# Let R be a commutative ring. Prove that Hom_R(R, M) and M are isomorphic R-modules

Let R be a commutative ring. Prove that $Ho{m}_{R}\left(R,M\right)$ and M are isomorphic R-modules
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Define $F:Hom\left(R,M\right)\to M$ by $F\left(\varphi \right)=\varphi \left(1\right)\in M$
First, we proved that F is an R-module homomorphism.
Let $\varphi ,\rho \in Hom\left(R,M\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}r\in R$. Then,
$F\left(r\varphi -\rho \right)=\left(r\varphi -\rho \right)\left(1\right)=r\varphi \left(1\right)-\rho \left(1\right)=rF\left(r\right)-F\left(r\right)$
Hence, F is an R-module homomorphism.
Now, to prove that it is an isomorphic R-module, we must prove that F is bijective.
Suppose $F\left(\varphi \right)=0$ for some $\varphi \in Hom\left(R,M\right)$. Then, $\varphi \left(1\right)=0.$
For any $r\in R,\varphi \left(r\right)=r\varphi \left(1\right)=r\cdot 0=0$. Hence, $\varphi =0$. Thus, F is injective.
Now, suppose $m\in M$.
Define a map $\varphi :R\to M$ by $\varphi \left(r\right)=rm$ for any $r\in R.$
Let $r,s,t\in R.$
$⇒\varphi \left(rs-t\right)=\left(rs-t\right)m=r\left(sm\right)-tm=r\varphi \left(s\right)-\varphi \left(t\right)$
Hence, $\varphi \in Hom\left(R,M\right).$
Futher, F $\left(\varphi \right)=\varphi \left(1\right)=m.$
Thus, F is surjective.
F is homomorphism and bijective. Therefore, $Hom\left(R,M\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}M$ are isomorphic R-module.