I am reading Adkins and Weintraub's book : Algebra: An Approach Through Module Theory. On page 119 (3.1) Theorem states:

Let M be an R-module and let , , ....., be submodules of M such that

(1) M = + + ..... + , and

(2) for 1 i n,

( + + ... + + + ..... + ) = 0

Then M .....

Proof:

Let : M be the inclusion map, that is, (x) = x for all s and define

f : ..... M

by f( , ......, ) = + .....+

f is an R-Module homomorphism and it follows from condition (1) that f is surjective.

Now suppose that ( , ......, ) Ker(f) .

Then + .....+ = 0 so that for 1 i n, we have

= -( + .....+ + + ....+ ).

Therefore ( + + ... + + + ..... + ) = 0

so that ( , ......, ) = 0 and f is an isomorphism

My questions are as follows:

1. why exactly is f surjective

2. why exactly does showing that ( , ......, ) Ker(f) implies ( , ......, ) = 0 demonstrate that f is an isomorphism?

Can anyone help?

I have attached the relevant page of Adkins and Weintraub since the paragraph at the top of the page may be relevant.

Peter