Notations:(adsbygoogle = window.adsbygoogle || []).push({});

σ denotes an R-map

M denotes an abelian group under addtion

R denotes a commutative ring with identity

hom_{R}(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∈hom_{R}(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!

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# About the idempotent R-homomorphism

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