Endo-Greatly Bounded Modules

Remarks:

Then ${\mathit{ann}}_z\left(N\right)=2Z$ and ${\mathit{ann}}_z$ ( $\left(\mathit{\phi o\psi }\right)(\overline{0},\overline{1}$ ))= ${\mathit{ann}}_z(\overline{0},\overline{0})=Z$

Therefor $\mathrm{{\rm H}}$ is not Endo-G.B module, but $\mathit{ann}\left(\overline{x}\right)=\mathit{ann}(\overline{0},\overline{1})=2Z$ = $\mathit{ann}\left(\mathrm{{\rm H}}\right)=2Z$ hence $\mathrm{{\rm H}}$ is bounded module.

Proposition 3.5

Let $\mathrm{{\rm H}}$ be an Endo-G.B.S-module and $B\le \mathrm{{\rm H}}.$ Then $\mathrm{{\rm H}}∕B$ is an Endo-G.B.S-module.

Proof:

Let $A,B\le \mathrm{{\rm H}}$ and $A$ containing $B$ such that $A/B\le \mathrm{{\rm H}}/B.$

Since $\mathrm{{\rm H}}$ is an Endo-G.B. module. Then $A$ is an End-G.B. submodule that is, there exists $\phi ,\psi \in \mathit{End}\left(\mathrm{{\rm H}}\right)$ such that either $\phi \left(x\right)\in N$ or $\psi \left(x\right)\in N,$ and ${\mathit{ann}}_S\left(\mathit{\phi o\psi }\left(x\right)\right)={\mathit{ann}}_S\left(A\right).$

Let $\phi ,\psi :\frac{\mathrm{{\rm H}}}{B}\to \frac{\mathrm{{\rm H}}}{B}$ by $\phi (x+B)=\phi \left(x\right)+B,\forall x\in \mathrm{{\rm H}}$ and $\psi (x+B)=\psi \left(x\right)+B,\forall x\in \mathrm{{\rm H}},$ then ${\mathrm{it}}^{,}\mathrm{s}$ clear that $\phi ,\psi$ are well –defined and an endomorphism of $\mathrm{{\rm H}}/B.$

Obviously, $\psi (x+B)\in A/B$ so ${\mathit{ann}}_R\left(\frac{A}{B}\right)\subseteq {\mathit{ann}}_S\left(\mathit{\phi o}\psi (x+B)\right).$

Now, let $r\in {\mathit{ann}}_S\left(\mathit{\phi o\psi }(x+B)\right)$ that means

Since $\mathit{r\phi \psi }\left(x\right)\in \mathit{rA},\forall x\in \mathrm{{\rm H}},$ therefor $\mathit{rN}+B=B$

We conclude that $r\in {\mathit{ann}}_S\left(\frac{A}{B}\right).$

Proposition 3.6

Let ${\mathrm{{\rm H}}}_1,{\mathrm{{\rm H}}}_2$ be two $S$ -modules such that ${\mathit{ann}}_S\left({\mathrm{{\rm H}}}_1\right)+{\mathit{ann}}_S\left({\mathrm{{\rm H}}}_2\right)=S.$

Then ${\mathrm{{\rm H}}}_1⨁{\mathrm{{\rm H}}}_2$ is an Endo-G.B.S-module if and only if ${\mathrm{{\rm H}}}_1$ and ${\mathrm{{\rm H}}}_2$ are Endo-G.B. S-modules.

Proof:

Assume that ${\mathrm{{\rm H}}}_1⨁{\mathrm{{\rm H}}}_2$ is an Endo-G.B, the $A_1⨁A_2\le {\mathrm{{\rm H}}}_1⨁{\mathrm{{\rm H}}}_2$ where

$A_1\le {\mathrm{{\rm H}}}_1$ and $A_2\le {\mathrm{{\rm H}}}_2.$

Let $\phi ,\psi :{\mathrm{{\rm H}}}_1⨁{\mathrm{{\rm H}}}_2\to {\mathrm{{\rm H}}}_1⨁{\mathrm{{\rm H}}}_2$ be an endomorphism of ${\mathrm{{\rm H}}}_1⨁{\mathrm{{\rm H}}}_2$ define as

either $\phi (a,b)=r(a,b),$ like that $\phi (a,b)\in A_1⨁A_2$

or $\psi (a,b)=r(a,b),$ like that $\psi (a,b)\in A_1⨁A_2$

So that $,(\mathit{ra},\mathit{rb})\in A_1⨁A_2$ where $\mathit{ra}\in A_1,\mathit{rb}\in A_2$ and

Consider the composition

Let $a\in {\mathrm{{\rm H}}}_1,\mathrm{\text{then}}\left(\mathit{\text{joψoφoi}}\right)\left(a\right)=\mathit{\text{joψoφ}}\left(i\left(a\right)\right)=\mathit{\text{joψoφ}}(a,0)=\mathit{jo\psi }\left(\phi (a,o)\right)$

