New Class of S-Pseudo Bounded Modules With Some Related Concepts

Definision 2.1

¹A $\mathcal{F}-\mathit{\text{module}}$ $\mathcal{M}$ is bounded module if there exists an element $x\in \mathcal{M}$ such that ${\mathit{ann}}_{\mathcal{F}}\left(\mathcal{M}\right)={\mathit{ann}}_{\mathcal{F}}\left(x\right).$

Definision 2.2

⁶A $\mathcal{F}-\mathit{\text{module}}$ $\mathcal{M}$ is said to be scalar if for every $\phi \in \mathit{End}\left(\mathcal{M}\right)$ there exists $r\in \mathcal{F}$ such that $\phi \left(x\right)=\mathit{rx},\forall x\in \mathcal{M}.$

Definision 2.3

⁷A $\mathcal{F}-\mathit{\text{module}}$ $\mathcal{M}$ is called monoform if $\forall \mathcal{\aleph }\ne 0$ of $\mathcal{M}$ is dense where a submodule $\mathcal{\aleph }$ of $\mathcal{M}$ is dense if for any $x,y\in \mathcal{M},x\ne 0$ there exists $t\in \mathcal{F}$ such that $\mathit{ty}\in \mathcal{\aleph }$ and $\mathit{tx}\ne 0$ .

Equivalently, A $\mathcal{F}-\mathit{\text{module}}$ $\mathcal{M}$ is said to be monoform if $\forall \mathcal{\aleph }\ne 0$ of $\mathcal{M}$ and $\forall 0\ne \phi \in \mathit{Hom}(\mathcal{\aleph },\mathcal{M})$ then $\phi$ is monomorphism.⁷ Also , if $\mathcal{M}$ is monoform module, then it is uniform and prime and hence we deduce that ${\mathit{ann}}_{\mathcal{F}}\left(\mathcal{M}\right)$ is prime ideal of $\mathcal{F}.$

Definision 2.4

⁸A $\mathcal{F}-\mathit{\text{module}}$ $\mathcal{M}$ is said to be finitely annihilated $\mathcal{F}-\mathit{\text{module}}$ if there exists a finitely generated submodule $\mathcal{\aleph }$ of $\mathcal{M}$ such that ${\mathit{ann}}_{\mathcal{F}}\left(\mathcal{M}\right)={\mathit{ann}}_{\mathcal{F}}\left(\mathcal{\aleph }\right).$

Definision 2.5

⁹A $\mathcal{F}-\mathit{\text{module}}$ $\mathcal{M}$ is called quasi-Dedekind if $\forall \mathcal{\aleph }\ne 0$ of $\mathcal{M}$ is quasi-invertible where $\mathcal{\aleph }$ is quasi-invertible if $\mathit{Hom}(\mathcal{M}/\mathcal{\aleph },\mathcal{M})=0.$

Equivalently, a $\mathcal{F}-\mathit{\text{module}}$ $\mathcal{M}$ is said to be quasi-Dedekind if for each non-zero endomorphism of $\mathcal{M}$ is a $\mathcal{F}-\mathit{\text{monomorphism}}.$

Definision 2.6

¹⁰ A $\mathcal{F}-\mathit{\text{module}}$ $\mathcal{M}$ is called polyform if every essential submodule $\mathcal{\aleph }$ of $\mathcal{M}$ is dense. Note that every monoform is polyform.

Definision 2.7

¹⁰A $\mathcal{F}-\mathit{\text{module}}$ $\mathcal{M}$ is said to be fully polyform if every P-essential submodule of $\mathcal{M}$ is dense where $\mathcal{\aleph }$ is called P-essential if every pure submodule $k$ of $\mathcal{M}$ such that $\mathcal{\aleph }\cap k=\left(0\right)$ implies that $k=\left(0\right).$

Definision 2.8

¹⁰A $\mathcal{F}-\mathit{\text{module}}$ $\mathcal{M}$ is said to be fully retractable if for every non-zero submodule $\mathcal{\aleph }$ of $\mathcal{M}$ and each non-zero homomorphism $f\in \mathit{Hom}(\mathcal{\aleph },\mathcal{M})$ implies that $\mathit{Hom}(\mathcal{\aleph },\mathcal{M})\ne 0.$

Definision 2.9

¹¹A $\mathcal{F}-\mathit{\text{module}}$ $\mathcal{M}$ is called coprime if ${\mathit{ann}}_{\mathcal{F}}\left(\mathcal{M}\right)={\mathit{ann}}_{\mathcal{F}}(\mathcal{M}/\mathcal{\aleph })$ for every proper submodule $\mathcal{\aleph }$ of $\mathcal{M}.$

Corollary 2.10

⁶If $\mathcal{M}$ is a multiplication finitely generated $\mathcal{F}-\mathit{\text{module}},$ then $\mathcal{M}$ is a scalar $\mathcal{F}-\mathit{\text{module}}.$

Remark 2.11

⁶Let $\mathcal{M}$ be an injective scalar $\mathcal{F}-\mathit{\text{module}}$ . Then $\mathcal{\aleph }$ is a scalar submodule of $\mathcal{M}.$

Remark 2.12

¹²Every finitely generated $\mathcal{F}-\mathit{\text{module}}$ is finitely annihilated.

Proposition 2.13

¹²Let $\mathcal{M}$ be a multiplication $\mathcal{F}-\mathit{\text{module}}\mathit{.}$ Then $\mathcal{M}$ is finitely generated if and only if $\mathcal{M}$ is finitely annihilated.

