Exercises in Classical Ring TheorySpringer-Verlag, 1995 - 287 pages Based in large part on the comprehensive "First Course in Ring Theory" by the same author, this book provides a comprehensive set of problems and solutions in ring theory that will serve not only as a teaching aid to instructors using that book, but also for students, who will see how ring theory theorems are applied to solving ring-theoretic problems and how good proofs are written. The author demonstrates that problem-solving is a lively process: in "Comments" following many solutions he discusses what happens if a hypothesis is removed, whether the exercise can be further generalized, what would be a concrete example for the exercise, and so forth. The book is thus much more than a solution manual. |
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Page 82
... contradiction . Ex . 8.12 . Let k be any field of characteristic 2 , and let G = S4 . Let M be the kG - module given by ke1 ... ke4 / k ( e1 + ··· + € 4 ) , ... on which G acts by permuting the e's . Compute the kG - composition fac ...
... contradiction . Ex . 8.12 . Let k be any field of characteristic 2 , and let G = S4 . Let M be the kG - module given by ke1 ... ke4 / k ( e1 + ··· + € 4 ) , ... on which G acts by permuting the e's . Compute the kG - composition fac ...
Page 202
... contradiction . If σ ( r ) < r , a similar contradiction results . Ex . 17.14 . Let ( R , P ) be an ordered ring for which P is a well - ordered set , i.e. every nonempty subset of P has a smallest element . Show that ( R , P ) is order ...
... contradiction . If σ ( r ) < r , a similar contradiction results . Ex . 17.14 . Let ( R , P ) be an ordered ring for which P is a well - ordered set , i.e. every nonempty subset of P has a smallest element . Show that ( R , P ) is order ...
Page 244
... contradiction . Therefore , det M = 0 , and the Cayley - Hamilton Theorem gives M2 = tM , where ttr ( M ) Є Z. But then M = nM2 = ntM implies nt = 1 , a final contradiction . Ex . 21.21 . Let R be a semilocal ring whose radical is nil ...
... contradiction . Therefore , det M = 0 , and the Cayley - Hamilton Theorem gives M2 = tM , where ttr ( M ) Є Z. But then M = nM2 = ntM implies nt = 1 , a final contradiction . Ex . 21.21 . Let R be a semilocal ring whose radical is nil ...
Table des matières
2 Semisimplicity | 16 |
Jacobson Radical Theory 35 1981 | 35 |
6 Group rings and the Jsemisimplicity problem | 57 |
Droits d'auteur | |
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a₁ abelian artinian ring assume automorphism B₁ central idempotents char commutative ring conjugate consider constructed contradiction decomposition Dedekind-finite defined division ring domain element endomorphism equation Exercise exists fact finite group finite-dimensional follows group G hopfian idempotent identity implies indecomposable integer inverse irreducible isomorphism J-semisimple Jacobson radical k-algebra kG-module left ideal left primitive ring Lemma Let G linear local ring M₁ matrix maximal ideal maximal left ideal maximal subfield minimal left module multiplication Neumann regular ring nil ideal nilpotent ideal noetherian ring noncommutative nonzero polynomial prime ideal primitive idempotents primitive rings proof prove quasi-regular R-module R/rad rad kG representation resp right ideal right R-module ring theory semilocal ring semiprime semisimple ring show that rad simple left R-module simple ring soc(RR Solution stable range strongly regular subdirect product subgroup subring Theorem unit-regular zero