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

[rachellcb]

Homework Statement

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 $M=ker(\varphi)\oplus im(\varphi)$

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?

 

[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)?

 

[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?

 

[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'?