Proposition 2.14

⁶Let $\mathcal{M}$ be a scalar torsion-free $\mathcal{F}-\mathit{\text{module}}$ with ( $\mathcal{F}$ is an integral domain). Then every $\phi \in \mathit{End}\left(\mathcal{M}\right)$ is $\mathcal{F}-\mathit{\text{monomorphism}}.$

Proposition 2.15

¹³Let $\mathcal{M}$ be a quasi-Dedekind retractable $\mathcal{F}-\mathit{\text{module}}.$ If every $0\ne \phi \in \mathit{End}\left(\mathcal{M}\right)$ is a monomorphism, then $\mathcal{M}$ is compressible module.

Proposition 2.16

¹⁴Let $\mathcal{M}$ be a retractable $\mathcal{F}-\mathit{\text{module}}$ such that End( $\mathcal{M})$ is a domain. Then $\mathcal{M}$ is critically compressible module if and only if it is polyform.

Proposition 2.17

¹⁴Let $\mathcal{M}$ be a retractable $\mathcal{F}-\mathit{\text{module}}.$ Then $\mathcal{M}$ is critically compressible module if and only if every non-zero partial endomorphism of $\mathcal{M}$ is monomorphism.

Proposition 2.18

¹⁴Let $\mathcal{M}$ be a fully retractable $\mathcal{F}-\mathit{\text{module}}$ with End( $\mathcal{M})$ is a domain. Then $\mathcal{M}$ is polyform.

Definition 2.19

¹⁵A $\mathcal{F}-\mathit{\text{module}}$ $\mathcal{M}$ is called Rickart $\mathcal{F}-\mathit{\text{module}}$ if and only if $\mathit{\text{Kerφ}}$ is a direct summand of $\mathcal{M}.$

Proposition 2.20

¹⁵If $\mathcal{M}$ is an injective prime $\mathcal{F}-\mathit{\text{module}},$ then $\mathcal{M}$ is Rickart module.

Definition 2.21

¹⁶A $\mathcal{F}-\mathit{\text{module}}$ $\mathcal{M}$ is called N-Rickart if for every homomorphism $g:\mathcal{M}\to \mathcal{\aleph },\mathit{ker}g$ is a summand of $\mathcal{M}.$

Corollary 2.22

¹⁴If $\mathcal{M}$ is uniform module, then $\mathcal{M}$ is fully polyform if and only if $\mathcal{M}$ is monoform module.

Corollary 2.23

¹⁷If $\mathcal{M}$ is finitely generated $\mathcal{F}-\mathit{\text{module}}$ , then $\mathcal{M}$ is compressible $\mathcal{F}-\mathit{\text{module}}$ if and only if $\mathcal{M}$ is uniform prime module.

Definition 3.1

A proper submodule $\mathcal{\aleph }$ of a $\mathcal{F}-\mathit{\text{module}}\mathcal{M}$ is called S-pseudo bounded $\mathcal{F}-\mathit{\text{submodule}}$ (symbolically S-PS.B. $\mathcal{F}$ -submodule) if there exists $\phi \left(x\right)\in S=\mathit{End}\left(\mathcal{M}\right)$ such that $\phi \left(x\right)\in \mathcal{\aleph }$ for some $x\in \mathcal{M}$ implies that $\sqrt{{\mathit{ann}}_{\mathcal{F}}\left(\mathcal{\aleph }\right)}=\sqrt{{\mathit{ann}}_{\mathcal{F}}\left(\phi \left(x\right)\right)}.$

Examples 3.2

1- Let $\mathcal{M}={\mathbb{Z}}_2⨁{\mathbb{Z}}_4$ as a ${\mathbb{Z}}_8-\mathit{\text{module}}$ and $\mathcal{\aleph }={\mathbb{Z}}_2⨁⟨\overline{2}⟩.$ Define $\phi :\mathcal{M}⟶\mathcal{M}$ by $\phi (\overline{a},\overline{b})=(\overline{a},\overline{2}),\forall (\overline{a},\overline{b})\in \mathcal{M}.$ If $(\overline{0},\overline{2})\in \mathcal{M}$ then $\phi (\overline{0},\overline{2})=(\overline{0},\overline{2})\in \mathcal{\aleph }$ . Therefore, $\sqrt{{\mathit{ann}}_{{\mathbb{Z}}_8}\left(\mathcal{\aleph }\right)}=\sqrt{{\mathit{ann}}_{{\mathbb{Z}}_8}\phi (\overline{0},\overline{2})},$ and hence $\mathcal{\aleph }$ is S-PS.B. ${\mathbb{Z}}_8-$ submodule of $\mathcal{M}.$

2- Consider $\mathcal{M}=\mathbb{Z}⨁{\mathbb{Z}}_2$ as a $\mathbb{Z}-\mathit{\text{module}}$ and $\mathcal{\aleph }=2\mathbb{Z}⨁⟨\overline{0}⟩$ , then there exists $\phi :\mathcal{M}⟶\mathcal{M}$ defined by $\phi (a,\overline{b})=(a,\overline{0}),\forall (a,\overline{b})\in \mathcal{M}$ clearly, $\phi \in \mathit{End}\left(\mathcal{M}\right).$ Now, suppose that $x=(2,\overline{1})\in \mathcal{M}$ . Sequently, $\phi \left(x\right)=\phi (2,\overline{1})=(2,\overline{0})\in \mathcal{\aleph }$ . Thus $\sqrt{{\mathit{ann}}_{\mathbb{Z}}\left(\mathcal{\aleph }\right)}=\sqrt{{\mathit{ann}}_{\mathbb{Z}}\phi (2,\overline{1})}=\sqrt{⟨\overline{0}⟩}.$ Therefore, $\mathcal{\aleph }$ is S-PS.B. $\mathbb{Z}-$ submodule of $\mathcal{M}.$

