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

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Example text

Applying the functor Hom z (-, Q) to this sequence we obtain an exact sequence 0-*A**-+F* in <€\ and, by Theorem 13, F* is A-injective. Let aeA. e. for a l l / i n Homz(A, Q). Clearly ^ e H o m z ( i * , Q) = ^ * * . We therefore have a mapping 0:^4^,4** in which (a) = a. Next is additive since ai+a2 = ^ a i + ^ a 2 . If AeA, then (A0a) (/) = ^O(/A) = (/A) (a). But (/A)(a)=/(Aa). Consequently (A^a) (/) =f(Aa) = ^ Ao (/) and therefore A^a = ^ A a . This shows that 0:^4 ^^4** is a homomorphism of left A-modules.

Definition. B is said to be an ' essential extension' of A if every nonzero submodule of B has a non-zero intersection with A. For example, a module is always an essential extension of itself. An extension of a A-module A which is different from A will be said to be non-trivial. Lemma 4. Let A be a submodule of a A-module B and let {CJ ie/ be a non-empty family of submodules of B such that if iv i2£l, then either Cti c: Gx or Ci2 <= Gi . Denote by C the set-theoretic union of the Ct. Then C is a submodule of B.

We first prove that Z(pco) is divisible and hence, by Theorem 12, injective. Let a == j 0 belong to Z(pQO) and let m #= 0 be an integer. r Then a = {a\p ) + Z, where r > 0, aeZ and p does not divide a; also m can be expressed in the form m = psny where s ^ 0 and n is an 50 THE HOM FUNCTOR integer not divisible by p. We can now find integers h, k such that hpr + kn = a. Thus hpr+s + km = aps and therefore Accordingly Z{pco) is injective. Next let C be the Z-submodule of Z(pco) generated by (\jpv) + Z.