\subseteq \left(0{:}_{M}ab\right)\cup Nil\left(M\right)·M$.

(iii) 

For all nonunits $a,b\in R$ with $ab\notin \left(N{:}_{R}M\right)$, either $\left(N{:}_{M}ab\right)=\left(0{:}_{M}ab\right)$ or $\left(N{:}_{M}ab\right)\subseteq Nil\left(M\right)·M$.

(iv) 

For all nonunits $a,b\in R$ and every submodule $K\le M$ with $0\ne abK\subseteq N$, either $ab\in \left(N{:}_{R}M\right)$ or $K\subseteq Nil\left(M\right)·M$.

(v) 

For every nonunit $a\in R$, every proper ideal J of R, and every submodule $K\le M$ with $0\ne aJK\subseteq N$, either $aJ\subseteq \left(N{:}_{R}M\right)$ or $K\subseteq Nil\left(M\right)·M$.

(vi) 

For all proper ideals $I,J$ of R and every submodule $K\le M$ with $0\ne IJK\subseteq N$, either $IJ\subseteq \left(N{:}_{R}M\right)$ or $K\subseteq Nil\left(M\right)·M$.

Proof. (i)⇒(ii). Assume that N is a weakly $(1,n)$-submodule. Let $a,b\in R$ be nonunits with $ab\notin \left(N{:}_{R}M\right)$, and let $m\in \left(N{:}_{M}ab\right)$, so that $abm\in N$. If $abm=0$, then $ab·m=0$, so $m\in \left(0{:}_{M}ab\right)$ by definition. If $abm\ne 0$, then the weakly $(1,n)$-condition gives $ab\in \left(N{:}_{R}M\right)$ or $m\in Nil\left(M\right)·M$; since $ab\notin \left(N{:}_{R}M\right)$, we get $m\in Nil\left(M\right)·M$. In both cases $m\in \left(0{:}_{M}ab\right)\cup Nil\left(M\right)·M$.

(ii)⇒(iii). By (ii), the submodule $\left(N{:}_{M}ab\right)$ is contained in the union of the two submodules $\left(0{:}_{M}ab\right)$ and $Nil\left(M\right)·M$. By the prime avoidance lemma for submodules, $\left(N{:}_{M}ab\right)$ must be contained in one of them. If $\left(N{:}_{M}ab\right)\subseteq \left(0{:}_{M}ab\right)$, then since the reverse inclusion $\left(0{:}_{M}ab\right)\subseteq \left(N{:}_{M}ab\right)$ always holds (as $abm=0\in N$ implies $m\in \left(N{:}_{M}ab\right)$), we obtain the equality $\left(N{:}_{M}ab\right)=\left(0{:}_{M}ab\right)$. Otherwise, $\left(N{:}_{M}ab\right)\subseteq Nil\left(M\right)·M$.

(iii)⇒(iv). Suppose $0\ne abK\subseteq N$ for nonunits $a,b$ and a submodule $K\le M$ with $ab\notin \left(N{:}_{R}M\right)$. Since every $k\in K$ satisfies $abk\in abK\subseteq N$, we have $K\subseteq \left(N{:}_{M}ab\right)$ by definition of the colon submodule. By (iii), either $\left(N{:}_{M}ab\right)=\left(0{:}_{M}ab\right)$ or $\left(N{:}_{M}ab\right)\subseteq Nil\left(M\right)·M$. In the first case $K\subseteq \left(0{:}_{M}ab\right)$, giving $abK=0$, a contradiction. Hence $K\subseteq \left(N{:}_{M}ab\right)\subseteq Nil\left(M\right)·M$.

(iv)⇒(v). Suppose $0\ne aJK\subseteq N$ for a nonunit $a\in R$, a proper ideal J, and a submodule $K\le M$ with $aJ⊈\left(N{:}_{R}M\right)$ and $K⊈Nil\left(M\right)·M$. Choose $b\in J$ with $ab\notin \left(N{:}_{R}M\right)$. If $abK\ne 0$, then (iv) gives $ab\in \left(N{:}_{R}M\right)$, a contradiction. Hence $abK=0$. Since $aJK\ne 0$, there exists $c\in J$ with $acK\ne 0$. Applying (iv) to $0\ne acK\subseteq N$ gives $ac\in \left(N{:}_{R}M\right)$ (since $K⊈Nil\left(M\right)·M$). Since $abK=0$, we have $a(b+c)K=abK+acK=0+acK=acK\ne 0$. Since $b,c\in J$ we also have $b+c\in J$, so $a(b+c)K\subseteq aJK\subseteq N$. Applying (iv) to this nonzero inclusion, we obtain $a(b+c)\in \left(N{:}_{R}M\right)$. Therefore $ab=a(b+c)-ac\in \left(N{:}_{R}M\right)$, contradicting the choice of b.

(v)⇒(vi). Suppose $0\ne IJK\subseteq N$ for proper ideals $I,J$ and a submodule $K\le M$ with $IJ⊈\left(N{:}_{R}M\right)$ and $K⊈Nil\left(M\right)·M$. Choose $a\in I$ with $aJ⊈\left(N{:}_{R}M\right)$. If $aJK\ne 0$, then (v) gives $aJ\subseteq \left(N{:}_{R}M\right)$ or $K\subseteq Nil\left(M\right)·M$, both of which contradict our assumptions.