3- Suppose that ${\mathbb{Z}}_6$ as a $\mathbb{Z}-\mathit{\text{module}}$ and $\mathcal{M}={\mathbb{Z}}_6,\mathcal{\aleph }=⟨\overline{3}⟩$ . An endomorphism $\phi :\mathcal{M}⟶\mathcal{M}$ defined by $\phi \left(\overline{x}\right)=\overline{0},\forall \overline{x}\in {\mathbb{Z}}_6,\phi \left(\overline{x}\right)\in \mathcal{\aleph }$ if $\overline{x}=\overline{2}\in {\mathbb{Z}}_6$ , we have $\mathbb{Z}=\sqrt{{\mathit{ann}}_{\mathbb{Z}}\phi \left(\overline{2}\right)}=\sqrt{{\mathit{ann}}_{\mathbb{Z}}\left(\overline{0}\right)}\ne \sqrt{{\mathit{ann}}_{\mathbb{Z}}⟨\overline{3}⟩}=2\mathbb{Z},\mathcal{\aleph }$ is not S-PS.B. $\mathbb{Z}-$ submodule.

We establish that every Endo-R.B. is S.PS.B. submodule but the converse is not necessary true in general. If $\mathcal{\aleph }$ is an Endo-R.B. then there exists $\phi \in \mathit{End}\left(\mathcal{M}\right)$ such that $\phi \left(x\right)\in \mathcal{\aleph },\mathit{\text{for some}}x\in \mathcal{M}$ implies that ${\mathit{ann}}_{\mathcal{F}}\left(\mathcal{\aleph }\right)={\mathit{ann}}_{\mathcal{F}}\phi \left(x\right)$ by Ref. 7. But $\sqrt{{\mathit{ann}}_{\mathcal{F}}\phi \left(x\right)}=\{r\in \mathcal{F}:r^n\in {\mathit{ann}}_{\mathcal{F}}\phi \left(x\right),n\in {\mathbb{Z}}^{+}\}$ and from above equality we get $\sqrt{{\mathit{ann}}_{\mathcal{F}}\phi \left(x\right)}=\sqrt{{\mathit{ann}}_{\mathcal{F}}\left(\mathcal{\aleph }\right).}$ Hence, $\mathcal{\aleph }$ is S-PS.B. $\mathcal{F}-$ submodule. But the converse is not true in general for example:

Let $\mathcal{M}={\mathbb{Z}}_2⨁{\mathbb{Z}}_4$ as $\mathbb{Z}-\mathit{\text{module}}$ . Define $\phi :\mathcal{M}⟶\mathcal{M}$ as: $\phi (\overline{a},\overline{b})=(\overline{0},\overline{b}),\forall (\overline{a},\overline{b})\in \mathcal{M}$ if we take $\mathcal{\aleph }=⟨\overline{0}⟩⨁{\mathbb{Z}}_4,(\overline{1},\overline{2})\in \mathcal{M}$ then $\phi (\overline{1},\overline{2})=(\overline{0},\overline{2})\in \mathcal{\aleph },$ implies that $\sqrt{{\mathit{ann}}_{\mathbb{Z}}\phi (\overline{1},\overline{2})}=\sqrt{{\mathit{ann}}_{\mathbb{Z}}(\overline{0},\overline{2})}$ = $\sqrt{2\mathbb{Z}}=2\mathbb{Z}$ and $\sqrt{{\mathit{ann}}_{\mathbb{Z}}(⟨\overline{0}⟩⨁{\mathbb{Z}}_4)}=\sqrt{4\mathbb{Z}}=2\mathbb{Z}$ .Thus $\mathcal{\aleph }$ is S-PS.B. $\mathbb{Z}-$ submodule, but ${\mathit{ann}}_{\mathbb{Z}}(⟨\overline{0}⟩⨁{\mathbb{Z}}_4)=4\mathbb{Z}$ is not equal to ${\mathit{ann}}_{\mathbb{Z}}\phi (\overline{1},\overline{2})={\mathit{ann}}_{\mathbb{Z}}(\overline{0},\overline{2})=2\mathbb{Z}$ . Therefore $\mathcal{\aleph }$ is not Endo-R.B. $\mathbb{Z}-$ submodule.

Definition 3.3

A $\mathcal{F}-\mathit{\text{module}}\mathcal{M}$ 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.

Examples 3.4

Proposition 4.1

If $\mathcal{M}$ is a multiplication torsion-free $\mathcal{F}-\mathit{\text{module}}$ with ( $\mathcal{F}$ is an integral domain), then the following statements are equivalent

$\left(\mathit{i}\right)$ $\mathcal{M}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}}.$

$\left(\mathit{ii}\right)$ $\mathcal{M}$ is monoform module.

Proof:

$\left(\mathit{i}\right)\mathit{⟹}\left(\mathit{ii}\right)$ Suppose that $\mathcal{M}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}}.$ Since $\mathcal{M}$ is a multiplication torsion-free, then $\mathcal{M}$ is finitely generated, by Remark (2.12) and Proposition (2.13). Therefore $\mathcal{M}$ is scalar by Corollary (2.10). By Proposition (2.14), every $\phi \in \mathit{End}\left(\mathcal{M}\right)$ is monomorphism. Suppose that $\phi :\mathcal{M}⟶\mathcal{M}$ be an $\mathcal{F}-\mathit{\text{homomorphism}}$ and $i:\mathcal{\aleph }⟶\mathcal{M}$ is inclusion map, then $\phi \circ i:\mathcal{\aleph }⟶\mathcal{M}$ is monomorphism and $\mathcal{M}$ is monoform module.

