S-Pseudo Bounded Radical of Submodules

Definition 2.1

A proper submodule $\mathcal{\aleph }$ of an $\mathcal{F}-\mathit{\text{module}}\mathcal{M}$ is said to be S-pseudo bounded submodule if there exists $\phi \in S=\mathit{End}\left(\mathcal{M}\right)$ such that $\phi \left(x\right)\in \mathcal{\aleph }$ , for all $x\in \mathcal{M}$ implies that $\sqrt{{\mathit{ann}}_{\mathcal{F}}\left(\mathcal{\aleph }\right)}=\sqrt{{\mathit{ann}}_{\mathcal{F}}\phi \left(x\right)},$ for some $x\in \mathcal{M}$ , where $\mathit{End}\left(\mathcal{M}\right)$ is the set of endomorphism maps over an $\mathcal{F}-\mathit{\text{module}}$ $\mathcal{M}.$

Examples 1

Definition 2.2

An $\mathcal{F}-\mathit{\text{module}}\mathcal{M}$ is said to be S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ if every proper submodule of $\mathcal{M}$ is S-PS.B $\mathcal{F}-$ submodule.

Lemma 2.3

If $\mathcal{\aleph }$ is a prime submodule of a scalar $\mathcal{F}-\mathit{\text{module}}$ $\mathcal{M},$ then $\mathcal{\aleph }$ is S-PS.B. $\mathcal{F}-\mathit{\text{submodule}}.$

Proof:

Let $\phi \in \mathit{End}\left(\mathcal{M}\right),x\in \mathcal{M}.$ Since $\mathcal{M}$ is scalar $\mathcal{F}-\mathit{\text{module}}$ , then for all $\phi \in \mathit{End}\left(\mathcal{M}\right),\exists 0\ne r\in \mathcal{F}$ such that $\phi \left(x\right)=\mathit{rx},\mathit{\text{for}}\mathit{all}x\in \mathcal{M}.$ Now, we must show that $\sqrt{{\mathit{ann}}_{\mathcal{F}}\phi \left(x\right)}=\sqrt{{\mathit{ann}}_{\mathcal{F}}\mathcal{\aleph }}.$ Let $a\in \sqrt{{\mathit{ann}}_{\mathcal{F}}\phi \left(x\right)},x\in \mathcal{M}.$ Then $a^n\in {\mathit{ann}}_{\mathcal{F}}\phi \left(x\right),\mathit{\text{for some}}n\in {\mathbb{Z}}^{+}$ and $a^n.\phi \left(x\right)=0,$ so $a^n.\mathit{rx}=0$ implies that $a^n\in {\mathit{ann}}_{\mathcal{F}}\left(\mathit{rx}\right),\mathit{\text{for}}\mathit{all}\mathit{rx}\in \mathcal{\aleph }.$ Since $\mathcal{\aleph }$ is a prime submodule, then either $x\in \mathcal{\aleph }\mathit{\text{or}}r\mathcal{M}\subseteq \mathcal{\aleph }$ so that $a^n\in {\mathit{ann}}_{\mathcal{F}}\mathcal{\aleph }$ and $a\in \sqrt{{\mathit{ann}}_{\mathcal{F}}\mathcal{\aleph }}.$ Similarly, $\sqrt{{\mathit{ann}}_{\mathcal{F}}\mathcal{\aleph }}\subseteq \sqrt{{\mathit{ann}}_{\mathcal{F}}\phi \left(x\right)}.$ Hence, $\mathcal{\aleph }$ is S-PS.B. $\mathcal{F}-\mathit{\text{submodule}}.$

Lemma 2.4

If $\mathcal{\aleph }$ is S-PS.B. $\mathcal{F}-\mathit{\text{submodule}}$ of a scalar $\mathcal{F}-\mathit{\text{module}}$ $\mathcal{M},$ then $\mathcal{\aleph }$ is S-prime submodule.

Proof:

Let $\phi \in \mathit{End}\left(\mathcal{M}\right)$ and define $\phi :\mathcal{M}$ ⟶ $\mathcal{M}$ be $\phi \left(x\right)=\mathit{rx}$ ,∀x ∈ $\mathcal{M}$ . Let x ∉ ℵ, then we have to prove φ( $\mathcal{M}$ ) ⊆ ℵ. Since $\mathcal{M}$ is scalar and by definition of S-PS.B submodule, we have φ(x) ∈ ℵ,∀x ∈ $\mathcal{M}$ , which means that φ( $\mathcal{M}$ ) ⊆ ℵ.

