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Let $f : M \to M$ be an endomorphism of a module $M$ such that $f^2=f$.Then prove that, $M = \text{Im}\ f \oplus \ker f$.

I was thinking that this can probably be proved if I can find an exact sequence and prove that it is also split exact because then it would follow from the definition of split exact sequence that $M = \text{Im}\ f \oplus \ker f$ since $\ker f$ is a submodule of $M$.

How can I find such an exact sequence ?

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  • $\begingroup$ I don't think that's necessary or even helpful. Just see whether any element of M can be written as a sum of something in the image and something in the kernel and argue that the intersection of the two submodules is 0. $\endgroup$ – John Brevik Mar 1 '15 at 9:34
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Well, why not?

The exact sequence

$$0 \to Ker(f) \to M \to Im(f) \to 0$$ is split by the inclusion $Im(f) \hookrightarrow M$, since $Im(f) \hookrightarrow M \to Im(f)$ is the identity.

Just for completion: The other splitting (whose existence is a formal consequence in general, but can be easily given here, too) is the map $\operatorname{id-f}: M \to Ker(f)$. One easily checks, that this map indeed maps into the kernel, and that $Ker(f) \hookrightarrow M \to Ker(f)$ is the identity.

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For a more down-to-earth solution, you can note that given $x\in M$, you may write $x=x-f(x)+f(x)$ and $f(x-f(x))=0$, so $M={\rm im}\, f+\ker f$. The sum is direct since if $y=f(x)$ is in the kernel of $f$; $f(y)=f^2(x)=f(x)=y=0$.

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