On elongations of QTAG-modules

Abstract

Mehdi studied (ω+k)-projective QTAG-modules with the help of their submodules contained in H^k(M) (the submodule generated by the elements of exponents at most k). These modules contain nice submodules N contained in H^k(M) such that M/N is a direct sum of uniserial modules. Here, we investigate the class

of QTAG-modules, containing nice submodules N⊆H^k(M) such that M/N is totally projective. We also study strong ω-elongation of totally projective QTAG-modules by (ω+k)-projective QTAG-modules.

Introduction

Throughout this paper, all rings will be associative with unity, and modules M are unital QTAG-modules. An element x∈M is uniform, if xR is a non-zero uniform (hence uniserial) module and for any R-module M with a unique composition series, d(M) denotes its composition length. For a uniform element x∈M, e(x)=d(x R) and $H_M\left(x\right)=\text{sup}\left\{d\left(\frac{\mathit{\text{yR}}}{\mathit{\text{xR}}}\right)\mid y\in M,x\in \mathit{\text{yR}}\text{and}y\text{uniform}\right\}$ are the exponent and height of x in M, respectively. H _k (M) denotes the submodule of M generated by the elements of height at least k and H^k(M) is the submodule of M generated by the elements of exponents at most k.M is h-divisible if $M=M^1=\bigcap _{k=0}^{\infty }{H}_k\left(M\right)$ and it is h-reduced if it does not contain any h-divisible submodule. In other words, it is free from the elements of infinite height. A h-reduced QTAG-module M is called totally projective if it has a nice system. A submodule N of M is h-pure in M if N∩H _k (M)=H _k (N), for every integer k≥0. For a limit ordinal α, $H_{\alpha }\left(M\right)=\bigcap _{\rho <\alpha }{H}_{\rho }\left(M\right),$ for all ordinals ρ<α and it is α-pure in M if H _σ (N)=H _σ (M)∩N for all ordinals σ<α. A submodule N⊂M is nice [1] Definition 2.3 in M, if H _σ (M/N)=(H _σ (M)+N)/N for all ordinals σ, i.e. every coset of M modulo N may be represented by an element of the same height. A QTAG-module M is said to be separable, if M¹=0. The cardinality of the minimal generating set of M is denoted by g(M). For all ordinals α, f _M (α) is the α th- Ulm invariant of M and it is equal to g(Soc(H _α (M))/Soc(H_α+1(M))). For a QTAG-module M, there is a chain of submodules M⁰⊃M¹⊃M²⋯⊃M^τ=0, for some ordinal τ. M^σ+1=(M^σ)¹, where M^σ is the σ th- Ulm submodule of M. Singh [2] proved that the results which hold for TAG-modules also hold good for QTAG-modules. Notations and terminology are followed from [3, 4]

Elongations of totally projective QTAG-modules by (ω+k)-projective QTAG-modules

Recall that a QTAG-module M is (ω+1)-projective if there exists submodule N⊂H¹(M) such that M/N is a direct sum of uniserial modules and a QTAG module M is (ω+k)-projective if there exists submodule N⊂H^k(M) such that M/N is a direct sum of uniserial modules [5]. A QTAG-module is an ω-elongation of a totally projective QTAG-module by a (ω+k)-projective QTAG-module if and only if H _ω (M) is totally projective and M/H _ω (M) is (ω+k)-projective. Suppose ${\mathcal{A}}_k$ denotes the family of QTAG-modules M which contain nice submodules N⊆H^k(M) free from the elements of infinite height, such that M/N is totally projective. The main goal of this section is to find a condition for the modules of the family ${\mathcal{A}}_k$ to be isomorphic. To achieve this goal we need some results. We start with the following:

Lemma 1

Let M be a QTAG-module and N⊆M such that N∩H _ω (M)=0, then N is nice in M if and only if N⊕H _ω (M) is nice in M.