Now suppose $aJK=0$. Since $IJK\ne 0$, there exists $a^{\prime }\in I$ with $a^{\prime }JK\ne 0$. If $a^{\prime }J⊈\left(N{:}_{R}M\right)$, then applying (v) to $0\ne a^{\prime }JK\subseteq N$ yields a contradiction. Hence $a^{\prime }J\subseteq \left(N{:}_{R}M\right)$. Since $aJK=0$, we have $(a+a^{\prime })JK=aJK+a^{\prime }JK=a^{\prime }JK\ne 0$, and $a+a^{\prime }\in I$ since I is an ideal. If $(a+a^{\prime })J⊈\left(N{:}_{R}M\right)$, then (v) again yields a contradiction. If $(a+a^{\prime })J\subseteq \left(N{:}_{R}M\right)$, then $aJ=(a+a^{\prime })J-a^{\prime }J\subseteq \left(N{:}_{R}M\right)$, contradicting the choice of a.

(vi)⇒(i). Let $0\ne abm\in N$ for nonunits $a,b\in R$ and $m\in M$. Define ideals $I=\left(a\right)$ and $J=\left(b\right)$ of R, and set $K=Rm$. Since a and b are nonunits, both I and J are proper ideals. Since $abm=\left(a\right)\left(b\right)m\in IJK$ and $abm\ne 0$, the product $IJK$ is nonzero. Since $IJK=\left(a\right)\left(b\right)\left(Rm\right)=abRm\subseteq RN=N$, we have $0\ne IJK\subseteq N$. By (vi), either $IJ=\left(ab\right)\subseteq \left(N{:}_{R}M\right)$ or $K=Rm\subseteq Nil\left(M\right)·M$, that is, $ab\in \left(N{:}_{R}M\right)$ or $m\in Nil\left(M\right)·M$. □

The structure theorems that follow require several standard notions from multiplicative module theory, which we now recall for the reader’s convenience.

An R-module M is called a multiplication module if every submodule of M has the form $IM$ for some ideal I of R; in this case $\left(IM{:}_{R}M\right)=I$. When M is, in addition, faithful (${ann}_R\left(M\right)=0$) and finitely generated, the cancellation law $IM=JM\Rightarrow I=J$ holds. It is well known that every faithful finitely generated multiplication module is a finitely generated projective module of rank one; in particular, over a quasi-local ring such a module is free of rank one and hence cyclic (see, e.g. [3]).

A ring R is called quasi-local if it possesses a unique maximal ideal. For a faithful finitely generated multiplication module M, one has $Nil\left(M\right)=N\left(R\right)$ and $Nil\left(M\right)·M=N\left(R\right)M$.

Lemma 2.

Let R be a quasi-local ring with maximal ideal $\mathfrak{m}$ and let M be a faithful finitely generated multiplication R-module. Then M is a cyclic module. More precisely, for every element $g\in M\setminus \mathfrak{m}M$, one has $M=Rg$. In particular, $M\cong R$ as R-modules.

Proof. 

Since M is a faithful finitely generated multiplication R-module, it is a finitely generated projective module of rank one (as recalled above). Every finitely generated projective module over a quasi-local ring is free, so $M\cong R$ as R-modules. In particular, M is cyclic, say $M=R{g}_0$ for some $g_0\in M$.

It remains to show that every $g\in M\setminus \mathfrak{m}M$ is a generator of M. Since M is a multiplication module, $Rg=IM$ for some ideal I of R. Suppose for contradiction that $I\subseteq \mathfrak{m}$. Then $Rg=IM\subseteq \mathfrak{m}M$, which gives $g\in \mathfrak{m}M$, contradicting the choice of g. Hence $I⊈\mathfrak{m}$. Since R is quasi-local with unique maximal ideal $\mathfrak{m}$, the only ideal not contained in $\mathfrak{m}$ is R itself, so $I=R$ and therefore $Rg=RM=M$.

It remains to confirm that such an element g always exists. Since M is finitely generated and $\mathfrak{m}=Jac\left(R\right)$, Nakayama’s lemma guarantees that $\mathfrak{m}M\ne M$ (otherwise $M=\mathfrak{m}M=Jac\left(R\right)·M$ would force $M=0$, a contradiction), so $M\setminus \mathfrak{m}M\ne \varnothing$.

Finally, the isomorphism $M\cong R$ follows from $M=Rg$. The surjective R-module homomorphism $R\to M$ defined by $r\mapsto rg$ has kernel ${ann}_R\left(g\right)={ann}_R\left(M\right)=0$ (since M is faithful and $M=Rg$), hence it is an isomorphism. □

The following lemma is the central technical tool of this section: it reduces questions about weakly $(1,n)$-submodules of a faithful finitely generated multiplication module to the ring-level theory of weakly $(1,n)$-ideals via the order-isomorphism between the ideal lattice of R and the submodule lattice of M.

Lemma 3

(Reduction Lemma). Let M be a faithful finitely generated multiplication R-module, let N be a proper submodule of M, and set $K=\left(N{:}_{R}M\right)$. The following two statements hold.

(i) 

The submodule N is a weakly $(1,n)$-submodule of M if and only if the ideal K is a weakly $(1,n)$-ideal of R.

(ii) 

$N⊈N\left(R\right)M$ if and only if $K⊈N\left(R\right)$.

Proof. 

Since M is a faithful finitely generated multiplication module, the map $I\mapsto IM$ is an order-isomorphism between the ideal lattice of R and the submodule lattice of M, with inverse map $L\mapsto \left(L{:}_{R}M\right)$. This isomorphism satisfies:

$$\begin{array}{c}L\subseteq N\left(R\right)M\iff \left(L{:}_{R}M\right)\subseteq N\left(R\right)(\mathrm{cancellation}\mathrm{law}),\\ L=0\iff \left(L{:}_{R}M\right)=0\left(M\mathrm{faithful}\right).\end{array}$$

(ii) Applying the first equivalence with $L=N=KM$, we obtain $N\subseteq N\left(R\right)M\iff K\subseteq N\left(R\right)$, which gives (ii) by negation.

