prove that $HOM_R(\bigoplus _{s\in S}R_s ,M)$ and $\prod_{s\in S}M_s$ are isomorphic as $R$ modules.

Question

Let $R$ be a ring, $M$ module over $R$ and $S$ a non empty set.
Let $\mathrm{Hom}_R(M,N)$ be the group of $R$ module homomorphisms from $M$ to $N$.
Denote $M_s:=M$ and $R_s:=R$.

Prove that the modules $\mathrm{Hom}_R(\bigoplus _{s\in S}R_s ,M)$ and $\prod_{s\in S}M_s$ are isomorphic as $R$ modules.
I am pretty clueless about this one.

Answer 1

A map $f$ from $\bigoplus _{s\in S}R_s$ into $M$, by the universal property of the co-product, is the same as a collection of maps $f_s:R_s\rightarrow M$, that is $$ \mathrm{Hom}\left( \bigoplus _{s\in S}R_s,M\right) \cong \prod _{s\in S}\mathrm{Hom}(R_s,M). $$ However, $\mathrm{Hom}\, (R,M)\cong M$, the isomorphism given by $x\in M$ being sent to the map $\phi :R\rightarrow M$ defined by $\phi (1):=x$. Thus, $$ \mathrm{Hom}\left( \bigoplus _{s\in S}R_s,M\right) \cong \prod _{s\in S}M_s. $$