# If $M=\bigoplus_{i\in I}M_i$ and $f:M\to M$ is idempotent then $f$ induces an isomorphism between $M_i$ and $f(M_i)$.

Let $$R$$ be a ring, $$M=\bigoplus_{i\in I} M_i$$ an $$R$$-module with $$\text{End}(M_i)$$ local for all $$i\in I$$. Then I want to show that if $$f:M\to M$$ is idempotent (i.e. $$f^2=f$$) and nonzero, then there exists at least one $$i\in I$$ such that $$f$$ induces an isomorphism of $$M_i$$ onto $$f(M_i)$$ and $$f(M_i)$$ is a direct summand of $$M$$.

I'm not really sure how to start this proof, so any help is appreciated.

Edit: I think I have a proof of this (which is a reformulation of Lemmas 1.1 and 1.2 in Chapter 5 of Abelian Categories with Applications to Rings and Modules by N. Popescu, which prove a more general result), but I'm not very confident in it.

A note on notation: for $$I'\subseteq I$$, $$M(I')=\bigoplus_{i\in I'} M_i$$.

Lemma 1: Let $$M=\bigoplus_{i\in I} M_i$$, where $$M_i$$ have local endomorphism rings. Let $$f,g\in\text{End}(M)$$ such that $$f+g=1_M$$. Then for any finite $$I'\subseteq I$$ there exist submodules $$N_i\subseteq M, i\in I'$$ such that

1. For each $$i\in I'$$ either $$f$$ or $$g$$ induces an isomorphism of $$N_i$$ with $$M_i$$.
2. $$M=N(I')\oplus M(I\setminus I')$$

Proof. Let $$\varepsilon_i:M_i\to M$$ be the inclusion maps and $$\pi_i:M\to M_i$$ the associated projection maps. Then $$\pi_i=\pi_i(f+g)=\pi_if+\pi_ig$$ and so $$1_{M_i}=\pi_i\varepsilon_i=\pi_if\varepsilon_i+\pi_ig\varepsilon_i.$$ Since $$\text{End}(M_i)$$ is local it follows that one of $$\pi_if\varepsilon_i$$ and $$\pi_ig\varepsilon_i$$ is an isomorphism. Fix $$i_1\in I'$$, suppose that $$\pi_{i_1} f\varepsilon_{i_1}$$ is an isomorphism and let $$N_1=\text{im }f\varepsilon_{i_1}$$. Since $$\pi_{i_1} f\varepsilon_{i_1}$$ is an isomorphism, it follows that $$\pi_{i_1}\vert_{N_1}:N_1\to M_{i_1}$$ is an isomorphism and so it follows that $$M=N_1\oplus M(I\setminus\{i_1\})$$. Repeating the argument for each $$i\in I'$$ we obtain $$N_1,N_2,\dots, N_n$$ with the desired properties.

So to prove the statement, consider the decomposition $$M=\bigoplus_{i\in I}M_i=\text{im } f\oplus\text{im }(1-f).$$ Since $$\text{im }f\neq 0$$, there is a finite $$I'\subseteq I$$ such that $$\text{im }f\cap M(I')\neq 0$$. Applying Lemma 1 with $$f$$ and $$g=1-f$$ we get the submodules $$N_i, i\in I'$$ such that $$M=N(I')\oplus M(I\setminus I'),$$ where $$f$$ or $$1-f$$ induces an isomorphism $$N_i\to M_i$$ for each $$i\in I'$$. Since $$\text{im }f\cap M(I')\subseteq \ker(1-f)$$ it can't be the case that $$1-f$$ induces the isomorphism $$M_i\to N_i$$ for all $$i\in I'$$ and so $$f$$ induces an isomorphism $$M_i\to N_i$$ for at least one $$i\in I'$$.

Any feedback is appreciated :)

• I don't have an answer but a few things of the top are 1) for local rings the only idempotent elements are $0$ and $1$. Also you have that $M=img(f)\oplus img(1-f)$ May 16 at 12:05
• Is it correct to say you are not assuming anything about commutativity? May 16 at 14:19
• @rschwieb Yes, that would be correct May 16 at 14:22