Direct Sums of Modules

**Bernhard** · August 30th 2011, 02:17 AM

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

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

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

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

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

Proof:

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

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

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

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

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

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

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

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

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

My questions are as follows:

1. why exactly is f surjective

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

**girdav** · August 30th 2011, 02:55 AM

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

**Bernhard** · August 30th 2011, 03:03 AM

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

**girdav** · August 30th 2011, 03:09 AM

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

Thread: Direct Sums of Modules

LinkBack

Thread Tools

Display

Direct Sums of Modules

Re: Direct Sums of Modules

Re: Direct Sums of Modules

Re: Direct Sums of Modules

Similar Math Help Forum Discussions

Confusion about Direct Limit of R-Modules

Direct Sums of Modules

Modules, Direct Sums and Annihilators

Direct sum of free modules

Direct Sums

Search Tags