By Lopez-Permouth S., Huynh D.V. (eds.)

This quantity contains refereed examine and expository articles by way of either plenary and different audio system on the overseas convention on Algebra and functions held at Ohio collage in June 2008, to honor S.K. Jain on his seventieth birthday. The articles are on a wide selection of parts in classical ring idea and module idea, resembling jewelry pleasurable polynomial identities, earrings of quotients, team earrings, homological algebra, injectivity and its generalizations, and so on. incorporated also are functions of ring conception to difficulties in coding thought and in linear algebra.

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Analogously (C ∨ D)⊥sext ≤ D⊥sext , and then (C ∨ D)⊥sext ≤ C⊥sext ∧ D⊥sext always happens. On the other side, suppose there exists 0 = M ∈ C⊥sext ∧ D⊥sext \ (C ∨ D) ⊥sext . Then her (M ) ∩ C = {0}, her (M ) ∩ D = {0} and her (M ) ∩ (C ∨ D) = {0}. Then, there exists 0 = N ≤ M such that N ∈ (C ∨ D), and hence there exists an exact sequence 0→C N L→0 with 0 = C ∈ C ∪ D. Since C ∈ her (M ) too, we have her (M ) ∩ C = {0} or ⊥ her (M ) ∩ D = {0}, a contradiction, so (C ∨ D) sext = C⊥sext ∧ D⊥sext .

2. If U is a class of modules, then class in R-qext generated by U. n∈N n∈N E(Q, Q)n ∈ R-quot. E(quot (U) , quot (U))n is the Denoting qext (U) = n∈N E(quot (U) , quot (U))n we have that R-qext is a complete big lattice where for each family {Cα }α∈X in R-qext: {Cα }α∈X = {Cα }α∈X = {Cα }α∈X qext {Cα }α∈X . 3. If Q ∈ R-qext, then Q has a unique pseudocomplement in R-qext given by Q⊥qext = {M ∈ R- mod | quot (M ) ∩ Q = {0}} . Example. qext (R) = {R M | M is ﬁnitely generated} . Proof. qext (R) is the class closed under extensions generated by the cyclic modules, thus it is clear that qext (R) contains just ﬁnitely generated modules.

So there exists an exact sequence 0 → I −→ S −→ K −→ 0 where I ∈ L(R) and K ∈ E(L(R), L(R))n−1 . As S is simple this implies that S ∼ = I. Conversely, assume that R is left artinian thus it is noetherian and semiartinian. As usual, let us deﬁne soc1 (R) = soc (R) and socn+1 (R) / socn (R) = soc (R/ socn (R)) . It follows that socn (R) ∈ sext(R-simp), for each n. Also there exists an m such that socm+1 (R) = socm (R) because R is left noetherian. Thus soc (R/ socm (R)) = 0 which implies that R/ socm (R) = 0 because R is semiartinian.