QTAG-modules isomorphic to their fully invariant submodules

A QTAG-module is Ihp-module provided it is isomorphic to a proper h-pure submodule and it is an Id-module in case it is isomorphic to its proper direct summand.

If a QTAG-module,M is h-reduced and also $\frac{M}{H_1\left(M\right)}$ is finitely generated then M is not an Id-module.

A QTAG-module is said to be an If-module provided it is isomorphic to a proper submodule which is fully invariant.

Let $M=\underset{i}{⨁}{M}_i$ , where each $M_i$ is the DSUM of length i. Then N is fully invariant submodule of M if and only if $N=\underset{i}{⨁}{H}_{n_i}\left(M_i\right)$ where $n_i⩽i$ , for every $i\in {\mathbb{Z}}^{+}$ and $n_i⩽n_{i+j}⩽n_i+j$ for $i,j\in {\mathbb{Z}}^{+}$ . A fully invariant submodule N is large in M if and only if $N=\underset{i}{⨁}{H}_{n_i}\left(M_i\right)$ , the above conditions hold and the sequence $<1-n_1,2-n_2,3-n_3,\dots .>$ is unbounded if M itself is not bounded.

Let N be a fully invariant submodule of M. Then $N=N\cap M=⨁(M_i\cap N)=⨁H_{n_i}\left(M_i\right)$ . Now $n_i⩽i$ for all $i\in {\mathbb{Z}}^{+}$ and the first condition holds.

If $N=0$ , then $H_{n_i}\left(M_i\right)=0$ for every i, therefore $n_i=i$ for every i and the second condition holds.

If $N\ne 0$ , then a least positive integer k exists such that $H_{n_k}\left(M_k\right)\ne 0$ . We claim that $H_{n_i}\left(M_i\right)\ne 0$ for all $i⩾k$ where $M_i\ne 0$ . Since $\mathit{Soc}\left(M_k\right)=\mathit{Soc}\left(H_{k-1}\right(M_k\left)\right)\subseteq N$ implies that $\mathit{Soc}\left(H_{k-1}\right(M\left)\right)\subseteq \mathit{Soc}\left(N\right)$ . Also $\mathit{Soc}\left(M_i\right)=\mathit{Soc}\left(H_{k-1}\right(M_i\left)\right)$ for all $i⩾k$ , we have ${\mathit{Soc}}_{(}{M}_i)\subseteq N\cap M_i=H_{n_i}(M_i)$ and the assertion follows.

Now suppose $N\ne 0$ and $M_i\ne 0\ne M_{i+j}$ . If $H_{n_{i+j}}\left(M_{i+j}\right)=0$ then $H_{n_i}\left(M_i\right)=0$ and $n_i=i,n_{i+j}=i+j=n_i+j$ and the second condition holds.

Therefore we assume that $H_{n_{i+j}}\left(M_{i+j}\right)\ne 0$ . Consider $x\in M_i$ such that $H\left(x\right)⩾n_{i+j}$ . $y\in H_{n_{i+j}}\left(M_{i+j}\right)$ such that $H\left(y\right)=n_{i+j}$ . Now there exists an endomorphism f of M which maps y onto x. Hence $x\in N$ and $H_{n_{i+j}}\left(M_i\right)\subseteq N\cap M_i=H_{n_i}\left(M_i\right)$ . Thus, $n_i⩽n_{i+j}$ .

Now suppose $H_{n_i}\left(M_i\right)=0$ . Then $n_i=i$ so $n_{i+j}⩽i+j=n_i+j$ .

If $H_{n_i}\left(M_i\right)\ne 0$ and $y\in M_{i+j}$ such that $H\left(y\right)⩾n_i+j$ . We may choose $x\in M_i$ such that $H\left(x\right)=n_i$ . Then $e\left(x\right)=i-n_i$ and $e\left(y\right)⩽i+j-(n_i+j)=i-n_i$ .

Again there exists an endomorphism f of M with $f\left(x\right)=y$ . Thus $y\in N$ and we have $H_{n_{i+j}}\left(M_{i+j}\right)\subseteq N\cap M_{i+j}=H_{n_{i+j}}\left(M_{i+j}\right)$ , therefore $n_{i+j}⩽n_i+j$ .