Lemma 2.5

If $\mathcal{\aleph }$ is a submodule of a scalar $\mathcal{F}-\mathit{\text{module}}$ $\mathcal{M},$ then $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right)=\mathit{rad}\left(\mathcal{\aleph }\right).$

Proof:

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

Definition 3.1

A S-pseudo bounded radical of a submodule $\mathcal{\aleph }$ of an $\mathcal{F}-$ module $\mathcal{M}$ is the intersection of all S-pseudo bounded submodules of $\mathcal{M}$ containing $\mathcal{\aleph }$ and denotes by $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right).$

If there is no S-PS.B. submodule of $\mathcal{M}$ contains $\mathcal{\aleph }$ , then $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right)=\mathcal{M}.$

Definition 3.2

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

Note that S-pseudo bounded radical submodule is proper submodule of $\mathcal{M}.$

Remark 3.3

If $\mathcal{\aleph }$ is S-PS.B. submodule of an $\mathcal{F}-$ module $\mathcal{M},$ then $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right)$ is S-PS.B. submodule too.

Proof:

Using the Definition (3.1), we have $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right)={\cap }_{j\in J}\{k_j:k_j\mathit{\text{is}}S-\mathit{PS}.B.\mathit{\text{submodule}},\mathcal{\aleph }\subseteq k_j\},$ for $j\in J$ and $J$ be an index set. Then by using induction and properties of S-PS.B. submodule we get $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right)$ is S-PS.B. submodule.

Proposition 3.4

If $\mathcal{\aleph },\mathcal{K}$ are two submodules of an $\mathcal{F}-$ module $\mathcal{M}$ . Then

Proof:

Proposition 3.5

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

Proof:

Hence $\sum _{j\in J}S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left({\mathcal{\aleph }}_j\right)\subseteq S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\sum _{j\in J}{\mathcal{\aleph }}_j\right).$

Since $\sum _{j\in J}{\mathcal{\aleph }}_j\subseteq \sum _{j\in J}S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left({\mathcal{\aleph }}_j\right),$ then it is easy to show that $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\sum _{j\in J}{\mathcal{\aleph }}_j\right)=S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}(\sum _{j\in J}S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left({\mathcal{\aleph }}_j\right)).$

Proposition 3.6

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

Proof:

Assume that $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right)=\mathcal{M}$ and $\mathcal{\aleph }\ne \mathcal{M}$ . Since $\mathcal{M}$ is a finitely generated module, then there exists a maximal submodule $\mathcal{K}$ of $\mathcal{M}$ such that $\mathcal{\aleph }\subseteq \mathcal{K}$ and hence $\mathcal{K}$ is prime. Since $\mathcal{M}$ is a finitely generated multiplication $\mathcal{F}-$ module, then $\mathcal{M}$ is scalar¹¹ and thus $\mathcal{K}$ is S-PS.B. submodule Lemma (2.34). Thus $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right)\subseteq \mathcal{K}$ so that $\mathcal{M}=\mathcal{K}$ which is contradicting the assumption. If $\mathcal{\aleph }=\mathcal{M},$ the other side is clear.

Corollary 3.7

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

Proof:

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

Hence by previous proposition, $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right)+S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{K}\right)=\mathcal{M}.$

Proposition 3.8

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

Proof:

Since $\mathcal{\aleph }\subseteq \mathcal{\aleph }+\sqrt{(\mathcal{\aleph }:\mathcal{M})}\mathcal{M},$ then $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right)\subseteq S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}(\mathcal{\aleph }+\sqrt{(\mathcal{\aleph }:\mathcal{M})}\mathcal{M})$ . Assume that $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right)={\cap }_{j\in J}{V}_j$ where $V_j$ is S-PS.B. submodule of $\mathcal{M}$ containing $\mathcal{\aleph }.$ Since $\mathcal{M}$ is a finitely generated multiplication module, then $\mathcal{M}$ is scalar¹¹ and thus $\mathcal{\aleph }$ is prime Lemma (2.4) so that $V_j$ is prime submodule and $(V_j:\mathcal{M})$ is prime ideal which implies that $\sqrt{(\mathcal{\aleph }:\mathcal{M})}\subseteq (V_j:\mathcal{M})$ and hence $\mathcal{\aleph }+\sqrt{(\mathcal{\aleph }:\mathcal{M})}\mathcal{M}\subseteq V_j.$ Therefore $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}(\mathcal{\aleph }+\sqrt{(\mathcal{\aleph }:\mathcal{M})}\mathcal{M})\subseteq {\cap }_{j\in J}{V}_j=S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right)$ . Hence $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right)=S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}(\mathcal{\aleph }+\sqrt{(\mathcal{\aleph }:\mathcal{M})}\mathcal{M}).$

Corollary 3.9

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

Proof:

