Direct Sums of Modules

**Bernhard** · August 29th 2011, 05:21 AM

I am reading Dummit and Foote to gain an understanding of Modules. In section 10.3 Proposition 5 we read:

Proposition 5

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

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

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

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

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

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

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

Then aj = $a_1$ + $a_2$ + .... + aj-1 + aj+1 + .... + $a_k$ for some $a_i$ $\in$ Ni and ( $a_1$ , $a_2$ , ....., aj-1, -aj, aj+1, ...., [LaTeX ERROR: Convert failed] ) would be a nonzero element of Ker $\pi$

My question is this: Why is a non-zero element of Ker $\pi$ a contradiction. Surely you can have non-zero elements of Ker $\pi$ ? 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

**girdav** · August 29th 2011, 09:15 AM

Since $\pi$ is supposed to be an isomorphism, the kernel is the set which consists of the vector $0$ .

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