Because $M$ is a finite-dimensional <semisimple module>, it has an <isotypic decomposition>
$$
M\simeq\bigoplus_{i=1}^t S_i^{m_i},
$$
where the $S_i$ are pairwise nonisomorphic <simple module>[simple right $R$-modules]. By <Schur lemma>,
$$
D_i=\operatorname{End}_R(S_i)
$$
is a <division ring>, while $\operatorname{Hom}_R(S_i,S_j)=0$ for $i\ne j$. Consequently every endomorphism preserves the isotypic summands and is a matrix of entries from $D_i$ on each one. Therefore
$$
\boxed{\operatorname{End}_R(M)\simeq
\prod_{i=1}^t M_{m_i}(D_i).}
$$
If the ground field is algebraically closed and the $S_i$ are finite-dimensional over it, <Schur lemma> gives $D_i=k$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 102 / 3 / b / Solution

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A division ring is a nonzero <ring> in which every nonzero element has a multiplicative inverse. Its multiplication need not be <commutative ring>[commutative].

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