My question is as simple as the title. Is it true that $\text{Hom}_R(R,M)$ is isomorphic to $M$? In which $R$ is a commutative ring with unit, and $M$ is an $R$-module. I actually made a proof, I just want to confirm it's right and I'm not missing anything, I don't know why but at first I had the intuition it would be false.
Since $R$ is always a free module over itself, generated by the unit $1$, all elements of $\text{Hom}_R(R,M)$ are of the form \begin{align} f_m:R &\rightarrow M\\ 1 &\mapsto m \end{align} for each $m\in M$.
We clearly have $$(rf_m+f_n)(1) = m+n = f_{rm+n}(1) $$ which implies $rf_m+f_n = f_{rm+n}$ for any $m,n\in M$ and $r\in R$.
This implies that the following map is a homomorphism (and since it is also clealy bijective, it is an isomorphism) \begin{align} M &\rightarrow \text{Hom}_R(R,M)\\ m &\mapsto f_m \end{align} showing that $M\cong \text{Hom}_R(R,M)$. Is this correct?
