By Gregory Karpilovsky

Enable N be a regular subgroup of a finite workforce G and allow F be a box. a tremendous technique for developing irreducible FG-modules contains the appliance (perhaps repeated) of 3 easy operations: (i) limit to FN. (ii) extension from FN. (iii) induction from FN. this can be the `Clifford concept' constructed by way of Clifford in 1937. some time past two decades, the speculation has loved a interval of energetic improvement. the rules were reinforced and reorganized from new issues of view, particularly from the perspective of graded jewelry and crossed items. the aim of this monograph is to tie jointly a variety of threads of the advance on the way to provide a accomplished photograph of the present kingdom of the topic. it truly is assumed that the reader has had the similar of a typical first-year graduate algebra direction, i.e. familiarity with uncomplicated ring-theoretic, number-theoretic and group-theoretic strategies, and an realizing of common houses of modules, tensor items and fields.

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Re,. e, = e;el + .. + e;en Hence 37 The radical of modules and rings proving that e,ej = S;ie; for al! i,j E ( 1 , . . ,n}. Moreover, R e ; C Vi and R R = Re1 @ . . @ R e , which implies that R e ; = V; for all i E {I,.. ,n}. Conversely, suppose that { e l , . . ,en} is a set of pairwise or- +. + thogonal idempotents and put e = el .. e,. Then e' = e and ee, = e;e = e; for each i. Thus R e = Re;. If E r i e ; = 0 , then multiplication on the right by ej implies that r j e j = 0 for all xi"=, j .

Ren and each Re, is irreducible (in particular, e i is a primitive idempotent of R ) . We may assume that R e l , . . ,Rer, 1 5 r 5 n, are all nonisomorphic modules of the set { R e l , . ,Re,}. 1,V E Re, for some i E {1,2,. . ,r } . This proves (i) and (ii). (iii) Put Di = EndR(Vi)", 1 5 i R" EndR(RR) 5 r . 21, EndR(Vi) E (eiRe,)" we have D; E eiRei, proving (iii). ) where each D;. is a division ring. Then there exist central idempotents u 1 , . . ,us of R such that and R u j E kj Mkj (Di) is a simple ring.

Clearly V, is a left ideal in R. Because V is irreducible, V = Rs Z R/Vz and so V. is a maximal left ideal in R. 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 .