# To show a module M = direct sum of Image and kernal of automorphism

1. Apr 18, 2012

### rachellcb

1. The problem statement, all variables and given/known data

Let R be a unital commutative ring. Let M be an R-module and $\varphi : M \rightarrow M$ a homomorphism.

To show: if $\varphi \circ \varphi = \varphi$ then $M=ker(\varphi)\oplus im(\varphi)$

3. The attempt at a solution

I have already shown that $M=ker(\varphi)\cap im(\varphi) = {0}$, so now I am trying to show that if $m \in M$ then m=m1+m2 where m1 $\in ker(\varphi)$ and m2 $\in im(\varphi)$

So far I have shown that $\varphi$ acts as the identity function on elements of $im(\varphi)$ and that $\exists$ m1, m2 with $\varphi$(m) = m2 then $\varphi$(m) = $\varphi$(m1 + m2) but I can not see a way to show that $\varphi$ is injective so this does not necessarily mean that m=m1+m2.

Not sure if I am using a good approach or not... Any hints or suggestions?

Last edited: Apr 18, 2012
2. Apr 18, 2012

### Dick

φ is injective on im(φ). You've already figured out that m2 must be φ(m). Doesn't that mean m1 must equal m-φ(m)?

3. Apr 19, 2012

### rachellcb

Thanks, I see that φ(m) = φ(m1) + φ(m2)
⇔ φ(m1) = φ(m) - φ(m2)
⇔ φ(m1) = φ(m - m2)
⇔ φ(m1) = φ(m - φ(m))

But neither m1 nor m need be in the image of φ, so injectivity does not necessarily apply, right?

4. Apr 19, 2012

### Dick

Suppose m has two decomposition m=m1+m2=m1'+m2'. That means φ(m2)=φ(m2') where both m2 and m2' are both in im(φ). What does that say about m2 and m2'?