1. Since , we have by definition of :
by hypothesis.
2. You can show that a morphism is injective if and only if its kernel only contains .
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