Read e-book online Algebras, Rings and Modules: Volume 1 PDF

By Michiel Hazewinkel

ISBN-10: 1402026900

ISBN-13: 9781402026904

ISBN-10: 1402026919

ISBN-13: 9781402026911

From the reports of the 1st edition:

"This is the 1st of 2 volumes which goal to take the idea of associative jewelry and their modules from primary definitions to the examine frontier. The e-book is written at a degree meant to be available to scholars who've taken usual easy undergraduate classes in linear algebra and summary algebra. … has been written with massive cognizance to accuracy, and has been proofread with care. … a truly welcome characteristic is the sizeable set of bibliographic and old notes on the finish of every chapter." (Kenneth A. Brown, Mathematical stories, 2006a)

"This publication follows within the footsteps of the dear paintings performed through the seventies of systematizing the research of homes and constitution of earrings through the use of their different types of modules. … A awesome novelty within the current monograph is the examine of semiperfect earrings via quivers. … one other stable notion is the inclusion of the examine of commutative in addition to non-commutative discrete valuation jewelry. each one bankruptcy ends with a few illustrative ancient notes." (José Gómez Torrecillas, Zentralblatt MATH, Vol. 1086 (12), 2006)

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Extra info for Algebras, Rings and Modules: Volume 1

Example text

F is both a monomorphism and an epimorphism, hence f is an isomorphism. 2 (Wedderburn-Artin). The following conditions are equivalent for a ring A: (a) A is right semisimple; (b) A is isomorphic to a direct sum of a finite number of full matrix rings over division rings; (c) A is left semisimple. Proof. (a) ⇒ (b). By definition, a ring A as the regular right A-module decomposes into a finite direct sum of simple right modules. Grouping isomorphic modules together we can consider that the decomposition has the form A = P1n1 ⊕ ...

Let L ⊂ N . Consider the restriction of π to N . We obtain a homomorphism π : L → M/L with the kernel Ker(π) = L and the image Im(π) = π(N ). Obviously, N ⊂ π −1 (π(N )). We now prove the converse inclusion. Let m ∈ π −1 (π(N )), then π(m) = π(x), where x ∈ N . , m − x = y ∈ Ker(π) = L. Since L ⊂ N , we have m = x + y ∈ N . As a result, π −1 (π(N )) = N and, by theorem N/Ker(π) = N/L. 1, π(N ) = Im(π) lemma. 4. Let L be a submodule of M and π : M → M/L be the natural projection. For any submodule N ⊂ M and any submodule N ⊂ M/L we have 1) π(N ) is a submodule of M/L; 2) π −1 (N ) is a submodule of M ; 3) π(π −1 (N )) = N ; 4) if L ⊂ N then π −1 (π(N )) = N .

If M0 = M , then, by assumption, M = M0 ⊕ M1 where M1 = 0. As we have shown above the submodule M1 contains a simple submodule, that contradicts the definition of M0 . So M is a sum of simple submodules. Let N be a submodule of M and let N0 be the sum of all simple submodules of N . By assumption M = N0 ⊕ M1 . Any element n ∈ N can be unique written as n = n0 + m1 , where n0 ∈ N0 and m1 ∈ M1 . Since m1 ∈ N , we obtain N = N0 ⊕ (N ∩ M1 ). The submodule N ∩ M1 = 0; otherwise it contains a simple submodule, that contradicts the definition of the module N0 .

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Algebras, Rings and Modules: Volume 1 by Michiel Hazewinkel


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