By Gregory Karpilovsky
Enable N be an ordinary subgroup of a finite crew G and allow F be a box. a major strategy for developing irreducible FG-modules involves the appliance (perhaps repeated) of 3 uncomplicated operations: (i) limit to FN. (ii) extension from FN. (iii) induction from FN. this can be the `Clifford idea' constructed by way of Clifford in 1937. some time past two decades, the idea has loved a interval of full of life improvement. the rules were reinforced and reorganized from new issues of view, in particular from the perspective of graded jewelry and crossed items. the aim of this monograph is to tie jointly numerous threads of the improvement in an effort to supply a accomplished photo of the present country of the topic. it really is assumed that the reader has had the identical of a customary first-year graduate algebra path, i.e. familiarity with easy ring-theoretic, number-theoretic and group-theoretic strategies, and an realizing of user-friendly homes of modules, tensor items and fields.
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18. Corollary. Let R be a f i n i t e direct product of full matrix The radical of modules and rings 35 rings over division rings. Then R is semisimple artinian. Proof. 17. To prove that R is an artinian, we may assume that R = Mn(D) for some n 2 1 and some division ring D. 3. rn To establish the converse of the above result, we need some preliminary observations. 19. Lemma. (Schur’s lemma). Let V be an irreducible module. Then EndR(v) is a division ring. Proof. Let f : V + V be a nonzero R-homomorphism.
Then the following conditions are equivalent: (i) V is artinian. (ii) V is noetherian. (iii) V is both artinian and noetherian. Proof. Assume that V is either artinian or noetherian. It suffices t o verify that V is both artinian and noetherian. 13, J(R)" = 0 for some n 2 1, hence it suffices t o show that .. each of the modulesV/J(R)V, J ( R ) V / J ( R ) 2 V , . 6). However these modules are annihilated by J ( R ) and so can be regarded as R / J ( R ) modules. 6, each of them is either artinian or noetherian.
Since I is the intersection of all Vz with 0 # s E V, (ii) is established. (iii) This is a direct consequence of (i) and (ii). rn We now proceed to examine J ( V ) in detail. 4. Proposition. (i) Let f :W +V be a homomorphism of Preliminaries 28 c R-modules. Then f ( J ( W ) ) J ( V ) with equality i f f is surjective and Kerf J(W). c (ii) If W is a submodule of an R-module V , then J ( W ) J ( V ) and + J ( V / W )2 ( J ( V ) W ) / W . (iii) If W is a submodule of an R-module V with W J ( V / W )= J ( V ) / W .