If $M_i\ne 0\ne M_{i+j}$ , then $n_i⩽n_{i+j}\ne n_i+j$ but if $M_i=0$ , we may define $n_i$ so that this inequality holds for all i. Thus all fully invariant submodules of M are the direct sums of $H_{n_i}\left(M_i\right)$ . If N is a large submodule of M and M is unbounded, N is also unbounded. Therefore $<1-n_1,2-n_2,3-n_3,\dots ..>$ must be unbounded.

For the converse, suppose $N=⨁H_{n_i}\left(M_i\right)$ where $n_i⩽i$ for all $i\in {\mathbb{Z}}^{+}$ and $n_i⩽n_{i+j}⩽n_i+j$ for all $i,j\in {\mathbb{Z}}^{+}$ . To establish the fully invariance of N, we consider any $i\in {\mathbb{Z}}^{+}$ and $x\in H_{n_i}\left(M_i\right)$ . We have to show that for any endomorphism f of $M,f\left(x\right)\in N$ .

Consider $x\ne 0$ , such that $f\left(x\right)=x_1+x_2+\dots ..+x_3$ where $x_r\in M_r$ and $H\left(x\right)⩽H\left(f\right(x\left)\right)=\mathit{min}\left(H\right(x_k\left)\right),1⩽k⩽l,e\left(x\right)⩾e\left(f\right(x\left)\right)$ =max $\left(e\right(x_k\left)\right|1⩽k⩽l)$ . If $k⩽i$ , then $H\left(x_k\right)⩾H\left(x\right)⩾n_i$ so $x_k\in H_{n_i}\left(M_k\right)\subseteq H_{n_k}\left(M_k\right)$ , because $n_k⩽n_i$ , hence $x_k\in N$ . If $k=i+j$ then $e\left(x_k\right)⩽e\left(x\right)⩽i-n_i=i+j-(n_i+j)⩽i+j-n_{i+j}$ because $n_{i+k}⩽n_i+k$ . Thus $x_k\in H^{(i+j-n_{i+j})}\left(M_{i+j}\right)=H_{n_{i+j}}\left(M_{i+j}\right)\subseteq N$ .

This implies that N is a fully invariant submodule of M. If M is unbounded and $<1-n_1,2-n_2,3-n_3,\dots .>$ is also unbounded, then N is unbounded and is therefore a large submodule of M.

A bounded QTAG-module can not be an If-module.

Let M be a bounded QTAG-module with $H\left(x\right)<k$ for every element x in M. Then M can be expressed as $\underset{i=1}{\overset{k}{⨁}}{M}_i$ where each $M_i$ is a DSUM of length i. If N is a fully invariant submodule of M, then by Theorem 1, $N=H_{n_1}\left(M_1\right)\oplus H_{n_2}\left(M_2\right)\oplus ...\oplus H_{n_k}\left(M_k\right)$ such that $n_j⩽j,n_j⩽n_{j+l}⩽n_j+l$ , for all $j,l\in N$ .

If $n_k=0$ then $n_1⩽n_2⩽....⩽n_k=0$ implies $N=M_1\oplus M_2\oplus \dots ..\oplus M_k=M$ , thus N is not a proper submodule of M. With no loss of generality, we assume that $n_k⩾1$ . Now $H_{n_k}\left(M_k\right)$ is a direct sum of uniserial modules of length $k-n_k$ which implies that no direct summand of N is a uniserial module of length k. Therefore N is not isomorphic to M.

Let $M=\underset{i\in I}{⨁}{M}_i$ be a QTAG-module. If $N=\underset{i\in I}{⨁}{N}_i,K=\underset{i\in I}{⨁}{K}_i$ such that $N_i$ and $K_i$ are the submodules of $M_i,\forall i$ then $N_i=K_i$ for all $i\in I$ .