$\mathrm{Now},$ to show that

Let $t\in {\mathit{ann}}_S\left(\mathit{\text{joψoφoi}}\left(a\right)\right),\mathit{\text{then}}t\left(\mathit{\text{joψoφoi}}\left(a\right)\right)=0,$ that is $t\left(\mathit{ra}\right)=0,$

Where $t\in S.$ This implies that $t\in {\mathit{ann}}_S\left(A_1\right).$

Hence ${\mathit{ann}}_S\left(\mathit{\text{joψoφoi}}\left(a\right)\right)={\mathit{ann}}_S\left(A_1\right).$

So ${\mathrm{{\rm H}}}_1$ is an End. G.B. Similarly,we can prove that ${\mathrm{{\rm H}}}_2$ is the Endo-GB S-module.

Conversely, suppose that ${\mathrm{{\rm H}}}_1\text{and}{\mathrm{{\rm H}}}_2$ are Endo-GB S-modules. Let $A_1$ and $A_2$ are two submodule of ${\mathrm{{\rm H}}}_1,{\mathrm{{\rm H}}}_2$ respectively. Then $\exists$ ${\phi }_1,{\psi }_1,{\phi }_2,{\psi }_2\in \text{Endo}\left({\mathrm{{\rm H}}}_1\right),$

$\mathit{\text{Endo}}$ $\left({\mathrm{{\rm H}}}_2\right)$ respectively defined as:

${\phi }_1,{\psi }_1:{\mathrm{{\rm H}}}_{1\to }{\mathrm{{\rm H}}}_1$ such that ${\phi }_1\left(a\right)=\mathit{ra},{\psi }_1\left(a\right)=a\forall a\in {\mathrm{{\rm H}}}_1$

${\phi }_2,{\psi }_2:{\mathrm{{\rm H}}}_{2\to }{\mathrm{{\rm H}}}_2$ such that ${\phi }_2\left(b\right)=\mathit{rb},{\psi }_2\left(b\right)=b\forall b\in {\mathrm{{\rm H}}}_2$

Also, ${\phi }_1\left(a\right)=\mathit{ra}\in A_1,{\psi }_1\left(a\right)=a\in A_1$ and ${\phi }_2\left(b\right)=\mathit{rb}\in A_2,{\psi }_2\left(b\right)=b\in A_2$

where ${\mathit{ann}}_S\left(\right({\phi }_{1}o{\psi }_1\left)\left(a\right)\right)={\mathit{ann}}_S\left(A_1\right),{\mathit{ann}}_S(\left({\phi }_{2}o{\psi }_2\right)\left(b\right))={\mathit{ann}}_S\left(A_2\right)$

Now, let $f({\phi }_1,$ ${\phi }_2)$ and $g({\psi }_1,{\psi }_2)$ , where $f,g:{\mathrm{{\rm H}}}_1⨁{\mathrm{{\rm H}}}_2\to {\mathrm{{\rm H}}}_1⨁{\mathrm{{\rm H}}}_2,$ defined as $\mathit{fog}=({\phi }_{1}o{\psi }_1,{\phi }_{2}o{\psi }_2)(a,b)=({\phi }_{1}o{\psi }_1\left(a\right),{\phi }_{2}o{\psi }_2\left(b\right))={\phi }_1\left({\psi }_1\left(a\right)\right),{\phi }_2\left({\psi }_2\left(b\right)\right)$

$=({\phi }_1\left(a\right),{\phi }_2\left(b\right))=(\mathit{ra},\mathit{rb})\in A_1⨁A_2$ where $A_1⨁A_2\le {\mathrm{{\rm H}}}_1⨁{\mathrm{{\rm H}}}_2.$

We claim that ${\mathit{ann}}_S\left(\mathit{fog}({\phi }_{1}o{\psi }_1,{\phi }_{2}o{\psi }_2)\right)={\mathit{ann}}_S(A_1⨁A_2),$

Let $t\in {\mathit{ann}}_S\left(\mathit{fog}({\phi }_{1}o{\psi }_1,{\phi }_{2}o{\psi }_2)\right)\text{implies that}t\left(\mathit{fog}({\phi }_{1}o{\psi }_1,{\phi }_{2}o{\psi }_2)(a,b)\right)=(0,0)$

$=t({\phi }_{1}o{\psi }_1\left(a\right),{\phi }_{2}o{\psi }_2\left(b\right))=t(\mathit{ra},\mathit{rb})$ .

