# Show that $\operatorname{Hom}_{R} (R,M) \cong M$

Let $R$ be a commutative ring with unit and $M$ a right $R$-module. Consider the assignment $\varphi: \operatorname{Hom}_{R} (R,M) \rightarrow M$, $f \mapsto \varphi(f) = f(1)$. Show that $\varphi$ is an isomorphism.

My approach: We have to prove $\varphi$ is a morphism, then $\varphi$ is a injection, surjection, hence $\varphi$ is an isomorphism.

$\forall f,g \in \operatorname{Hom}_{R} (R,M)$

$\varphi(fg) = (fg)(1) = f(1)g(1) = \varphi(f) \varphi(g)$

$\Rightarrow \varphi$ is a morphism

$\forall f,g \in \operatorname{Hom}_{R} (R,M)$

$\varphi(f) = \varphi(g) \Rightarrow f(1) = g(1) \Rightarrow f = g$

$\Rightarrow \varphi$ is injective $\Rightarrow \varphi$ is a homomorphism

How can I prove $\varphi$ is surjective to conclude $\varphi$ is an isomorphism? Sorry for my poor English.

First of all, for $$\varphi$$ to be a homomorphism, it means that $$\varphi(fr + gs) = \varphi(f)r + \varphi(g)s$$ for $$r, s \in R$$ and $$f, g \in \operatorname{Hom}_R(R,M)$$. Notice that multiplication is not an operation on modules, only addition and scalar multiplication.
Secondly, it is not clear to me why $$f(1) = g(1) \implies f = g$$. But we can handle this and surjectivity at the same time with the following realization.
Given $$m \in M$$ there is a unique map $$f : R \to M$$ with $$f(1) = m$$.
To see this, define $$f(r) = mr$$. Then clearly $$f(1) = m$$. If there is another $$R$$-linear map $$g$$ with $$g(1) = m$$, then $$mr = g(1)r = g(r)$$ for all $$r \in R$$ so $$g = f$$ .
• I see. We need to define a map $f$ with $f(r) = rm$ to see a surjection – Minh Nguyễn Hoàng May 28 '17 at 14:59
• What $rm$ means if $M$ is a right $R$-module? Sorry for the silly question but I am confused – Cornelius Mar 20 at 20:19