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

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!

• No, you are missing 2 homs! generally morphisms between direct sums can be thought of as matrixes, hence you get here actually 4 copies of $\mathrm{Hom} (M,M)$, your proof only considers the "diagonal" maps. For example, if you have $M\oplus M'$, you get $$\mathrm{Hom} (M\oplus M',M\oplus M') \cong \mathrm{Hom}(M,M') \oplus \mathrm{Hom}(M,M)^{\oplus 2} \oplus \mathrm{Hom}(M',M)$$ Also $M\cong M\oplus M$ seems kind of strong, nealry as if your object is the zero object or you will get an Eilenberg swindle – Enkidu Jul 23 '19 at 14:43

$$S=\text{Hom}_R(M,M) \cong \text{Hom}_R(M, M \oplus M) \cong \text{Hom}_R(M,M) \oplus \text{Hom}_R(M,M)=S \oplus S.$$
• Well that one is just univ property of $\oplus$. – Enkidu Jul 23 '19 at 14:47
• Do you know of any places that I could learn about the universal property of $\oplus$, what exactly that means and why it explains the middle isomorphism? – HaKuNa MaTaTa Jul 23 '19 at 14:59
• @MathematicalMushroom, Try to show this one $\text{Hom}_R(M, M \oplus M) \cong \text{Hom}_R(M,M) \oplus \text{Hom}_R(M,M)$. – Why Jul 23 '19 at 15:03