Question

Let E = End(M) be the ring of endomorphisms of an R-module M, and let nM denote the direct sum of n copies of M. Show that End (nM) is isomorphic to Mn(E).

Answer 1

Say $M$ is a right $R$-module, and we write endomorphisms on the left. Let $\varepsilon_j: M \to nM$ be the $j$-th inclusion, and $\pi_i: nM \to M$ be the $i$th projection. For any endomorphism $F: nM \to nM$, let $f_{ij}$ be the composition $\pi_i F \varepsilon_j \in E$. Define a map $\alpha: \operatorname{End}R(nM) \to \mathbf{M}n(E)$ by $\alpha(F) = (f_{ij})$. Routine calculations show that $\alpha$ is an isomorphism of rings.