S-Pseudo Bounded Radical of Submodules

Amal Madhi Rashid; Buthyna Najad Shihab

doi:10.12688/f1000research.172188.1

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Research Article

S-Pseudo Bounded Radical of Submodules

[version 1; peer review: awaiting peer review]

Amal Madhi Rashid ^1,2, Buthyna Najad Shihab^1,2

PUBLISHED 09 Dec 2025

Author details Author details

¹ Mathematics, University of Baghdad College of Education for Pure Science Ibn Al-Haitham, Baghdad, Baghdad Governorate, 31001, Iraq
² Mathematics, University of Baghdad College of Education for Pure Science Ibn Al-Haitham, Baghdad, Baghdad Governorate, 10001, Iraq

Amal Madhi Rashid
Roles: Writing – Review & Editing

Buthyna Najad Shihab
Roles: Writing – Original Draft Preparation

OPEN PEER REVIEW

REVIEWER STATUS AWAITING PEER REVIEW

This article is included in the Fallujah Multidisciplinary Science and Innovation gateway.

Abstract

In this paper, every module M is unitary and every ring F is commutative with identity. Some properties of S-Pseudo bounded radical are studied. A proper submodule ℵ of an F − module M is said to be S-pseudo bounded submodule if there exists , φ ∈ End ( M ) , x ∈ M such that φ ( x ) ∈ ℵ implies ann F ( ℵ ) = ann F φ ( x ) making use of endomorphism map over an F − module . The characterization of S-Pseudo bounded radical for finitely generated and multiplication module is given. It must be emphasized that scalar module and prime submodule played a major role in achieving new results and to study the relationship between the radical of submodules and the radical of S-pseudo bounded submodule. In our work many properties and corollaries will be proved which explain the idea of the radical of S-pseudo bounded submodule and a new class of F − module as well as F − submodule provided with some examples that illustrate and clarify in a nice way this type of module (submodule). We used the symbol End ( M ) which means the set of all endomorphism maps of F − module M and S − rad M PS . B . ( ℵ ) refers to the intersection of all S-pseudo bounded submodules of M containing ℵ .

Keywords

S-pseudo bounded radical, Finitely generated module, Prime submodule, Multiplication modules, Scalar module.

Corresponding author: Amal Madhi Rashid

Competing interests: No competing interests were disclosed.

Grant information: The author(s) declared that no grants were involved in supporting this work.

Copyright: © 2025 Madhi Rashid A and Najad Shihab B. This is an open access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

How to cite: Madhi Rashid A and Najad Shihab B. S-Pseudo Bounded Radical of Submodules [version 1; peer review: awaiting peer review]. F1000Research 2025, 14:1378 (https://doi.org/10.12688/f1000research.172188.1) First published: 09 Dec 2025, 14:1378 (https://doi.org/10.12688/f1000research.172188.1) Latest published: 09 Dec 2025, 14:1378 (https://doi.org/10.12688/f1000research.172188.1)

1. Introduction

The concept of the radical of an ideal plays important role in the study of rings and it was generalized to modules over commutative rings.^1,2 A submodule $ℵ$ of $M$ is called prime submodule if $ℵ \neq M$ and given $t \in F, m \in M, tm \in ℵ$ implies $m \in ℵ$ or $t \in (ℵ : M)$ where $(ℵ : M) = {t \in F / t M \subseteq ℵ}$ , if $ℵ$ is prime then $(ℵ : M)$ is prime ideal of $F$ .^3–5 The $M - radical$ of a submodule $ℵ$ was defined as the intersection of all prime submodules of $M$ containing $ℵ$ , if there is no prime submodule containing $ℵ$ then $rad (ℵ) = M$ .^6,7 For a proper submodule $ℵ$ of an $F - module M,$ the intersection of all S-PS.B. submodules of $M$ containing $ℵ$ is called the radical of $ℵ$ and denoted by $S - {rad}_{M}^{PS . B .} (ℵ)$ , if there is no S-PS.B. submodule of $M$ containing $ℵ$ , then $S - {rad}_{M}^{PS . B .} (ℵ) = M$ . A submodule $ℵ$ of $M$ is called a radical submodule if $S - {rad}_{M}^{PS . B .} (ℵ) = ℵ .$ The radical of an ideal I of $F$ has the characterization $\sqrt{I} = {x \in F / x^{m} \in I for some m > 0} .$ ⁸

The primary objective of this study is to present novel and diverse results concerning the S-pseudo bounded radical of submodules, and to establish a characterization for this concept, which is regarded as a new notion. We explore the relationship between the radical of submodules and that of S-pseudo bounded submodules. This connection yields significant results regarding their radicals. In the following section, we provide definitions, examples, and lemmas to further clarify our study.

2. S-Pseudo bounded submodules

At the beginning of this section, it is essential to introduce the definition of an S-pseudo bounded submodule along with illustrative examples. Additionally, it is necessary to present several lemmas that play a significant role in the study of the S-pseudo bounded radical of submodules.

Definition 2.1

A proper submodule $ℵ$ of an $F - module M$ is said to be S-pseudo bounded submodule if there exists $φ \in S = End (M), x \in M$ such that $φ (x) \in ℵ$ implies $\sqrt{{ann}_{F} (ℵ)} = \sqrt{{ann}_{F} φ (x)}, End (M)$ the set of endomorphism maps over an $F - module$ $M .$

Examples 2.2

1- Let $M = ℤ_{2} ⨁ ℤ_{4}$ as a $ℤ - module$ . Define $φ : M ⟶ M$ as $φ (\bar{a}, \bar{b}) = (\bar{0,} \bar{b}), \forall (\bar{a}, \bar{b}) \in M$ . Then if we take $ℵ = ⟨ \bar{0} ⟩ ⨁ ℤ_{4}$ , then $φ (\bar{a}, \bar{b}) \in ℵ .$ Hence $ℵ$ is S-PS.B. $ℤ -$ submodule of $M$ , since when $(\bar{a}, \bar{b}) = (\bar{0}, \bar{2}) \in M$ implies that $\sqrt{{ann}_{ℤ} (ℵ)} = \sqrt{{ann}_{ℤ} (\bar{0}, \bar{2})} = 2 ℤ .$
2- Consider $M = ℤ ⨁ ℤ_{3}$ as a $ℤ - module .$ Define $φ : M ⟶ M$ as $φ (a, \bar{b}) = (0, \bar{0}), \forall (a, \bar{b}) \in M,$ then if we take $ℵ = 3 ℤ ⨁ ⟨ \bar{0} ⟩$ , it is clear $φ \in End (M) .$ Let $(a, \bar{b}) = (3, \bar{0}) \in M,$ hence $\sqrt{{ann}_{ℤ} (ℵ)} = \sqrt{{ann}_{ℤ} (3 ℤ ⨁ ⟨ \bar{0} ⟩)} = ⟨ \bar{0} ⟩,$ but $\sqrt{{ann}_{ℤ} φ (3, \bar{0})} = \sqrt{{ann}_{ℤ} (0, \bar{0})} = ℤ .$ Thus $ℵ$ is not S-PS.B. $ℤ -$ submodule of $M .$

Definition 2.3

An $F - module M$ is said to be S-PS.B. $F - module$ if every proper submodule of $M$ is S-PS.B $F -$ submodule.