(i) By Theorem 2.16(vi), N is weakly $(1,n)$ if and only if the following condition holds: for all proper ideals $I,J\le R$ and every submodule $L\le M$ with $0\ne IJL\subseteq N$, one has $IJ\subseteq K$ or $L\subseteq N\left(R\right)M$.

We pass from the submodule L to its colon ideal by setting $L^{\prime }=\left(L{:}_{R}M\right)$, so that $L=L^{\prime }M$ under the isomorphism. Applying the order-isomorphism and its listed properties, we obtain the equivalences

$$\begin{array}{c}IJL\subseteq N\iff IJ{L}^{\prime }\subseteq K,L\subseteq N\left(R\right)M\iff L^{\prime }\subseteq N\left(R\right),\\ IJL\ne 0\iff IJ{L}^{\prime }\ne 0.\end{array}$$

Substituting these translations, we see that N is a weakly $(1,n)$-submodule of M if and only if, for all proper ideals $I,J,L^{\prime }$ of R satisfying $0\ne IJ{L}^{\prime }\subseteq K$, one has $IJ\subseteq K$ or $L^{\prime }\subseteq N\left(R\right)$. By Theorem 2.16(vi) applied with $M=R$ (using Remark 2.2), this is precisely the weakly $(1,n)$-ideal condition on the ideal K of R. □

We can now transfer the ring-theoretic structure theorem for weakly $(1,n)$-ideals ([1], Theorem 2) to the module setting. We state that theorem here for the reader’s convenience.

Theorem 2

([1], Theorem 2). Let K be a weakly $(1,n)$-ideal of R with $K⊈N\left(R\right)$. Then one of the following holds.

(i) 

R is a quasi-local ring with unique maximal ideal $\mathfrak{m}$ such that ${\mathfrak{m}}^2\subseteq K\subseteq \mathfrak{m}$.

(ii) 

$R=R_1\times R_2$, where $(R_i,{\mathfrak{m}}_i)$ are quasi-local rings with ${\mathfrak{m}}_i^2=\left(0\right)$ and $K=R_1\times \left(0\right)$ or $K=\left(0\right)\times R_2$. Furthermore, $R_1$ or $R_2$ (at least one of the two) is a field.

Theorem 3.

Let M be a faithful finitely generated multiplication module and let N be a weakly $(1,n)$-submodule with $N⊈N\left(R\right)M$. Then one of the following holds:

(i) 

R is quasi-local with maximal ideal $\mathfrak{m}$ and ${\mathfrak{m}}^{2}M\subseteq N$; or

(ii) 

$R\cong R_1\times R_2$ where $(R_1,{\mathfrak{m}}_1)$ and $(R_2,{\mathfrak{m}}_2)$ are quasi-local rings with ${\mathfrak{m}}_1^2={\mathfrak{m}}_2^2=\left(0\right)$, and $N=M_1\times \left\{0\right\}$ or $N=\left\{0\right\}\times M_2$ under $M\cong M_1\times M_2$. Furthermore, $R_1$ or $R_2$ (at least one of the two) is a field.

Proof. 

Set $K=\left(N{:}_{R}M\right)$. By the Reduction Lemma (Lemma 2.18), K is a weakly $(1,n)$-ideal of R with $K⊈N\left(R\right)$. By the theorem above (the ring-theoretic result of [1] recalled just before this proof), one of the following holds.

Case (a): R is quasi-local with maximal ideal $\mathfrak{m}$ and ${\mathfrak{m}}^2\subseteq K$. Multiplying by M gives ${\mathfrak{m}}^{2}M\subseteq KM=N$, establishing (i).

Case (b): $R\cong R_1\times R_2$ with $(R_i,{\mathfrak{m}}_i)$ quasi-local, ${\mathfrak{m}}_i^2=\left(0\right)$, $K=R_1\times \left\{0\right\}$ or $K=\left\{0\right\}\times R_2$, and $R_1$ or $R_2$ a field. Since M is a faithful finitely generated multiplication R-module and $R\cong R_1\times R_2$, we have $M\cong M_1\times M_2$, where $M_i$ is a faithful finitely generated multiplication $R_i$-module. Since each $R_i$ is quasi-local, Lemma 2.17 gives $M_i\cong R_i$ (this holds regardless of whether $R_i$ is a field). If $K=R_1\times \left\{0\right\}$, then $N=KM$ corresponds to $M_1\times \left\{0\right\}$; if $K=\left\{0\right\}\times R_2$, then N corresponds to $\left\{0\right\}\times M_2$. This establishes (ii), including the assertion that $R_1$ or $R_2$ is a field, which is inherited directly from the cited theorem. □

Remark 6.

One might expect part (ii) of Theorem 2.19 to assert the stronger claim thatboth$R_1$ and $R_2$ are fields. This is false in general: for example, if $D=k\left[\epsilon \right]/\left({\epsilon }^2\right)$ (a quasi-local ring with ${\mathfrak{m}}_D^2=0$ that is not a field) and $F=k$ is a field, then $R=D\times F$ (with $M=R$) admits the weakly $(1,n)$-submodule $N=D\times \left\{0\right\}⊈N\left(R\right)M$, yet $R\ncong k\times k$. The correct statement, matching ([1], Theorem 2), requires only thatoneof $R_1,R_2$ be a field.

A ring is a UN-ring if every element is a unit or nilpotent, equivalently quasi-local with maximal ideal $N\left(R\right)$.

