Since is supposed to be an isomorphism, the kernel is the set which consists of the vector .
I am reading Dummit and Foote to gain an understanding of Modules. In section 10.3 Proposition 5 we read:
Proposition 5
Let , , ....., Nk be submodules of the R-module M. Then the following are equivalent:
(1) The map : X X ...X Nk + + ...+ Nk defined by
( , , ..... , ) = + + .... + is an isomorphism of R-modules
i.e. + + .... + Nk is isomorphic to X X .... X Nk
(2) Nj ( + + Nj-1 + Nj+1 .... + Nk ) = 0 for all j = {1,2, ...... , k}
(3) Every x + + .... + Nk can be written uniquely in the form + + .... +
Proof: To prove (1) implies (2), suppose that for some j that (2) fails to hold and let aj ( + + Nj-1 + Nj+1 + ....+ Nk) Nj, with aj not equal to 0 .
Then aj = + + .... + aj-1 + aj+1 + .... + for some Ni and ( , , ....., aj-1, -aj, aj+1, ...., [LaTeX ERROR: Convert failed] ) would be a nonzero element of Ker
My question is this: Why is a non-zero element of Ker a contradiction. Surely you can have non-zero elements of Ker ? WHy not?
{Since I had real trouble with subscripts in Latex, I have attached the relevant page from Dummit and Foote showing the relevant proposition and proof}
Peter