Lemma 2.4

If $ℵ$ is a prime submodule of a scalar $F - module$ $M,$ then $ℵ$ is S-PS.B. $F - submodule .$

Proof:

Let $φ \in End (M), x \in M .$ Since $M$ is scalar $F - module$ , then for all $φ \in End (M), \exists 0 \neq r \in F$ such that $φ (x) = rx, for all x \in M .$ Now, we must show that $\sqrt{{ann}_{F} φ (x)} = \sqrt{{ann}_{F} ℵ} .$ Let $a \in \sqrt{{ann}_{F} φ (x)}, x \in M .$ Then $a^{n} \in {ann}_{F} φ (x), for some n \in ℤ^{+}$ and $a^{n} . φ (x) = 0,$ so $a^{n} . rx = 0$ implies that $a^{n} \in {ann}_{F} (rx), for all rx \in ℵ .$ Since $ℵ$ is a prime submodule, then either $x \in ℵ or r M \subseteq ℵ$ so that $a^{n} \in {ann}_{F} ℵ$ and $a \in \sqrt{{ann}_{F} ℵ} .$ In a same way, $\sqrt{{ann}_{F} ℵ} \subseteq \sqrt{{ann}_{F} φ (x)} .$ Hence, $ℵ$ is S-PS.B. $F - submodule .$

Lemma 2.5

If $ℵ$ is S-PS.B. $F - submodule$ of a scalar $F - module$ $M,$ then $ℵ$ is S-prime submodule.

Proof:

Assume that $ℵ$ be a proper submodule of $M$ and $φ (x) \in ℵ, φ \in End (M), x \in M .$ Let $x \notin ℵ,$ then we have to prove $φ (M) \subseteq ℵ .$ Since $ℵ$ is S-PS.B. $F - submodule$ and $M$ is scalar, then $φ (x) \in ℵ, \forall x \in M$ which means that $φ (M) \subseteq ℵ .$

Lemma 2.6

If $ℵ$ is a submodule of a scalar $F - module$ $M,$ then $S - {rad}_{M}^{PS . B .} (ℵ) = rad (ℵ) .$

Proof:

Note that every S-prime submodule is prime, then from previous lemmas we get the result.

3. Some results related to S-pseudo bounded radical of submodules

In this part, we have examined the S-pseudo bounded radical of submodules presenting various properties exploring several relationships and introducing new results.

Definition 3.1

A S-pseudo bounded radical of a submodule $ℵ$ of an $F -$ module $M$ is the intersection of all S-pseudo bounded submodules of $M$ containing $ℵ$ and denoted by $S - {rad}_{M}^{PS . B .} (ℵ) .$

If there is no S-PS.B. submodule of $M$ contains $ℵ$ , then $S - {rad}_{M}^{PS . B .} (ℵ) = M .$

Definition 3.2

A proper submodule $ℵ$ of an $F -$ module $M$ is called S-pseudo bounded radical submodule if $S - {rad}_{M}^{PS . B .} (ℵ) = ℵ .$

Note that S-pseudo bounded radical submodule is proper submodule of $M .$

Remark 3.3

If $ℵ$ is S-PS.B. submodule of an $F -$ module $M,$ then $S - {rad}_{M}^{PS . B .} (ℵ)$ is S-PS.B. submodule too.

Proof:

Using the Definition (3.1), we have $S - {rad}_{M}^{PS . B .} (ℵ) = \cap_{j \in J} {k_{j} : k_{j} is S - PS . B . submodule, ℵ \subseteq k_{j}},$ for $j \in J$ and $J$ be an index set. Then by using induction and properties of S-PS.B. submodule we get $S - {rad}_{M}^{PS . B .} (ℵ)$ is S-PS.B. submodule.

Proposition 3.4

If $ℵ, K$ are two submodules of an $F -$ module $M$ . Then

(i) $ℵ \subseteq S - {rad}_{M}^{PS . B .} (ℵ) .$
(ii) $If ℵ \subseteq K, then S - {rad}_{M}^{PS . B .} (ℵ) \subseteq S - {rad}_{M}^{PS . B .} (K) .$
(iii) $S - {rad}_{M}^{PS . B .} (S - {rad}_{M}^{PS . B .} (ℵ)) = S - {rad}_{M}^{PS . B .} (ℵ) .$

Proof:

(i) By the Definition (3.1), we have $S - {rad}_{M}^{PS . B .} (ℵ) = \cap_{j \in J} k_{j}$ , $j \in J$ and $J$ be an index set and S-PS.B. submodules $k_{j}$ of $M$ containing $ℵ$ so that $ℵ \subseteq S - {rad}_{M}^{PS . B .} (ℵ) .$
(ii) Let $L_{j}$ be an S-PS.B. submodule of an $F -$ module $M$ such that $K \subseteq L_{j},$ $j \in J$ and $J$ be an index set. Then $ℵ \subseteq K \subseteq L_{j}$ so that $ℵ \subseteq L_{j}$ and by Definition (3.1), we get $S - {rad}_{M}^{PS . B .} (ℵ) \subseteq S - {rad}_{M}^{PS . B .} (K) .$
(iii) From Definition (3.1), we have $S - {rad}_{M}^{PS . B .} (S - {rad}_{M}^{PS . B .} (ℵ)) = \cap_{j \in J} G_{j}$ where $G_{j}$ is an S-PS.B. submodules of $M$ such that $S - {rad}_{M}^{PS . B .} (ℵ) \subseteq G_{j}$ . By part $(i),$ we get $ℵ \subseteq S - {rad}_{M}^{PS . B .} (ℵ),$ thus $S - {rad}_{M}^{PS . B .} (S - {rad}_{M}^{PS . B .} (ℵ)) \subseteq S - {rad}_{M}^{PS . B .} (ℵ) .$ From part $(ii),$ we obtain $S - {rad}_{M}^{PS . B .} (ℵ) \subseteq S - {rad}_{M}^{PS . B .} (S - {rad}_{M}^{PS . B .} (ℵ)) .$

Proposition 3.5

If $ℵ_{j}$ is a submodule of an $F -$ module $M, J$ is an index set, $j \in J,$ then

(i) $S - {rad}_{M}^{PS . B .} (\cap_{j \in J} ℵ_{j}) \subseteq \cap_{j \in J} S - {rad}_{M}^{PS . B .} (ℵ_{j}) =$ $S - {rad}_{M}^{PS . B .} (\cap_{j \in J} S - {rad}_{M}^{PS . B .} (ℵ_{j})) .$
(ii) $\sum_{j \in J} S - {rad}_{M}^{PS . B .} (ℵ_{j}) \subseteq S - {rad}_{M}^{PS . B .} (\sum_{j \in J .} ℵ_{j}) =$ $S - {rad}_{M}^{PS . B .} (\sum_{j \in J .} S - {rad}_{M}^{PS . B .} (ℵ_{j})) .$

Proof:

(i) Suppose that $S - {rad}_{M}^{PS . B .} (\cap_{j \in J} ℵ_{j}) = \cap_{k \in K} V_{k}$ where $V_{k}$ is S-PS.B. submodule containing $\cap_{j \in J} ℵ_{j}$ for $k \in K, K$ be an index set. Let ${G_{kj}}$ be the set of S-PS.B. submodules containing $ℵ_{j} .$ Since $\cap_{j \in J} ℵ_{j} \subseteq ℵ_{j} \subseteq G_{kj},$ then $S - {rad}_{M}^{PS . B .} (\cap_{j \in J} ℵ_{j}) \subseteq S - {rad}_{M}^{PS . B .} (ℵ_{j}) .$ Since $\cap_{j \in J} S - {rad}_{M}^{PS . B .} (ℵ_{j}) \subseteq S - {rad}_{M}^{PS . B .} (ℵ_{j}),$ then $S - {rad}_{M}^{PS . B .} (\cap_{j \in J} S - {rad}_{M}^{PS . B .} (ℵ_{j})) \subseteq S - {rad}_{M}^{PS . B .} (S - {rad}_{M}^{PS . B .} (ℵ_{j})) = S - {rad}_{M}^{PS . B .} (ℵ_{j}), j \in J .$ Hence $S - {rad}_{M}^{PS . B .} (\cap_{j \in J} S - {rad}_{M}^{PS . B .} (ℵ_{j})) \subseteq \cap_{j \in J} S - {rad}_{M}^{PS . B .} (ℵ_{j}) .$ By previous proposition part $(i),$ $\cap_{j \in J} S - {rad}_{M}^{PS . B .} (ℵ_{j}) = S - {rad}_{M}^{PS . B .} (\cap_{j \in J} S - {rad}_{M}^{PS . B .} (ℵ_{j}))$ is clear.
(ii) Since $ℵ_{j} \subseteq \sum_{j \in J} ℵ_{j}$ for $j \in J,$ then from previous proposition part (ii) $S - {rad}_{M}^{PS . B .} (ℵ_{j}) \subseteq S - {rad}_{M}^{PS . B .} (\sum_{j \in J} ℵ_{j}) .$

Hence $\sum_{j \in J} S - {rad}_{M}^{PS . B .} (ℵ_{j}) \subseteq S - {rad}_{M}^{PS . B .} (\sum_{j \in J} ℵ_{j}) .$

Since $\sum_{j \in J} ℵ_{j} \subseteq \sum_{j \in J} S - {rad}_{M}^{PS . B .} (ℵ_{j}),$ then it is easily to show that $S - {rad}_{M}^{PS . B .} (\sum_{j \in J} ℵ_{j}) = S - {rad}_{M}^{PS . B .} (\sum_{j \in J} S - {rad}_{M}^{PS . B .} (ℵ_{j})) .$

Proposition 3.6

If $ℵ$ is a submodule of a multiplication finitely generated $F -$ module $M .$ Then $S - {rad}_{M}^{PS . B .} (ℵ) = M$ if and only if $ℵ = M .$

Proof:

Assume that $S - {rad}_{M}^{PS . B .} (ℵ) = M$ and $ℵ \neq M$ . Since $M$ is a finitely generated module, then there exists a maximal submodule $K$ of $M$ such that $ℵ \subseteq K$ and hence $K$ is prime. Since $M$ is a finitely generated multiplication $F -$ module, then $M$ is scalar⁹ and thus $K$ is S-PS.B. submodule Lemma (2.4). Thus $S - {rad}_{M}^{PS . B .} (ℵ) \subseteq K$ so that $M = K$ which is contradicting the assumption. If $ℵ = M,$ the other side is clear.

Corollary 3.7

Let $ℵ, K$ be two submodules of a multiplication finitely generated $F -$ module $M$ . Then $S - {rad}_{M}^{PS . B .} (ℵ) + S - {rad}_{M}^{PS . B .} (K) = M$ if and only if $ℵ + K = M .$

Proof:

Assume that $S - {rad}_{M}^{PS . B .} (ℵ) + S - {rad}_{M}^{PS . B .} (K) = M,$ then by proposition (3.4), (3.5) and previous proposition, $M = S - {rad}_{M}^{PS . B .} (ℵ + K)$ implies that $ℵ + K = M$ . On the other hand if $ℵ + K = M,$ then $M = S - {rad}_{M}^{PS . B .} (ℵ + K) = S - {rad}_{M}^{PS . B .} (S - {rad}_{M}^{PS . B .} (ℵ) + S - {rad}_{M}^{PS . B .} (K)) .$

Hence by previous proposition, $S - {rad}_{M}^{PS . B .} (ℵ) + S - {rad}_{M}^{PS . B .} (K) = M .$

Proposition 3.8

If $M$ be a multiplication finitely generated $F -$ module and $ℵ$ be a submodule of $M,$ then $S - {rad}_{M}^{PS . B .} (ℵ) = S - {rad}_{M}^{PS . B .} (ℵ + \sqrt{(ℵ : M)} M),$ where $\sqrt{(ℵ : M)}$ be a prime ideal of $F .$

Proof:

Since $ℵ \subseteq ℵ + \sqrt{(ℵ : M)} M,$ then $S - {rad}_{M}^{PS . B .} (ℵ) \subseteq S - {rad}_{M}^{PS . B .} (ℵ + \sqrt{(ℵ : M)} M)$ . Assume that $S - {rad}_{M}^{PS . B .} (ℵ) = \cap_{j \in J} V_{j}$ where $V_{j}$ is S-PS.B. submodule of $M$ containing $ℵ .$ Since $M$ is a finitely generated multiplication module, then $M$ is scalar⁹ and thus $ℵ$ is prime Lemma (2.5) so that $V_{j}$ is prime submodule and $(V_{j} : M)$ is prime ideal which implies that $\sqrt{(ℵ : M)} \subseteq (V_{j} : M)$ and hence $ℵ + \sqrt{(ℵ : M)} M \subseteq V_{j} .$ Therefore $S - {rad}_{M}^{PS . B .} (ℵ + \sqrt{(ℵ : M)} M) \subseteq \cap_{j \in J} V_{j} = S - {rad}_{M}^{PS . B .} (ℵ)$ . Hence $S - {rad}_{M}^{PS . B .} (ℵ) = S - {rad}_{M}^{PS . B .} (ℵ + \sqrt{(ℵ : M)} M) .$

Corollary 3.9

If $M$ is a multiplication finitely generated $F -$ module and $ℵ$ is a submodule of $M,$ then $S - {rad}_{M}^{PS . B .} (ℵ) = ℵ + \sqrt{(ℵ : M)} M,$ where $\sqrt{(ℵ : M)}$ is a maximal ideal of $F .$

Proof:

Since $ℵ + \sqrt{(ℵ : M)} M \subseteq S - {rad}_{M}^{PS . B .} (ℵ + \sqrt{(ℵ : M)} M) = S - {rad}_{M}^{PS . B .} (ℵ),$ by previous proposition and since $\sqrt{(ℵ : M)} \subseteq ((ℵ + \sqrt{(ℵ : M)} M) : M),$ then $((ℵ + \sqrt{(ℵ : M)} M) : M) = \sqrt{(ℵ : M)}$ or $((ℵ + \sqrt{(ℵ : M)} M) : M) = F .$ If $((ℵ + \sqrt{(ℵ : M)} M) : M) = F,$ then $FM = M \subseteq ℵ + \sqrt{(ℵ : M)} M \subseteq S - {rad}_{M}^{PS . B .} (ℵ)$ which implies that $M = S - {rad}_{M}^{PS . B .} (ℵ) .$ Since $M$ is finitely generated, then $M = ℵ$ by proposition (2.6) which is contradiction. So $((ℵ + \sqrt{(ℵ : M)} M) : M) = \sqrt{(ℵ : M)}$ , therefore $S - {rad}_{M}^{PS . B .} (ℵ) = ℵ + \sqrt{(ℵ : M)} M .$

Proposition 3.10

If $ℵ$ is a submodule of $M$ and $g : M \to M^{'}$ is an epiomorphism such that $Ker (g) \subseteq ℵ$ , then

(i) $g (S - {rad}_{M}^{PS . B .} (ℵ)) = S - {rad}_{M}^{PS . B .} g (ℵ) .$
(ii) $g^{- 1} (S - {rad}_{M}^{PS . B .} (ℵ^{'})) = S - {rad}_{M}^{PS . B .} g^{- 1} (ℵ^{'}) .$

Proof:

(i) By the Definition (3.1), we have $S - {rad}_{M}^{PS . B .} (ℵ) = \cap_{j \in J} V_{j}$ where $V_{j}$ is S-PS.B. submodule of $M$ containing $ℵ$ $, j \in J$ and $J$ be an index set. Therefore $g (S - {rad}_{M}^{PS . B .} (ℵ)) = g (\cap_{j \in J} V_{j})$ and $g (S - {rad}_{M}^{PS . B .} (ℵ)) = \cap_{j \in J} g (V_{j})$ with $g (ℵ) \subseteq g (V_{j}),$ since $Ker (g) \subseteq ℵ \subseteq V_{j}$ . Thus $g (S - {rad}_{M}^{PS . B .} (ℵ)) = S - {rad}_{M}^{PS . B .} g (ℵ) .$
(ii) Assume that $ℵ^{'}$ is a submodule of $M^{'},$ then by Definition (3.1), $S - {rad}_{M}^{PS . B .} (ℵ^{'}) = \cap_{j \in J} V_{j}^{'}$ the intersection is over all S-PS.B. submodules $V_{j}^{'}$ of $M^{'}$ containing $ℵ^{'} .$ By Ref. 10, we have $g^{- 1} (S - {rad}_{M}^{PS . B .} (ℵ^{'})) = g^{- 1} (\cap_{j \in J} V_{j}^{'}) = \cap_{j \in J} g^{- 1} (V_{j}^{'})$ the intersection is over all S-PS.B. submodules $g^{- 1} (V_{j}^{'})$ of $M$ containing $g^{- 1} (ℵ^{'})$ . Hence $g^{- 1} (S - {rad}_{M}^{PS . B .} (ℵ^{'})) = S - {rad}_{M}^{PS . B .} g^{- 1} (ℵ^{'}) .$

Proposition 3.11

If $ℵ, K$ are two submodules of an $F -$ module $M$ , then $S - {rad}_{M}^{PS . B .} (ℵ) \cap S - {rad}_{M}^{PS . B .} (K) = ℵ \cap K$ if and only if $ℵ \cap K$ is radical and $S - {rad}_{M}^{PS . B .} (ℵ \cap K) = S - {rad}_{M}^{PS . B .} (ℵ) \cap S - {rad}_{M}^{PS . B .} (K) .$

Proof:

Assume that $S - {rad}_{M}^{PS . B .} (ℵ) \cap S - {rad}_{M}^{PS . B .} (K) = ℵ \cap K .$ From Proposition (3.4) part $(i),$ we have $ℵ \cap K \subseteq S - {rad}_{M}^{PS . B .} (ℵ \cap K),$ then $S - {rad}_{M}^{PS . B .} (ℵ) \cap S - {rad}_{M}^{PS . B .} (K) \subseteq S - {rad}_{M}^{PS . B .} (ℵ \cap K)$ and from Proposition (3.5) the inequality $S - {rad}_{M}^{PS . B .} (ℵ \cap K) \subseteq S - {rad}_{M}^{PS . B .} (ℵ) \cap S - {rad}_{M}^{PS . B .} (K)$ satisfies. Therefore, $S - {rad}_{M}^{PS . B .} (ℵ \cap K) = S - {rad}_{M}^{PS . B .} (ℵ) \cap S - {rad}_{M}^{PS . B .} (K) .$ We get that $ℵ \cap K$ is $S - {rad}_{M}^{PS . B .}$ submodule. Since $S - {rad}_{M}^{PS . B .} (ℵ) \cap S - {rad}_{M}^{PS . B .} (K) = ℵ \cap K,$ thus we have the another result. The converse can be proved by the assumption together.

Proposition 3.12

If $ℵ, K$ are two submodules of a finitely generated multiplication $F -$ module $M$ such that $(ℵ : M)$ and $(K : M)$ are both radical ideals, then $(S - {rad}_{M}^{PS . B .} (ℵ \cap K) : M) = (ℵ \cap K : M) .$

Proof:

Clearly, $(ℵ \cap K : M) = (ℵ : M) \cap (K : M) = \sqrt{(ℵ : M)} \cap \sqrt{(K : M)} = \sqrt{(ℵ \cap K : M)}$ . Since $M$ is finitely generated, then by Ref. 11 we have $\sqrt{(ℵ \cap K : M)} = (S - {rad}_{M}^{PS . B .} (ℵ \cap K) : M) .$ Thus $(ℵ \cap K : M) = (S - {rad}_{M}^{PS . B .} (ℵ \cap K) : M) .$

Corollary 3.13

If $ℵ, K$ are two submodules of a multiplication finitely generated $F -$ module $M$ , then

(S - {rad}_{M}^{PS . B .} (ℵ \cap K) : M) = (S - {rad}_{M}^{PS . B .} (ℵ) \cap S - {rad}_{M}^{PS . B .} (K) : M) .