$\left(\mathit{ii}\right)\mathit{⟹}\left(\mathit{i}\right)$ Let $\mathcal{M}$ is monoform module implies that $\mathcal{M}$ is uniform prime module and every submodule $\mathcal{\aleph }\ne 0$ of $\mathcal{M}$ is dense. Suppose that $\phi \in \mathit{End}\left(\mathcal{M}\right)$ such that $\phi :\mathcal{M}⟶\mathcal{M}$ and $\phi \left(x\right)=\mathit{tx},0\ne x\in \mathcal{M},0\ne t\in \mathcal{F}.$ Since $\mathcal{M}$ is uniform, then $\mathit{tx}\in \mathcal{\aleph }.$ Hence ${\mathit{ann}}_{\mathcal{F}}\left(\mathcal{\aleph }\right)\subseteq {\mathit{ann}}_{\mathcal{F}}\phi \left(x\right)$ and thus $\sqrt{{\mathit{ann}}_{\mathcal{F}}\left(\mathcal{\aleph }\right)}\subseteq \sqrt{{\mathit{ann}}_{\mathcal{F}}\phi \left(x\right)}.$ Let $a\in \sqrt{{\mathit{ann}}_{\mathcal{F}}\phi \left(x\right)}$ implies that $a^n.\phi \left(x\right)=0,\mathit{\text{for some}}n\in {\mathbb{Z}}^{+}$ which implies $a^n.\left(\mathit{tx}\right)=0$ and $a^{n}t\left(x\right)=0.$ Since $\mathcal{\aleph }$ is dense submodule and ${\mathit{ann}}_{\mathcal{F}}\left(\mathcal{M}\right)$ is prime ideal of $\mathcal{F},$ then $t\notin {\mathit{ann}}_{\mathcal{F}}\left(x\right)$ and $a^n\in {\mathit{ann}}_{\mathcal{F}}\left(x\right)={\mathit{ann}}_{\mathcal{F}}\left(\mathcal{M}\right)\subseteq {\mathit{ann}}_{\mathcal{F}}\left(\mathcal{\aleph }\right).$ Thus $\sqrt{{\mathit{ann}}_{\mathcal{F}}\left(\mathcal{\aleph }\right)}=\sqrt{{\mathit{ann}}_{\mathcal{F}}\phi \left(x\right)}$ and $\mathcal{M}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}}.$

However, the condition torsion-free is suffice to prove $\mathcal{M}$ is monoform module.

Corollary 4.2

Let $\mathcal{M}$ be a torsion-free S-PS.B. $\mathcal{F}-\mathit{\text{module}}\mathit{.}$ Then $\mathcal{M}$ is monoform module.

Proof:

Suppose that $\mathcal{\aleph }$ be any non-zero submodule of $\mathcal{M}.$ Since $\mathcal{M}$ is a torsion-free module, then $\forall x\in \mathcal{M}$ there exists $0\ne t\in \mathcal{F}$ such that $\mathit{tx}=0,$ we obtain $x=0.$ Since $\mathcal{M}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ , then $\exists \phi \in \mathit{End}\left(\mathcal{M}\right)$ such that $\phi :\mathcal{M}⟶\mathcal{M}$ and define as $\phi \left(x\right)=\mathit{tx},x\in \mathcal{M}$ and hence $\mathit{tx}\in \mathcal{\aleph }.$ Now, assume that $0\ne x\in \mathcal{M},$ we have to show that $\mathit{tx}\ne 0.$ Let $\mathit{tx}=0$ implies that $x=0,$ since $\mathcal{M}$ is torsion-free module which is contradiction and thus $\mathit{tx}\ne 0.$ Since $\mathcal{\aleph }$ is an artibrary submodule, then $\mathcal{M}$ is monoform module.

Proposition 4.3

If $\mathcal{M}$ is quasi-Dedekind S-PS.B. $\mathcal{F}-\mathit{\text{module}},$ then $\mathcal{M}$ is monoform module.

Proof:

It is sufficient to show that every non-zero submodule of $\mathcal{M}$ is dense. Assume that $\mathcal{\aleph }$ is any non-zero submodule of $\mathcal{M}.$ Define $\phi :\mathcal{M}⟶\mathcal{M}$ by $\phi \left(x\right)=\mathit{tx},\forall x\in \mathcal{M}.$ Suppose that $0\ne x\in \mathcal{M}$ and $\mathit{tx}=0$ and since $\mathcal{M}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ implies that $\mathit{tx}\in \mathcal{\aleph }$ and $\phi \left(x\right)=0$ which means $x\in ker\phi .$ Since $\mathcal{M}$ is quasi-Dedekind module, then $\phi$ is monomorphism. Hence $x=0$ which is contradiction. Thus $\mathit{tx}\ne 0$ and $\mathcal{\aleph }$ is dense submodule.

Proposition 4.4

Every S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ is retractable $\mathcal{F}-\mathit{\text{module}}.$

Proof:

Assume that $\mathcal{\aleph }$ is a non-zero submodule of S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ $\mathcal{M},$ then there exists $0\ne \phi \in \mathit{End}\left(\mathcal{M}\right)$ such that $\phi :\mathcal{M}⟶\mathcal{M}$ and defined as $\phi \left(x\right)=\mathit{tx},0\ne x\in \mathcal{M},\phi \left(x\right)\in \mathcal{\aleph }$ and $\sqrt{{\mathit{ann}}_{\mathcal{F}}\left(\mathcal{\aleph }\right)}=\sqrt{{\mathit{ann}}_{\mathcal{F}}\phi \left(x\right)}.$ Let $\mathit{Hom}(\mathcal{M},\mathcal{\aleph })=0$ and $i:\mathcal{\aleph }⟶\mathcal{M}$ the inclusion map. Then $\phi =i\circ g$ where $g:\mathcal{M}⟶\mathcal{\aleph },g=0.$ Thus $\phi \left(x\right)=(i\circ g)\left(x\right)=i\left(g\left(x\right)\right)=0$ that means $\sqrt{{\mathit{ann}}_{\mathcal{F}}\left(\mathcal{\aleph }\right)}\ne \sqrt{{\mathit{ann}}_{\mathcal{F}}\phi \left(x\right)}$ which is a contradiction. Hence $\mathit{Hom}(\mathcal{M},\mathcal{\aleph })\ne 0\forall \mathcal{\aleph }\ne 0$ of $\mathcal{M}.$ Therefore $\mathcal{M}$ is retractable $\mathcal{F}-\mathit{\text{module}}.$

