A note on ADS∗ modules

We study the ADS∗ modules which are the dualizations of ADS modules studied by Alahmadi et al. (J Algebra 352:215–222, 2012). Mainly we prove that an amply supplemented module M is ADS∗ if and only if M1 and M2 are mutually projective whenever M = M1 ⊕ M2 if and only if for any direct summand S1 and a submodule S2 with M = S1 + S2, the epimorphism αi : M −→ Si/(S1 ∩ S2) with Ker(αi ) = S j (i = j = 1, 2) can be lifted to an idempotent endomorphism βi of M with βi (M) ⊆ Si .

If any module M is π -projective, then it is ADS * by [6, 41.14(2)]. Since every ring R is π -projective, R R and R R are ADS * . If any module M is both supplemented and π -projective then M is called quasi-discrete. Now we have that Note that the Z-module Q Z is ADS * but it is not π -projective and so not quasidiscrete. On the other hand, if M is ADS * and lifting, then it is quasi-discrete by [6, 41.15(b)]. Note that any module M is called lifting if it is amply supplemented and every supplement submodule of M is a direct summand. We should note that there is an amply supplemented ADS * module which is not π -projective (equivalently not quasi-discrete). For example; let R be a local artinian ring with radical W such that Note that U is 2-generated, ADS * and amply supplemented. On the other and, U cannot be lifting since it is not cyclic. Hence U is not quasi-discrete and so not π -projective.
In this note we provide equivalent conditions for a module to be ADS * . An amply supplemented module M is ADS * if and only if any decomposition of M into two modules, the summands are mutually projective (Theorem 2.1). We give an example showing that "amply supplemented" condition in Theorem 2.1 is necessary (Example 2.2). We call a module M completely ADS * if each of its subfactors is ADS * . Let M = ⊕ i∈I M i be an indecomposable decomposition of M. Suppose that M is amply supplemented completely ADS * . Then . We state that "amply supplemented" condition in Theorem 2.7 is necessary. Finally, we prove that a right semiartinian ring R is a right Vring if and only if every right R-module is ADS * if and only if every 2-generated right R-module is ADS * (Theorem 2.8).

Results
Theorem 2.1 For an amply supplemented module M the following are equivalent: (ii) For any direct summand S 1 and a submodule S 2 with M = S 1 + S 2 , the epi- is the natural epimorphism. Clearly, π 1 β 1 = α 1 . Therefore β 1 lifts α 1 . Similarly, the projection map β 2 : M −→ T lifts the epimorphism α 2 : We want to show that M = A ⊕ C. By hypothesis, there exists an idempotent endomorphism f : Note that there exists an ADS * module which is not amply supplemented with a decomposition M = A ⊕ B such that A is not B-projective: Example 2.2 Let R be a right V -domain which is not a division ring and S be any simple right R-module which is not R-projective. Put M = R ⊕ S. Since R is a right V -ring, every right R-module has a zero radical so M is (completely) ADS * . Note that M is not amply supplemented since R is not a division ring.

Proposition 2.3 Let A and B two submodules of an amply supplemented ADS * module M such that A is a direct summand of M and B is coclosed in M and M = A + B. Then A ∩ B is coclosed in M.
Proof Let C be a supplement of A contained in B. By hypothesis, M = A ⊕ C. Then Proof It is Lemma 3.6 in [1].
With the same notations as in Lemma 2.4 we give an interesting property of an ADS * module.  it is satisfied. (ii) Assume that i = 1, j = 2 and let σ :

Theorem 2.5 A module M is ADS
Since M is completely ADS * , M 1 /m 1 R is M 1 -projective. Hence m 1 R is a direct summand of M 1 and so m 1 R = M 1 . This means that M 1 is simple.
Note that Theorem 2.7 is not correct if M is not amply supplemented: If we consider Example 2.2, then Hom(R R , S) = 0 but R is not simple. Therefore Theorem 2.7(i) and (ii) are not satisfied.
Finally, we investigate that when a semiartinian ring is a right V -ring in terms of ADS * modules: (iii) ⇒ (i): Let S be a simple right R-module, and E be the injective hull of S. Assume S = E. Then by semiartinian assumption, we can find a cyclic submodule T of E containing S as a maximal submodule. Put X = T ⊕ (T /S) and A = {(t, t + S) | t ∈ T }. Clearly, A is a supplement of T ⊕ 0 in X but A ∩ (T ⊕ 0) = 0, contradicting the ADS * assumption. So R is a right V -ring.