Proof:

From Ref. 11, we get $(S - {rad}_{M}^{PS . B .} (ℵ \cap K) : M) = \sqrt{(ℵ \cap K : M)} = \sqrt{(ℵ : M)} \cap \sqrt{(K : M)} = (S - {rad}_{M}^{PS . B .} (ℵ) : M) \cap (S - {rad}_{M}^{PS . B .} (K) : M) = (S - {rad}_{M}^{PS . B .} (ℵ) \cap S - {rad}_{M}^{PS . B .} (K) : M) .$

Proposition 3.14

If $ℵ$ is a submodule of a multiplication finitely generated $F -$ module $M$ , then $S - {rad}_{M}^{PS . B .} (ℵ) = \sqrt{(ℵ : M)} M .$

Proof:

From, Ref. 12, we have $\sqrt{(ℵ : M)} M \subseteq S - {rad}_{M}^{PS . B .} (ℵ) .$ Now, let $L$ be an S-PS.B. submodule of $M$ such that $ℵ \subseteq L,$ then $(ℵ : M) \subseteq (L : M) .$ Since $M$ is a a finitely generated multiplication module, then $M$ is scalar⁹ and hence $ℵ$ is prime Lemma( 2.5) and $L$ is prime submodule so that $(L : M)$ is prime ideal implies that $\sqrt{(ℵ : M)} \subseteq (L : M) .$ Therefore $\sqrt{(ℵ : M)} M \subseteq (L : M) M \subseteq L .$ Since $L$ is an arbitrary S-PS.B. submodule containing $ℵ,$ then $S - {rad}_{M}^{PS . B .} (ℵ) \subseteq \sqrt{(ℵ : M)} M .$ Thus $S - {rad}_{M}^{PS . B .} (ℵ) = \sqrt{(ℵ : M)} M .$

Proposition 3.15

If $ℵ, K$ are two submodules of an $F - module$ $M$ such that $ℵ \subseteq K \subseteq M, ℵ$ is a direct summand of $M$ and rad( $M)$ be an essential submodule of $M .$ If $S - {rad}_{M}^{PS . B .} (ℵ) = S - {rad}_{M}^{PS . B .} (K)$ then $ℵ = K .$

Proof:

Suppose that $ℵ$ is a direct summand of $M,$ then $\exists ℵ^{'} < M$ such that $M = ℵ ⨁ ℵ^{'} .$ Thus $K = ℵ ⨁ (K \cap ℵ^{'})$ implies that $S - {rad}_{M}^{PS . B .} (K) = S - {rad}_{M}^{PS . B .} (ℵ) ⨁ S - {rad}_{M}^{PS . B .} (K \cap ℵ^{'}) .$ Therefore $S - {rad}_{M}^{PS . B .} (K \cap ℵ^{'}) = 0$ which can be written as rad( $M) \cap (K \cap ℵ^{'}) = 0$ and since rad( $M)$ is an essential submodule of $M$ implies that $(K \cap ℵ^{'}) = 0,$ thus $ℵ = K .$

Proposition 3.16

If $ℵ, K$ are two submodules of a multiplication finitely generated $F -$ module $M,$ then $\sqrt{(ℵ : M) + (K : M)} = (S - {rad}_{M}^{PS . B .} (ℵ + K) : M),$ where $ℵ, K$ are both radical submodules of $M .$

Proof:

By, Ref. 13, since $M$ is finitely generated multiplication module, then we have $\sqrt{(I M : M)} = \sqrt{I + (0 : M)}$ for all ideal I of $F$ . Consider the finitely generated $F -$ module $M / K$ and the ideal $(ℵ : M)$ instead of $M$ and I respectively, then

\sqrt{(ℵ : M) + (K : M)} = \sqrt{(ℵ : M) + (0 : \frac{M}{K})} = \sqrt{((ℵ : M) (\frac{M}{K}) : \frac{M}{K})} = \sqrt{((\frac{(ℵ : M) M + K}{K}) : \frac{M}{K})} = \sqrt{((ℵ : M) M + K : M)} = \sqrt{(ℵ + K : M)} = (S - {rad}_{M}^{PS . B .} (ℵ + K) : M) .

Corollary 3.17

Let $ℵ$ be a submodule of a finitely generated multiplication $F -$ module $M$ . Then $(S - {rad}_{M}^{PS . B .} ((ℵ : M) M) : M) = (ℵ : M),$ where $ℵ$ is radical submodules of $M .$

Proof:

Since $ℵ$ is radical submodules of $M,$ then $(ℵ : M) M \subseteq ℵ$ implies that $S - {rad}_{M}^{PS . B .} ((ℵ : M) M) \subseteq ℵ .$ Therefore $(S - {rad}_{M}^{PS . B .} ((ℵ : M) M) : M) \subseteq (ℵ : M) .$ On the other hand $(ℵ : M) \subseteq ((ℵ : M) M : M) \subseteq (S - {rad}_{M}^{PS . B .} ((ℵ : M) M) : M) .$ Therefore $(S - {rad}_{M}^{PS . B .} ((ℵ : M) M) : M) = (ℵ : M) .$

Corollary 3.18

If $M$ is a multiplication finitely generated $F -$ module, then $S - {rad}_{M}^{PS . B .} ((S - {rad}_{M}^{PS . B .} (I M) : M) M) = S - {rad}_{M}^{PS . B .} (I M),$ where I is radical ideal of $F .$

Proof:

Since $(S - {rad}_{M}^{PS . B .} (I M) : M) M \subseteq S - {rad}_{M}^{PS . B .} (I M),$

then $S - {rad}_{M}^{PS . B .} ((S - {rad}_{M}^{PS . B .} (I M) : M) M) \subseteq S - {rad}_{M}^{PS . B .} (I M) .$ On the other hand $I M \subseteq S - {rad}_{M}^{PS . B .} (I M)$ implies that $I \subseteq (S - {rad}_{M}^{PS . B .} (I M) : M)$ and thus $I M \subseteq (S - {rad}_{M}^{PS . B .} (I M) : M) M$ which means

S - {rad}_{M}^{PS . B .} (I M) \subseteq S - {rad}_{M}^{PS . B .} ((S - {rad}_{M}^{PS . B .} (I M) : M) M) .

Therefore $S - {rad}_{M}^{PS . B .} ((S - {rad}_{M}^{PS . B .} (I M) : M) M) = S - {rad}_{M}^{PS . B .} (I M) .$

Proposition 3.19

If $ℵ, K$ are two submodules of a scalar $F -$ module $M$ , then

(i) $If S - {rad}_{M}^{PS . B .} (ℵ + K) = M,$ then $\exists t \in F$ such that $t M \subseteq S - {rad}_{M}^{PS . B .} (ℵ)$ and $(1 - t) M \subseteq S - {rad}_{M}^{PS . B .} (K) .$
(ii) $If M$ is finitely generated module such that $(ℵ + K) = M,$ then $\exists t \in F$ such that $t M \subseteq ℵ$ and $(1 - t) M \subseteq K .$ Where $ℵ, K$ are radical submodules of $M .$

Proof:

(i) Since
$\begin{matrix} F = (M : M) = (S - {rad}_{M}^{PS . B .} (ℵ + K) : M) \\ = (S - {rad}_{M}^{PS . B .} (S - {rad}_{M}^{PS . B .} (ℵ) + S - {rad}_{M}^{PS . B .} (K)) : M) \\ = \sqrt{(S - {rad}_{M}^{PS . B .} (ℵ) : M) + (S - {rad}_{M}^{PS . B .} (K) : M)} . \end{matrix}$

Therefore $F = (S - {rad}_{M}^{PS . B .} (ℵ) : M) + (S - {rad}_{M}^{PS . B .} (K) : M) .$ Thus the required result is clear.

