On elongations of QTAG-modules

Mehdi studied (ω+k)-projective QTAG-modules with the help of their submodules contained in Hk(M) (the submodule generated by the elements of exponents at most k). These modules contain nice submodules N contained in Hk(M) such that M/N is a direct sum of uniserial modules. Here, we investigate the class of QTAG-modules, containing nice submodules N⊆Hk(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 For a QTAG-module M, there is a chain of submodules Singh [2] proved that the results which hold for TAGmodules 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 ⊂ H 1 (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 Suppose 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 A k to be isomorphic. To achieve this goal we need some results. We start with the following:

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
Proof. Since N is a nice submodule we have Again,

Lemma 3. Let M be a QTAG-module and N a submod-
Again, 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).
there is a monomorphism from M into a direct sum of uniserial modules that does not decrease heights.
Now, we consider the family 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 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. is also a direct sum of uniserial modules.

Theorem 1. A direct summand of a module in
Again,  H ω (M )).
Since f is height-preserving isomorphism, it maps Again, if we put As these modules are totally projective, there is an isomorphism g 2 : H ω (M) → H ω (M ), which is again height preserving. Now, the isomorphisms g 1 , g 2 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 for all α and M ∼ = M [6,7].
Here, we study strong ω-elongations and separate ωelongations. We start with the following:     After this, we recall some results from previous work, which are helpful in proving the next theorem: It is well known that each totally projective module is a -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: 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 -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 -module, if and only if it is a direct sum of countably generated modules with lengths at most ω + 1.