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'?