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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.

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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 $g = f$ .

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  • $\begingroup$ I see. We need to define a map $f$ with $f(r) = rm$ to see a surjection $\endgroup$ – Minh Nguyễn Hoàng May 28 '17 at 14:59
  • $\begingroup$ What $rm$ means if $M$ is a right $R$-module? Sorry for the silly question but I am confused $\endgroup$ – Cornelius Mar 20 at 20:19
  • $\begingroup$ @Cornelius Sorry, I'm just more used to left-modules. I will update the answer. $\endgroup$ – Trevor Gunn Mar 20 at 23:34

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