Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 7 / 2 / c / Solution

Here the Hom groups must be taken over $A$; the printed subscript $B$ in this part has no defined coefficient ring. For an arbitrary module $P$, form the free moduleand define a module homomorphism $\pi :F\to P$ by $e_p\mapsto p$. Each element of $F$ has finite support, so this prescription extends uniquely and is well-defined. It is surjective because every $p\in P$ is the image of $e_p$. Thus every module is a quotient of a free module.

Let $K=\ker \pi$. The short exact sequence $0\to K\to F\stackrel{\pi }{\to }P\to 0$ and the defining exactness condition for a projective module makesurjective. Lift $1_P$ to a map $s:P\to F$. Then $\pi s=1_P$, so $s$ is injective. Every $f\in F$ has the decompositionwhose first term lies in $K$ and second lies in $s(P)$. If $s(p)\in K$, applying $\pi$ gives $p=0$, so the intersection is zero. ConsequentlyThis is a split short exact sequence and proves that projective modules are direct summands of free modules, including modules that are not finitely generated.