Let $I$ be an ideal of $R$ such that the mapping $f:I\otimes_R\operatorname{Hom}_R (I,R)→R$ defined (on the generators) by $f(i\otimes α)=α(i)$ for all $i∈I$ and $α∈\operatorname{Hom}_R (I,R)$ is onto. Show that $I$ is a finitely generated projective $R$-module.
Here is the hint I have been given: Show there is a split exact sequence $0→K→F→I→0$, where $F$ is a finitely generated free $R$-module.
Any help is appreciated, thanks a lot.
Using the surjectivity of the evaluation map $\mathrm{Hom}_R(I,R)\otimes_R I\to R$, you can find $R$-linear maps $\alpha_1,\dots,\alpha_m:I\to R$ and elements $i_1,\dots,i_m\in I$ such that $$\alpha_1(i_1)+\cdots+\alpha_m(i_m)=1$$ Consider the map $\phi:R^{\oplus m}\to I,(r_1,\dots,r_m)\mapsto r_1i_1+\cdots+r_mi_m$. This map is surjective. Indeed, write $\rho_k=\alpha_k(i_k)$ for $k=1,\dots,m$. Then, if $i\in I$, $$i=i\cdot 1=\sum_{k=1}^m i\rho_k=\sum_{k=1}^m i\alpha_k(i_k)=\sum_{k=1}^m \alpha_k(i)i_k=\phi(\alpha_1(i),\dots,\alpha_m(i))$$ This furthermore shows that $$\psi=(\alpha_1,\dots,\alpha_m):I\to R^{\oplus m}$$ is a splitting: $\phi\circ\psi=\mathrm{id}_I$.