Theorem 4.

Let M be a faithful finitely generated multiplication module. The following statements are equivalent:

(i) 

Every proper principal submodule of M is a weakly $(1,n)$-submodule of M.

(ii) 

Every proper submodule of M is a weakly $(1,n)$-submodule of M.

(iii) 

R is a UN-ring, or $R\cong F_1\times F_2$ for fields $F_1,F_2$.

Proof. (i)⇒(ii). Let N be any proper submodule of M and suppose $0\ne abm\in N$ for some nonunits $a,b\in R$ and $m\in M$. Since $R\left(abm\right)\subseteq N⊊M$, the principal submodule $R\left(abm\right)$ is proper and is a weakly $(1,n)$-submodule of M by (i). The weakly $(1,n)$-condition applied to $0\ne abm\in R\left(abm\right)$ yields $ab\in \left(R\left(abm\right){:}_{R}M\right)$ or $m\in Nil\left(M\right)·M$. Since $R\left(abm\right)\subseteq N$, we have $\left(R\left(abm\right){:}_{R}M\right)\subseteq \left(N{:}_{R}M\right)$, so $ab\in \left(N{:}_{R}M\right)$ or $m\in Nil\left(M\right)·M$. Hence N is a weakly $(1,n)$-submodule of M.

(ii)⇒(iii). Suppose every proper submodule of M is a weakly $(1,n)$-submodule.

Case A: Every proper submodule of M is contained in $N\left(R\right)M$. For any nonunit $a\in R$, the submodule $aM=\left(a\right)M$ is proper: if $aM=M$, then $\left(aM{:}_{R}M\right)=\left(a\right)=R$ (since M is a multiplication module), which forces $a\in U\left(R\right)$, contradicting the assumption that a is a nonunit. Thus $aM⊊M$, so $aM\subseteq N\left(R\right)M$, giving $a\in \left(N\left(R\right)M{:}_{R}M\right)=N\left(R\right)$ by the cancellation property. Hence every nonunit of R is nilpotent, so R is a UN-ring.

Case B: Some proper submodule $N_0⊈N\left(R\right)M$. By Theorem 2.19, either R is quasi-local with maximal ideal $\mathfrak{m}$ and ${\mathfrak{m}}^{2}M\subseteq N_0$, or $R\cong R_1\times R_2$ where $(R_i,{\mathfrak{m}}_i)$ are quasi-local rings with ${\mathfrak{m}}_i^2=\left(0\right)$.

First subcase: R is quasi-local. By Lemma 2.17, $M\cong R$. Under this isomorphism, submodules of M correspond to ideals of R, and a weakly $(1,n)$-submodule of M corresponds to a weakly $(1,n)$-ideal of R (Remark 2.2 and Lemma 2.18(i) applied with $R=M$). Since every proper submodule of M is weakly $(1,n)$ by hypothesis (ii), every proper ideal of R is a weakly $(1,n)$-ideal. By ([1], Theorem 4), R is a UN-ring or $R\cong F_1\times F_2$ for fields $F_1,F_2$. Since R is quasi-local, it cannot be isomorphic to a product of two fields (such a product has two maximal ideals). Hence R is a UN-ring.

Second subcase: $R\cong R_1\times R_2$, $(R_i,{\mathfrak{m}}_i)$ quasi-local with ${\mathfrak{m}}_i^2=\left(0\right)$. We show that $R_1$ and $R_2$ are both fields. As in the first subcase, $M\cong R$ (via $M_i\cong R_i$ by Lemma 2.17 applied to each factor), so hypothesis (ii) transfers to: every proper ideal of $R\cong R_1\times R_2$ is weakly $(1,n)$.

We claim every proper ideal $J_1$ of $R_1$ equals $\left(0\right)$; the symmetric argument then shows $R_2$ is a field. Let $J_1⊊R_1$ and set $J=J_1\times R_2$, a proper ideal of R (proper since $J_1\ne R_1$). Since $R_2⊈{\mathfrak{m}}_2$, we have $J=J_1\times R_2⊈{\mathfrak{m}}_1\times {\mathfrak{m}}_2=N\left(R\right)$. By hypothesis, J is weakly $(1,n)$, and since R is not quasi-local (it has the two distinct maximal ideals ${\mathfrak{m}}_1\times R_2$ and $R_1\times {\mathfrak{m}}_2$), Theorem 2.19 forces J into the “product” alternative (ii) of that theorem applied to the ring $R\cong R_1\times R_2$ itself (as in the first subcase, with $M=R$): that is, $J=R_1\times \left\{0\right\}$ or $J=\left\{0\right\}\times R_2$. Since the second coordinate of $J=J_1\times R_2$ is all of $R_2\ne \left\{0\right\}$ (as $M\ne 0$), the first alternative $J=R_1\times \left\{0\right\}$ is impossible. Hence $J=\left\{0\right\}\times R_2$, which forces $J_1=\left(0\right)$.

Therefore every proper ideal of $R_1$ is zero, i.e. $R_1$ is a field; symmetrically, $R_2$ is a field. This establishes $R\cong F_1\times F_2$ for fields $F_1,F_2$, completing Case B and the proof of (ii)⇒(iii).

(iii)⇒(i). Let N be any proper principal submodule of M.

