By Gregory Karpilovsky

ISBN-10: 0444873775

ISBN-13: 9780444873774

Enable N be a standard subgroup of a finite crew G and enable F be a box. a tremendous procedure for developing irreducible FG-modules includes the appliance (perhaps repeated) of 3 uncomplicated operations: (i) limit to FN. (ii) extension from FN. (iii) induction from FN. this is often the `Clifford thought' constructed via Clifford in 1937. some time past two decades, the speculation has loved a interval of energetic improvement. the principles were reinforced and reorganized from new issues of view, specially from the perspective of graded jewelry and crossed items. the aim of this monograph is to tie jointly numerous threads of the advance with a view to provide a entire photo of the present nation of the topic. it truly is assumed that the reader has had the identical of a typical first-year graduate algebra direction, i.e. familiarity with uncomplicated ring-theoretic, number-theoretic and group-theoretic recommendations, and an knowing of user-friendly homes of modules, tensor items and fields.

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V, of V. 10(ii). 12. 2. Corollary. Let R be a ring. T h e n the following conditions are equivalent: Preliminaries 26 (i) RR is a finite direct s u m of irreducible modules. (ii) RR is completely reducible. (iii) R i s semisimple artinian. (iv) Every R-module i s completely reducible. Proof. 1 for V = R R. (ii+(iii): Assume RR is completely reducible. 1 (applied t o V = R R), it suffices to prove that R is artinian. Since 1 lies in the sum of finitely many irreducible submodules of RR, the same is true for RR.

Both the underlying set and the additive structure of R" are just those of R. But the multiplication, denoted by 0, is given by z o y = yz for all z,y E R. 11. Corollary. Let R be a ring. T h e n J ( R " ) = J ( R ) and J ( R ) is the intersection of all maximal right ideals of R. Proof. 10. 12. Corollary. If I I i s a left or right nil ideal of R , then J(R). Proof. Assume that I is a left nil ideal of R and let z E I . Then, for any r E R, rz E I and hence ( r ~=) 0 ~for some n 2 1. Applying the identity we see that 1 - r z is a unit.

Then J(R)V J ( V ) with equality i f R / J ( R ) is artinian. Proof. If V has no maximal submodules, then J ( V ) = V and hence J(R)V C J ( V ) . Let M be a maximal submodule of V . Then V I M is an irreducible R-module. 3(iii)), it follows that J ( R ) V M . Thus J(R)V C J ( V ) . Assume that R / J ( R ) is artinian. 4(v), it suffices to verify that J(V/J(R)V)= 0. 2). In particular, the R / J ( R ) module V / J ( R ) V is completely reducible. Hence V / J ( R ) V is a 31 The radical of modules and rings completely reducible R-module and therefore J ( V / J ( R ) V ) = 0.

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