(ii) Since $(ℵ + K) = M = S - {rad}_{M}^{PS . B .} (ℵ + K),$ by part $(i)$ we have $F = (S - {rad}_{M}^{PS . B .} (ℵ) : M) + (S - {rad}_{M}^{PS . B .} (K) : M) .$ By Ref. 14, since $M$ is finitely generated module we have $F = \sqrt{(ℵ : M)} + \sqrt{(K : M)}$ and hence $F = (ℵ : M) + (K : M) .$ Thus clearly result follows.
Recall that a submodule $ℵ$ of an $F -$ module $M$ is named completely irreducible if for any two submodules $ℵ_{1}, ℵ_{2}$ of $M$ such that $ℵ_{1} \cap ℵ_{2} \subseteq ℵ$ implies that $ℵ_{1} \subseteq ℵ$ or $ℵ_{2} \subseteq ℵ$ .^15,16

Proposition 3.20

If $ℵ_{1}, ℵ_{2}$ are two submodules of an $F -$ module $M$ and every S-PS.B. submodule $L$ of $M$ is completely irreducible such that $ℵ_{1} \cap ℵ_{2} \subseteq L$ , then $S - {rad}_{M}^{PS . B .} (ℵ_{1} \cap ℵ_{2}) = S - {rad}_{M}^{PS . B .} (ℵ_{1}) \cap S - {rad}_{M}^{PS . B .} (ℵ_{2}) .$

Proof:

By Proposition (3.5), we have $S - {rad}_{M}^{PS . B .} (ℵ_{1} \cap ℵ_{2}) \subseteq S - {rad}_{M}^{PS . B .} (ℵ_{1}) \cap S - {rad}_{M}^{PS . B .} (ℵ_{2}) .$ Let $L$ be an S-PS.B. submodule of $M$ such that $ℵ_{1} \cap ℵ_{2} \subseteq L .$ Since every S-PS.B. submodule of $M$ that contains $ℵ_{1} \cap ℵ_{2}$ is completely irreducible, then either $ℵ_{1} \subseteq L$ or $ℵ_{2} \subseteq L$ . Thus $S - {rad}_{M}^{PS . B .} (ℵ_{1}) \subseteq L$ or $S - {rad}_{M}^{PS . B .} (ℵ_{2}) \subseteq L$ . Therefore $S - {rad}_{M}^{PS . B .} (ℵ_{1}) \cap S - {rad}_{M}^{PS . B .} (ℵ_{2}) \subseteq L$ for any $L$ and $S - {rad}_{M}^{PS . B .} (ℵ_{1} \cap ℵ_{2}) .$ Therefore $S - {rad}_{M}^{PS . B .} (ℵ_{1}) \cap S - {rad}_{M}^{PS . B .} (ℵ_{2}) \subseteq S - {rad}_{M}^{PS . B .} (ℵ_{1} \cap ℵ_{2})$ .

Hence $S - {rad}_{M}^{PS . B .} (ℵ_{1} \cap ℵ_{2}) = S - {rad}_{M}^{PS . B .} (ℵ_{1}) \cap S - {rad}_{M}^{PS . B .} (ℵ_{2}) .$

Now, we give some notations:

G^{S - PS .} (ℵ) = {k : k \in {Spec}^{S ‐ PS .} (M), ℵ \subseteq k},

$S - {rad}_{M}^{PS . B .} (ℵ) = \cap_{k \in {Spec}^{S - PS .}} k,$ where ${Spec}^{S - PS .}$ denoted all S-PS.B. submodules of $M .$

Proposition 3.21

Let $M$ be an $F -$ module. Then

(i) $G^{S - PS .} (ℵ_{1}) \cap G^{S - PS .} (ℵ_{2}) = G^{S - PS .} (ℵ_{1} + ℵ_{2}) .$
(ii) $G^{S - PS .} (ℵ_{1}) \cup G^{S - PS .} (ℵ_{2}) \subseteq G^{S - PS .} (ℵ_{1} \cap ℵ_{2}) .$ Where $ℵ_{1}, ℵ_{2}$ are two submodules of $M .$

Proof:

(i) Since
$\begin{matrix} G^{S - PS .} (ℵ_{1}) = {k : k \in {Spec}^{S - PS .} (M), ℵ_{1} \subseteq k}, \\ G^{S - PS .} (ℵ_{2}) = {k : k \in {Spec}^{S - PS .} (M), ℵ_{2} \subseteq k}, \\ \begin{matrix} G^{S - PS .} (ℵ_{1}) \cap G^{S - PS .} (ℵ_{2}) = {k : k \in {Spec}^{S - PS .} (M), ℵ_{1} \subseteq k, ℵ_{2} \subseteq k}, \\ G^{S - PS .} (ℵ_{1} + ℵ_{2}) = {k : k \in {Spec}^{S - PS .} (M), (ℵ_{1} + ℵ_{2}) \subseteq k} . \end{matrix} \end{matrix}$

Hence $G^{S - PS .} (ℵ_{1}) \cap G^{S - PS .} (ℵ_{2}) = G^{S - PS .} (ℵ_{1} + ℵ_{2}) .$

(ii) Since
$\begin{matrix} G^{S - PS .} (ℵ_{1}) = {k : k \in {Spec}^{S - PS .} (M), ℵ_{1} \subseteq k}, \\ G^{S - PS .} (ℵ_{2}) = {k : k \in {Spec}^{S - PS .} (M), ℵ_{2} \subseteq k}, \\ G^{S - PS .} (ℵ_{1}) \cup G^{S - PS .} (ℵ_{2}) = {k : k \in {Spec}^{S - PS .} (M), ℵ_{1} \subseteq k or ℵ_{2} \subseteq k} . \end{matrix}$

Also $G^{S - PS .} (ℵ_{1} \cap ℵ_{2}) = {{k : k \in {Spec}^{S - PS .} (M), ℵ_{1} \cap ℵ_{2} \subseteq k} .$

Now, let $L \in G^{S - PS .} (ℵ_{1}) \cup G^{S - PS .} (ℵ_{2})$ then $L$ is S-PS.B. submodule such that $ℵ_{1} \subseteq L$ or $ℵ_{2} \subseteq L .$ Thus $ℵ_{1} \cap ℵ_{2} \subseteq ℵ_{1} \subseteq L$ and $ℵ_{1} \cap ℵ_{2} \subseteq ℵ_{2} \subseteq L$ and hence $L \in G^{S - PS .} (ℵ_{1} \cap ℵ_{2}) .$ Therefore $G^{S - PS .} (ℵ_{1}) \cup G^{S - PS .} (ℵ_{2}) \subseteq G^{S - PS .} (ℵ_{1} \cap ℵ_{2}) .$

Corollary 3.22

If $ℵ_{1}, ℵ_{2}$ are two submodules of an $F -$ module $M$ and every S-PS.B. submodule $k$ of $M$ is completely irreducible such that $ℵ_{1} \cap ℵ_{2} \subseteq k$ , then

G^{S - PS .} (ℵ_{1}) \cup G^{S - PS .} (ℵ_{2}) = G^{S - PS .} (ℵ_{1} \cap ℵ_{2}) .