Let $x_i\in N_i$ . Then $x_i\in N=⨁K_i$ and we may write $x_i=y_{i1}+y_{i2}+\dots +y_{\mathit{ik}}$ where $y_{\mathit{ij}}\in K_j,j=1,2,\dots ,k$ . Since $N_i$ and $K_i$ are the submodules of $M_i,x_i\in M_i,y_{\mathit{ij}}\in M_{\mathit{ij}}$ and M is a direct sum of $M_i$ ’s, $x_i=y_{\mathit{ij}}$ for some j. In fact $y_{\mathit{ij}}=x_i$ and $y_{\mathit{ik}}=0$ if $j\ne k$ . Therefore $x_i\in K_i$ or $N_i\subseteq K_i$ . Similarly $K_i\subseteq N_i$ and $N_i=K_i$ for all i.

Consider a QTAG-module $M=⨁M_i$ with each $M_i$ fully invariant. M is an If-module if and only if there exists at least one $M_i$ which is an If-module.

Let us consider N as a proper submodule of M which is fully invariant such that $M\simeq N$ . Since N is fully invariant, $N=⨁(N\cap M_i)=⨁N_i$ where $N_i=N\cap M_i$ . Let $f:M⟶N$ be an isomorphism. Then $f_i$ , the restriction of f on $M_i$ is an endomorphism of $M_i$ . For $x\in M,x=x_{i1}+\dots +x_{\mathit{ik}}$ where $x_{\mathit{ij}}\in M_{\mathit{ij}}$ and we have $f\left(x\right)=f_{i1}\left(x_{i1}\right)+\dots +f_{\mathit{ik}}\left(x_{\mathit{ik}}\right)$ . Thus $N=f\left(M\right)=\underset{i\in I}{⨁}{f}_i\left(M_i\right)=⨁N_i$ and by Lemma 1, $f_i\left(M_i\right)=N_i$ . Since $\mathit{Ker}\left(f\right)=0,\mathit{Ker}\left(f_i\right)=0\forall i\in I$ and each $f_i$ is also an isomorphism from $M_i$ onto $N_i$ . As $M\ne N,M_{\mathit{il}}\simeq N_{\mathit{il}}$ and $M_{\mathit{il}}\ne N_{\mathit{il}}$ for some il implying that $M_{\mathit{il}}$ is an If-module.

Conversely, consider $M=⨁M_i$ where each $M_i$ is a fully invariant submodule of M. Let $M_{i_0}$ be an If-module for some $i_0\in I$ . Therefore there exists a submodule $N_{i_0}$ of $M_{i_0}$ which is proper and invariant such that $N_{i_0}\simeq M_{i_0}$ . Consider $N=N_{i_0}\oplus \left(\underset{j\ne i_0}{⨁}{M}_j\right)$ . Since $N_{i_0}\ne M_{i_0},N$ is a proper submodule of M. As $N_{i_0}\simeq M_{i_0}$ , we have $N=N_{i_0}\oplus \left(\underset{j\ne i_0}{⨁}{M}_j\right)\simeq M_{i_0}\oplus \left(\underset{j\ne i_0}{⨁}{M}_j\right)=⨁M_i=M$ $i.e$ $N\simeq M$ . Now consider an arbitrary endomorphism f of M and $x\in N$ . Now $x=x_{i_0}+y_{i_1}+\dots +y_{i_k}$ where $y_{\mathit{ij}}\in M_{\mathit{ij}},x_{i_0}\in N_{i_0}$ . We have $f\left(x\right)=f(x_{i_0}+y_{i_1}+\dots +y_{i_k})=f\left(x_{i_0}\right)+f\left(y_{i_1}\right)+\dots +f\left(y_{i_k}\right)$ . Since $M_{\mathit{ij}}$ ’s are fully invariant $f\left(y_{\mathit{ij}}\right)\in M_{\mathit{ij}}$ . Also $N_{i_0}$ is fully invariant in $M_{i_0}$ which is fully invariant in M, thus $f\left(x_{i_0}\right)\in N_{i_0}$ . Now we have $f\left(x\right)\in N\forall x\in N$ , therefore N is fully invariant in M. Since N is a proper submodule of M with $N\simeq M,M$ is an If-module and we are done.

Let M be a separable QTAG-module. A strictly increasing sequence of positive integers $i_0,i_1,i_2,\dots ,i_n,\dots$ is said to be admissible for M if $f_M\left(k\right)={\sum }_{i=l_k}^{k+1}{f}_M\left(i\right),k<\omega$ .

