# Are there any characterisations of endomorphisms whose (a) kernels and (b) images are direct summands?

It has recently been explained to me on this site what direct summands really are. (Here.)

Now I think I have a more difficult question. There is a theorem (easy to prove) which says

For a module $M$ and its endomorphism $f,$ $f$ is a von Neumann regular element of $End(M)$ iff $\ker f$ is a direct summand of $M$ and at the same time $\operatorname{im} f$ is a direct summand of $M.$

I would like to know if there are any useful (or even less useful) necessary and/or sufficient conditions for

1. $\ker f$ is a direct summand of $M;$

2. $\operatorname{im} f$ is a direct summand of $M.$

I understand that this is technically open-ended and I can't know exactly what the scope of it is, but based on my (not that great) mathematical experience, I believe this is not a real issue here. However, if I am wrong and this is a very broad subject, please just give me some references and a short outline of the most important facts if it's possible.

I would be especially grateful for a condition that would be equivalent to (a) but didn't use the standard notion of a kernel, but used the notion of the kernel being an equivalence relation (congruence). I am trying to rewrite certain things that are true for modules in the language of semimodules where the module-like kernels simply don't exist.

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Please note that I'm looking for separate conditions for images and kernels. I'm not sure how clear it is from the question. – user23211 Jan 21 '12 at 11:22

The kernel of $f$ is a direct summand if and only if there is a projector $p: M \to M$ (i.e. an endomorphism $p$ such that $p^2 = p$) whose kernel is equal to the kernel of $M$. (We then have $M = ker(p) \oplus im(p) = ker(f) \oplus im(p)$.)

Thus the kernel of $f$ is a direct summand of $M$ if and only if there is an idempotent $p$ in $End(M)$ such that $f$ and $p$ have the same kernel.

Similarly, the image of $f$ is a direct summand of $M$ if and only if there is an idempotent $p$ in $End(M)$ such that $f$ and $p$ have the same image.

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This is very helpful, thank you. – user23211 Jan 22 '12 at 10:48