Does fiberwise exactness imply exactness?

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.]

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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 $C^{n-1}$.

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Thanks! $\quad$ – Charles Staats May 21 '11 at 20:44
You're welcome. I believe that for the converse (in fact, $C^\bullet$ exact implies $C^\bullet\otimes M$ exact for any $R$ and any $R$-module $M$ if $C^n$ flat) you can also get away without bounded below. But you do need bounded above. – Tom Goodwillie May 22 '11 at 0:07