Let M be an unbounded QTAG-module with M is a direct sum of uniserial modules and all Ulm-Kaplansky invariants of M are finite. Then M is not an If-module if and only if there exists only one admissible sequence for M, consisting of all non negative numbers.

We may express $M=\underset{k\in N}{⨁}{M}_k$ where each $M_k$ is a DSUM of length k. Thus $f_M\left(i\right)=g\left(M_{i+1}\right)$ for each $i\in {\mathbb{Z}}^{+}$ . Suppose M is not an If-module which has an admissible sequence $l_0,l_1,\dots$ different from $0,1,2,\dots$ . Since the Ulm-Kaplansky invariants of M are finite we have two cases. Either $l_0\ne 0$ or $l_0=0$ . If $l_0\ne 0$ consider the submodule N of M such that

$N=H_1\left(M_1\right)\oplus H_2\left(M_2\right)\oplus \dots \oplus H_{l_0}\left(M_{l_0+1}\right)\oplus H_{l_0+1}\left(M_{l_0+2}\right)\oplus \dots \oplus H_{l_1-1}\left(M_{l_1}\right)\oplus H_{l_1-1}\left(M_{l_1+1}\right)\oplus H_{l_1}\left(M_{l_1+2}\right)\oplus \dots \oplus H_{l_2-2}\left(M_{l_2}\right)\oplus H_{l_2-2}\left(M_{l_2+1}\right)\oplus \dots \oplus H_{i_2}\left(M_{l_2+3}\right)\oplus \dots \oplus H_{i_3-3}\left(M_{i_3}\right)\oplus \dots$ .

In short we may say that $N=⨁H_{n_k}\left(M_k\right)$ where $n_{l_j}=n_{l_{j+1}}=l_j-j,n_{l_{j+m}}=l_j-j+m-1$ . Now N is a proper submodule of M. By Theorem 1, N is fully invariant in M. Since Ulm-Kaplansky invariants of M and N are equal $M\simeq N$ because $f_N\left(m\right)=f_M\left(l_m\right)+f_M\left(l_{m+1}\right)+\dots .+f_M(l_{m+1}-1)=f_M\left(m\right)$ for each $m\in {\mathbb{Z}}^{+}$ . Therefore $N\simeq M$ but $N\ne M$ and thus M is an If-module. This contradiction proves our assertion. If $l_0=0$ , we have a least natural number j such that $l_{j+1}>j+1$ . Now $l_0=0,l_1=1,\dots l_j=j$ and the admissible sequence has the form $0,1\dots j,l_{j+1},l_{j+2}$ ….For these sequences we may write $$\begin{array}{cc}\hfill f_M\left(0\right)=& f_M\left(0\right)\hfill \\ \hfill f_M(j-1)=& f_M(j-1)\hfill \\ \hfill f_M\left(j\right)=& f_M\left(j\right)+f_M(j+1)+\dots +f_M(j_m-1)\hfill \\ \hfill f_M\left(k\right)=& f_M\left(l_k\right)+f_M\left(l_{k+1}\right)+\dots +f_M(l_{k+1}-1)\hfill \end{array}$$

Now the sum of the right hand side of the last inequality consists of more than one summand.

We consider $$\begin{array}{cc}\hfill N=& M_1\oplus M_2\oplus \dots \oplus M_{k+1}\oplus H_1\left(M_{k+2}\right)\oplus \dots \hfill \\ \hfill & H_{l_{j+1}-j-1}\left(M_{l_{j+1}}\right)\oplus H_{l_{j+1}-j-1}(M_{l_{j+1}}+1)\oplus \dots \hfill \end{array}$$

By Theorem 1, N is fully invariant submodule of M. Now $$\begin{array}{cc}\hfill f_N\left(0\right)=& f_M\left(0\right)\hfill \\ \hfill f_N\left(1\right)=& f_M\left(1\right)\hfill \\ \hfill f_N(j-1)=& f_M(j-1)\hfill \\ \hfill f_N\left(j\right)=& f_M\left(j\right)+f_M(j+1)+\dots +f_M(l_{j+1}-1)\hfill \\ \hfill f_N\left(r\right)=& f_M\left(l_r\right)+f_M\left(l_{r+1}\right)+\dots +f_M(l_{r+1}-1),r>j\hfill \end{array}$$

These two set of equalities ensure that $N\simeq M$ . Since $N\ne M,M$ is an If-module, again a contradiction.

