If an $R$-module $M \cong M \oplus M$ then if $S=\operatorname{Hom}_R(M,M)$ we have $S \cong S \oplus S$

Question

If an $R$-module $M \cong M \oplus M$ then if $S=\operatorname{Hom}_R(M,M)$ we have $S \cong S \oplus S$

$proof:$ So I need help with the details, but in general I was hoping something like this would work,

$\operatorname{Hom}_R(M,M) \cong \operatorname{Hom}_R(M \oplus M,M \oplus M) \cong \operatorname{Hom}_R(M,M) \oplus \operatorname{Hom}_R(M,M)$

Is this right? The first $\cong$ would mean that $\operatorname{Hom}_R$ is invariant under isomorphism classes, or at least in the case of endomorphisms as we have here, which makes perfect sense to me.

I could see the second $\cong$ holding as well, but wouldn't mind somebody helping me spell out the details. Thanks!

Answer 1

$S=\text{Hom}_R(M,M) \cong \text{Hom}_R(M, M \oplus M) \cong \text{Hom}_R(M,M) \oplus \text{Hom}_R(M,M)=S \oplus S.$