Thus, $t\in {\mathit{ann}}_S(A_1⨁A_2).$

Remark 3.7:

$\mathit{\text{If}}$ ${\mathrm{{\rm H}}}_1\text{and}{\mathrm{{\rm H}}}_2$ are two Endo-G.B.S-modules, then it is not necessary that ${\mathrm{{\rm H}}}_1⨁{\mathrm{{\rm H}}}_2$ be an Endo-G.B. module so the condition in Proposition (3.6) can not be dropped and the next example shows that:

Example 3.8:

Let ${\mathrm{{\rm H}}}_1=Z_6$ and $N_1=\left(\overline{3}\right)$ . Define ${\phi }_1,{\psi }_1:Z_6\to Z_6$ as ${\phi }_1\left(\overline{a}\right)=3a,{\psi }_1\left(\overline{a}\right)\mathit{=}\overline{\mathit{a}}$ . Then ${\mathit{ann}}_S\left({\phi }_{1}o{\psi }_1\right)\left(\overline{a}\right)={\mathit{ann}}_S\left(N_1\right),{\phi }_1\left(\overline{a}\right)\in N_1$ and

let ${\mathcal{M}}_2=Z_3,N_1=\left(\overline{0}\right).$

Define ${\phi }_2,{\psi }_2:Z_3\to Z_3$ as ${\phi }_2\left(\overline{a}\right)=\overline{0},{\psi }_2\left(\overline{a}\right)\mathit{=}\overline{\mathit{a}}$

Then ${\mathit{ann}}_S({\phi }_{2}O{\psi }_2\left(\overline{a}\right))={\mathit{ann}}_S\left(N_2\right),{\phi }_1\left(\overline{a}\right)\in N_2.$

Let $\mathrm{{\rm H}}=Z_6⨁Z_3$ and $N=\left(\overline{3}\right)⨁\left(\overline{0}\right).$ Define $\phi ,$ $\psi :Z_6⨁Z_3\to Z_6⨁Z_3$ as $\phi (\overline{a},\overline{b})=(\overline{a},0).$ Then $\mathrm{{\rm H}}$ is not Endo-G.B. modul since $\phi (\overline{a},\overline{b})\notin N.$

Corollary 3.9:

Let $n$ be a positive integer and ${\mathrm{{\rm H}}}_i(i\in I)$ be an S-module, such that ${\mathit{ann}}_S{\mathrm{{\rm H}}}_i+{\mathit{ann}}_S{\mathrm{{\rm H}}}_j=S,\forall 1\le i\le j\le n.$ Then $\mathrm{{\rm H}}={\mathrm{{\rm H}}}_i⨁{\mathrm{{\rm H}}}_j⨁\dots \dots \dots \dots .⨁{\mathrm{{\rm H}}}_n$ is an Endo-GB module if and only if ${\mathrm{{\rm H}}}_i$ is Endo-G.B. module $\forall 1\le i\le n.$

Proof:

The proof by induction and by using Proposition (3.6).

Remark 3.10:

The direct summand of the Endo-G.B. module $\mathrm{{\rm H}}$ is also Endo-G.B.

Proof:

Assume that $A$ is a direct summand of $\mathrm{{\rm H}}$ , then $\mathrm{{\rm H}}=A⨁B$ for some sub module $\mathrm{B}$ of $\mathrm{{\rm H}}.$

Since $A\le A⨁B,$ by using the hereditary property of an Endo-G.B. module we have That $A$ is an Endo-G.B. submodule.

Proposition 3.11:

Let $\mathrm{{\rm H}}$ = $A⨁B$ be an S-module, where $A$ and $B$ are two sub modules of $\mathrm{{\rm H}}$ such that ${\mathit{ann}}_S\left(A\right)+{\mathit{ann}}_S\left(B\right)=S.$ Then $A$ and $B$ are End-G.B. submodules if and only if $\mathrm{{\rm H}}$ is End-G.B.S-module.

Proof:

Suppose that $\mathrm{{\rm H}}$ is Endo-G.B.S-module. Since $A$ and $B$ are direct summands of $\mathrm{{\rm H}}$ , then by Remark (3.10) $A$ and $B$ are Endo-G.B. submodules.

Conversely, assume that $A$ and $B$ are Endo-GB S-submodules. Let $L⨁N$ be a submodule of $A⨁B$ where $L\le A$ and $N\le B$ and ${\mathit{ann}}_S\left(A\right)+{\mathit{ann}}_S\left(B\right)=S.$

Thus, from Proposition (3.6), we find that $A⨁B$ is an Endo-G.B.S-module.

Recall that a sub-module $A$ of S-module $\mathrm{{\rm H}}$ is said to be fulley invariant if $\phi \left(A\right)\subseteq A$ for every $\phi ,\psi \in \mathit{End}\left(H\right).$ ¹⁰

Proposition 3.12:

Let $X,\mathrm{Y}$ be two fully invariant sub modules of an S-module $\mathrm{{\rm H}}$ like that $\mathrm{{\rm H}}=X⨁Y.$ Then $X\text{and}Y$ are two Endo-G.B. submodules if and only if $\mathrm{{\rm H}}$ is End-G.B.S-module.