Proof:

It is sufficient to show $G^{S - PS .} (ℵ_{1} \cap ℵ_{2}) \subseteq G^{S - PS .} (ℵ_{1}) \cup G^{S - PS .} (ℵ_{2}) .$ Let $k \in G^{S - PS .} (ℵ_{1} \cap ℵ_{2})$ implies that $k$ is S-PS.B. submodule and $ℵ_{1} \cap ℵ_{2} \subseteq k$ . Since every S-PS.B. submodule of $M$ that contains $ℵ_{1} \cap ℵ_{2}$ is completely irreducible, then either $ℵ_{1} \subseteq k$ or $ℵ_{2} \subseteq k$ which means $k \in G^{S - PS .} (ℵ_{1})$ or $k \in G^{S - PS .} (ℵ_{2})$ so that $k \in G^{S - PS .} (ℵ_{1}) \cup G^{S - PS .} (ℵ_{2}) .$ Thus $G^{S - PS .} (ℵ_{1} \cap ℵ_{2}) \subseteq G^{S - PS .} (ℵ_{1}) \cup G^{S - PS .} (ℵ_{2}) .$

Corollary 3.23

Let $M$ be a finitely generated multiplication $F -$ module and I, J are an ideals of $F .$ Then

(i) $G^{S - PS .} (I M) = G^{S - PS .} (\sqrt{I} M)$ .
(ii) $G^{S - PS .} (I M + J M) = G^{S - PS .} (I M) \cap G^{S - PS .} (J M) .$
(iii) $G^{S - PS .} ((I \cap J) M) = G^{S - PS .} (I M \cap J M) = G^{S - PS .} (I M) \cup G^{S - PS .} (J M) .$

Proof:

(i) Assume that I is an ideal of $F .$ Since $I \subseteq \sqrt{I}$ then $I M \subseteq \sqrt{I} M$ and hence $G^{S - PS .} (\sqrt{I} M) \subseteq G^{S - PS .} (I M) .$ Let $ℵ \in G^{S - PS .} (I M),$ then $ℵ$ is S-PS.B. submodule and since $M$ is a finitely generated multiplication, then $M$ is scalar⁹ and thus $ℵ$ is prime Lemma (2.5). Since $I \subseteq (I M : M) \subseteq (ℵ : M),$ thus
$\sqrt{I} \subseteq (ℵ : M)$ and hence $(\sqrt{I} M : M) \subseteq ((ℵ : M) M : M) \subseteq (ℵ : M)$ it follows that $ℵ \in G^{S - PS .} (\sqrt{I} M)$ and thus $G^{S - PS .} (I M) \subseteq G^{S - PS .} (\sqrt{I} M) .$
(ii) Let $I$ and $J$ be two ideals of $F .$ Suppose that $ℵ \in G^{S - PS .} (I M) \cap G^{S - PS .} (J M),$ then $ℵ$ is S-PS.B. submodule and since $M$ is a finitely generated multiplication, then $M$ is scalar⁹ and thus $ℵ$ is prime Lemma (2.5). We have $I \subseteq (I M : M) \subseteq (ℵ : M)$ and $J \subseteq (J M : M) \subseteq (ℵ : M) .$ Thus $(I + J) M \subseteq (ℵ : M) M \subseteq ℵ .$ Therefore, $((I + J) M : M) \subseteq (ℵ : M)$ and thus $ℵ \in G^{S - PS .} (I M + J M) .$ Clearly, $G^{S - PS .} (I M + J M) \subseteq G^{S - PS .} (I M) \cap G^{S - PS .} (J M) .$ Therefore, $G^{S - PS .} (I M + J M) = G^{S - PS .} (I M) \cap G^{S - PS .} (J M) .$
(iii) Let $I$ and $J$ be two ideals of $F .$ Clearly, we have $G^{S - PS .} (I M) \cup G^{S - PS .} (J M) \subseteq G^{S - PS .} (I M \cap J M) \subseteq G^{S - PS .} ((I \cap J) M) .$ Let $ℵ \in G^{S - PS .} ((I \cap J) M),$ then $ℵ$ is S-PS.B. submodule and from⁹ and Lemma (2.5) hence $ℵ$ is prime, So that $I \cap J \subseteq ((I \cap J) M : M) \subseteq (ℵ : M) .$
Now, since $(ℵ : M)$ is prime ideal of $F,$ then $I \subseteq (ℵ : M)$ or $J \subseteq (ℵ : M) .$ Hence $(I M : M) \subseteq (ℵ : M)$ or $\subseteq (J M : M) \subseteq (ℵ : M) .$ It follows that $ℵ \in G^{S - PS .} (I M)$ or $ℵ \in G^{S - PS .} (J M)$ implies that $ℵ \in G^{S - PS .} (I M) \cup G^{S - PS .} (J M) .$

Corollary 3.24

If $ℵ$ is a submodule of a multiplication finitely generated $F -$ module $M$ , then $G^{S - PS .} (ℵ) = G^{S - PS .} ((ℵ : M) M)$ .

Proof:

Clearly, $G^{S - PS .} (ℵ) \subseteq G^{S - PS .} ((ℵ : M) M) .$ Let $L \in G^{S - PS .} ((ℵ : M) M)$ , then we have $((ℵ : M) M) \subseteq L .$ Since $(ℵ : M) \subseteq ((ℵ : M) M : M) \subseteq (L : M)$ so that $(ℵ : M) \subseteq (L : M)$ and hence $L \in G^{S - PS .} (ℵ) .$ Thus $G^{S - PS .} (ℵ) = G^{S - PS .} ((ℵ : M) M)$ .

4. Conclusion

In this study, we have presented several significant findings concerning the S-pseudo bounded radical submodule. Our results are supported by facts related to the prime submodule, prime ideal, and S-prime submodule. We examined the properties of the S-pseudo bounded radical submodule, particularly in the context of finitely generated and multiplication modules.

Ethical considerations

This research did not involve any studies with human participants or animals and therefore did not require ethical approval.

Data availability

No experimental data were generated or analyzed in this study. The research is entirely theoretical within the field of pure mathematics (abstract algebra); therefore, data sharing is not applicable. No datasets were generated or analyzed during the current study. All results are theoretical and derived analytically within the framework of abstract algebra.Therefore, data sharing is not applicable to this article as no datasets were created or used.

Reporting guidelines

Thank you for the provided guidelines regarding reporting standards. I will ensure compliance with the relevant guidelines applicable to pure mathematics research to maintain clarity and scientific rigor.

