# Download e-book for kindle: A First Course of Homological Algebra by D. G. Northcott

By D. G. Northcott

ISBN-10: 0511565887

ISBN-13: 9780511565885

ISBN-10: 0521201969

ISBN-13: 9780521201964

ISBN-10: 0521299764

ISBN-13: 9780521299763

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Additional resources for A First Course of Homological Algebra

Example text

Exercise 18. Let {A^UI be a family of modules in ^ A , and for each iel let Bi be an essential extension of A^ Show that © Bi is an essential iel extension of © A{. iel Solution. Let C be a non-zero submodule of © B{. Let ceC and iel suppose that c 4= 0. Then c = bx + b2 + ... + bn. , in are distinct elements of /. We shall SOLUTIONS TO EXERCISES 51 construct, in succession, a sequence A1? © Aif. This will ensure that Xnc is a non-zero member of © Ai and complete the solution. iel Since Bix is an essential extension of Ati, there exists fixeA such that fJL1b1 is a non-zero member of Ai .

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). Now Ai ^ Ax + A2 and A1f]A2^ A{. Thus F(A{) c F(A1 + A2) and F{AX n A2) c F(Ai). Accordingly F(A1) + F(A2) ^ F(A1 + A2) and Let X = A1 © J[2 and define ft:X->A by fi(ava2) = a{. Then ( / i + / 2 ) K ^ 2 ) = «i + «2Consequently Im/^ = Ai and Im (/x +/ 2 ) = A1 + A2. F7 is additive (Theorem 6). Hence, by Exercise 5, we have (as submodules of F(A)) = Im It follows that Let 7 = 4 / ^ x A\A2.

Since A is a (A, Y)-bimodule, Hom^ (A,B) is a Y-module. Since B is a (A, Y)-bimodule this also induces a Y-module structure on HomA (A,B). Show that these two structures coincide. Solution. Let fe HomA (A, B) and y e Y. If g is the product of y and / i n the case where A is considered as a (A, F)-bimodule, then y 40 THE HOM FUNCTOR is a commutative diagram. Now let In be the product of 7 and/when B is considered as a (A, F)-bimodule. In this case the diagram is commutative. However because/is a A-homomorphism, 7 is also commutative.