Conversely is not true in general, we have this example:

Consider ${\mathbb{Z}}_4$ as a $\mathbb{Z}-\mathit{\text{module}}$ and $\mathcal{M}={\mathbb{Z}}_4,\mathcal{\aleph }=⟨\overline{2}⟩.$ Define $\phi :{\mathbb{Z}}_4⟶{\mathbb{Z}}_4$ as $\phi \left(\overline{x}\right)=\overline{0},\forall \overline{x}\in {\mathbb{Z}}_4,\phi \in \mathit{End}\left({\mathbb{Z}}_4\right).$ Since $\phi \left(\overline{x}\right)\in \mathcal{\aleph },\forall \overline{x}\in {\mathbb{Z}}_4,\mathit{Im}\phi \subseteq \mathcal{\aleph }$ then $\mathcal{M}$ is retractable module. But $\mathcal{M}$ is not S-PS.B $\mathbb{Z}-\mathit{\text{module}}$ , since $2\mathbb{Z}=\sqrt{{\mathit{ann}}_{\mathbb{Z}}\left(\overline{2}\right)}\ne \sqrt{{\mathit{ann}}_{\mathbb{Z}}\phi \left(\overline{1}\right)}=\mathbb{Z},$ where $\overline{1}\in {\mathbb{Z}}_4.$

Proposition 4.5

If $\mathcal{M}$ is S-PS.B. torsion-free module, then $\mathcal{M}$ is critically compressible module.

Proof:

By Corollary (4.2) and Proposition (2.17) we get $\mathcal{M}$ is critically compressible.

Corollary 4.6

Let $\mathcal{M}$ be S-PS.B. $\mathcal{F}-\mathit{\text{module}},$ then $\mathcal{M}$ is critically compressible $\mathcal{F}-\mathit{\text{module}}$ if and only if every non-zero partial endomorphism of $\mathcal{M}$ is monomorphism.

Proposition 4.7

If $\mathcal{M}$ is a duo S-PS.B. $\mathcal{F}-\mathit{\text{module}},$ then $\mathcal{M}$ is fully retractable module.

Proof:

Since $\mathcal{M}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}},$ then there exists $\phi$ is a non-zero endomorphism of $\mathcal{M}$ such that $\phi \left(x\right)\in \mathcal{\aleph },x\in \mathcal{M}$ and $\sqrt{{\mathit{ann}}_{\mathcal{F}}\left(\mathcal{\aleph }\right)}=\sqrt{{\mathit{ann}}_{\mathcal{F}}\phi \left(x\right)}$ where $\mathcal{\aleph }$ is a submodule of $\mathcal{M}.$ For every $\phi \in \mathit{End}\left(\mathcal{M}\right)$ we have $\phi \left(\mathcal{\aleph }\right)\subseteq \mathcal{\aleph },$ since $\mathcal{M}$ is a duo. Thus the partial endomorphism of $\mathcal{M}$ is not zero and $0\ne \phi :\mathcal{\aleph }⟶\mathcal{M}.$ Therefore $\mathcal{M}$ is retractable module, by Proposition (4.4) so that there exists $0\ne \psi :\mathcal{M}⟶\mathcal{\aleph }$ a homomorphism. Hence $\psi \circ \phi \ne 0$ and thus $\mathcal{M}$ is fully retractable module.

Corollary 4.8

If $\mathcal{M}$ is a duo S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ and $\mathit{End}\left(\mathcal{M}\right)$ is a domain, then $\mathcal{M}$ is polyform module.

Proof:

Assume that $\mathcal{M}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ then by previous proposition, $\mathcal{M}$ is fully retractable module. Applying Proposition (2.18) we get the result.

Proposition 4.9

Let $\mathcal{M}$ be an uniform S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ and $\mathit{End}\left(\mathcal{M}\right)$ is a domain. Then the following statements are equivalent:

$\left(i\right)$ $\mathcal{M}$ is critically compressible module.

$\left(\mathit{ii}\right)$ $\mathcal{M}$ is polyform module.

Proof:

$\left(\mathit{i}\right)\mathit{⟹}\left(\mathit{ii}\right)$ Suppose that $\mathcal{M}$ is critically compressible module, then $\mathcal{M}$ is monoform, by Corollary(4.6) and thus $\mathcal{M}$ is polyform.

$\left(\mathit{ii}\right)\mathit{⟹}\left(\mathit{i}\right)$ Assume that $\mathcal{M}$ is polyform module, then Corollary (2.22), $\mathcal{M}$ is monoform, since $\mathcal{M}$ is uniform. Since $\mathcal{M}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ , then $\mathcal{M}$ is retractable. By Proposition (2.16), $\mathcal{M}$ is critically compressible module.

Corollary 4.10

If $\mathcal{M}$ is S-PS.B. uniform $\mathcal{F}-\mathit{\text{module}}$ where $\mathit{End}\left(\mathcal{M}\right)$ is a domain, then the following statements are equivalent:

$\left(\mathit{i}\right)$ $\mathcal{M}$ is fully polyform module.

$\left(\mathit{ii}\right)$ $\mathcal{M}$ is critically compressible module.

Proof:

If $\mathcal{M}$ is uniform, then polyform and fully polyform are equivalent, by Ref. 10.

