# Thread: Direct Sums of Modules

1. ## Direct Sums of Modules

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 $\displaystyle M_1$ , $\displaystyle M_2$ , ....., $\displaystyle M_n$ be submodules of M such that

(1) M = $\displaystyle M_1$ + $\displaystyle M_2$ + ..... + $\displaystyle M_n$ , and

(2) for 1 $\displaystyle \leq$ i $\displaystyle \leq$ n,

$\displaystyle M_i$ $\displaystyle \cap$ ($\displaystyle M_1$ + $\displaystyle M_2$ + ... + $\displaystyle M_{i-1}$ + $\displaystyle M_{i+1}$ + ..... + $\displaystyle M_n$) = 0

Then M $\displaystyle \cong$ $\displaystyle M_1$ $\displaystyle \oplus$ $\displaystyle M_2$ $\displaystyle \oplus$ ..... $\displaystyle \oplus$ $\displaystyle M_n$

Proof:

Let $\displaystyle f_i$ : $\displaystyle M_i$ $\displaystyle \rightarrow$ M be the inclusion map, that is, $\displaystyle f_i$ (x) = x for all s $\displaystyle \in$ $\displaystyle M_i$ and define

f : $\displaystyle M_1$ $\displaystyle \oplus$ $\displaystyle M_2$ $\displaystyle \oplus$ ..... $\displaystyle \oplus$ $\displaystyle M_n$ $\displaystyle \rightarrow$ M

by f( $\displaystyle x_1$ , ......, $\displaystyle x_n$) = $\displaystyle x_1$ + .....+ $\displaystyle x_n$

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

Now suppose that ($\displaystyle x_1$ , ......, $\displaystyle x_n$) $\displaystyle \in$ Ker(f) .

Then $\displaystyle x_1$ + .....+ $\displaystyle x_n$ = 0 so that for 1 $\displaystyle \leq$ i $\displaystyle \leq$ n, we have

$\displaystyle x_i$ = -($\displaystyle x_1$ + .....+ $\displaystyle x_{i-1}$ +$\displaystyle x_{i+1}$ + ....+ $\displaystyle x_n$).

Therefore $\displaystyle x_i$ $\displaystyle \in$ $\displaystyle M_i$ $\displaystyle \cap$ ($\displaystyle M_1$ + $\displaystyle M_2$ + ... + $\displaystyle M_{i-1}$ + $\displaystyle M_{i+1}$ + ..... + $\displaystyle M_n$) = 0

so that ( $\displaystyle x_1$ , ......, $\displaystyle x_n$) = 0 and f is an isomorphism

My questions are as follows:

1. why exactly is f surjective

2. why exactly does showing that ($\displaystyle x_1$ , ......, $\displaystyle x_n$) $\displaystyle \in$ Ker(f) implies ( $\displaystyle x_1$ , ......, $\displaystyle x_n$) = 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

2. ## Re: Direct Sums of Modules

1. Since $\displaystyle M_1+\ldots + M_n=\left\{m_1+\ldots+m_n,m_1\in M_1,\ldots,m_n\in M_n\right\}$, we have by definition of $\displaystyle f$: $\displaystyle f(M_1\oplus\ldots\oplus M_n)=\left\{m_1+\ldots+m_n,m_1\in M_1,\ldots,m_n\in M_n\right\}=M_1+\ldots + M_n=M$
by hypothesis.
2. You can show that a morphism is injective if and only if its kernel only contains $\displaystyle 0$.

3. ## Re: Direct Sums of Modules

Thanks girdav

Can you give me some guidance on how to show that a morphism is injective if and only if its kernel only contains 0

- or let me know where to find a proof

Peter

4. ## Re: Direct Sums of Modules

Let $\displaystyle M$ and $\displaystyle M'$ two modules, and $\displaystyle \varphi\colon M\to M'$ a morphism.
If $\displaystyle \ker\varphi =\left\{0\right\}$ and if $\displaystyle x_1, x_2\in M$ are such that $\displaystyle \varphi(x_1)=\varphi(x_2)$, then $\displaystyle \varphi(x_1-x_2)=0$ and $\displaystyle x_1-x_2\in\ker\varphi$.
If $\displaystyle \varphi$ is injective, then for $\displaystyle x\in \ker\varphi$ we have $\displaystyle \varphi(x)=\varphi(0)$ hence $\displaystyle x=0$.