Proof

Suppose N is nice in M. Since a submodule K is nice in M if M/K is separable, it is sufficient to show that M/(N⊕H _ω (M)) is separable. If $H\left(\bar{x}\right)$ is infinite in M/(N⊕H _ω (M)), where $\bar{x}=x+N\oplus H_{\omega }\left(M\right),$ then there exist a sequence {x _k } in N⊕H _ω (M) such that H(x+x _k )≥k, for every k∈Z⁺. If x _k =y _k +z _k where y _k ∈N, z _k ∈H _ω (M); then H(x+y _k )≥k and the coset x+N has infinite height in M/N. Now for some u∈N, H(x+u)≥ω and x=−u+(x+u)∈N⊕H _ω (M), thus N⊕H _ω (M) is nice in M. For the converse suppose N⊕H _ω (M) is nice in M. Since H _ω (M)⊆N⊕H _ω (M), M/(N⊕H _ω (M)) must be separable. By the previous argument, an element x+N has height ω in M/N if and only if it can be represented by an element of H _ω (M) and the result follows. □

Lemma 2

If N is nice submodule of H^k(M)⊆M which is bounded by k such that N∩H _ω (M)=0 and M/N is totally projective, then

• (i) M/(N⊕H _ω (M)) is a direct sum of uniserial modules and

• (ii) M/H _ω (M) is (ω+k)-projective.

Proof

Since N is a nice submodule we have H _ω (M/N)=(H _ω (M)+N)/N. Now, M/(N⊕H _ω (M))≅(M/N)/H _ω (M/N) and M/N is totally projective; therefore, (M/N)/H _ω (M/N) is a direct sum of uniserial modules. Thus, M/(N⊕H _ω (M)) is also a direct sum of uniserial modules. Again, (N⊕H _ω (M))/H _ω (M) is a submodule of M/H _ω (M), which is bounded by k. Thus, M/H _ω (M) is (ω+k)-projective module. □

Lemma 3

Let M be a QTAG-module and N a submodule of H^k(M)⊆M such that N∩H _ω (M)=0. If H _ω (M) is totally projective and M/(N⊕H _ω (M)) is a direct sum of uniserial modules, then M/N is totally projective.

Proof

Now, N⊕H _ω (M) is nice in M; therefore, by Lemma 1, N is a nice submodule of M. This implies that H _ω (M/N)=(N⊕H _ω (M))/N≅H _ω (M) because N∩H _ω (M)=0. Again,

$$\begin{array}{c}(M/N)/H_{\omega }(M/N)=(M/N)/\left[\right(N\oplus H_{\omega }\left(M\right))/N]\\ \cong M/(N\oplus H_{\omega }(M\left)\right)\end{array}$$

is a direct sum of uniserial modules implying that M/N is totally projective. □

Lemma 4

Let N be a submodule of H^k(M)⊆M such that N∩H _ω (M)=0. Then the Ulm-invariants of N⊕H _ω (M) with respect to M can be determined by H^k(M).

Proof

The σ th Ulm-invariant of N⊕H _ω (M) with respect to M is

$$\begin{array}{l}g\left(\right.\text{Soc}\left(H_{\sigma }\right(M\left)\right)/\\ \left(\right(H_{\sigma +1}\left(M\right)+(N\oplus H_{\omega }(M\left)\right))\cap \text{Soc}(H_{\sigma }\left(M\right)\left)\right)\left.\right)\left[5\right].\end{array}$$

If σ is an integer, then H_σ+1(M)+N⊕H _ω (M)=H_σ+1(M)+N and if x∈H_σ+1(M), y∈N such that x+y∈Soc(H_σ+1(M)+N), then there exist x^′, y^′ such that d(x^′R/x R)=k−1=d(y^′R/y R). This implies that x ∈ H_σ+1 (H^k(M)) and Soc (H_σ+1(M)+N+H _ω (M))=Soc [ H_σ+1(H^k(M))+N] and if σ≥ω, then H _σ (M)⊆N+H _ω (M) and the σ th relative Ulm-invariant is zero. □

Definition 1

A QTAG-module M is h-distinctive if there is a monomorphism from M into a direct sum of uniserial modules that does not decrease heights.

Remark 1

Let M be a QTAG-module and N a submodule of M such that M/N is a direct sum of uniserial modules. If N is h-distinctive, then M is also a direct sum of uniserial modules.

Now, we consider the family ${\mathcal{A}}_k$ of QTAG-modules M which contains nice submodules N⊆H^k(M) free from the elements of infinite height, such that M/N is totally projective. In fact, any module in ${\mathcal{A}}_k$ is an extension of a totally projective module H _ω (M) by a separable (ω+k)-projective module M/H _ω (M) or M is a ω-elongation of a totally projective module by a separable (ω+k)-module.

