By Michiel Hazewinkel

From the stories of the 1st edition:

"This is the 1st of 2 volumes which objective to take the idea of associative jewelry and their modules from basic definitions to the learn frontier. The publication is written at a degree meant to be obtainable to scholars who've taken general simple undergraduate classes in linear algebra and summary algebra. … has been written with significant awareness to accuracy, and has been proofread with care. … a really welcome function is the immense set of bibliographic and historic notes on the finish of every chapter." (Kenneth A. Brown, Mathematical experiences, 2006a)

"This booklet follows within the footsteps of the dear paintings performed through the seventies of systematizing the research of houses and constitution of earrings through the use of their different types of modules. … A extraordinary novelty within the current monograph is the examine of semiperfect earrings by way of quivers. … one other solid suggestion is the inclusion of the learn of commutative in addition to non-commutative discrete valuation earrings. each one bankruptcy ends with a few illustrative old notes." (José Gómez Torrecillas, Zentralblatt MATH, Vol. 1086 (12), 2006)

**Read Online or Download Algebras, Rings and Modules: Volume 1 PDF**

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**Additional info for Algebras, Rings and Modules: Volume 1**

**Example text**

Since N is a submodule and π is a homomorphism, π(x1 ) + π(x2 ) = π(x1 + x2 ) ∈ π(N ) and π(x)a = π(xa) ∈ N for all x1 , x2 ∈ N , a ∈ A. Therefore π(N ) is a submodule of M/L. If N is a submodule of M/L we deﬁne π −1 (N ) = {m ∈ M : π(m) ∈ N }. Since π(y) = 0 ∈ N for any y ∈ L, we have L ⊂ N . Let m1 , m2 ∈ π −1 (N ) and a ∈ A, then π(m1 + m2 ) = π(m1 ) + π(m2 ) ∈ N and π(m1 a) = π(m1 )a ∈ N . Hence π −1 (N ) is a submodule of M containing L. Furthermore, every element m ∈ N is of the form π(m), where m ∈ M , and also m ∈ π −1 (N ) because π(m) = m ∈ N .

If A is a commutative ring, and F is a free A-module, then any two free bases of F have the same cardinal number. Proof. By Zorn’s lemma, A has a maximal ideal M . Let { fi ∈ F : i ∈ I} A/M . be a free basis of F and denote by Fi an A/M -module fi A/fi AM If πi : fi A → Fi is a natural projection, then we denote πi (fi ) = f¯i . Deﬁne the homomorphism σi : Fi → F/F M by σi (f¯i ) = fi + F M and the projection Fi → Fi . Then it is easy to show that the homomorphism σ = σ i τi τi : i∈I i∈I gives an isomorphism of A/M -modules Fi and F/F M .

Mn } be a set of generators of an A-module M . , an ) = n mi ai . It is easy to see that ϕ i=1 is an epimorphism and by the homomorphism theorem M An /Ker(ϕ). Let F be a free A-module and α : F → ⊕ AA be an isomorphism of A-modules. i∈I Consider the elements fi for which α(fi ) = ei are elements of ⊕ AA having the i∈I identity of A at the i-th position and zeroes elsewhere. Then any element f ∈ F fi ai , where ai ∈ A and only a ﬁnite number of ai are can be written as f = i∈I ei ai = 0, all ai = 0.