Let the sequence $$0\to A\to B\to C\to 0$$ be a split exact sequence of $R$-modules over a ring $R$. The ring $R$ is a commutative ring with identity. Let $A,B$ be free and finitely generated. Is $C$ also free and finitely generated? If it is not, then under which conditions on the ring $R$ we can say that $C$ is free and finitely generated?
I know that without splitting it is certainly not true, due to the common example: $0\to Z\to 2Z\to Z/2Z\to 0$. However, I have no idea on how to approach this…
Best Answer
Split exact implies that $B$ is isomorphic to the direct sum of $A$ and $C$. Thus, $C$ is a direct summand of the free and finitely generated module $B$ (and thus $C$ must be a projective module). So, for example, the structure theorem for finitely generated modules over principal ideal domains implies that a sufficient condition for this result to hold is that $R$ be a principal ideal domain.
Also, note that $C$ is always finitely generated, being a surjective image of a finitely generated module. The only issue is whether $C$ need be free.
Without the f.g., conditions, a module $C$ that can occur in this split-exact sequence is called stably free. (Note that not all projective modules are stably free, since $A$ is not required to be free in order for $C$ to be projective.) Thus, $C$ in your question is a finitely generated stably free module. There are commutative rings $R$ where not all such modules are free. For example, let $R=\Bbb{R}[x,y,z]/(x^2+y^2+z^2-1)$ and let $T=\{(f,g,h)\in R^3 | xf+yg+zh=0\}$. Then $T$ is not free, but $T\oplus R\cong R^3$. See here for a proof, and a discussion of stably free modules generally.
Also, note that not all projective modules are stably free. For example, if $R$ is a Dedekind domain that is not a principal ideal domain, then its non-principal ideals are projective modules that are not stably free. In fact, I think you could conclude from the structure theory for f.g. modules over Dedekind domains that $R$ being a Dedekind domain is sufficient to make $C$ free.