UN-ring case. Every nonunit of R lies in $N\left(R\right)=Nil\left(M\right)$. Suppose $0\ne abm\in N$ for nonunits $a,b\in R$ and $m\in M$. If $m\in Nil\left(M\right)·M$, the second alternative holds. Now suppose $m\notin Nil\left(M\right)·M=N\left(R\right)M$. Let $J=\left(Rm{:}_{R}M\right)$. Since M is a multiplication module, $Rm=JM$. If $J\subseteq N\left(R\right)$, then $Rm\subseteq N\left(R\right)M$, contradicting $m\notin N\left(R\right)M$. Hence $J⊈N\left(R\right)$, so J contains a unit, giving $J=R$ and hence $Rm=JM=M$. Since $Rm=M$, every element $m^{\prime }\in M$ has the form $m^{\prime }=rm$ for some $r\in R$, so $ab·m^{\prime }=r\left(abm\right)\in N$ (since $abm\in N$ and N is closed under R-multiplication). Therefore $ab·M\subseteq N$, giving $ab\in \left(N{:}_{R}M\right)$.

$F_1\times F_2$ case. Since $R\cong F_1\times F_2$, we have $M\cong F_1\times F_2$ and $Nil\left(M\right)·M=\left\{\right(0,0\left)\right\}$. The proper submodules of M are precisely $\left\{\right(0,0\left)\right\}$, $F_1\times \left\{0\right\}$, and $\left\{0\right\}\times F_2$.

We verify the weakly $(1,n)$-condition for $N=F_1\times \left\{0\right\}$; the case $N=\left\{0\right\}\times F_2$ is symmetric and $N=\left\{\right(0,0\left)\right\}$ is vacuous. Let $0\ne abm\in N$ with nonunits $a,b$. Since $abm\ne 0$, both a and b are nonzero. Write $a=(a_1,a_2)$, $b=(b_1,b_2)$, and $m=(m_1,m_2)$. Then $abm=(a_1{b}_1{m}_1,a_2{b}_2{m}_2)\in F_1\times \left\{0\right\}$ forces $a_2{b}_2{m}_2=0$. Since $(a_1,a_2)\ne (0,0)$ is a nonunit in $F_1\times F_2$, exactly one of $a_1,a_2$ is zero; the same holds for b.

Case $a_2=0$: Then $a=(a_1,0)$ with $a_1\ne 0$. For any $(x_1,x_2)\in M$, $ab·(x_1,x_2)=(a_1{b}_1{x}_1,0)\in F_1\times \left\{0\right\}=N$. Hence $ab\in \left(N{:}_{R}M\right)$.

Case $a_1=0$: Then $a=(0,a_2)$ with $a_2\ne 0$. If $b_2=0$, then $b=(b_1,0)$, so $ab=(0,0)$, giving $abm=0$, a contradiction. If $b_1=0$, then $b=(0,b_2)$ with $b_2\ne 0$, so $ab=(0,a_2{b}_2)$ with $a_2{b}_2\ne 0$; then $abm=(0,a_2{b}_2{m}_2)\in F_1\times \left\{0\right\}$ forces $m_2=0$, so $abm=(0,0)$, again a contradiction. Hence this case cannot occur.

In all non-contradictory cases, $ab\in \left(N{:}_{R}M\right)$, so N is a weakly $(1,n)$-submodule of M. □

The corresponding classification for weakly n-submodules takes a simpler form: it characterises exactly the UN-rings, with no product-of-fields alternative.

Theorem 5.

Let M be a faithful finitely generated multiplication module. The following statements are equivalent:

(i) 

R is a UN-ring.

(ii) 

Every proper submodule of M is a weakly n-submodule of M.

(iii) 

Every proper cyclic submodule of M is a weakly n-submodule of M.

Proof. (i)⇒(ii). Since R is a UN-ring, it is quasi-local with maximal ideal $N\left(R\right)$, and $Nil\left(M\right)=N\left(R\right)$ for the faithful module M. Moreover, every proper ideal of R is contained in $N\left(R\right)$, so every proper submodule $N=IM$ satisfies $N\subseteq N\left(R\right)M=Nil\left(M\right)·M$.

Let N be a proper submodule of M and suppose $0\ne am\in N$ for some $a\in R$ and $m\in M$ with $a\notin \left(N{:}_{R}M\right)$. We show $m\in Nil\left(M\right)·M$.

Case $a\in U\left(R\right)$: Since a is a unit, $m=a^{-1}\left(am\right)\in N$, and $N\subseteq Nil\left(M\right)·M$ gives $m\in Nil\left(M\right)·M$.

Case $a\in N\left(R\right)=Nil\left(M\right)$: Suppose for contradiction that $m\in U\left(R\right)$. Since m is a unit, $Rm=M$. Then $aM=a\left(Rm\right)=R\left(am\right)\subseteq N$, giving $a\in \left(N{:}_{R}M\right)$, contradicting our assumption. Therefore $m\notin U\left(R\right)$, so $m\in N\left(R\right)=Nil\left(M\right)$, giving $m=m·1\in Nil\left(M\right)·M$.

In both cases $m\in Nil\left(M\right)·M$, so N is a weakly n-submodule.

(ii)⇒(iii). Every proper cyclic submodule is a proper submodule, so (ii) implies (iii) directly.

