Show an ideal is a finitely generated projective module via a split exact sequence

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.

• Have you tried something along this line: $1=\alpha_1(i_1)+\cdots+\alpha_m(i_m)$, and see whether the morphism $\oplus^mR\to I, r_1\oplus\cdots\oplus r_m\mapsto\sum_{k=1}^m r_ki_k$ is onto and split? – Olivier Bégassat Mar 13 '15 at 1:17
• Thank you, I haven't tried that, but what you have wrote will imply that $1\in I$, and so $I=R$? Shouldn't $I$ be arbitrary? – user138017 Mar 13 '15 at 1:29
• If $f$ is onto, you do have such a relation. On another hand, there's no reason why $\alpha(i)$ should belong to $I$. – Bernard Mar 13 '15 at 1:30

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: $\phi\circ\psi=\mathrm{id}_I$.
• Thanks a lot for answering. can I ask about your notation $R^{⊕m}$? is that means the $m$ copies direct sum of $R$, $⊕_m R$ ? – user138017 Mar 13 '15 at 4:09
• Yes, it means $R\oplus R\oplus\cdots\oplus R$, $m$ times. – Olivier Bégassat Mar 13 '15 at 4:14
• I guess the free module $F$ in the exact in the hint provided is $F=⊕_m R$, so may I ask if I need to find the $K$ that listed in the hint? Thanks a lot. – user138017 Mar 13 '15 at 4:24
• $K$ is the kernel of $\phi$. – Olivier Bégassat Mar 13 '15 at 4:48
• But you do not really have to find $K$. The surjection $F \to I$ splits, hence $I$ is a direct summand of a free module and thus projective. – MooS Mar 13 '15 at 6:48