Trying to show $\mathrm{End}_R(U_1 \oplus U_2) \cong \mathrm{End}_R(U_1) \times \mathrm{End}_R(U_2)$ I'm trying to prove
$\mathrm{End}_R(U_1 \oplus U_2) \cong \mathrm{End}_R(U_1) \times \mathrm{End}_R(U_2)$
Where $U_1, U_2$ are simple $R$-modules s.t $U_1 \not\simeq U_2$
So far I have constructed a mapping $\phi \mapsto (\pi_1 \phi i_1 , \pi_2 \phi i_2)$
Where $$\phi:U_1 \oplus U_2 \mapsto U_1 \oplus U_2$$
$$\pi_1:(u_1,u_2) \mapsto u_1 $$
$$\pi_2:(u_1,u_2) \mapsto u_2 $$
$$ i_1 : u  \mapsto (u,0) $$
$$ i_2 : u  \mapsto (0,u) $$
I feel I just need to need to use Schur's lemma now, although not quite sure how
 A: Endomorphisms of finite direct sums are most conveniently described with matrices. Suppose $M$ and $N$ are modules; write the elements of $M\oplus N$ as columns; then an endomorphism of $M\oplus N$ is described by a matrix
$$
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix},
$$
where $a\colon M\to M$, $b\colon N\to M$, $c\colon M\to N$ and $d\colon N\to N$, with the obvious action
$$
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\begin{bmatrix} x\\ y\end{bmatrix}
=
\begin{bmatrix}
ax+by \\ cx+dy
\end{bmatrix}.
$$
This is just a different way to put in your maps $i_1$, $i_2$, $\pi_1$ and $\pi_2$.
Note that composition of endomorphisms corresponds to (formal) matrix multiplication. In the case where
$$
\operatorname{Hom}_R(M,N)=0,
\quad
\operatorname{Hom}_R(N,M)=0,
$$
we obviously have
$$
\operatorname{End}_R(M\oplus N)\cong
\operatorname{End}_R(M)\times\operatorname{End}_R(N)
$$
and this is the case when $M$ and $N$ are nonisomorphic simple modules. Indeed, a nonzero homomorphism $f$ from a simple module $M$ to a simple module $N$ must be injective (because $\ker f\ne M$, so $\ker f=\{0\}$) and surjective (because the image of $f$ is not $0$, so it is $N$), hence an isomorphism.
It is definitely not true without the assumption that the Hom groups are zero. In particular,
$$
\operatorname{End}_R(M\oplus M)\cong
M_2(\operatorname{End}_R(M))
$$
