By John Clark, Christian Lomp, N. Vanaja, Robert Wisbauer

Extending modules are generalizations of injective modules and, dually, lifting modules generalize projective supplemented modules. there's a yes asymmetry during this duality. whereas the speculation of extending modules is definitely documented in monographs and textual content books, the aim of our monograph is to supply a radical research of vitamins and projectivity stipulations had to examine sessions of modules concerning lifting modules.The textual content starts with an advent to small submodules, the novel, diversifications on projectivity, and hole size. the following chapters ponder preradicals and torsion theories (in specific with regards to small modules), decompositions of modules (including the alternate estate and native semi-T-nilpotency), vitamins in modules (with particular emphasis on semilocal endomorphism rings), completing with a protracted bankruptcy on lifting modules, major up their use within the conception of ideal earrings, Harada jewelry, andquasi-Frobenius rings.Most of the fabric within the monograph seems in publication shape for the 1st time. the most textual content is augmented by means of a ample provide of workouts including reviews on additional comparable fabric and on how the speculation has advanced.

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**Example text**

A)⇒(b). Let K be strongly coclosed in M and f : M → X such that (K)f = 0. For L = Ke f , f induces an isomorphism f : M/L → Im f . Denoting by p : M → M/L the projection map, we have (K)f = (K)pf = ((K + L)/L)f Im f. (b)⇒(c). Suppose f : M → X is a homomorphism such that (K)f = 0. If L ⊆ (K)f is cosmall in Im f , then (K)f pL Im f pl , where pL : Im f → Im f /L denotes the projection. This contradicts our hypothesis. Hence (K)f is coclosed in Im f . (c)⇒(a). Let L ⊆ M with K ⊆ L and let pL : M → M/L denote the projection.

Nn } in M such that M/Ni is hollow for all i ≤ n and N1 ∩ · · · ∩ Nn is small in M . Then a submodule K is small in M if and only if Ni + K = M for every i ∈ {1, . . , n}. Proof. The necessity is clear. Conversely assume that Ni + K = M for all i ∈ {1, . . , n}. Then (K + Ni )/Ni is small in M/Ni since M/Ni is hollow, that is, cs cs n n Ni →(K + Ni ) for all i ≤ n. 3, it follows that i=1 Ni → i=1 (K + Ni ). 2 (2) implies that K ⊆ n i=1 (K M + Ni ) M. 5. Lemma. Let {N1 , . . , Nn } be a family of coindependent submodules of M such n that i=1 Ni M and each M/Ni is hollow.

Nn , L} is also coindependent, contradicting the maximality of X. Hence N1 ∩ · · · ∩ Nn is small in M , as required. (d)⇒(c). Let {L1 , . . , Lk } be a coindependent family of M . 5, k ≤ n. Thus every coindependent family of M has at most n elements. (c)⇒(b). If (b) is not satisfied, then there is a chain M = N1 ⊃ N2 ⊃ · · · of submodules of M such that for any j ≥ 1 there exists an integer k(j) > j such that Nj /Nk(j) is not small in M/Nk(j) . Let {jm }m∈N be the sequence of indices defined as follows: j1 := 1 and jm := k(jm−1 ) for all m > 1.