Proof:

Assume that $\mathrm{{\rm H}}$ is an Endo-GB S-module. The direct summands of the Endo-GB module are also Endo-G.B.

Therefore, we conclude that $X$ and $Y$ are Endo-GB S-submodules.

Conversely, suppose that X and $Y$ are Endo-G.B.S-submodules of an S-module $\mathrm{{\rm H}}$ .

Then in order to show that $\mathrm{{\rm H}}$ is an Endo-G.B.S-module, Let $\phi ,\psi :\mathrm{{\rm H}}\to \mathrm{{\rm H}}$ be an endomorphism and $N\le \mathrm{{\rm H}}.$ Firstly, we have to prove that either $\phi \left(x\right)\in N$ or $\psi \left(x\right)\in N,$ where $x\in \mathrm{{\rm H}}.$ Since $X,Y$ are fully invariant S-modules, then $N=(N\cap X)⨁(N\cap Y)$ either $\psi \left(x\right)\in (N\cap X)⨁(N\cap Y),$ then $\psi \left(x\right)={\phi }_1\left(c\right)+{\phi }_2\left(d\right)$ or $\phi \left(x\right)\in (N\cap X)⨁(N\cap Y),$ then $\phi \left(x\right)={\psi }_1\left(c\right)+{\psi }_2\left(d\right)$ .

Where ${\phi }_1,{\psi }_1,{\phi }_2,{\psi }_2\in \mathit{End}\left(\mathrm{{\rm H}}\right)$ either ${\phi }_1\left(c\right)\text{or}{\psi }_1\left(c\right)\in N\cap X,\text{either}{\phi }_2\left(d\right)\text{or}$ ${\psi }_2\left(d\right)\in N\cap Y$ implies that ${\phi }_1\left(c\right),{\psi }_1\left(c\right)\in N,{\phi }_1\left(c\right),{\psi }_1\left(c\right)\in X$ and ${\phi }_2\left(d\right),{\psi }_2\left(d\right)\in N,{\phi }_2\left(d\right),{\psi }_2\left(d\right)\in Y.$

So, ${\phi }_1\left(c\right),{\psi }_1\left(c\right),{\phi }_2\left(d\right),{\psi }_2\left(d\right)\in N$ which implies $\left({\phi }_{1}o{\psi }_1\right)\left(c\right)+\left({\phi }_{2}o{\psi }_2\right)\left(d\right)\in N,$ it is clear that ${\mathit{ann}}_S\left(N\right)\subseteq {\mathit{ann}}_S\left(\mathit{\phi o}\psi \left(x\right)\right).$

Now, let $r\in {\mathit{ann}}_S\left(\mathit{\phi o}\psi \right)\left(x\right),$ then $r\left(\mathit{\phi o}\psi \right)\left(x\right)=0$

Thus ${\mathit{ann}}_S\left(\left(\mathit{\phi o}\psi \right)\left(x\right)\right)={\mathit{ann}}_S\left(N\right).$

Proposition 3.13:

The homomorphic image of the Endo-GB module is also an Endo-GB module.

Proof:

Let $\mathrm{{\rm H}},{\mathrm{{\rm H}}}^{/}$ be two S-modules, and $f:\mathrm{{\rm H}}\to {\mathrm{{\rm H}}}^{/}$ be an S-homomorphism.

Let $\mathrm{{\rm H}}$ is an Endo-G.B. module, we claim that $f\left(\mathrm{{\rm H}}\right)$ is an Endo-G.B. module.

Since $\mathrm{{\rm H}}$ is an Endo-G.B. there exist $\phi ,\psi \in \mathit{End}\left(\mathrm{{\rm H}}\right)$ and define as $\phi \left(x\right)=\mathit{rx},\text{and}\psi \left(x\right)=\mathit{rx},\forall x\in \mathrm{{\rm H}},$ for all proper submodule $N$ of $\mathrm{{\rm H}}$ such that:

${\mathit{ann}}_S\left(\left(\mathit{\phi O\psi }\right)\left(x\right)\right)={\mathit{ann}}_S\left(N\right),$ we know that $f\left(\mathrm{{\rm H}}\right)\le {\mathrm{{\rm H}}}^{/}$ and $f\left(N\right)\le f\left(\mathrm{{\rm H}}\right)\le {\mathrm{{\rm H}}}^{/}$ ${\mathrm{it}}^{,}\mathrm{s}$ enough to prove that ${\mathrm{{\rm H}}}^{/}$ is an Endo-G.B. module.

