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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
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1 Answer

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+50

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
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