Theorem 1

A direct summand of a module in ${\mathcal{A}}_k$ is again in ${\mathcal{A}}_k.$

Proof

Let $M\in {\mathcal{A}}_k$, such that M=T⊕K and N⊆H^k(M) a nice submodule of M, N∩H _ω (M)=0 and M/N totally projective. We define

$$M_1=T\cap (N\oplus H_{\omega }(M\left)\right)\text{and}{M}_2=K\cap (N\oplus H_{\omega }(M\left)\right).$$

Now, by Lemma 2, M/(N⊕H _ω (M)) is a direct sum of uniserial modules; therefore

$$\begin{array}{c}T/M_1\cong (T+(N\oplus H_{\omega }\left(M\right)\left)\right)/(N\oplus H_{\omega }(M\left)\right)\\ \subseteq M/(N\oplus H_{\omega }(M\left)\right)\hfill \end{array}$$

is also a direct sum of uniserial modules. Again, H _ω (M)⊆M₁⊕M₂⊆N⊕H _ω (M), therefore

$$M_1\oplus M_2=H_{\omega }\left(M\right)\oplus (N\cap (M_1\oplus M_2\left)\right).$$

Since H _ω (M)=H _ω (T)⊕H _ω (K),

$$M_1=H_{\omega }\left(T\right)\oplus [M_1\cap \left(H_{\omega }\left(K\right)\oplus (N\cap (M_1\oplus M_2\left)\right)\right)].$$

Now, the submodule M₁∩(H _ω (K)⊕(N∩(M₁⊕M₂))) is contained in H^k(M) and free from the elements of infinite height. Since H _ω (T) is a summand of the totally projective module H _ω (M), by applying Lemma 3, on T and $M_1\cap \left(\right(N\cap (M_1\oplus M_2))\oplus H_{\omega }(K\left)\right),T\in {\mathcal{A}}_k,$ which completes the proof. □

Theorem 2

Let $M,M^{\prime }\in {\mathcal{A}}_k.$ Then M is isomorphic to M^′ if and only if there is a height-preserving isomorphism f:H^k(M)→H^k(M^′).

Proof

Consider the height-preserving isomorphism f: H^k(M)→H^k(M^′). Since $M,M^{\prime }\in {\mathcal{A}}_k,$ there are nice submodules N⊆H^k(M)⊆M and N^′⊆H^k(M^′)⊆M^′ such that N∩H _ω (M)=0, N^′∩H _ω (M^′)=0 and M/N, M^′/N^′ are totally projective. By Lemma 2, M/(N⊕H _ω (M)) and M^′/(N^′⊕H _ω (M^′)) are direct sums of uniserial modules. We put

$$K=(N\oplus H_{\omega }(M\left)\right)\cap \left(f^{-1}\right(N^{\prime })\oplus H_{\omega }(M\left)\right)$$

and consider the exact sequence

$$0\to (N\oplus H_{\omega }(M\left)\right)/K\to M/K\to M/(N\oplus H_{\omega }(M\left)\right)\to 0.$$

Let x∈N,y∈H _ω (M) such that H(x+y+K)≥m. Since y∈K, x+y+K=x+K and H(x+K)≥m, and there exists some z∈M such that d[ (z+K)R/(x+K)R]=m. Now there is some z^′∈(x+K)R such that z^′−x∈K. Therefore, (z^′−x)∈(f⁻¹(N^′)⊕H _ω (M)) and for some u^′∈N^′, z^′′=x+f⁻¹(u^′) where $H_{M^{\prime }}\left(f\right(x)+u^{\prime })=H_M\left(z^{\mathrm{\prime \prime }}\right)\ge m$. This implies that the height of the coset f(x)+u^′+(N^′⊕H _ω (M^′)) is greater than equal to m in M^′/(N^′⊕H _ω (M^′)). The map $\bar{f}:\left((N\oplus H_{\omega }(M\left)\right)/K\right)\to M^{\prime }/(N^{\prime }\oplus H_{\omega }(M\left)\right)$ is a monomorphism which does not decrease heights; thus, (N⊕H _ω (M))/K is h-distinctive, and by Remark 1, M/K is a direct sum of uniserial modules. Similarly, M^′/K^′ is a direct sum of uniserial modules, where

$$K^{\prime }=\left(f\right(N)\oplus H_{\omega }(M^{\prime }\left)\right)\cap (N^{\prime }\oplus H_{\omega }(M^{\prime }\left)\right).$$

