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Let $M$ be a free right $R$-module. When $M_R\cong R_R^n$ with $n\in \mathbb{Z}_{\geq 1}$, then we know that the endomorphism ring $E={\rm End}(M_R)$ is isomorphic to $\mathbb{M}_n(R)$. We also know that $\mathbb{M}_{n}(R)$ is generated as a ring by its idempotents, when $n\geq 2$. Is $E$ generated by its idempotents if $M_R$ is not finitely generated?

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Yes it's true: every element can be written as $tu+vw+x+y-4z$ with each of $t,\dots, z$ idempotent.

More generally, this holds for an arbitrary module $M$ that is isomorphic to $N\times N$ for some module $N$.

Indeed, in this setting, every endomorphism of $M$ can be written as block matrix $\begin{pmatrix} A & B\\ C & D\end{pmatrix}$. Then

$$\begin{pmatrix} A & B\\ C & D\end{pmatrix}=\begin{pmatrix} I & A\\ 0 & 0\end{pmatrix}\begin{pmatrix} I & 0\\I & 0\end{pmatrix}+\begin{pmatrix} 0 & 0\\ D & I\end{pmatrix}\begin{pmatrix} 0 & I\\ 0 & I\end{pmatrix}$$ $$+\begin{pmatrix} I & B\\ 0 & 0\end{pmatrix}+\begin{pmatrix} 0 & 0\\ C & I\end{pmatrix}-4\begin{pmatrix} I & 0\\ 0 & I\end{pmatrix}.$$

(Although unnecessary to the question, the proof can be adapted to $M$ that is isomorphic to the $n$-power of a given module for $n\ge 2$, with a different writing.)

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  • $\begingroup$ Of course! In other words, when $M\cong N\times N$, then ${\rm End}(M)\cong \mathbb{M}_2({\rm End}(N))$. Great observation. $\endgroup$
    – Pace Nielsen
    Commented Oct 2, 2020 at 18:35
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    $\begingroup$ Blast, I spent a long time editing, improving this to any module of the form $M^2\oplus N^3$ (which uniformly covers all modules $P^n$ for $n\ge 3$, and hence all ultraproducts of such endomorphism algebras)... and when I refreshed the page after the other answer was deleted, all has disappeared. The argument is not hard but it takes time to write down all these matrices... So well, let's stick to have it as a remark. $\endgroup$
    – YCor
    Commented Oct 2, 2020 at 18:44
  • $\begingroup$ i think it must be true for any cardinal $\kappa$ such that $M= R^{\kappa}$ $\endgroup$
    – Cary
    Commented Oct 4, 2020 at 23:13
  • $\begingroup$ @Cary yes, this is covered by my answer (at least when $\kappa$ is infinite, assuming ZFC, or finite even) $\endgroup$
    – YCor
    Commented Oct 5, 2020 at 6:18
  • $\begingroup$ Thank you so much @Ycor $\endgroup$
    – Cary
    Commented Oct 5, 2020 at 9:54

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