By Carl Faith

E-book through religion, Carl

**Read Online or Download Algebra II. Ring Theory: Ring Theory PDF**

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**Extra info for Algebra II. Ring Theory: Ring Theory**

**Sample text**

From this the assertion follows. (5) Choose an idempotent e ∈ End(M ) with M1 = M e and let p : M → M/N be the canonical projection. By self-projectivity of M there exists α ∈ End(M ) with p = α e p|M1 . Then d = e + (1 − e)αe is an idempotent in End(M ), M d = M e = M1 and M2 = Ke d = M (1 − e)(1 − αe) ⊆ N is a direct summand of M satisfying our claim. 11. Self-projectives with noetherian self-injective hull. (1) Let M be a self-projective left R-module whose self-injective hull M is noetherian.

Kjm , . } is coindependent. This contradicts (c). (b)⇒(a). If (a) is not satisﬁed, then M contains an inﬁnite coindependent family of submodules {N1 , N2 , . }. Then N1 ⊃ N1 ∩ N2 ⊃ N1 ∩ N2 ∩ N3 ⊃ · · · and, for any k ∈ N, (N1 ∩ · · · ∩ Nk ) + Nk+1 = M. From this it follows that (N1 ∩ · · · ∩ Nk )/(N1 ∩ · · · ∩ Nl ) is not small in M/(N1 ∩ · · · ∩ Nl ) for all l > k, contradicting (b). (d)⇒(e). Let N1 , . . , Nn be given as in (d). Then the projection yields G M/ M n i=1 Ni n i=1 M/Ni , where each factor module M/Ni is hollow.

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 M and each M/Ni is hollow.