# Direct Sums of Modules

• Aug 29th 2011, 05:21 AM
Bernhard
Direct Sums of Modules
I am reading Dummit and Foote to gain an understanding of Modules. In section 10.3 Proposition 5 we read:

Proposition 5

Let $\displaystyle N_1$ , $\displaystyle N_{2}$ , ....., Nk be submodules of the R-module M. Then the following are equivalent:

(1) The map $\displaystyle \pi$ : $\displaystyle N_1$ X $\displaystyle N_{2}$ X ...X Nk $\displaystyle \rightarrow$ $\displaystyle N_1$ + $\displaystyle N_{2}$ + ...+ Nk defined by

$\displaystyle \pi$ ( $\displaystyle a_1$ , $\displaystyle a_2$ , ..... , $\displaystyle a_k$ ) = $\displaystyle a_1$ + $\displaystyle a_2$ + .... + $\displaystyle a_k$ is an isomorphism of R-modules

i.e. $\displaystyle N_1$ + $\displaystyle N_{2}$ + .... + Nk is isomorphic to $\displaystyle N_1$ X $\displaystyle N_{2}$ X .... X Nk

(2) Nj $\displaystyle \cap$ ( $\displaystyle N_1$ + $\displaystyle N_{2}$ + Nj-1 + Nj+1 .... + Nk ) = 0 for all j = {1,2, ...... , k}

(3) Every x $\displaystyle \in$ $\displaystyle N_1$ + $\displaystyle N_{2}$ + .... + Nk can be written uniquely in the form $\displaystyle a_1$ + $\displaystyle a_2$ + .... + $\displaystyle a_k$

Proof: To prove (1) implies (2), suppose that for some j that (2) fails to hold and let aj $\displaystyle \in$ ( $\displaystyle N_1$ + $\displaystyle N_{2}$ + Nj-1 + Nj+1 + ....+ Nk) $\displaystyle \cap$ Nj, with aj not equal to 0 .

Then aj = $\displaystyle a_1$ + $\displaystyle a_2$ + .... + aj-1 + aj+1 + .... +$\displaystyle a_k$ for some $\displaystyle a_i$ $\displaystyle \in$ Ni and ( $\displaystyle a_1$ , $\displaystyle a_2$ , ....., aj-1, -aj, aj+1, ...., $\displaystyle a_k$ ) would be a nonzero element of Ker $\displaystyle \pi$

My question is this: Why is a non-zero element of Ker $\displaystyle \pi$ a contradiction. Surely you can have non-zero elements of Ker $\displaystyle \pi$? 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
• Aug 29th 2011, 09:15 AM
girdav
Re: Direct Sums of Modules
Since $\displaystyle \pi$ is supposed to be an isomorphism, the kernel is the set which consists of the vector $\displaystyle 0$.