References

1. McCasland RL, Moore ME: On radicals of submodules. Communications in Algebra. 1991; 19(5): 1327–1341. Publisher Full Text
2. Abdul-Al-Kalik AJ: Semi–Bounded Modules. Baghdad Sci. J. 2012; 9(4): 20. Publisher Full Text
3. Al-Mothafar NS, Abdil-Khalik AJ: End-Ψ-Prime Submodules. Ibn Al-Haitham J. Pure Appl. Sci. 2016; 29(2). Reference Source
4. Murad MS, Shihab BN: Endo-restricted bounded submodules related to prime submodules and scalar modules. AIP Conference Proceedings. AIP Publishing LLC; 2025; Vol. 3264(1). Publisher Full Text
5. Al-Quraishi AD, Shihab BN: Pu-visible Submodule with their most prominent characteristics. IOP Conference Series: Materials Science and Engineering. IOP Publishing; 2020; Vol. 871(1). Reference Source
6. Hadi IM, Ali SNA-A, Shyaa FD: T-Essentially Coretractable and Weakly T-Essentially Coretractable Modules. Baghdad Sci. J. 2021; 18(1): 0156. Publisher Full Text
7. Ramadhan HA, Al Mothafar NS: Weakly Small Semiprime Submodules. J. Phys. Conf. Ser. 2021; 1879(3): 032128. IOP Publishing. Publisher Full Text
8. Abbas MR, Ahmed MA: Modules Whose St-Closed Submodules are Fully Invariant. Baghdad Sci. J. 2025; 22(3): 988–996. Publisher Full Text
9. Shihab BN: Scalar Reflexive Modules. Iraq: University of Baghdad; 2004. Doctoral dissertation, PhD. Thesis. Reference Source
10. Kasch F: Modules and rings. Vol. 17. .Academic Press; 1982. Publisher Full Text
11. McCasland RL, Moore ME, Smith PF: Modules with bounded spectra. Commun. Algebra. 1998; 26(10): 3403–3417. Publisher Full Text
12. McCasland RL, Moore ME: On radicals of submodules of finitely generated modules. Can. Math. Bull. 1986; 29(1): 37–39. Publisher Full Text
13. Faith C: Algebra II Ring Theory: Vol. 2: Ring Theory. Vol. 191. . Springer Science & Business Media; 2012. Publisher Full Text
14. McCasland RL, Moore ME: Prime submodules. Commun. Algebra. 1992; 20(6): 1803–1817. Publisher Full Text
15. Ajeel AS, Mohammad HK, Ali.: Approximaitly Prime Submodules and Some Related Concepts. Ibn Al-Haitham J. Pure Appl. Sci. 2019; 32(2): 103–113. Publisher Full Text
16. Hakeem A, Hashim MB, Al-Mothafar NS, et al.: Semi-Small Compressible Modules and Semi-Small Retractable Modules. Ibn Al-Haitham J. Pure Appl. Sci. 2023; 36(4): 407–413. Publisher Full Text

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¹ Mathematics, University of Baghdad College of Education for Pure Science Ibn Al-Haitham, Baghdad, Baghdad Governorate, 31001, Iraq
² Mathematics, University of Baghdad College of Education for Pure Science Ibn Al-Haitham, Baghdad, Baghdad Governorate, 10001, Iraq

Amal Madhi Rashid
Roles: Writing – Review & Editing

Buthyna Najad Shihab
Roles: Writing – Original Draft Preparation

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No competing interests were disclosed.

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The author(s) declared that no grants were involved in supporting this work.

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Madhi Rashid A and Najad Shihab B. S-Pseudo Bounded Radical of Submodules [version 1; peer review: awaiting peer review]. F1000Research 2025, 14:1378 (https://doi.org/10.12688/f1000research.172188.1)

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[1] 1. McCasland RL, Moore ME: On radicals of submodules. Communications in Algebra. 1991; 19(5): 1327–1341. Publisher Full Text

[2] 2. Abdul-Al-Kalik AJ: Semi–Bounded Modules. Baghdad Sci. J. 2012; 9(4): 20. Publisher Full Text

[3] 3. Al-Mothafar NS, Abdil-Khalik AJ: End-Ψ-Prime Submodules. Ibn Al-Haitham J. Pure Appl. Sci. 2016; 29(2). Reference Source

[4] 4. Murad MS, Shihab BN: Endo-restricted bounded submodules related to prime submodules and scalar modules. AIP Conference Proceedings. AIP Publishing LLC; 2025; Vol. 3264(1). Publisher Full Text

[5] 5. Al-Quraishi AD, Shihab BN: Pu-visible Submodule with their most prominent characteristics. IOP Conference Series: Materials Science and Engineering. IOP Publishing; 2020; Vol. 871(1). Reference Source

[6] 6. Hadi IM, Ali SNA-A, Shyaa FD: T-Essentially Coretractable and Weakly T-Essentially Coretractable Modules. Baghdad Sci. J. 2021; 18(1): 0156. Publisher Full Text

[7] 7. Ramadhan HA, Al Mothafar NS: Weakly Small Semiprime Submodules. J. Phys. Conf. Ser. 2021; 1879(3): 032128. IOP Publishing. Publisher Full Text

[8] 8. Abbas MR, Ahmed MA: Modules Whose St-Closed Submodules are Fully Invariant. Baghdad Sci. J. 2025; 22(3): 988–996. Publisher Full Text

[9] 9. Shihab BN: Scalar Reflexive Modules. Iraq: University of Baghdad; 2004. Doctoral dissertation, PhD. Thesis. Reference Source

[10] 10. Kasch F: Modules and rings. Vol. 17. .Academic Press; 1982. Publisher Full Text

[11] 11. McCasland RL, Moore ME, Smith PF: Modules with bounded spectra. Commun. Algebra. 1998; 26(10): 3403–3417. Publisher Full Text

[12] 12. McCasland RL, Moore ME: On radicals of submodules of finitely generated modules. Can. Math. Bull. 1986; 29(1): 37–39. Publisher Full Text

[13] 13. Faith C: Algebra II Ring Theory: Vol. 2: Ring Theory. Vol. 191. . Springer Science & Business Media; 2012. Publisher Full Text

[14] 14. McCasland RL, Moore ME: Prime submodules. Commun. Algebra. 1992; 20(6): 1803–1817. Publisher Full Text

[15] 15. Ajeel AS, Mohammad HK, Ali.: Approximaitly Prime Submodules and Some Related Concepts. Ibn Al-Haitham J. Pure Appl. Sci. 2019; 32(2): 103–113. Publisher Full Text

[16] 16. Hakeem A, Hashim MB, Al-Mothafar NS, et al.: Semi-Small Compressible Modules and Semi-Small Retractable Modules. Ibn Al-Haitham J. Pure Appl. Sci. 2023; 36(4): 407–413. Publisher Full Text

S-Pseudo Bounded Radical of Submodules

Abstract

Keywords

1. Introduction

2. S-Pseudo bounded submodules

Definition 2.1

Examples 2.2

Definition 2.3

Lemma 2.4

Proof:

Lemma 2.5

Proof:

Lemma 2.6

Proof:

3. Some results related to S-pseudo bounded radical of submodules

Definition 3.1

Definition 3.2

Remark 3.3

Proof:

Proposition 3.4

Proof:

Proposition 3.5

Proof:

Proposition 3.6

Proof:

Corollary 3.7

Proof:

Proposition 3.8

Proof:

Corollary 3.9

Proof:

Proposition 3.10

Proof:

Proposition 3.11

Proof:

Proposition 3.12

Proof:

Corollary 3.13

Proof:

Proposition 3.14

Proof:

Proposition 3.15

Proof:

Proposition 3.16

Proof:

Corollary 3.17

Proof:

Corollary 3.18

Proof:

Proposition 3.19

Proof:

Proposition 3.20

Proof:

Proposition 3.21

Proof:

Corollary 3.22

Proof:

Corollary 3.23

Proof:

Corollary 3.24

Proof:

4. Conclusion

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References

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