1. May 11, 2009

### sanctifier

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. May 11, 2009

### matt grime

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

$$1=1-\sigma+\sigma$$

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).

3. May 20, 2009

### sanctifier

Thanks! matt grime