(iii)⇒(i). Let $a\in R$ be a nonunit. If $aM=0$, then $a\in {ann}_R\left(M\right)=0$ since M is faithful, so $a=0\in N\left(R\right)$. Hence we may assume $aM\ne 0$. Furthermore, $aM⊊M$: if $aM=M$, then $\left(aM{:}_{R}M\right)=\left(a\right)=R$, forcing $a\in U\left(R\right)$, a contradiction. Let $m\in M$ be arbitrary. The cyclic submodule $R\left(am\right)$ is proper since $R\left(am\right)\subseteq aM⊊M$. By hypothesis (iii), $R\left(am\right)$ is a weakly n-submodule of M. By Lemma 2.11, $R\left(am\right)\subseteq Nil\left(M\right)·M=N\left(R\right)M$, so $am\in N\left(R\right)M$. Since $m\in M$ was arbitrary, $aM\subseteq N\left(R\right)M$. Hence $a\in \left(N\left(R\right)M{:}_{R}M\right)$. Since M is a faithful finitely generated multiplication module, the cancellation law gives $\left(N\left(R\right)M{:}_{R}M\right)=N\left(R\right)$, so $a\in N\left(R\right)$. Therefore every nonunit of R is nilpotent, and R is a UN-ring. □

3. Behaviour under Module Operations

Having established the definitions, basic properties, and structural characterisations of weakly $(1,n)$-submodules, we now investigate how these properties behave under the standard module-theoretic constructions of localisation, quotient modules, and module homomorphisms.

Proposition 5.

Let M be finitely generated and let S be a multiplicatively closed subset of R with $S\cap \left(N{:}_{R}M\right)=\varnothing$. If N is a weakly $(1,n)$-submodule of M, then $S^{-1}N$ is a weakly $(1,n)$-submodule of $S^{-1}M$.

Proof. 

Since M is finitely generated, ${ann}_{S^{-1}R}\left(S^{-1}M\right)=S^{-1}{ann}_R\left(M\right)$. Taking radicals and using $\sqrt{S^{-1}I}=S^{-1}\sqrt{I}$, we obtain $Nil\left(S^{-1}M\right)=S^{-1}Nil\left(M\right)$, hence $Nil\left(S^{-1}M\right)·S^{-1}M=S^{-1}(Nil\left(M\right)·M)$.

The submodule $S^{-1}N$ is proper in $S^{-1}M$: using $\left(S^{-1}N{:}_{S^{-1}R}{S}^{-1}M\right)=S^{-1}\left(N{:}_{R}M\right)$, one has $S^{-1}N=S^{-1}M$ iff $1/1\in S^{-1}\left(N{:}_{R}M\right)$, iff $S\cap \left(N{:}_{R}M\right)\ne \varnothing$. Since this fails by hypothesis, $S^{-1}N\ne S^{-1}M$.

Let $0\ne (a/s)(b/t)(m/u)\in S^{-1}N$ for nonunits $a/s,b/t\in S^{-1}R$ with $m/u\notin Nil\left(S^{-1}M\right)·S^{-1}M$. Write $abm/\left(stu\right)=n/v$ for some $n\in N$, $v\in S$. There exists $w\in S$ with $w(v·abm-stu·n)=0$, so $s^{\prime }=wv\in S$ satisfies $s^{\prime }abm=wstu·n\in N$. Since $abm/\left(stu\right)\ne 0$ in $S^{-1}M$, no element of S annihilates $abm$ in M, so $s^{\prime }abm\ne 0$.

Since $a/s$ is a nonunit in $S^{-1}R$, a cannot be a unit in R (else $a/s\in U\left(S^{-1}R\right)$); similarly $b\notin U\left(R\right)$.

We claim $s^{\prime }m\notin Nil\left(M\right)·M$. If $s^{\prime }m={\sum }_j{c}_j{n}_j$ with $c_j\in Nil\left(M\right)$, $n_j\in M$, then in $S^{-1}M$, $m/u=\left(s^{\prime }m\right)/\left(s^{\prime }u\right)\in S^{-1}Nil\left(M\right)·S^{-1}M=Nil\left(S^{-1}M\right)·S^{-1}M$, a contradiction. Hence $s^{\prime }m\notin Nil\left(M\right)·M$.

Applying the weakly $(1,n)$-condition to $0\ne a·b·\left(s^{\prime }m\right)\in N$, we obtain $ab\in \left(N{:}_{R}M\right)$, i.e. $ab·M\subseteq N$. For any $m^{\prime }/u^{\prime }\in S^{-1}M$, $(a/s)(b/t)(m^{\prime }/u^{\prime })=ab{m}^{\prime }/\left(st{u}^{\prime }\right)\in S^{-1}N$, so $(a/s)(b/t)\in \left(S^{-1}N{:}_{S^{-1}R}{S}^{-1}M\right)$. Therefore $S^{-1}N$ is a weakly $(1,n)$-submodule of $S^{-1}M$. □

The converse of Proposition 3.1 holds when S consists of regular elements.

Corollary 1.

Let M be a finitely generated R-module. Let $\mathrm{reg}\left(M\right)=\{r\in R:rm=0\Rightarrow m=0\}$ denote the set of non-zero-divisors on M, and define $\mathrm{reg}(M/N)$ similarly. Let $S\subseteq \mathrm{reg}\left(M\right)\cap \mathrm{reg}(M/N)$ be a multiplicatively closed subset of R satisfying:

(a) 

for every $s\in S$ and $m\in M$, $sm\in Nil\left(M\right)·M$ implies $m\in Nil\left(M\right)·M$;

(b) 

for every $a\in R$, if $a\notin U\left(R\right)$ then $a/1\notin U\left(S^{-1}R\right)$.

If $S^{-1}N$ is a weakly $(1,n)$-submodule of $S^{-1}M$ and $S^{-1}N\ne S^{-1}M$, then N is a weakly $(1,n)$-submodule of M.

Proof. 

Since $S^{-1}N\ne S^{-1}M$, N is proper in M.

Since M is finitely generated, as before $Nil\left(S^{-1}M\right)·S^{-1}M=S^{-1}(Nil\left(M\right)·M)$.

