Let $R$ be a commutative ring. Prove that $\text{Hom}_R(R,M)$ and $M$ are isomorphic as left $R$-mod. Question Let $R$ be a commutative ring. Prove that $\text{Hom}_R(R,M)$ and $M$ are isomorphic as left $R$-modules.
Attempt
- Define a map $\phi:\text{Hom}_R(R,M)\to M$ by $\phi(f)=f(1), \forall f\in \text{Hom}_R(R,M) $. This makes sense as any $R$-module map $R\to M$ is determined by where it sends $1$.
- Define a map $\psi:M\to \text{Hom}_R(R,M)$ by $\psi(m)=f_m$, makes sense as $f_m:R\to M$ $r\mapsto rm$ is a well defined homomorphism.
$$\text{Hom}_R(R,M)\longrightarrow^\phi M\longrightarrow^\psi \text{Hom}_R(R,M) .$$ $\phi o \psi(m)=\phi(\psi(m))=\phi(f_m)=f_m(1)=m=\text{id}_M(m).$
$$M\longrightarrow^\psi \text{Hom}_R(R,M)\longrightarrow^\phi M.$$ $\psi\phi(f)=\psi(f(1))=f_{f(1)}=f=\text{id}_{\text{Hom}_R(R,M)}(f).$
I just need a clarification whether this solution is correct. I don't need alternative solution I already done.
$$\psi\phi(f) = \psi(f(1)) = f_{f(1)} \stackrel{!}= f = \text{id}_{\text{Hom}_R(R,M)}(f)$$ where $(!)$ is true because $(\forall r \in R) \ f_{f(1)}(r) = rf(1) = f(r)$.