Given commutative rings $A \subseteq B \subseteq C$ with $C$ a smooth $A$-algebra, I am interested to know if there are some "mild" conditions that make $B$ a smooth $A$-algebra.

**For example:** Assume $A \subseteq B \subseteq C$ are commutative noetherian domains of zero characteristic, $C$ is a f.g. $B$-module, $B$ is a f.p. $A$-algebra ($B$ is not necessarily a f.g. $A$-module), $C$ is a flat and separable $A$-algebra (hence, $C$ is a smooth $A$-algebra) and $B$ is a flat $A$-algebra.

Is it true that $B$ is a smooth $A$-algebra?; equivalently, is it true that $fd_{B \otimes_A B}(B) < \infty$? (please see Corollary 2).

I have tried to show formal smoothness of $B$ over $A$ (since $B$ is f.p. over $A$, smoothness is equivalent to formal smoothness), namely: For each $A$-algebra $E$, and each ideal $J$ in $E$ with $J^2=0$, the canonical homomorphism $Hom_{A−alg}(B,E) \to Hom_{A−alg}(B,E/J)$ is surjective. However, I had a problem when, given an $A$-algebras homomorphism $f: B \to E/J$ to extend it to an $A$-algebras homomorphism $F: C \to E/J$ (since, if we have such $F$, then $F$ has a preimage $G: C \to E$, and then we can restrict $G$ to $B$, and get a preimage for $f$, if I am not wrong).

I suspect there exists a counterexample.

**Remarks:**
(1) This answer is not applicable in the current "example", since there $B$ is not a domain. (Also, I prefer to assume that $A$ is not a field, and here $C$ is not assumed to be a finitely generated $A$-module).
(2) This question asks about smoothness of $C$ over $B$.
(3) See also this question and comment, *which is now almost answered*.