Let $0\ne abm\in N$ for some nonunits $a,b\in R$ and $m\in M$ with $m\notin Nil\left(M\right)·M$. Since every $s\in S$ is a non-zero-divisor on M and $abm\ne 0$, $\left(abm\right)/1\ne 0$ in $S^{-1}M$, and $\left(abm\right)/1\in S^{-1}N$.

By hypothesis (b), $a/1,b/1\notin U\left(S^{-1}R\right)$.

We claim $m/1\notin Nil\left(S^{-1}M\right)·S^{-1}M=S^{-1}(Nil\left(M\right)·M)$. If $m/1\in S^{-1}(Nil\left(M\right)·M)$, then $sm\in Nil\left(M\right)·M$ for some $s\in S$; by (a), $m\in Nil\left(M\right)·M$, a contradiction.

Applying the weakly $(1,n)$-condition to $0\ne (a/1)(b/1)(m/1)\in S^{-1}N$, we obtain $\left(ab\right)/1\in \left(S^{-1}N{:}_{S^{-1}R}{S}^{-1}M\right)$ (the alternative $m/1\in Nil\left(S^{-1}M\right)·S^{-1}M$ is excluded). Thus for any $m^{\prime }\in M$, $\left(ab{m}^{\prime }\right)/1\in S^{-1}N$, meaning $s\left(ab{m}^{\prime }\right)\in N$ for some $s\in S$. Since $S\subseteq \mathrm{reg}(M/N)$, $s\left(ab{m}^{\prime }\right)\in N$ implies $ab{m}^{\prime }\in N$. As $m^{\prime }\in M$ was arbitrary, $ab\in \left(N{:}_{R}M\right)$. Therefore N is a weakly $(1,n)$-submodule of M. □

Remark 7.

The hypotheses of Corollary 3.2 deserve comment. Condition (a) together with $S\subseteq \mathrm{reg}\left(M\right)$ ensures that the nilpotency locus descends from $S^{-1}M$ to M. Condition (b) is needed because the weakly $(1,n)$ property requires nonunit factors in $S^{-1}R$, not merely in R; it fails in general, but holds whenever $S\subseteq U\left(R\right)$ (in which case $S^{-1}R\cong R$ and the corollary reduces to Proposition 3.1). The condition $S\subseteq \mathrm{reg}(M/N)$ ensures S-saturation of N. Note that $\mathrm{reg}\left(R\right)⊈\mathrm{reg}\left(M\right)$ in general: $2\in \mathrm{reg}\left(\mathbb{Z}\right)$ but $2\notin \mathrm{reg}(\mathbb{Z}/2\mathbb{Z})$. Condition (a) holds automatically when $Nil\left(M\right)·M=0$ or $S\subseteq U\left(R\right)$.

Proposition 6.

Let $K\subseteq N$ be submodules of M with $N\ne M$. Then the following hold.

(i) 

If N is a weakly $(1,n)$-submodule of M, then $N/K$ is a weakly $(1,n)$-submodule of $M/K$.

(ii) 

Suppose that:

• $N/K$ is a weakly $(1,n)$-submodule of $M/K$;
• K is a weakly $(1,n)$-submodule of M;
• for every $m\in M$, if $m+K\in Nil(M/K)·(M/K)$, then $m\in Nil\left(M\right)·M$.

Then N is a weakly $(1,n)$-submodule of M.

(iii) 

If N is a weakly $(1,n)$-submodule of M and $\left\{0\right\}$ is a $(1,n)$-submodule of M, then N is a $(1,n)$-submodule of M.

Proof. (i) Let $0\ne ab(m+K)\in N/K$ for some nonunits $a,b\in R$ and $m\in M$ with $m+K\notin Nil(M/K)·(M/K)$. Since $ab(m+K)\ne 0$ in $M/K$, we have $abm\notin K$, so $0\ne abm\in N$. We claim $m\notin Nil\left(M\right)·M$: if $m\in Nil\left(M\right)·M$, then since $Nil\left(M\right)\subseteq Nil(M/K)$, the coset $m+K\in Nil(M/K)·(M/K)$, a contradiction. Applying the weakly $(1,n)$-condition to $0\ne abm\in N$, we obtain $ab\in \left(N{:}_{R}M\right)$. Since $ab·M\subseteq N$, we have $ab·(M/K)\subseteq N/K$, giving $ab\in (N/K{:}_{R}M/K)$.

(ii) Let $0\ne abm\in N$ for some nonunits $a,b\in R$ and $m\in M$ with $m\notin Nil\left(M\right)·M$.

Case 1: $abm\in K$. Since $abm\ne 0$ and K is a weakly $(1,n)$-submodule and $m\notin Nil\left(M\right)·M$, we obtain $ab\in \left(K{:}_{R}M\right)\subseteq \left(N{:}_{R}M\right)$.

Case 2: $abm\notin K$. Then $0\ne ab(m+K)\in N/K$. By the third hypothesis, $m+K\notin Nil(M/K)·(M/K)$. Since $N/K$ is a weakly $(1,n)$-submodule of $M/K$, we obtain $ab\in (N/K{:}_{R}M/K)=\left(N{:}_{R}M\right)$.

(iii) Let $abm\in N$ for nonunits $a,b\in R$ and $m\in M$ with $m\notin Nil\left(M\right)·M$. If $abm\ne 0$, then the weakly $(1,n)$-condition gives $ab\in \left(N{:}_{R}M\right)$. If $abm=0$, then since $\left\{0\right\}$ is a $(1,n)$-submodule and $m\notin Nil\left(M\right)·M$, we get $ab\in {ann}_R\left(M\right)\subseteq \left(N{:}_{R}M\right)$. Hence N is a $(1,n)$-submodule of M. □

Proposition 7.

