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 :)

  • 1
    $\begingroup$ 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 $M=img(f)\oplus img(1-f)$ $\endgroup$
    – MIO
    May 16, 2022 at 12:05
  • $\begingroup$ Is it correct to say you are not assuming anything about commutativity? $\endgroup$
    – rschwieb
    May 16, 2022 at 14:19
  • 1
    $\begingroup$ @rschwieb Yes, that would be correct $\endgroup$
    – 14159
    May 16, 2022 at 14:22


You must log in to answer this question.

Browse other questions tagged .