Let $R$ be a commutative ring $M$ and $N$: $R$-modules Let $f : M \rightarrow N$ is a homomorphism,
I have shown that the map $b: R\times M \rightarrow N$ given by $b(r,m) = r \cdot f(m)$ is R-bilinear. I am trying to show that
$f \mapsto b$ gives an isomorphism of $R$-modules $\Phi_N : \text {Hom}_R(M,N) \rightarrow \text {Bilin}_R(R×M,N)$
Injectivity: Let $f_1,f_2\in \text {Hom}_R(M,N)$
$\Phi(f_1)=\Phi(f_2)\implies b_1=b_2$ whith $b_1(r,m)=rf_1(m)$ and $b_2(r,m)=rf_2(m)$
then $rf_1(m)=rf_2(m) \forall m \in M$.. How do I conclude that $f_1=f_2$ ? In a commutative ring elements are not necessarily invertible , so would it suffice to take $r=1$ to get rid of that r and conclude injectivity ?
surjectivity: Let $b \in \text {Bilin}_R(R×M,N)$ I need to find an $f\in \text {Hom}_R(M,N)$ s.t. $\Phi_N(f)=b$, how do I do that? I am stuck here. Any help is welcome