Proof that the map Hom$_R(R,M)\rightarrow M$ given by $f\mapsto f(1)$ is an isomorphism of $R$-modules Let $M$ be a module over a commutative ring $R$ and let Hom$_R(R,M)$ be an $R$-module as well. How do I prove that the map Hom$_R(R,M)\rightarrow M$ given by $f\mapsto f(1)$ is an isomorphism of $R$-modules?
Edit: It's not that I haven't tried anything, it's just that I have no clue what to try.
 A: $\phi: \mathrm{Hom}_R(R,M)\rightarrow M$ is given by $\phi(f)=f(1)$ for all $f\in \mathrm{Hom}_R(R,M)$

*

*To check it is an $R$-module map

(i) $\phi(f+g)=(f+g)(1)=f(1)+g(1)=\phi(f)+\phi(g)$
(ii) $\phi(rf)=rf(1)=r\cdot f(1)=r\cdot\phi(f)$


*To check it is bijective:

(i) injective: If $\phi(f)=0 \implies f(1)=0$
Now, $f$ is R-module homomorphism, i.e., $f(r)=f(r\cdot 1)=rf(1)=r\cdot 0=0$ i.e., $f$ is a zero-homomorphism.
(ii) Surjective: Let $x\in M$. Consider the function $g:R\rightarrow M $ defined by $g(1)=x$, check that $g$ is a $R$-module homomorphism from $R\rightarrow M$, then $\phi(g)=x$. Therefore, $\phi$ is bijective.
Hence, $\phi$ is an isomorphism of $R$-modules and $\mathrm{Hom}_R(R,M)\cong M$
A: First, you want to check that $\phi: \text{Hom}_R(R,M) \to M$ given by $\phi(f) = f(1_R)$ is indeed $R$-linear.
So, first, we have to show that $\phi(f+g) = \phi(f) + \phi(g)$.
Recall that we define $f+g$ "point-wise" that is:
$(f+g)(r) = f(r) + g(r)$ (the right-hand sum takes place in $M$).
So $\phi(f+g) = (f+g)(1_R) = f(1_R) + g(1_R) = \phi(f) + \phi(g)$. That was easy.
Second, we verify that for any $a \in R$, we have: $\phi(af) = a\phi(f)$.
Again, $af$ is defined "point-wise" by $(af)(r) = a(f(r))$ (the latter is the $R$-action in $M$, the former the $R$-action in $\text{Hom}_R(R,M)$).
Thus $\phi(af) = (af)(1_R) = a(f(1_R)) = a\phi(f)$. Evidently, then $\phi$ is $R$-linear (technically, $R$ is "left $R$-linear", but $R$ is commutative, so we don't really care).
So we at least have a homomorphism of $R$-modules.
To show that $\phi$ is surjective, we need to exhibit, for any $m \in M$, an element $f \in \text{Hom}_R(R,M)$ with $f(1_R) = m$.
The obvious candidate is $f_m: R \to M$ given by $f_m(r) = rm$. It is straight-forward to show this is indeed an element of $\text{Hom}_R(R,M)$, and I will not do so here.
We then have that $M \cong \text{Hom}_R(R,M)/\text{ker }\phi$.
Suppose that $\phi(f) = 0_M$. It then follows that $f(1_R) = 0_M$, and thus for any $r \in R$:
$f(r) = f(r1_R) = rf(1_R) = r0_M = 0_M$, that is: $f$ is the $0$-map of $\text{Hom}_R(R,M)$. QED.
