Exercises in Classical Ring TheorySpringer Science & Business Media, 29 juin 2013 - 288 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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... submodule N C M is a direct summand of M , ( 2 ) M is the sum of a family of simple submodules , or ( 3 ) M is the direct sum of a family of simple submodules . Here , a simple module means a nonzero R - module M such that the only ...
... submodule N C M is a direct summand of M , ( 2 ) M is the sum of a family of simple submodules , or ( 3 ) M is the direct sum of a family of simple submodules . Here , a simple module means a nonzero R - module M such that the only ...
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... submodule of M. Show that N ≈ i = i - ni Vi for suitable ni ≤ mi , and that M / N ; ( mi — ni ) Vi . = Solution . Write N , Ni , where the N's are the isotypic components of N. Then N , CM ;, so we have NiniV ; for some ni mi . By the ...
... submodule of M. Show that N ≈ i = i - ni Vi for suitable ni ≤ mi , and that M / N ; ( mi — ni ) Vi . = Solution . Write N , Ni , where the N's are the isotypic components of N. Then N , CM ;, so we have NiniV ; for some ni mi . By the ...
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... submodule N CM is certainly not sufficient to determine the isomorphism type of M / N . ( For N = M , M / N = 0 ; but for N'00 V11 , = M / N'≈ V1 . And yet N≈ N ' . ) For another cancellation result not requiring any semisimplicity ...
... submodule N CM is certainly not sufficient to determine the isomorphism type of M / N . ( For N = M , M / N = 0 ; but for N'00 V11 , = M / N'≈ V1 . And yet N≈ N ' . ) For another cancellation result not requiring any semisimplicity ...
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Table des matières
1 | |
16 | |
Jacobson Radical Theory 4 The Jacobson radical 123535 | 35 |
5 Jacobson radical under change of rings | 52 |
Introduction to Representation Theory 689 | 69 |
Linear groups | 98 |
Prime and Primitive Rings 103 | 102 |
11 Structure of primitive rings the Density Theorem | 119 |
Introduction to Division Rings | 151 |
15 Tensor products and maximal subfields | 176 |
Ordered Structures in Rings | 191 |
Local Rings Semilocal Rings | 211 |
21 The theory of idempotents | 229 |
22 Central idempotents and block decompositions | 251 |
24 Homological characterizations of perfect | 265 |
Subject Index | 281 |
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Expressions et termes fréquents
a₁ abelian artinian ring assume automorphism B₁ central idempotents char commutative ring conjugate 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 matrix maximal ideal maximal left ideal maximal subfield minimal left Mn(R 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