Suppose that $h,g:{\mathrm{{\rm H}}}^{/}\to {\mathrm{{\rm H}}}^{/}$ and define as

Then $h\left(b\right)=\mathit{rf}\left(a\right)=f\left(\mathit{ra}\right)\in f\left(N\right)$ where $f\left(N\right)<f\left(H\right)$ and $g\left(b\right)=\mathit{rf}\left(a\right)=f\left(\mathit{ra}\right)\in f\left(N\right)$ where $f\left(N\right)<f\left(H\right),$ and $\left(\mathit{hog}\right)\left(b\right)=\left(\mathit{\text{foφoψ}}\right)\left(a\right),\forall b\in {\mathrm{{\rm H}}}^{/}\mathit{\text{and}}a\in \mathrm{{\rm H}}$

Then $f\left(b\right)=\phi \left(\psi \left(a\right)\right)=\phi \left(\mathit{ra}\right)=\mathit{ra},$

Now, we show that ${\mathit{ann}}_S\left(\left(\mathit{hog}\right)\left(b\right)\right)={\mathit{ann}}_S\left(f\left(N\right)\right).$

Let $t\in {\mathit{ann}}_S\left(\mathit{hog}\right)\left(b\right),$ then $t\in {\mathit{ann}}_S\left(\mathit{\text{foφoψ}}\right)\left(\mathit{ra}\right)\Rightarrow t\left(\mathit{\text{foφoψ}}\right)\left(\mathit{ra}\right)=0$

Thus, $t\in {\mathit{ann}}_S\left({\mathrm{{\rm H}}}^{/}\right)\subseteq {\mathit{ann}}_S\left(f\left(N\right)\right)$ so ${\mathit{ann}}_S\left(\left(\mathit{hog}\right)\left(b\right)\right)\subseteq {\mathit{ann}}_S\left(f\left(N\right)\right),$ and ${\mathrm{it}}^{,}\mathrm{s}$ clear that ${\mathit{ann}}_{S}f\left(N\right)\subseteq {\mathit{ann}}_S\left(\mathit{hog}\right)\left(b\right).$ Therefore, ${\mathrm{{\rm H}}}^{/}$ is an End-G.B. module implies that $f\left(\mathrm{{\rm H}}\right)$ is End-G.B.

Remark 4.1:

Every Endo-G.B.S-module is bounded module.

Proof:

Assume that $\phi ,\psi :\mathrm{{\rm H}}\to \mathrm{{\rm H}}$ as $\phi \left(a\right)=a,\psi \left(a\right)=a,\forall a\in \mathrm{{\rm H}},$ either $\phi \left(a\right)\in N$ or $\psi \left(a\right)\in N$ , where $N$ is a proper submodule of $\mathrm{{\rm H}}.$

By use the definition of an Endo-G.B. we have ${\mathit{ann}}_S\left(\left(\mathit{\phi O\psi }\right)\left(a\right)\right)={\mathit{ann}}_S\left(N\right)$ and implies that ${\mathit{ann}}_S\left(a\right)={\mathit{ann}}_S\left(N\right)$ for all proper sub module $N$ of $\mathrm{{\rm H}}.$ Hence $N$ is bounded submodule and conclude that every proper submodule is bounded then $\mathrm{{\rm H}}$ is fully bounded S-module and $\mathrm{{\rm H}}$ is bounded S-module.

The opposite of (Remark 4.1) is generally not true in the following example:

Example 4.2:

Let $\mathrm{{\rm H}}=Z_2⨁Z_2$ as Z-module and let $N=Z_2⨁\left(\overline{0}\right)$ .

Let $\phi ,\psi :\mathrm{{\rm H}}\to \mathrm{{\rm H}}$ by $\phi (\overline{c},d)=(\overline{0},\overline{0}),\psi (\overline{c},\overline{d})=(\overline{0},\overline{0}),$

Since $\psi (\overline{c},\overline{d})=(\overline{0},\overline{0})\in N$ but $2Z={\mathit{ann}}_Z\left(N\right)\ne {\mathit{ann}}_Z(\overline{0},\overline{0})=Z.$

Therefore, $\mathrm{{\rm H}}$ is not Endo-G.B. while $\mathrm{{\rm H}}$ is bounded S-module since there exist an element $x=(\overline{0},\overline{1})\in Z_2⨁Z_2$ such that $2Z={\mathit{ann}}_Z\left(\mathrm{{\rm H}}\right)={\mathit{ann}}_z(\overline{0}$ , $\overline{1})=2Z$ .

Recall an S-module is called cyclic over $\mathit{End}\left(\mathrm{{\rm H}}\right)$ if there exist an element $x\in \mathrm{{\rm H}}$ like that for every element $h\in \mathrm{{\rm H}}$ can be writen $h=f\left(x\right)$ for some $f\in \mathit{End}\left(H\right).$ ¹¹

Proposition 4.3:

Let $\mathrm{{\rm H}}$ be a bounded S-module and $\mathrm{SX}$ is an essential sub module of $\mathrm{{\rm H}}.$ Then $\mathrm{{\rm H}}$ is End-G.B.