Since $\mathcal{\aleph }+\sqrt{(\mathcal{\aleph }:\mathcal{M})}\mathcal{M}\subseteq S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}(\mathcal{\aleph }+\sqrt{(\mathcal{\aleph }:\mathcal{M})}\mathcal{M})=S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right),$ by previous proposition and since $\sqrt{(\mathcal{\aleph }:\mathcal{M})}\subseteq ((\mathcal{\aleph }+\sqrt{(\mathcal{\aleph }:\mathcal{M})}\mathcal{M}):\mathcal{M}),$ then $((\mathcal{\aleph }+\sqrt{(\mathcal{\aleph }:\mathcal{M})}\mathcal{M}):\mathcal{M})=\sqrt{(\mathcal{\aleph }:\mathcal{M})}$ or $((\mathcal{\aleph }+\sqrt{(\mathcal{\aleph }:\mathcal{M})}\mathcal{M}):\mathcal{M})=\mathcal{F}.$ If $((\mathcal{\aleph }+\sqrt{(\mathcal{\aleph }:\mathcal{M})}\mathcal{M}):\mathcal{M})=\mathcal{F},$ then $\mathcal{FM}=\mathcal{M}\subseteq \mathcal{\aleph }+\sqrt{(\mathcal{\aleph }:\mathcal{M})}\mathcal{M}\subseteq S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right)$ which implies that $\mathcal{M}=S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right).$ Since $\mathcal{M}$ is finitely generated, then $\mathcal{M}=\mathcal{\aleph }$ by Proposition (3.6) which is contradiction. So $((\mathcal{\aleph }+\sqrt{(\mathcal{\aleph }:\mathcal{M})}\mathcal{M}):\mathcal{M})=\sqrt{(\mathcal{\aleph }:\mathcal{M})}$ , therefore $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right)=\mathcal{\aleph }+\sqrt{(\mathcal{\aleph }:\mathcal{M})}\mathcal{M}.$

Proposition 3.10

If $\mathcal{\aleph }$ is a submodule of $\mathcal{M}$ and $g:\mathcal{M}\to {\mathcal{M}}^{\prime }$ is an epimorphism such that $\mathit{Ker}\left(g\right)\subseteq \mathcal{\aleph }$ , then

Proof:

Proposition 3.11

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

Proof:

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

Proposition 3.12

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

Proof:

Clearly, $(\mathcal{\aleph }\cap \mathcal{K}:\mathcal{M})=(\mathcal{\aleph }:\mathcal{M})\cap (\mathcal{K}:\mathcal{M})=\sqrt{(\mathcal{\aleph }:\mathcal{M})}\cap \sqrt{(\mathcal{K}:\mathcal{M})}=\sqrt{(\mathcal{\aleph }\cap \mathcal{K}:\mathcal{M})}$ . Since $\mathcal{M}$ is finitely generated, then by Ref. 13 Theorem 4.4, we have $\sqrt{(\mathcal{\aleph }\cap \mathcal{K}:\mathcal{M})}=(S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}(\mathcal{\aleph }\cap \mathcal{K}):\mathcal{M}).$ Thus $(\mathcal{\aleph }\cap \mathcal{K}:\mathcal{M})=(S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}(\mathcal{\aleph }\cap \mathcal{K}):\mathcal{M}).$

Corollary 3.13

If $\mathcal{\aleph },\mathcal{K}$ are two submodules of a multiplication finitely generated $\mathcal{F}-$ module $\mathcal{M}$ , then

Proof:

From Ref. 13 Theorem 4.4, we get $(S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}(\mathcal{\aleph }\cap \mathcal{K}):\mathcal{M})=\sqrt{(\mathcal{\aleph }\cap \mathcal{K}:\mathcal{M})}=\sqrt{(\mathcal{\aleph }:\mathcal{M})}\cap \sqrt{(\mathcal{K}:\mathcal{M})}=(S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right):\mathcal{M})\cap (S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{K}\right):\mathcal{M})=(S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right)\cap S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{K}\right):\mathcal{M}).$

Proposition 3.14

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

Proof:

From, Ref. 14, we have $\sqrt{(\mathcal{\aleph }:\mathcal{M})}\mathcal{M}\subseteq S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right).$ Now, let $L$ be an S-PS.B. submodule of $\mathcal{M}$ such that $\mathcal{\aleph }\subseteq L,$ then $(\mathcal{\aleph }:\mathcal{M})\subseteq (L:\mathcal{M}).$ Since $\mathcal{M}$ is a finitely generated multiplication module, then $\mathcal{M}$ is scalar¹¹ and hence $\mathcal{\aleph }$ is prime Lemma (2.4) and $L$ is prime submodule so that $(L:\mathcal{M})$ is prime ideal implies that $\sqrt{(\mathcal{\aleph }:\mathcal{M})}\subseteq (L:\mathcal{M}).$ Therefore $\sqrt{(\mathcal{\aleph }:\mathcal{M})}\mathcal{M}\subseteq (L:\mathcal{M})\mathcal{M}\subseteq L.$ Since $L$ is an arbitrary S-PS.B. submodule containing $\mathcal{\aleph },$ then $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right)\subseteq \sqrt{(\mathcal{\aleph }:\mathcal{M})}\mathcal{M}.$ Thus $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right)=\sqrt{(\mathcal{\aleph }:\mathcal{M})}\mathcal{M}.$

