By D. G. Northcott

In line with a chain of lectures given at Sheffield in the course of 1971-72, this article is designed to introduce the scholar to homological algebra keeping off the frilly equipment frequently linked to the topic. This booklet provides a couple of very important subject matters 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 the ultimate bankruptcy includes a few formerly unpublished fabric and should supply extra curiosity either for the prepared pupil and his teach. a few simply confirmed effects and demonstrations are left as workouts for the reader and extra routines are integrated to extend the most issues. options are supplied to all of those. a quick bibliography presents references to different courses during which the reader may perhaps persist with up the topics taken care of within the ebook. Graduate scholars will locate this a useful direction textual content as will these undergraduates who come to this topic of their ultimate yr.

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**Example text**

Next, since Kerf-> A -> B is exact, so too is F(Kerf) -> F(A) -> F(B) and, moreover, F(Kerf)->F(A) is monic. Hence KevF(f) = F(Kevf) provided that the latter is regarded as a submodule of F(A). SOLUTIONS TO EXERCISES ( 19 ( Exercise 6. ,An are A-submodules of A. Show that, as submodules of F(A), and F(AX n A2 n... n An) = F(AJ n F(A2) n... n F(An). Solution. We may suppose that n = 2. Let BVB2 be submodules of A with BX^B2. On applying i^7 to the commutative diagram B1- A formed by the various inclusion mappings, we see that F(BX) g F{B2).

Also V is isomorphic to V and therefore it inherits property (b). Accordingly E' is not an essential extension of V and so there exists a submodule C, of E, strictly containing B and such that V n (GjB) = 0. It follows that V n C = 0 and therefore Ce 2. This however contradicts the maximal property of B and now the theorem is proved. Definition. Let A be a A-module. An 'injective envelope' of A is an injective, essential extension of A. Theorem 17. Let A be a A-module and E an injective extension module of A.

Show that a Z-module is injective if and only if it is divisible. Solution. Suppose that E is Z-injective and let m be a non-zero integer. Let eeE and consider the diagram E where / is the ideal of Z generated by m and / is the homomorphism defined by f(nm) = ne. Since E is injective there is a homomorphism h:Z->E extending/. Accordingly e = f(m) = h(m) = mh(l). Whence eemE. It follows that E = mE and the divisibility of E is proved. Conversely, suppose E is divisible and let / be any non-zero ideal of Z.