In Category of R-Module, Hom(R,M) is isomorphic to M

I met an interesting exercise when I am learning module theory

View $R$$R$ is the module over itself, proving that ${\mathrm{Hom}}_{R-Mod}(R,M)\cong M$$\mathrm{Hom}_{R-Mod}(R,M)\cong M$

$\mathrm{Proof}.$$\mathrm{Proof}.$

Firstly, I need to prove that ${\mathrm{Hom}}_{R-Mod}(R,M)$$\mathrm{Hom}_{R-Mod}(R,M)$ is a module

We already know that in $\mathrm{Ab}$$\mathrm{Ab}$, $\mathrm{Hom}(H,G)$$\mathrm{Hom}(H,G)$ is an Abelian group

We define

$(f+g)(r)=f(r)+g(r),(\lambda f)(r)=\lambda f(r)$$(f+g)(r)=f(r)+g(r),(\lambda f)(r)=\lambda f(r)$

And observe that $r(f):=f(r)$$r(f):=f(r)$ give an homomorphism $r:{\mathrm{Hom}}_{R-Mod}(R,M)\to M$$r:\mathrm{Hom}_{R-Mod}(R,M)\to M$

It is natural to consider $1(f):=f(1)$$1(f):=f(1)$

Easy to see it is injective because $1(f)=1(g)\Rightarrow f(1)=g(1)\Rightarrow \mathrm{\forall }r\in R,rf(1)=rg(1)\Rightarrow \mathrm{\forall }r\in R,f(r)=g(r)$$1(f)=1(g)\Rightarrow f(1)=g(1)\Rightarrow\forall r\in R,rf(1)=rg(1)\Rightarrow\forall r\in R,f(r)=g(r)$

To see it is surjective, we only need to consider $\mathrm{\forall }m\in M,f_m(r)=rm,1(f_m):=f_m(1)=m$$\forall m\in M,f_m(r)=rm,1(f_m):=f_m(1)=m$

And we need to prove that $f_m(r)$$f_m(r)$ is a Module homomorphism

$f_m(r_1+r_2)=(r_1+r_2)m=r_{1}m+r_{2}m=f_m(r_1)+f_m(r_2)$$f_m(r_1+r_2)= (r_1+r_2)m=r_1m+r_2m=f_m(r_1)+f_m(r_2)$

$f_m(r_1{r}_2)=r_1{r}_{2}m=r_1(r_{2}m)=r_{1}f(r_2)$$f_m(r_1r_2)=r_1r_2m=r_1(r_2m)=r_1 f(r_2)$

Thus we show that $1$$1$ is an isomorphism.

And every ${\mathrm{Hom}}_{R-Mod}(R,M)$$\mathrm{Hom}_{R-Mod}(R,M)$ has a form like $f(r)=rm$$f(r)=rm$