# Finitely generated free module is projective.

Call a $$R$$-module projective if every short exact sequence $$0 \to A\stackrel{f} \to B\stackrel{g} \to C \to 0$$ of $$R$$-modules splits.

Call a short exact sequence as above split, if it admits a section. i.e. an $$R$$-linear map $$h:C\rightarrow B$$ such that $$g\circ h= \text{id}_C$$.

I wish to show that a finitely generated free module is projective. So I need to produce a section for the above short exact sequence where $$C$$ is finitely generated and free.

My thoughts:

1. I can write down an $$R$$-linear map $$h:B/A\rightarrow B$$. But no information is given about the map $$g$$ (except I know that it is surjective). How could I verify that the required composition is the identity on $$C$$?

2. Perhaps, to show that the sequence splits, I can show that $$B\cong A \oplus C$$. I'm not sure how I would proceed to do this though. I know that $$C$$ has a finite basis. Maybe this helps?

This is a homework question. Please don't provide complete solutions. Hints are appreciated. Thanks!

You want a homomorphism $$h:C\to B$$ with certain properties. As $$C$$ is finitely generated projective it has a basis $$c_1,\ldots,c_n$$. Given $$b_1,\ldots,b_n\in B$$, there is a unique homomorphism $$h:C\to B$$ with $$h(c_i)=b_i$$ for all $$i$$.
If you can choose the $$b_i$$ so that $$g(b_i)=c_i$$, then $$(h\circ g)(b_i)=c_i$$. That would imply $$h\circ g=\text{id}_C$$.