Is there a complete local analogue of the Artin-Tate lemma?

Question

The Artin-Tate lemma states that if $A \subseteq B \subseteq C$ are commutative rings where $A$ and $C$ are Noetherian, $C$ is finitely generated as an $A$-algebra, and $C$ is finitely generated as a $B$-module, then $B$ is finitely-generated as an $A$-algebra.

I am wondering about a kind of "complete local" analogue of the above. Specifically, let $(C,\mathfrak m)$ be a complete Noetherian local ring, let $A$ be a coefficient ring (as in the Cohen structure theorem) [or perhaps just a complete Noetherian local subring with topology compatible with that of $C$?], and let $B$ be ring between $A$ and $C$ such that $C$ is finite as a $B$-module. Does it follow that $B$ is $(\mathfrak m \cap B)$-adically complete?

(Recall that Eakin's theorem guarantees that $B$ is Noetherian.)

Answer 1

Yes. In fact, if $B \subset C$ is a module-finite extension of noetherian rings with $C$ local and complete then $B$ is local and complete. Indeed, by standard prime-lifting stuff with module-finite extensions we see that $B$ is local, so the issue is just whether $B$ is complete. Since $\widehat{B} \otimes_B C$ is identified with the completion of $C$, yet $C$ is complete, the natural map $C \rightarrow \widehat{B} \otimes_B C$ is an isomorphism. But the natural map $\widehat{B} \rightarrow \widehat{C}=C$ is injective (as $\widehat{B}$ is $B$-flat and $B \rightarrow C$ is injective), so it follows that $\widehat{B}$ is $B$-finite.

Now we can forget about $C$ and instead are reduced to showing that if $B$ is a local noetherian ring such that $\widehat{B}$ is $B$-finite then $B$ is itself complete. But this is immediate from Nakayama's Lemma: the module-finite inclusion $B \rightarrow \widehat{B}$ becomes surjective modulo the maximal ideal of $B$, whence it is surjective and so an isomorphism.