Since f is height-preserving isomorphism, it maps H^k(K) onto H^k(K^′), where

$$H^k\left(K\right)=\left(N\oplus H_{\omega }\left(H^k\right(M\left)\right)\right)\cap \left(f^{-1}\left(N^{\prime }\right)\oplus H_{\omega }\left(H^k\right(M\left)\right)\right).$$

Again, if we put

$$\begin{array}{l}T=N\cap \left(f^{-1}\left(N^{\prime }\right)\oplus H_{\omega }\left(H^k\right(M\left)\right)\right),\\ T^{\prime }=N^{\prime }\cap \left(f\left(N\right)\oplus H_{\omega }\left(H^k\right(M^{\prime }\left)\right)\right),\end{array}$$

then K=T⊕H _ω (M), K^′=T^′⊕H _ω (M^′). From Lemma 3, M/T and M^′/T^′ are totally projective. Now f(T)⊕H _ω (M^′)=T^′⊕H _ω (M^′); therefore, f induces a height-preserving isomorphism g₁:T→T^′. The Ulm-invariants of H _ω (M) and H _ω (M^′) are determined by the cardinality of the minimal generating sets of their socles and f is height preserving therefore these are equal for H _ω (M) and H _ω (M^′). As these modules are totally projective, there is an isomorphism g₂:H _ω (M)→H _ω (M^′), which is again height preserving. Now, the isomorphisms g₁, g₂ help us to define an isomorphism ϕ:K→K^′, where K and K^′ are nice in M and M^′, respectively. Since the submodules T and T^′ have elements of finite heights only and the modules H _ω (M) and H _ω (M^′) have elements of height ≥ω only, ϕ must be height preserving. Therefore, by Lemma 4, the Ulm-invariants of K with respect to M can be determined with the help of H^k(M). As

$$f\left(H^k\right(K\left)\right)=H^k\left(K^{\prime }\right),f_{\alpha }(K,M)=f_{\alpha }(K^{\prime },M^{\prime })$$

for all α and M≅M^′[6, 7]. □

Remark 2

Thus, the isomorphic modules M in ${\mathcal{A}}_k$ can be identified by H^k(M).

Strong ω-elongations of totally projective QTAG-modules by (ω+k)-projective QTAG-modules

In the last section, we studied ω-elongations of a totally projective module by (ω+k)-projective module where H _ω (M) is totally projective and M/H _ω (M) is (ω+k)-projective. Here, we study strong ω-elongations and separate ω-elongations. We start with the following:

Definition 2

A QTAG-module M is a strong ω-elongation of a totally projective module by a (ω+k)-projective module when H _ω (M) is totally projective and there is a submodule N⊆H^k(M) such that M/(N+H _ω (M)) is a direct sum of uniserial modules.

Definition 3

A QTAG-module M is a separate strong ω-elongation of a totally projective module by a separable (ω+k)-projective module if there is a submodule N⊆H^k(M), with N∩H _ω (M)=0, H _ω (M) is totally projective and M/(N⊕H _ω (M)) is a direct sum of uniserial modules.

Remark 3

For the separable modules, M/(N+H _ω (M))≅(M/N)/(N+H _ω (M))/N is a direct sum of uniserial modules, we have H _ω (M/N)=(H _ω (M)+N)/N and these are separate strong ω-elongations.

Now, we prove some basic results:

Proposition 1

A direct summand of a strong ω-elongation of a totally projective module by a (ω+k)- projective module is again a strong ω-elongation of a totally projective module by a (ω+k)-projective module.

Proof

Let M=T ⊕ K and N⊆M such that N⊆H^k(M) and M/(N+H _ω (M)) is a direct sum of uniserial modules. We put M₁=T∩(N+H _ω (M)) to get

$$\begin{array}{c}T/M_1\cong (T+(N+H_{\omega }\left(M\right))/(N+H_{\omega }\left(M\right))\\ \subseteq M/(N+H_{\omega }(M\left)\right),\hfill \end{array}$$

which is a direct sum of uniserial modules. Since H _ω (M) is totally projective and H _ω (M)=H _ω (T)⊕H _ω (K), H _ω (T) is also totally projective. Again,