Proof:

Let $\phi ,\psi \in \mathit{End}\left(\mathrm{{\rm H}}\right)$ and define $\phi ,\psi :\mathrm{{\rm H}}\to \mathrm{{\rm H}}$ as $\phi \left(a\right)=\mathit{ra},$ $\psi \left(a\right)=a,\forall a\in \mathrm{{\rm H}}.$ Let $N$ be any proper sub-module of $H,$ then $\phi \left(a\right)\in N.$

Now, let $t\in {\mathit{ann}}_S\left(\left(\phi 0\psi \right)\left(a\right)\right),$ then $t\left(\phi 0\psi \right)\left(a\right)=0,$ implse that $\mathit{t\phi }\left(a\right)=0$ so $t\left(\mathit{ra}\right)=0,\forall r\in S.$ Thus $t\in {\mathit{ann}}_S\left(\mathit{Sa}\right).$ Since S is commutative ring with identity, then by Lemma (1.2) Alni, we get $t\in {\mathit{ann}}_S\left(a\right)={\mathit{ann}}_S\left(\mathrm{{\rm H}}\right)\subseteq {\mathit{ann}}_S\left(N\right).$

Proposition 4.4:

Every Endo-GB S-module is an Endo-GB E-module, where $E=\mathit{End}\left(\mathrm{{\rm H}}\right).$

Proof:

Suppose that $\mathrm{{\rm H}}$ is Endo-G.B.S-module, and let $N\le \mathrm{{\rm H}}.$

Let $\phi ,\psi \in \mathit{End}\left(H\right)$ and as $\phi \left(a\right)=b,\psi \left(a\right)=b,\forall a\in \mathrm{{\rm H}},$ so we have $\phi \left(a\right)\in N$ and ${\mathit{ann}}_S\left(\left(\mathit{\phi O\psi }\right)\left(a\right)\right)={\mathit{ann}}_S\left(N\right).$ Since $N$ as E-submodule is S-submodule, then $\phi \left(a\right)\in N$ as E-submodule.

Now, let $\delta \in {\mathit{ann}}_E\left(\left(\mathit{\phi o\psi }\right)\left(a\right)\right)$ where $\delta \in \mathit{End}\left(\mathrm{{\rm H}}\right),$ then $\delta \left(\left(\mathit{\phi O\psi }\right)\left(a\right)\right)=0.$

But $\phi \left(a\right)=b.$ Thus $\delta \left(b\right)=0,\forall b\in N$ implies that $\delta \left(N\right)=0,$ therefore, $\delta \in {\mathit{ann}}_E\left(N\right).$

Proposition 4.5:

Every cyclic submodule of the bounded module is an endo-GB submodule.

Proof:

Let $A$ be a cyclic sub module of a bounded S-module $\mathrm{{\rm H}}$ , then there exist $x\in \mathrm{{\rm H}}$ such that $A=\mathit{Sx}.$ Let $\phi ,\psi \in \mathit{\text{Endo}}\left(H\right)$ and as $\phi \left(x\right)=\mathit{rx},\psi \left(x\right)=x,\text{for every}x\in \mathrm{{\rm H}},$ $\phi \in \mathit{End}\left(\mathrm{{\rm H}}\right).$ Then $\phi \left(x\right)\in A$ and ${\mathit{ann}}_S\left(A\right)\subseteq {\mathit{ann}}_S\left(\left(\mathit{\phi O\psi }\right)\left(x\right)\right).$

Now, let $t\in {\mathit{ann}}_S\left(\left(\mathit{\phi o\psi }\right)\left(x\right)\right),$ then $t\left(\mathit{\phi O\psi }\right)\left(x\right)=0$

$t\phi \left(\mathit{rx}\right)=0$ , $t\left(\mathit{rx}\right)=0,\forall r\in S,t\left(\mathit{Sx}\right)=0.$

Thus $t\in {\mathit{ann}}_S\left(\mathit{Sx}\right)=\mathit{ann}\left(x\right)$ since S has the identity and using Lemma (1.2 bounded paper).

Therefore $t\in {\mathit{ann}}_S\left(\mathrm{{\rm H}}\right)\subseteq {\mathit{ann}}_S\left(A\right).$

Proposition 4.6:

Every cyclic bounded module is an Endo-G.B.

Corollary 4.7:

Let $\mathrm{{\rm H}}$ be a cyclic torsion-free S-module(where S is an integral domain), then $\mathrm{{\rm H}}$ is Endo-G.B.

Proof:

From proposition (1.5 bounded paper), $\mathrm{{\rm H}}$ is bounded S-module and using Proposition 4.6 we conclude that $\mathrm{{\rm H}}$ is an End-G.B.

Corollary 4.8:

Let $S$ be an integral domain and $\mathrm{{\rm H}}$ be a cyclic divisible multipilication S-module. Then $\mathrm{{\rm H}}$ is an End-G.B.

Proposition 4.9:

Every fully stable bounded S-module is an endo-GB.

Proof:

Let $\mathrm{{\rm H}}$ be a fully stable bounded S-module, then by [coro 1.2 bounded] $\mathrm{{\rm H}}$ is cyclic and using(prop 3.7), the result is as follows.

