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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!

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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 $h\circ g=\text{id}_C$.

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