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 $M$ is not a finitely generated free $R$-module?

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?

Let $M$ be a free $R$-module we knew that the endomorphism ring of $M$ is isomorphic to the ring of matrices over $R$. and also we know that $M_{n}(R)=I(M_{n}(R))$, where $I(M_{n}(R))$ is the subring of $M_{n}(R)$ generated by idempotents of $M_{n}(R)$. Does endomorphism ring of any free $R$-module equal to $I(M_{n}(R))$? or $M$ must be finitely generated free $R$-module?

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 $M$ is not a finitely generated free $R$-module?