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About the idempotent R-homomorphism

  1. May 11, 2009 #1
    Notations:
    σ denotes an R-map
    M denotes an abelian group under addtion
    R denotes a commutative ring with identity
    homR(M, M) is an R-endomorphism
    ker(?) denotes the kernel of a linear transformation "?"
    im(?) denotes the image of "?"
    ⊕ denotes the inner direct sum

    Terms:
    idempotent: σσ = σ
    R-module: a module whose base ring is R
    R-homomorphism or R-map: a function which preserves the module operations (similar to linear transformation)
    R-endomorphism: an R-homomorphism from M to itself

    Question:
    Let M be an R-module and let σ∈homR(M, M). if σ is idempotent show that M=ker(σ)⊕im(σ).

    My idea:
    Above all, since σσ=σ, im(σ)∩ker(σ)={0}.
    Let v=s+t where s,t∈M.
    1) If s∈ker(σ) and t=0, then σσ=σ holds;
    2) If t∈im(σ) and s=0, I realized that σ is idempotent only when t=σt, just like a projection of a vector space. I wonder wether it's right, or there's a better explanation.
    3) If s∈ker(σ) and t∈im(σ), σ is idempotent if 2) holds.
    4) s doesn't belong to ker(σ) and t doesn't belong to im(σ), it couldn't happen, every v∈M will be mapped by σ, as M is the domain of σ.

    Thanks for any help!
     
  2. jcsd
  3. May 11, 2009 #2

    matt grime

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    Science Advisor
    Homework Helper

    Assuming that R contains 1, then the best way to see this is to note that

    [tex]1=1-\sigma+\sigma[/tex]

    Hence for any r in R consider 1r - it clearly decomposes as something in Im(\sigma) plus something in ker(\sigma) (note that sigma annihilates (1-sigma)r for any r).
     
  4. May 20, 2009 #3
    Thanks! matt grime
     
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