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?



Endomorphism rings of infinitely generated free modules generated by idempotents?
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.)