Proposition 4.11

If $\mathcal{M}$ is a compressible $\mathcal{F}-\mathit{\text{module}},$ then $\mathcal{M}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}}.$

Proof:

Assume that $\phi \in \mathit{End}\left(\mathcal{M}\right)$ where $\phi :\mathcal{M}⟶\mathcal{M}$ define as $\phi \left(x\right)=\mathit{tx},x\in \mathcal{M}.$ For each submodule $\mathcal{\aleph }\ne 0$ of $\mathcal{M}$ there exists $h:\mathcal{M}⟶\mathcal{\aleph }$ a monomorphism map, since $\mathcal{M}$ is a compressible module. If $\phi =i\circ h$ where $i$ is the inclusion map, then $\phi \left(x\right)=(i\circ h)\left(x\right)=i\left(h\left(x\right)\right)=h\left(x\right)\in \mathcal{\aleph }.$ Suppose that $a\in \sqrt{{\mathit{ann}}_{\mathcal{F}}\phi \left(x\right)}$ implies that $a^n.\phi \left(x\right)=0,\mathit{\text{for some}}n\in {\mathbb{Z}}^{+},\forall x\in \mathcal{M}.$ Thus $a^n.h\left(x\right)=0,\forall x\in \mathcal{M}$ so that $h\left(a^{n}x\right)=0$ and $h\left(a^{n}x\right)=h\left(0\right)$ which means $a^{n}x=0,\forall x\in \mathcal{M}.$ Hence $a\in \sqrt{{\mathit{ann}}_{\mathcal{F}}\left(\mathcal{M}\right)}$ and $a^n\in {\mathit{ann}}_{\mathcal{F}}\left(\mathcal{M}\right)={\mathit{ann}}_{\mathcal{F}}\left(\mathcal{\aleph }\right),$ since $\mathcal{M}$ is prime module. Therefore $\sqrt{{\mathit{ann}}_{\mathcal{F}}\phi \left(x\right)}=\sqrt{{\mathit{ann}}_{\mathcal{F}}\left(\mathcal{\aleph }\right)}$ and $\mathcal{M}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}}.$

Proposition 4.12

Let $\mathcal{M}$ be a torsion-free $$ multiplication $\mathcal{F}-\mathit{\text{module}}$ with ( $\mathcal{F}$ is an integral domain).Then $\mathcal{M}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ if and only if $\mathcal{M}$ is a compressible module.

Proof:

Let $\mathcal{M}$ be S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ . By Corollary (2.10) $\mathcal{M}$ is a scalar module and hence every $\phi \in \mathit{End}\left(\mathcal{M}\right)$ is monomorphism. Since $\mathcal{M}$ is retractable module by Proposition (4.4), then there exists a homomorphism $0\ne f:\mathcal{M}⟶\mathcal{\aleph }$ for every submodule $\mathcal{\aleph }\ne 0$ of $\mathcal{M}.$ Hence $\phi =i\circ f$ is monomorphism where $i$ is an inclusion map. Thus $f$ is a monomorphism and hence $\mathcal{M}$ is a compressible module.

Conversely, applying previous proposition.

Corollary 4.13

If $\mathcal{M}$ is a quasi-Dedekind retractable $\mathcal{F}-\mathit{\text{module}},$ then $\mathcal{M}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ .

Proof:

Every $\mathcal{F}-\mathit{\text{homomorphism}}$ $\phi \in \mathit{End}\left(\mathcal{M}\right)$ is monomorphism, since $\mathcal{M}$ is a quasi-Dedekind. Hence $\mathcal{M}$ is a compressible module, by Proposition (2.15) and thus $\mathcal{M}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ , by Proposition (4.11).

Corollary 4.14

If $\mathcal{M}$ is a finitely generated, then every uniform prime $\mathit{\text{module}}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ .

Proof:

Since $\mathcal{M}$ is a finitely generated implies that $\mathcal{M}$ is a compressible module, by (2.23). The result is obtained by Proposition (4.11).

Proposition 4.15

If $\mathcal{M}$ is a quasi-Dedekind $\mathcal{F}-\mathit{\text{module}}$ and $\phi \left(\mathcal{M}\right)⊈{\cap }_{\psi \in \mathit{Hom}(\mathcal{F},\mathcal{\aleph })}\mathit{\text{kerψ}}$ where $\phi \in \mathit{Hom}(\mathcal{M},\mathcal{F})$ and $\mathcal{\aleph }$ is a submodule of $\mathcal{M},$ then $\mathcal{M}$ is S-PS.B. $\mathit{\text{module}}.$

Proof:

Suppose that $\phi \left(\mathcal{M}\right)⊈{\cap }_{\psi \in \mathit{Hom}(\mathcal{F},\mathcal{\aleph })}\mathit{\text{kerψ}},$ then $\exists {\psi }^{\ast }:\mathcal{F}\to \mathcal{\aleph }$ such that $0\ne {\psi }^{\ast }\circ \phi \in \mathit{Hom}(\mathcal{M},\mathcal{\aleph }).$ Thus $\mathcal{M}$ is retractable module. Since $\mathcal{M}$ is a quasi-Dedekind module, then $\mathcal{M}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ by proposition (4.13).