To prove it conversely, consider a fully invariant submodule N of M. Now by Theorem 1, $N=\underset{k\in N}{⨁}{H}_{n_k}\left(M_k\right)$ such that $n_k⩽k,n_k⩽n_{k+l}⩽n_k+l$ for all $k,l\in N$ .

Now $f_N\left(n\right)=g\left(\underset{n\in N}{⨁}\right(H_{n_k}\left(M_k\right)\left)\right)$ , where $H_{n_k}\left(M_k\right)$ is a DSUM of length $(n+1)$ .

Therefore $$\begin{array}{ccc}{f}_N\left(n\right)\hfill & =\hfill & g\left(\oplus H_{n_k}\right(M_k\left)\right)\mathrm{where}k-n_k=n+1\hfill \\ \hfill & =\hfill & {\displaystyle \sum _{k\in N}}{f}_M(k-1)\mathrm{suchthat}k-n_k-1=n\hfill \\ \hfill \end{array}$$

Again by Theorem 1,

$(k+1)-n_{k+1}-1⩾(k+1)-(n_k+1)-1=k-n_k-1$

$(k+1)-n_{k+1}-1⩽(k+1)-n_k-1=(k-n_k-1)+1$

On putting   $j_n=\underset{k\in N}{\min }\{k-1|k-n_k-1=n\}$ , we have.

$f_N\left(n\right)={\sum }_{j=j_n}^{j_{n+1}-1}{f}_M\left(j\right)$

If $N\simeq M$ , then $f_M\left(n\right)=f_N\left(n\right)={\sum }_{j=j_n}^{j_{n+1}-1}{f}_M\left(j\right)$   for any $n\in {\mathbb{Z}}^{+}$ .

Now the sequence $i_0,i_1,\dots .,i_n,\text{.}$ . is admissible for M, therefore $j_n=n$ for any n. Since $j_n=\underset{k\in N}{\min }\{k-1|k-n_k-1=n\}$ we have $n_k=0$ for any k or $M=N$ implying that M is not an If-module. Hence proved.

A separable QTAG-module M is not an If-module if its basic submodule is not an If-module.

Let B be the basic submodule of the separable QTAG-module such that B is not an If-module. Without loss of generality we assume that M is h-reduced. If M is bounded then by Theorem 2, M is not an If-module.

Suppose M is unbounded If-module. Then there exists a proper fully invariant submodule N of M such that $N\simeq M$ . Since M is h-reduced, so there is no element whose height is infinite. N is unbounded fully invariant submodule of M, therefore $N\cap B$ is a basic submodule of N. If $N\cap B=0$ , then N is h-divisible because $\frac{N+B}{B}$ is a submodule of a h-divisible module. Since M is h-reduced, it is not possible.

If $N\cap B=B$ then N is a large submodule of M and $B\subset N$ implies $N+B=N$ which is not possible as N is a proper submodule of M. Now we may conclude that $N\subset B$ is a proper submodule of M. Since $N\simeq M$ , the basic submodule of M and N are isomorphic (Mehdi and Khan, 1984) and $N\cap B\simeq B$ . As N is a large submodule of $M,N\cap B$ is a large submodule of B (Mehdi et al., 2014). Thus we obtain that the basic submodule B of M has a proper fully invariant submodule $N\cap B$ , is isomorphic to B which is a contradiction proving the result.

Let M be a separable unbounded QTAG-module with finite $\mathit{Ulm}-\mathit{Kaplansky}$ invariants. Then it is not an If-module if there exists only one admissible sequence for it consisting of all non negative integers.

Let B be the basic submodule of a separable QTAG-module with finite Ulm-Kaplansky invariants. Let $0,1,2\dots$ be the only admissible sequence for M. We have to show that B is not an If-module. Since B is a DSUM and $f_M\left(k\right)=f_B\left(k\right)$ , by Theorem 4, B is not an If-module and the result follows.

An unbounded separable QTAG- module is not an If-module if its $\mathit{Ulm}-\mathit{Kaplansky}$ invariants are finite and form an increasing sequence.

