# A theorem characterizing von Neumann regular endomorphisms.

I have completed a proof of a theorem characterizing von Neumann regularity of endomorphisms but I think my proof is imperfect. The theorem is stated in the language of modules but they could be groups just as well, and probably many other things.

My question is:

The theorem below states the equivalence of three conditions. How can I prove $(3)\Longrightarrow (1)$ without proving $(2)$ in between?

The theorem says what follows.

Let $M$ be an $R$-module. Let $\mathrm{E}:=\operatorname{End}(M),$ and $\rho\in\mathrm E$. The following conditions are equivalent.

$(1)$ $(\exists \sigma\in\mathrm E) \;\rho\sigma\rho=\rho;$

$(2)$ $\ker\rho$ and $\operatorname{im}\rho$ are direct summands in $M.$

$(3)$ There exist idempotent $\pi_1,\pi_2\in\mathrm E$ such that $\ker \pi_1=\ker\rho$ and $\operatorname{im}\pi_2=\operatorname{im}\rho.$

Proof.

$(1)\Longrightarrow(2).$ Let $\rho\sigma\rho=\rho$ for some $\sigma\in\mathrm E$.

Suppose $m\in\ker\rho\cap\operatorname{im}\rho\sigma.$ Then $m\rho=0$ and $m=x\rho\sigma.$ Therefore $$x\rho=x\rho\sigma\rho=0$$ and $$m=x\rho\sigma=0.$$ Thus

$$\ker\rho\cap\operatorname{im}\rho\sigma=0.$$

Now let $m\in M.$ We have $$(m-m\rho\sigma)\rho=m\rho-m\rho\sigma\rho=m\rho-m\rho=0,$$ so $(m-m\rho\sigma)\in\ker\rho$ and since $m\rho\sigma\in\operatorname{im}\rho\sigma,$ we have $$M=\ker\rho+\operatorname{im}\rho\sigma.$$ Thus $$M=\ker\rho\oplus\operatorname{im}\rho\sigma.$$

Suppose $m\in\operatorname{im}\rho\cap\ker\sigma\rho.$ Then $m\sigma\rho=0$ and $m=x\rho.$ Therefore, $$m=x\rho=x\rho\sigma\rho=0.$$ Thus

$$\operatorname{im}\rho\cap\ker\sigma\rho=0.$$

Now let $m\in M.$ We have $$(m-m\sigma\rho)\sigma\rho=m\sigma\rho-m\sigma\rho\sigma\rho=m\sigma\rho-m\sigma\rho=0,$$ and since $m\sigma\rho\in\operatorname{im}\rho,$ we have $$M=\operatorname{im}\rho+\ker\sigma\rho.$$ Thus $$M=\operatorname{im}\rho\oplus\ker\sigma\rho.$$

$(2)\Longrightarrow (1).$ Suppose $M=\ker\rho\oplus K=\operatorname{im}\rho\oplus L.$ From the first isomorphism theorem, we have an isomorphism $$i :(\ker\rho\oplus K)/\ker\rho \longrightarrow\operatorname{im}\rho$$ such that for any $m\in M$

$$(m+\ker\rho)i=m\rho.$$

We can write uniquely any $m\in M$ as a sum $m=a_{\ker\rho}+a_K.$ Let $m,n\in M$ with decompositions $m=a_1+a_2,$ $n=b_1+b_2$ where $a_1,b_1\in\ker\rho$ and $a_2,b_2\in K.$ Then the residue classes of $m,n$ in $(\ker\rho\oplus K)/\ker\rho$ are equal iff $a_2=b_2.$ Therefore, the map $$\sigma'':(\ker\rho\oplus K)/\ker\rho\longrightarrow \ker\rho\oplus K$$ such that

$$(a_{\ker\rho}+a_K+\ker\rho)\longmapsto (0+a_K)$$

is well-defined and it is easily seen to be a homomorphism. We define $\sigma':\operatorname{im}\rho\longrightarrow M$ by the formula $\sigma'=i^{-1}\sigma''.$ We can define $\sigma$ by putting $\sigma|_{\operatorname{im}\rho}=\sigma'$ and $\sigma|_L=0.$ Now $$\rho=\rho\sigma\rho.$$

$(2)\Longrightarrow (3).$ Suppose $M=\ker\rho\oplus K=\operatorname{im}\rho\oplus L.$ We define $\pi_1,\pi_2$ by $$\pi_1|_{\ker\rho}=0 \text{ and }\pi_1|_K=\operatorname{id}_K,$$ and $$\pi_2|_{\operatorname{im}\rho}=\operatorname{id}_{\operatorname{im}\rho}\text{ and } \pi_2|_L=0.$$

$(3)\Longrightarrow (2)$. It is easy to see that for an idempotent map $\pi:M\longrightarrow M,$ we have $M=\operatorname{im}\pi\oplus\ker \pi.$ The implication follows.

