Download A First Course of Homological Algebra by D. G. Northcott PDF

By D. G. Northcott

According to a chain of lectures given at Sheffield in the course of 1971-72, this article is designed to introduce the scholar to homological algebra warding off the frilly equipment often linked to the topic. This publication offers a few very important themes and develops the required instruments to deal with them on an advert hoc foundation. the ultimate bankruptcy includes a few formerly unpublished fabric and should supply extra curiosity either for the prepared pupil and his train. a few simply confirmed effects and demonstrations are left as routines for the reader and extra workouts are integrated to extend the most subject matters. strategies are supplied to all of those. a quick bibliography offers references to different courses within which the reader may well keep on with up the topics handled within the publication. Graduate scholars will locate this a useful direction textual content as will these undergraduates who come to this topic of their ultimate 12 months.

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We have / = Zm for some non-zero integer m. Put e — f(m). Since mE = E there is an element e' in E such that e = me'. Define a homomorphism h:Z-+E by h(n) = ne'. Then h extends/. That E is injective now follows from Theorem 11. Exercise 14. Show that every left A-module is protective if and only if every left A-module is injective. Solution. First suppose that every left A-module is injective and that A is a left A-module. where P is projective. But K is injective. It follows, from Theorem 10, that the sequence splits.

Let Q be the additive group of the rational numbers. Clearly Q is divisible. Put Q, = Q\Z. The divisibility of Q is inherited by Q; consequently (Theorem 12) Q. is Z-injective. For each A in ^ A , put A* = Homz(A,Cl). 1) Then A* is a A-module of the opposite type to A. Also, by Theorem 13, Hom z (-, £}) is a contravariant exact functor from ^ A to ^ A and it converts projectives into injectives. Likewise it may be regarded as a contravariant exact functor from ^ A to ^\ again converting projective modules into injective modules.

Evidently B #= 0. Let B' be a non-zero submodule of B. Since B is a submodule of M, we have 5 = Bf © JB" for a certain module 5". Since A <^ A+B' ^ N we see that i r e ^ + I ? ' ) . On the other hand (A+B')n(A+B") = A and therefore x$(A+B"). " = 0 and B = B'. This shows that i? is simple and establishes our claim. Let {Sj}jeJ be the family consisting of all simple submodules of M. Then M = ( £ £,,) © JV for some submodule N of If. Now and therefore N cannot contain any simple submodule. It follows that N = 0 and therefore M = 2 £,.

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