Proposition 4.16

Let $\mathcal{M}$ is a quasi-Dedekind S-PS.B. $\mathcal{F}-\mathit{\text{module}}.$ Then ${\mathit{ann}}_{\mathcal{F}}\left(\mathcal{M}\right)\ne {\mathit{ann}}_{\mathcal{F}}\left(\frac{\mathcal{M}}{\mathcal{\aleph }}\right),\mathcal{\aleph }\le \mathcal{M}.$

Proof:

Assume that $\mathcal{\aleph }\le \mathcal{M}$ and ${\mathit{ann}}_{\mathcal{F}}\left(\mathcal{M}\right)={\mathit{ann}}_{\mathcal{F}}\left(\frac{\mathcal{M}}{\mathcal{\aleph }}\right).$ Hence $\left[\mathcal{\aleph }{:}_{\mathcal{F}}\mathcal{M}\right]={\mathit{ann}}_{\mathcal{F}}\left(\mathcal{M}\right)={\mathit{ann}}_{\mathcal{F}}\left(x\right),\forall x\in \mathcal{M},$ since $\mathcal{M}$ is prime. Thus for every $t$ such that $t\mathcal{M}\subseteq \mathcal{\aleph }$ and $\mathit{ty}\in \mathcal{\aleph }$ so that $\mathit{tx}=0,x\ne 0,x,y\in \mathcal{M}$ which is contradiction since $\mathcal{M}$ is monoform module by Proposition (2.18). Thus ${\mathit{ann}}_{\mathcal{F}}\left(\mathcal{M}\right)\ne {\mathit{ann}}_{\mathcal{F}}\left(\frac{\mathcal{M}}{\mathcal{\aleph }}\right).$

Corollary 4.17

Let $\mathcal{M}$ is a quasi-Dedekind S-PS.B. $\mathcal{F}-\mathit{\text{module}}.$ Then $\mathcal{M}$ is not coprime $\mathcal{F}-\mathit{\text{module}}.$

Proposition 4.18

Let $\mathcal{\aleph }\le \mathcal{M}$ and $\mathcal{M}/\mathcal{\aleph }$ be a quasi-Dedekind $\mathcal{F}-\mathit{\text{module}}$ . Then $\mathcal{M}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ .

Proof:

Suppose that $\phi \in \mathit{End}\left(\mathcal{M}\right)$ and $\phi :\mathcal{M}⟶\mathcal{M}$ defined as $\phi \left(n\right)=\mathit{tn},n\in \mathcal{M}.$ Since $\mathcal{M}/\mathcal{\aleph }$ be a quasi-Dedekind module so if we define $\psi :\mathcal{M}/\mathcal{\aleph }\to \mathcal{M}/\mathcal{\aleph }$ as $\psi (x+\mathcal{\aleph })=\mathit{tx}+\mathcal{\aleph },\forall x\in \mathcal{M}$ then either $\psi =0$ or $\psi$ is a monomrphism. Let $\psi \ne 0$ and $n+\mathcal{\aleph }\in \mathit{Ker}\psi$ then $\psi (n+\mathcal{\aleph })=\mathcal{\aleph }$ which means $\mathit{tn}+\mathcal{\aleph }=\mathcal{\aleph }.$ Therefore, $n+\mathcal{\aleph }=\mathcal{\aleph }$ so that $n\in \mathcal{\aleph }$ and $\mathit{tn}\in \mathcal{\aleph },\forall n\in \mathcal{\aleph }.$ Hence $\sqrt{{\mathit{ann}}_{\mathcal{F}}\left(\mathcal{\aleph }\right)}\subseteq \sqrt{{\mathit{ann}}_{\mathcal{F}}\phi \left(n\right)}$ . If $\psi =0$ we obtain $\mathit{tn}+\mathcal{\aleph }=\mathcal{\aleph }$ so $\mathit{tn}\in \mathcal{\aleph }$ and also $\sqrt{{\mathit{ann}}_{\mathcal{F}}\left(\mathcal{\aleph }\right)}\subseteq \sqrt{{\mathit{ann}}_{\mathcal{F}}\phi \left(n\right)}.$ Now, let $a\in \sqrt{{\mathit{ann}}_{\mathcal{F}}\phi \left(n\right)}$ so $a^n.\phi \left(n\right)=0,\mathit{\text{for some}}n\in {\mathbb{Z}}^{+}$ and $a^n\left(\mathit{tn}\right)=0$ then $a\in \sqrt{{\mathit{ann}}_{\mathcal{F}}\left(\mathit{tn}\right)},\forall \mathit{tn}\in \mathcal{\aleph }.$ Thus $a\in \sqrt{{\mathit{ann}}_{\mathcal{F}}\left(\mathcal{\aleph }\right)}$ and we get $\sqrt{{\mathit{ann}}_{\mathcal{F}}\left(\mathcal{\aleph }\right)}=\sqrt{{\mathit{ann}}_{\mathcal{F}}\phi \left(n\right)}.$

Proposition 4.19

If $\mathcal{M}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ , then $\mathcal{M}\ne t\mathcal{M}.$

Proof:

Assume that $\mathcal{M}=t\mathcal{M}$ and $\exists \phi \in \mathit{End}\left(\mathcal{M}\right)$ such that $\phi :\mathcal{M}⟶\mathcal{M}$ and defined as $\phi \left(x\right)=\mathit{tx},\forall x\in \mathcal{M},$ since $\mathcal{M}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}}.$ Thus $\mathcal{M}$ is retractable by (4.4) which means $\mathit{Im\phi }\subseteq \mathcal{\aleph }$ for every submodule $\mathcal{\aleph }\ne 0$ of $\mathcal{M}.$ Therefore $\phi \left(\mathcal{M}\right)\subseteq \mathcal{\aleph }$ so that $t\mathcal{M}\subseteq \mathcal{\aleph }$ and since $\mathcal{M}=t\mathcal{M}$ implies $\mathcal{M}\subseteq \mathcal{\aleph }$ which is contradiction. Thus $\mathcal{M}\ne t\mathcal{M}.$

Proposition 4.20

Let $\mathcal{M}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ such that $0\ne \phi \in \mathit{End}\left(\mathcal{M}\right).$ Then $\phi$ is not epimorphism.