Let M be an unbounded separable QTAG-module. Suppose the sequence of Ulm-Kaplansky invariants of M is increasing and all the invariants are finite. Consider an admissible sequence $i_o,i_1,\dots ,i_n$ for M. Now

$f_M\left(k\right)={\sum }_{i=i_k}^{i_{k+1}-1}{f}_M\left(i\right)=f_M\left(i_k\right)+f_M(i_k+1)+\dots +f_M(i_{k+1}-1)$ , where k is an arbitrary non negative integer. Since the sequence $\left\{f_M\right(k\left)\right\}$ is increasing, $f_M\left(k\right)=f_M\left(i_k\right)$ and $i_k=k$ for every $k\in {\mathbb{Z}}^{+}$ . Therefore the admissible sequence $i_o,i_1,\dots ,i_n$ coincides with $0,1,2,\dots$ and by Theorem 6, M is not an If-module.

A closed QTAG-module M is an If-module if and only if its basic submodule B is an If-module.

Let M be a closed QTAG-module and B its basic submodule which is an If-module. Now there exists a proper fully invariant submodule N of B such that $B\simeq N$ . Now N is a proper large submodule of B. This N can be extended to $N\prime$ , a proper large submodule of M such that $N\prime \cap B=N$ . Now N is a basic submodule of $N\prime$ . Since $N\prime$ is a large submodule of a closed module it is also closed. Now M has a proper fully invariant submodule $N\prime$ such that the basic submodule B of M is isomorphic to the basic submodule N of $N\prime$ .

Now M and $N\prime$ , both are closed therefore $M\simeq N\prime$ and M is an If-module. The converse is trivial.

Let M be a closed QTAG-module with finite Ulm-Kaplansky invariants. Then the following conditions are equivalent.

• $\left(i\right)$ M is not an If-module.

• $\left(\mathit{ii}\right)$ No basic submodule of M is an If-module.

• $\left(\mathit{iii}\right)$ There exists only one admissible sequence for M consisting of all nonnegative integers.

Theorem 7 ensures that $\left(i\right)\Rightarrow \left(\mathit{ii}\right)$ .

$\left(\mathit{ii}\right)\Rightarrow \left(\mathit{iii}\right)$ . Let B be a basic submodule of M, and B is not an If-module. If M is bounded then $M=B$ . Otherwise M is unbounded and B is also unbounded. Since $f_M\left(k\right)=f_B\left(k\right)$ , by Theorem 4, we have that M has only one admissible sequence of the form $0,1,2,\dots$

$\left(\mathit{iii}\right)\Rightarrow \left(i\right)$ . If M is bounded then it is not an If-module by Theorem 2. If M is unbounded for which there exists only one admissible sequence of the form $0,1,2,\dots$ then its basic submodule has the same property. Therefore by Theorem 4, B is not an If-module. Since $\left(\mathit{ii}\right)$ and $\left(i\right)$ are equivalent, M is not an If-module.

Let M be a closed QTAG-module with finite Ulm-Kaplansky invariants. If for all $n\in {\mathbb{Z}}^{+},\exists k\in \mathbb{N}$ such that $f_M\left(n\right)=f_M(n+k),M$ is an If-module.

Let M be a closed QTAG-module with finite Ulm-Kaplansky invariants. Suppose there exists $k<\omega$ such that for all $n\in {\mathbb{Z}}^{+},f_M\left(n\right)=f_M(n+k)$ . Then the sequence $k,k+1,k+2,\dots$ is admissible for this module and by Theorem 8, M is an If-module.

Let M be a closed QTAG-module such that there exists a $m\in \mathbb{N}$ such that $f_M\left(n\right)=m\forall n\in {\mathbb{Z}}^{+}$ , then M is an If-module.

Let M be a closed QTAG-module and $f_M\left(n\right)=m$ for every $n\in {\mathbb{Z}}^{+}$ where $m<\omega$ . Now $f_M\left(n\right)=f_M(n+1)\forall n\in {\mathbb{Z}}^{+}$ and the sequence ${\left\{f_M\right(n\left)\right\}}_{n<\omega }$ is admissible for M. Therefore by Corollary 2.,  M is an If-module.