$$\begin{array}{l}{M}_1=T\cap (N+H_{\omega }(T)+H_{\omega }(K\left)\right)\\ =H_{\omega }\left(T\right)+(T\cap (N+H_{\omega }\left(K\right)\left)\right);\end{array}$$

thus,

$$H_k(T\cap (N+H_{\omega }\left(K\right)\left)\right)\subseteq H_k\left(T\right)\cap H_{\omega }\left(K\right)=0$$

as H _k (N)=0. Consequently, the result follows. □

Remark 4

Direct sums of strong ω-elongations of a totally projective module by a (ω+k)-projective module is a strong ω-elongations of a totally projective module by (ω+k)-projective module.

After this, we recall some results from previous work, which are helpful in proving the next theorem:

Result 1

A QTAG-module M is a Σ-module if and only if $\text{Soc}\left(M\right)=\bigcup _{k<\omega }{M}_k,M_k\subseteq M_{k+1}$ and for every k∈Z⁺, M _k ∩H _k (M)= Soc (H _ω (M)).

Result 2

Let N be a submodule of a QTAG-module M such that M/N is a direct sum of uniserial modules. Then M is a direct sum of uniserial modules if and only if $N=\bigcup _{k<\omega }{N}_k,N_k\subseteq N_{k+1}$ and N _k ∩H _k (M)=0. Equivalently if Soc$\left(N\right)=\bigcup _{k<\omega }\left(S_k\right),S_k\subseteq S_{k+1}$ and S _k ∩H _k (M)=0 for every k∈Z⁺. It is well known that each totally projective module is a $\sum$-module. The next statement answers under what conditions the converse holds. These additional conditions include the new elongations of totally projective modules by (ω+1)-projective modules.

Now we are in the state to prove the following:

Theorem 3

A QTAG-module M which is a strong ω-elongation of a totally projective module by a (ω+1)-projective module, is a Σ-module if and only if M is a totally projective module.

Proof

Suppose M is a Σ- module. Since H _ω (M) is totally projective, in order to prove that M is totally projective, we have to show that M/H _ω (M) is a direct sum of uniserial modules. By the structure of M, there exists a submodule N⊆ Soc (M), such that M/(N+H _ω (M)) is a direct sum of uniserial modules. Also

$$(M/H_{\omega }(M\left)\right)/(N+H_{\omega }(M)/H_{\omega }(M\left)\right)\cong M/(N+H_{\omega }(M\left)\right).$$

Since M is a Σ-module, by Result 1, $\text{Soc}\left(M\right)=\bigcup _{k<\omega }{M}_k$, M _k ⊆M_k+1 and M _k ∩H _k (M)⊆H _ω (M) for every k∈Z⁺. As N⊆ Soc (M), $N=\bigcup _{k<\omega }{N}_k,N_k=N\cap M_k,N_k\subseteq N_{k+1}$ and N _k ∩H _k (M)⊆H _ω (M). Therefore,

$$(N+H_{\omega }(M\left)\right)/H_{\omega }\left(M\right)=\bigcup _{k<\omega }\left[\right(N_k+H_{\omega }\left(M\right))/H_{\omega }(M\left)\right]\text{and}$$

$$\begin{array}{l}[N_k+H_{\omega }(M)/H_{\omega }(M\left)\right]\cap H_k(M/H_{\omega }(M\left)\right)\\ =\left[\right(N_k+H_{\omega }\left(M\right))\cap H_k(M\left)\right]/H_{\omega }\left(M\right)\\ =\left[H_{\omega }\right(M)+(N_k\cap H_k\left(M\right)\left)\right]/H_{\omega }\left(M\right)\\ =0.\end{array}$$

Now, by Result 2, M/H _ω (M) is a direct sum of uniserial modules, and the result follows. The converse is trivial. □

Corollary 1

A module M is summable and a strong ω-elongation of a totally projective module by a (ω+1)-projective module if and only if M is a totally projective module of length ≤ω+1. In other words M is a direct sum of countably generated modules.

Proof

Every summable module M is a Σ-module and every totally projective module of length ω+1 is a direct sum of countably generated modules. Therefore M is summable. □

We end this paper with the following remark:

Remark 5

Now we may say that a QTAG-module M is a (ω+1)-projective Σ-module, if and only if it is a direct sum of countably generated modules with lengths at most ω+1.