Proof:

Suppose that $\phi \in \mathit{End}\left(\mathcal{M}\right).\phi \ne 0$ Since $\mathcal{M}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}},$ then $\mathcal{M}$ is retractable module ad there exists a homomorphism $0\ne f:\mathcal{M}⟶\mathcal{\aleph },$ for every submodule $\mathcal{\aleph }\ne 0$ of $\mathcal{M}.$ Put $\phi =i\circ f:\mathcal{M}⟶\mathcal{M}$ where $i:\mathcal{\aleph }⟶\mathcal{M}$ is the inclusion map. Therefore $\phi$ is not epimorphism.

Proposition 4.21

Let $\mathcal{M}$ be S-PS.B. torsion-free $\mathcal{F}-\mathit{\text{module}}$ . Then $\mathcal{M}$ is an injective.

Proof:

Since $\mathcal{M}$ a torsion-free and S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ , so that $\mathcal{M}$ is a monoform module by Corollary (4.2) and thus there exists a non-zero monmorphism $g:\mathcal{\aleph }⟶\mathcal{M}$ defined by $g\left(y\right)=y,\forall y\in \mathcal{\aleph }.$ Since $\mathcal{M}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}},$ then $\mathcal{M}$ is retractable and there exists a homomorphism $0\ne f:\mathcal{M}⟶\mathcal{\aleph }$ defined by $f\left(y\right)=y,\forall y\in \mathcal{M}.$ Moreover we consider the diagram

Suppose that $y\in \mathcal{M},$ then $(g\circ f)\left(y\right)=g\left(f\left(y\right)\right)=g\left(y\right)=I_{\mathcal{M}}$ . Thus $g\circ f=I_{\mathcal{M}}$ and the diagram is commutative so $\mathcal{M}$ is an injective $\mathcal{F}-\mathit{\text{module}}.$

Proposition 4.22

If $\mathcal{M}$ is a torsion-free and S-PS.B. $\mathcal{F}-\mathit{\text{module}},$ then $\mathit{End}\left(\mathcal{M}\right)$ has no zero-divisors.

Proof:

Suppose that $\phi ,\psi \in \mathit{End}\left(\mathcal{M}\right)$ are two non-zero homomorphism implies that $\exists n,n^{\prime }\in \mathcal{M}$ such that $\phi \left(n\right)=x\ne 0,\psi \left(n^{\prime }\right)=y\ne 0$ where $x,y\in \mathcal{M}.$ Since $\mathcal{M}$ is monoform by (4.2) and thus $\mathcal{M}$ is quasi-Dedekind module. Hence $\phi ,\psi \in$ are two $\mathcal{F}-$ monomorphisms. Thus $(\phi \circ \psi )\left(n^{\prime }\right)=\phi \left(y\right)\ne 0$ and $(\psi \circ \phi )\left(n\right)=\psi \left(x\right)\ne 0$ which means $\mathit{End}\left(\mathcal{M}\right)$ has no zero-divisors.

Proposition 4.23

If $\mathcal{M}$ is torsion-free S-PS.B., then $\mathcal{M}$ is a Rickart $\mathcal{F}-\mathit{\text{module}}.$

Proof:

Since $\mathcal{M}$ is an injective, by Proposition (4.21) and hence $\mathcal{M}$ is prime module since every torsion-free S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ is monoform. By Proposition (2.20) we obtain that $\mathcal{M}$ is a Rickart.

Proposition 4.24

Let $\mathcal{M}$ be an indecomposable S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ with $\mathcal{M}$ is N-Rickart. Then $\mathcal{M}$ is quasi-Dedekind.

Proof:

Suppose that $\mathcal{M}$ is S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ , then $\mathcal{M}$ is a retractable module, by (4.4) and thus $\mathit{Hom}(\mathcal{M},\mathcal{\aleph })\ne 0$ for every non-zero submodule $\mathcal{\aleph }$ of $\mathcal{M}.$ Suppose that $\phi :\mathcal{M}\to \mathcal{M}$ is an endomorphism of $\mathcal{M}.$ Since $\mathcal{M}$ is N-Rickart, then $\mathit{\text{kerφ}}$ is direct summand of $\mathcal{M}.$ Since $\mathcal{M}$ be an indecomposable so that $\mathit{\text{kerφ}}=0.$ Thus $\phi$ is a $\mathcal{F}-\mathit{\text{monomorphism}}$ and $\mathcal{M}$ is quasi-Dedekind module.

Proposition 4.25

Let $\mathcal{M}$ be a torsion-free S-PS.B. Then for each $\phi \in \mathit{End}\left(\mathcal{M}\right)$ there exists once $\phi$ is splits in $\mathcal{M}.$

Proof:

From (4.23) we get $\mathcal{M}$ is a Rickart $\mathcal{F}-\mathit{\text{module}}$ , Consider the following short exact sequence $0\to \mathit{\text{kerφ}}={\mathit{ann}}_{\mathcal{M}}\phi \left(x\right)\to \mathcal{M}\to \phi \mathcal{M}\to 0$ .

Consequently, $\mathit{\text{kerφ}}$ is direct summand of $\mathcal{M}$ , since $\mathcal{M}$ is Rickart module. But $\mathit{\text{kerφ}}={\mathit{ann}}_{\mathcal{M}}\phi \left(x\right)$ which means $\phi$ is splits in $\mathcal{M}.$

Proposition 4.26

Let $\mathcal{M}$ be a S-PS.B. $\mathcal{F}-\mathit{\text{module}}$ . Then $\mathit{End}\left(\mathcal{M}\right)$ is a Rickart ring if and only if $\mathcal{M}$ is a Rickart module.

Proof:

Since $\mathcal{M}$ is a S-PS.B. $\mathcal{F}-\mathit{\text{module}},$ which means $\mathcal{M}$ is retractable module so that we get the result by (proposition (3.3), Ref. 16).