Remark 4.10:

Every finitely annihilated S-module is an End-GB S-module.

Proof:

Because every bounded S-module is finitely annihilated and every End-GB S-module is bounded, the result is as follows.

Proposition 4.11:

Let $\mathrm{{\rm H}}$ be a multiplication End-G.B.S-module, then $\mathrm{{\rm H}}$ is finite generated of S-module.

Proof:

Since every End-GB is finitely annihilated, then by [proposition 3.1 f.ann paper] finitely annihilated with multiplication gives the result.

Corollary 4.12:

Let $\mathrm{{\rm H}}$ be an End-G.B. fully stable semi-simple S-module, then $\mathrm{{\rm H}}$ is finitely generated.

Recall that an S-module $\mathrm{{\rm H}}$ is said that finendo if it is finitely generated over $\mathit{End}\left(\mathrm{{\rm H}}\right).$ ¹²

Proposition 4.13:

Let $\mathrm{{\rm H}}$ be a semi-simple Endo-G.B.S-module, then $\mathrm{{\rm H}}$ is a finendo S-module.

Proof:

Let $\mathrm{{\rm H}}$ be End-G.B, then $\mathrm{{\rm H}}$ is finite annihilated and by using [coro 2.3 f.ann paper] implies that $\mathrm{{\rm H}}$ is finendo.

Proposition 4.14:

Let $\mathrm{{\rm H}}$ be a divisible Noetherian S-module, then $\mathrm{{\rm H}}$ is a finite annihilated if and only if $\mathrm{{\rm H}}$ is an Endo-G.B.

Proof:

Suppose that $\mathrm{{\rm H}}$ is finite annihilated and let $A\le \mathrm{{\rm H}}.$ Then, $N$ is a finitely generated submodule, and by the hypotheses, we have ${\mathit{ann}}_S\left(\mathrm{{\rm H}}\right)={\mathit{ann}}_S\left(A\right)$

Now, let $\phi ,\psi \in \mathrm{End}\left(\mathrm{H}\right)$ and as $\phi \left(x\right)=r{x}_1+{\mathit{rx}}_2+\dots +{\mathit{rx}}_n,\forall x\in \mathrm{{\rm H}},$ since $\mathrm{{\rm H}}$ is finitely generated, $\phi \left(x\right)\in \mathrm{{\rm H}}$ and $\psi \left(x\right)=x_1+x_2+\dots +x_n,\forall x\in \mathrm{{\rm H}},$ for some $n\in Z^{+}$ . Then $\phi \left(x\right)\in A$ so ${\mathit{ann}}_S\left(A\right)\subseteq {\mathit{ann}}_S\left(\left(\mathit{\phi o\psi }\right)\left(x\right)\right).$

Let $t\in {\mathit{ann}}_S\left(\left(\mathit{\phi o\psi }\right)\left(x\right)\right),$ then $t\left(\mathit{\phi O\psi }\right)\left(x\right)=0$ implies that $t(r{x}_1+{\mathit{rx}}_2+\dots +{\mathit{rx}}_n)=0$ so that $\mathit{tr}(x_1+x_2+\dots +x_n)=0.$ Hence $\mathit{tr}\in {\mathit{ann}}_S(x_1+x_2+\dots +x_n)$ and $\mathit{tr}\in {\mathit{ann}}_S\left(A\right)={\mathit{ann}}_S\left(\mathrm{{\rm H}}\right).$ Therefore $\mathit{tr}\left(m\right)=0,\forall m\in \mathrm{{\rm H}}.$ Thus, $t\in {\mathit{ann}}_S\left(r\mathrm{{\rm H}}\right)={\mathit{ann}}_S\left(\mathrm{{\rm H}}\right)\subseteq {\mathit{ann}}_S\left(A\right).$

Remark 5.1:

Every prime S-module is Bounded.

Proposition 5.2:

Let $\mathrm{{\rm H}}$ be a cyclic prime module, then $\mathrm{{\rm H}}$ is an End-G.B.

Poof:

Since every prime module is bounded, then by Proposition (4.6) $\mathrm{{\rm H}}$ is an End-G.B.

