Endomorphism ring of a simple module is a division ring

Remarks

$\End_R(M)$ is nonzero, since $M \ne 0$ gives $\id_M \ne 0$. By Schur's lemma every nonzero endomorphism of $M$ is an isomorphism, so its inverse map is again an endomorphism and is its inverse in $\End_R(M)$. Example: $\End_\Z(\Z/p\Z) \cong \Z/p\Z$, a field, for $p$ prime. Simplicity is needed: for the semisimple $\Z$-module $M = \Z/p\Z \oplus \Z/p\Z$, the ring $\End_\Z(M) \cong M_2(\Z/p\Z)$ has the nonzero non-invertible element $E_{11}$. If $R$ is an algebra over an algebraically closed field $k$ and $M$ has finite dimension over $k$, then $\End_R(M) = k$.

Going deeper

For a student

This is how division rings enter the structure theory of rings. In the Artin–Wedderburn theorem, a simple Artinian ring is a matrix ring over the division ring $\End_R(M)^{\mathrm{op}}$ for a simple module $M$. In representation theory over $\C$, it says that endomorphisms of a finite-dimensional irreducible representation are scalars.