Proposition 3.15

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

Proof:

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

Proposition 3.16

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

Proof:

By, Ref. 15, since $\mathcal{M}$ is finitely generated multiplication module, then we have $\sqrt{(I\mathcal{M}:\mathcal{M})}=\sqrt{I+(0:\mathcal{M})}$ for all ideal I of $\mathcal{F}$ . Consider the finitely generated $\mathcal{F}-$ module $\mathcal{M}/\mathcal{K}$ and the ideal $(\mathcal{\aleph }:\mathcal{M})$ instead of $\mathcal{M}$ and I respectively, then

Corollary 3.17

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

Proof:

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

Corollary 3.18

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

Proof:

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

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

Therefore $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left((S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(I\mathcal{M}\right):\mathcal{M})\mathcal{M}\right)=S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(I\mathcal{M}\right).$

Proposition 3.19

Let $\mathcal{\aleph }$ and $\mathcal{K}$ be two submodules of a scalar $\mathcal{F}-$ module $\mathcal{M}$ , then

Proof:

Therefore $\mathcal{F}=(S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right):\mathcal{M})+(S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{K}\right):\mathcal{M}).$ Thus the required result is clear.

Proposition 3.20

If ${\mathcal{\aleph }}_1,{\mathcal{\aleph }}_2$ are two submodules of an $\mathcal{F}-$ module $\mathcal{M}$ and every S-PS.B. submodule $\mathcal{L}$ of $\mathcal{M}$ is completely irreducible such that ${\mathcal{\aleph }}_1\cap {\mathcal{\aleph }}_2\subseteq \mathcal{L}$ , then $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}({\mathcal{\aleph }}_1\cap {\mathcal{\aleph }}_2)=S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left({\mathcal{\aleph }}_1\right)\cap S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left({\mathcal{\aleph }}_2\right).$

Proof:

By Proposition (3.5), we have $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}({\mathcal{\aleph }}_1\cap {\mathcal{\aleph }}_2)\subseteq S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left({\mathcal{\aleph }}_1\right)\cap S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left({\mathcal{\aleph }}_2\right).$ Let $\mathcal{L}$ be an S-PS.B. submodule of $\mathcal{M}$ such that ${\mathcal{\aleph }}_1\cap {\mathcal{\aleph }}_2\subseteq \mathcal{L}.$ Since every S-PS.B. submodule of $\mathcal{M}$ that contains ${\mathcal{\aleph }}_1\cap {\mathcal{\aleph }}_2$ is completely irreducible, then either ${\mathcal{\aleph }}_1\subseteq \mathcal{L}$ or ${\mathcal{\aleph }}_2\subseteq \mathcal{L}$ . Thus $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left({\mathcal{\aleph }}_1\right)\subseteq \mathcal{L}$ or $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left({\mathcal{\aleph }}_2\right)\subseteq \mathcal{L}$ . Therefore $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left({\mathcal{\aleph }}_1\right)\cap S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left({\mathcal{\aleph }}_2\right)\subseteq \mathcal{L}$ for any $\mathcal{L}$ and $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}({\mathcal{\aleph }}_1\cap {\mathcal{\aleph }}_2).$ Therefore $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left({\mathcal{\aleph }}_1\right)\cap S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left({\mathcal{\aleph }}_2\right)\subseteq S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}({\mathcal{\aleph }}_1\cap {\mathcal{\aleph }}_2)$ .

Hence $S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}({\mathcal{\aleph }}_1\cap {\mathcal{\aleph }}_2)=S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left({\mathcal{\aleph }}_1\right)\cap S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left({\mathcal{\aleph }}_2\right).$

Definition 3.21

Now, we give some notations:

$S-{\mathit{rad}}_{\mathcal{M}}^{\mathit{PS}.B.}\left(\mathcal{\aleph }\right)={\cap }_{k\in {\mathit{\text{Spec}}}^{S-\mathit{PS}.}}k,$ where ${\mathit{\text{Spec}}}^{S-\mathit{PS}.}$ denoted all S-PS.B. submodules of $\mathcal{M}.$

Proposition 3.22

Let $\mathcal{M}$ be an $\mathcal{F}-$ module. Then

Proof:

Hence $G^{S-\mathit{PS}.}\left({\mathcal{\aleph }}_1\right)\cap G^{S-\mathit{PS}.}\left({\mathcal{\aleph }}_2\right)=G^{S-\mathit{PS}.}({\mathcal{\aleph }}_1+{\mathcal{\aleph }}_2).$