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

  1. Apr 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Let R be a unital commutative ring. Let M be an R-module and [itex]\varphi : M \rightarrow M [/itex] a homomorphism.

    To show: if [itex]\varphi \circ \varphi = \varphi [/itex] then [itex]M=ker(\varphi)\oplus im(\varphi) [/itex]

    3. The attempt at a solution

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

    So far I have shown that [itex]\varphi[/itex] acts as the identity function on elements of [itex]im(\varphi)[/itex] and that [itex]\exists [/itex] m1, m2 with [itex]\varphi[/itex](m) = m2 then [itex]\varphi[/itex](m) = [itex]\varphi[/itex](m1 + m2) but I can not see a way to show that [itex]\varphi[/itex] 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. jcsd
  3. Apr 18, 2012 #2

    Dick

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    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)?
     
  4. Apr 19, 2012 #3
    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?
     
  5. Apr 19, 2012 #4

    Dick

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    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'?
     
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