By D. G. Northcott

ISBN-10: 0511565887

ISBN-13: 9780511565885

ISBN-10: 0521201969

ISBN-13: 9780521201964

ISBN-10: 0521299764

ISBN-13: 9780521299763

In accordance with a sequence of lectures given at Sheffield in the course of 1971-72, this article is designed to introduce the coed to homological algebra warding off the frilly equipment frequently linked to the topic. This e-book provides a couple of vital themes and develops the mandatory instruments to address them on an advert hoc foundation. the ultimate bankruptcy comprises a few formerly unpublished fabric and may offer extra curiosity either for the willing pupil and his teach. a few simply confirmed effects and demonstrations are left as workouts for the reader and extra workouts are integrated to extend the most issues. ideas are supplied to all of those. a quick bibliography presents references to different guides during which the reader may perhaps keep on with up the topics taken care of within the e-book. Graduate scholars will locate this a useful path textual content as will these undergraduates who come to this topic of their ultimate yr.

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**Additional resources for A First Course of Homological Algebra**

**Sample text**

In other words, (b) is equivalent to the assumption that Hom A (P,-) preserves epimorphisms. The lemma now follows because, in any case, Hom A (P, -) is left exact. We recall that a A-module is called free if it is isomorphic to a direct sum of copies of A or (equivalently) if it has a base, that is a linearly independent system of generators. Theorem 3. Every free A-module is A-projective. This is a simple application of Lemma 1. Exercise 3. Prove Theorem 3. Theorem 4. Let {Ai}ieI be an arbitrary family of A-modules and put A = ® A{.

Exercise 7. Give an example of a projective module which is not free. 44 THE HOM FUNCTOR Solution. We have 2Z + 3Z = Z and 2Z n 3Z = 6Z. Hence if R is the ring Z/6Z, A = 2Z/6Z, and 5 = SZ/6Z, then i? = A ® B and therefore A is 12-projective. But R contains six elements. Consequently a free 12-module contains either an infinite number of elements or a finite number k of elements, where k is a multiple of six. However the number of elements in A lies between zero and six. Consequently A is not free.

Then the sequence splits. Proof. By Lemma 2, there exists a A-homomorphism h:A->E such that the diagram E ^A INJECTIVE MODULES 31 is commutative. That the sequence splits now follows from (Chapter 1, Theorem 4). Theorem 10 shows that whenever an injective module E is a submodule of a module A, then E is a direct summand of A. We shall see later (Theorem 15) that this is a characteristic property of injective modules. We can greatly strengthen Lemma 2. Theorem 11. Let E be a left A-module. A-^-E such that the diagram is commutative.

### A First Course of Homological Algebra by D. G. Northcott

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