Let $R$ be a local Noetherian domain with fraction field $K$ and residue field $\Bbbk$. Let $C^{\bullet}$ be a bounded complex of free, finitely generated $R$-modules. Suppose that $C^{\bullet} \otimes\_R K$ and $C^{\bullet} \otimes\_R \Bbbk$ are both exact. Does it follow that $C^{\bullet}$ is exact?


[Note: The converse holds, since tensoring anything by an exact sequence of flat modules (with zeros on both ends) gives an exact sequence.]



Does fiberwise exactness imply exactness?
Yes, it follows from Nakayama's Lemma. You can get by with the weaker set of hypotheses:


$R$ is a local ring with residue field $\mathbb k$.


$C^\bullet$ is a complex of finitely generated projective $R$-modules, bounded above.


$C^\bullet\otimes\_R\mathbb k$ is exact. 


For the key step, note that if $C^{n-1}\to C^n\to 0$ becomes exact when tensored with $\mathbb k$ then it is already exact, so that $C^n$ can be split off $C^{n-1}$.