Let $\phi :M\to M^{\prime }$ be an R-module homomorphism. Then the following hold.

(i) 

If φ is surjective, N is a weakly $(1,n)$-submodule of M, and $ker\left(\phi \right)\subseteq N$, then $\phi \left(N\right)$ is a weakly $(1,n)$-submodule of $M^{\prime }$.

(ii) 

If φ is injective, $\phi \left(M\right)⊈N^{\prime }$, and $N^{\prime }$ is a weakly $(1,n)$-submodule of $M^{\prime }$ satisfying ${\phi }^{-1}(Nil\left(M^{\prime }\right)·M^{\prime })\subseteq Nil\left(M\right)·M$, then ${\phi }^{-1}\left(N^{\prime }\right)$ is a weakly $(1,n)$-submodule of M.

In particular, the condition ${\phi }^{-1}(Nil\left(M^{\prime }\right)·M^{\prime })\subseteq Nil\left(M\right)·M$ holds when φ is an isomorphism.

Proof. (i) Since $ker\left(\phi \right)\subseteq N⊊M$ and $\phi$ is surjective, $\phi \left(N\right)\ne M^{\prime }$, so $\phi \left(N\right)$ is proper. Since $ker\left(\phi \right)\subseteq N$, $\phi$ induces an isomorphism $M/ker\left(\phi \right)\cong M^{\prime }$ under which $N/ker\left(\phi \right)$ corresponds to $\phi \left(N\right)$. By Proposition 3.4(i), $N/ker\left(\phi \right)$ is a weakly $(1,n)$-submodule of $M/ker\left(\phi \right)$, hence $\phi \left(N\right)$ is a weakly $(1,n)$-submodule of $M^{\prime }$.

(ii) Since $\phi \left(M\right)⊈N^{\prime }$, there exists $m\in M$ with $\phi \left(m\right)\notin N^{\prime }$, so ${\phi }^{-1}\left(N^{\prime }\right)\ne M$; hence ${\phi }^{-1}\left(N^{\prime }\right)$ is proper.

Let $0\ne abm\in {\phi }^{-1}\left(N^{\prime }\right)$ for some nonunits $a,b\in R$ and $m\in M$ with $m\notin Nil\left(M\right)·M$. Then $\phi \left(abm\right)=ab\phi \left(m\right)\in N^{\prime }$, and since $\phi$ is injective and $abm\ne 0$, $ab\phi \left(m\right)\ne 0$.

If $\phi \left(m\right)\in Nil\left(M^{\prime }\right)·M^{\prime }$, then $m\in {\phi }^{-1}(Nil\left(M^{\prime }\right)·M^{\prime })\subseteq Nil\left(M\right)·M$, a contradiction. Hence $\phi \left(m\right)\notin Nil\left(M^{\prime }\right)·M^{\prime }$.

Applying the weakly $(1,n)$-condition to $0\ne ab\phi \left(m\right)\in N^{\prime }$, we obtain $ab\in \left(N^{\prime }{:}_R{M}^{\prime }\right)$. For any $m^{\prime \prime }\in M$, $\phi (ab·m^{\prime \prime })=ab·\phi \left(m^{\prime \prime }\right)\in ab·M^{\prime }\subseteq N^{\prime }$, so $ab·m^{\prime \prime }\in {\phi }^{-1}\left(N^{\prime }\right)$. Hence $ab\in \left({\phi }^{-1}\left(N^{\prime }\right){:}_{R}M\right)$, and ${\phi }^{-1}\left(N^{\prime }\right)$ is a weakly $(1,n)$-submodule of M. □

4. Conclusions

We have introduced weakly n- and weakly $(1,n)$-submodules as the module-theoretic counterparts of the weakly n- and $(1,n)$-ideals of [1], established the strict hierarchy

$$\mathrm{weakly}n\Rightarrow \mathrm{weakly}(1,n)\Rightarrow \mathrm{weakly}1-\mathrm{absorbing}\mathrm{primary}$$

(Proposition 2.8, Examples 2.5 and 2.10), and given a six-fold characterisation (Theorem 2.16). The structural containment $N\subseteq Nil\left(M\right)·M$ was identified as the feature separating the weakly n-class from the weakly $(1,n)$-class (Lemma 2.11, Remark 2.12). Via the Reduction Lemma 2.18, the ring-level structure theory of [1] was lifted to faithful finitely generated multiplication modules: every proper submodule of such a module is weakly $(1,n)$ if and only if R is a UN-ring or a product of two fields (Theorem 2.21), while the weakly n analogue isolates exactly the UN-rings (Theorem 2.22). The behaviour under localisations, quotient modules, and homomorphisms was treated in Section 3.

We emphasise that the substantive module-theoretic content of Theorem 2.19 lies in the Reduction Lemma 2.18 itself — a translation between the module and ring settings that has no counterpart in [1], which is purely ring-theoretic — rather than in the ring-theoretic classification of ([1], Theorem 2), which is cited directly rather than re-derived. As noted in the Introduction, this reliance on [1] is confined to five of the paper’s twenty-seven numbered results; every other result is proved for a general module M with no ring-theoretic counterpart in [1].

Three natural directions remain open: (i) removing the faithful finitely generated multiplication hypothesis from the structure theorems and describing the rings and modules over which every proper submodule is weakly $(1,n)$ (resp. weakly n) in full generality; (ii) investigating the two classes in trivial extensions and amalgamated modules along a homomorphism, paralleling the ring-level study in [1]; and (iii) defining weakly $(m,n)$-submodules for $m\ge 1$ and clarifying their relation to the weakly classical 1-absorbing prime submodules of [11].