This ends the proof. However, I'm not satisfied with it. I have proven four implications, whereas only three are usually proven in such cases. I am able to fix it by proving $(1)\Longrightarrow (3),$ thus obtaining a cycle $(1)\Longrightarrow (3)\Longrightarrow (2)\Longrightarrow (1).$ But this is not good enough for me. I need $(3)\Longrightarrow (1)$ because I want to be able to see whether it is possible to generalize this theorem to structures in which the kernel cannot be defined as a substructure but has to be defined as a congruence. It is easy to see that the theorem generalizes to groups and perhaps it generalizes to rings too (although I haven't thought about it). That's because there is nothing specific to modules in the proof. The only important thing is that we can meaningfully say that the kernel is a direct summand. That's impossible I think when the kernel is just a relation, as in the plethora of various "semi-structures". If I were able to discard $(2)$ in the theorem, I would have a theorem that translated easily to the language of such structures. Could you please help me find a proof of $(3)\Longrightarrow (1)$ that doesn't use direct sums in any way and uses the kernels as congruences? Is it possible?

I suspect this must have a lot to do with category theory, but I know very little about it. Please, if you have to use it, do it bearing in mind that I will not understand anything other than the most elementary explanation.

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In fact, it is a lot neater to prove $(1) \implies (3)$ as follows :

Suppose we have a $\sigma$ such that $\rho \sigma \rho = \rho$.

Pick $\pi_1 = \rho \sigma$ : $\ker \rho \subset \ker \rho \sigma \subset \ker \rho \sigma \rho = \ker \rho$, so $\ker \rho \sigma = \ker \rho$. It is an idempotent because $(\rho\sigma)(\rho\sigma) = (\rho\sigma\rho)\sigma = \rho\sigma$.

Pick $\pi_2 = \sigma \rho$ : $\operatorname{im} \rho = \operatorname{im} \rho \sigma \rho \subset \operatorname{im} \sigma \rho \subset \operatorname{im} \rho$, so $\operatorname{im} \rho \sigma = \operatorname{im} \rho$. It is an idempotent because $(\sigma \rho)(\sigma \rho) = \sigma(\rho\sigma\rho) = \sigma\rho$.

One good thing about this it that it explains why you mysteriously chose $(3)$ and not

$(3')$ : There exist idempotent $\pi_1, \pi_2 \in E$ such that $\ker \pi_1 = \operatorname{im} \rho$ and $\operatorname{im} \pi_2 = \ker \rho$

about which it is equally easy to prove $(3') \implies (2)$ or $(2) \implies (3')$.

Proving $(3) \implies (1)$ without going through $(2)$ seems hopeless, you really need a way to "invert" $\rho$, and you can only do this via the isomorphism theorem and using the direct sums, not really by looking at some idempotents $\pi_1, \pi_2$ that behave a bit like $\rho$.

Edit : in fact you can reformulate what you are doing in $(3) \implies (2) \implies (1)$ without explicitly mentioning the direct sums :

Since $\operatorname{im} \pi_2 \subset \operatorname{im} \rho$, let $\rho' : M \to \operatorname{im} \rho$ and $\pi_2' : M \to \operatorname{im} \rho$ be the corestrictions of $\rho$ and $\pi_2$ : Forall $x\in M$, $x\rho = x\rho'$ and $x\pi_2 = x\pi_2'$

The isomorphism theorem says there is an isomorphism $\iota : \operatorname{im} \rho \to M / \ker \rho$ such that $x (\rho'\iota) = (x \rho) \iota = x + \ker \rho$, i.e. $\rho' \iota : M \to M / \ker \rho$ is the canonical surjection.

Since $\ker \rho \subset \ker \pi_1$, there is a map $\tilde{\pi_1} : M/ \ker \rho \to M$ such that $\rho'\iota\tilde{\pi_1} = \pi_1$. Now pick $\sigma = \pi_2' \iota \tilde{\pi_1}$.

Since $\operatorname{im} \rho \subset \operatorname{im} \pi_2$, for all $x \in M$, there exists $y\in M$ such that $x \rho = y \pi_2$. Then $x (\rho \pi_2') = (x \rho) \pi_2' = (y \pi_2) \pi_2 = y (\pi_2 \pi_2) = y \pi_2 = x \rho = x \rho'$, and so : $\rho \sigma \rho = (\rho \pi_2') (\iota \tilde{\pi_1} \rho) = \rho' (\iota \tilde{\pi_1} \rho) = (\rho' \iota \tilde{\pi_1}) \rho = \pi_1 \rho$.

Now since $\pi_1\pi_1 = \pi_1$, $(\pi_1 - id)\pi_1 = 0$ Since $\ker \pi_1 \subset \ker \rho$, $(\pi_1 - id)\rho = 0$, so we have $\pi_1 \rho = \rho$, and we have proved that $\rho \sigma \rho = \rho$.

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Thank you very much. Would it be possible to give some kind of argument for the impossibility of such a proof? – user23211 Mar 18 '12 at 13:24
I think you meant $M$ instead of $E$ everywhere. Thanks again. – user23211 Mar 18 '12 at 16:27