Proposition 5.3:

Let $\mathrm{{\rm H}}$ be an Endo-G.B.S-module, $0\ne x\in \mathrm{{\rm H}}$ such that the following conditions hold:

Proof:

Since $\mathrm{{\rm H}}$ is an End-G.B. module, then $\mathrm{{\rm H}}$ is bounded so that there exist $x\in \mathrm{{\rm H}}$

${\mathit{ann}}_S\left(x\right)={\mathit{ann}}_S\left(\mathrm{{\rm H}}\right).$ Let $N\subseteq \mathrm{{\rm H}},$ there exist $0\ne t\in S,0\ne \mathit{tx}\in N.\mathrm{It}\text{is}$ obvious that ${\mathit{ann}}_S\left(\mathrm{{\rm H}}\right)\subseteq {\mathit{ann}}_S\left(N\right)\mathrm{\text{and}}{\mathit{ann}}_S\left(N\right)\subseteq {\mathit{ann}}_S\left(\mathit{tx}\right).$

Let $r\in {\mathit{ann}}_S\left(\mathit{tx}\right)$ imply that $r\left(\mathit{tx}\right)=0$ then $\mathit{rt}\in {\mathit{ann}}_S\left(x\right)$ and ${\mathit{ann}}_S\left(x\right)$ be prime ideals. Therefore, $r\in {\mathit{ann}}_S\left(x\right)={\mathit{ann}}_S\left(\mathrm{{\rm H}}\right).$ Therefore, ${\mathit{ann}}_S\left(N\right)\subseteq {\mathit{ann}}_S\left(\mathit{tx}\right)\subseteq {\mathit{ann}}_S\left(x\right)={\mathit{ann}}_S\left(\mathrm{{\rm H}}\right).$

Hence, ${\mathit{ann}}_S\left(N\right)={\mathit{ann}}_S\left(\mathrm{{\rm H}}\right)$ means that $\mathrm{{\rm H}}$ is prime modules.

Recall, a submodule A of an S-module $\mathrm{{\rm H}}$ is called pure if $I\mathrm{{\rm H}}\cap A=\mathit{IA},$ for every ideal I of $S.$ ¹⁴

Proposition 5.4:

Let $S$ be a principle ideal ring and $\mathrm{{\rm H}}$ is uniform F-regular S-module. Then $\mathrm{{\rm H}}$ is a prime module.

Proof:

Let $A\subseteq \mathrm{{\rm H}}\mathit{,}$ then $A$ be a pure submodule and $r\mathrm{{\rm H}}\cap A=\mathit{rA}.$ It is clear that ${\mathit{ann}}_S\left(\mathrm{{\rm H}}\right)\subseteq {\mathit{ann}}_S\left(A\right).$

Now, let $r\in {\mathit{ann}}_S\left(A\right),$ then $\mathit{rA}=0.$ Thus, $r\mathrm{{\rm H}}\cap A=0$ and since $\mathrm{{\rm H}}$ is uniform module, then $A$ is an essential submodule. Therefore, $r\mathrm{{\rm H}}=0$ and $r\in {\mathit{ann}}_S\left(\mathrm{{\rm H}}\right).$

Corollary 5.5:

Let $\mathrm{{\rm H}}$ be a cyclic F-regular uniform S-module, then $\mathrm{{\rm H}}$ is an End-G.B.module.

Proof:

By previous proposition $\mathrm{{\rm H}}$ is prime module implies that $\mathrm{{\rm H}}$ is bounded module and since $\mathrm{{\rm H}}$ is a cyclic, then by (prop. 3.7) $\mathrm{{\rm H}}$ is an End-G.B.

There is a relationship between the scalar S-module and an End-GB, so we present some connections in the next properties and corollaries.

Remember that, let $S$ be an integral domain and let $\mathrm{{\rm H}}$ be an S-module. An element $h\in \mathrm{{\rm H}}$ is called torsion if there is a nonzero element $r\in S$ such that $\mathit{rh}=0,$ denoted by $T\left(H\right).$ ^15,16

Proposition 5.6:

Let $\mathrm{{\rm H}}$ be a scalar torsion-free S-module, where(S is an integral domain) and every submodule is cyclic. Then $\mathrm{{\rm H}}$ is an End-G.B.

Proof:

Let $\phi ,\psi \in \mathit{End}\left(\mathrm{{\rm H}}\right)$ and as $\phi \left(x\right)=\mathit{rx}\mathrm{\text{and}}\psi \left(x\right)=x,\forall x\in \mathrm{{\rm H}}.$ Let $N=\mathit{SX}$ be a sub- module of $\mathrm{{\rm H}}.$ Then $\phi \left(x\right)\in N$ so ${\mathit{ann}}_S\left(N\right)\subseteq {\mathit{ann}}_S\left(\left(\mathit{\phi O}\psi \right)\left(x\right)\right).$

Let $t\in {\mathit{ann}}_S\left(\left(\mathit{\phi O}\psi \right)\left(x\right)\right).$ Since $\mathrm{{\rm H}}$ is scalar torsion-free, then by coro[1.21 scalar thesis] $\mathrm{{\rm H}}$ is quasi-dedekind module.

Thus, $t\left(\left(\mathit{\phi O}\psi \right)\left(x\right)\right)=0$ and $t(\phi \left(\psi \left(x\right)\right)=o$ means that $t\left(\phi \left(x\right)\right)=0$ and $\phi \left(\mathit{tx}\right)=0$ hence $\mathit{tx}=0$ since $\phi$ is monomorphism.