Smoothness of a subalgebra 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.
My question: 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 question, 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) Please see this question.
 A: First let me come back to my initial example $A = k$, $B=k[x,y]/(y^2-x^3-x^2)$, $C$ the integral closure of $B$ in its quotient field $K(B)$.
Then $B/A$ is flat and of finite type, $C$ is nonsingular, therefore $C/k$ smooth, and $C/B$ is finite. The properties "finite", "of finite type", "flat" and "smooth" are preserved under base extension. Therefore it is possible to extend the base $A=k$ to every $k$--scheme $S$. Of course this gives only utterly trivial examples, as every fiber is the same.
To get something a little more interesting one can use the slightly changed original setup:
$$
\begin{align}
A & =\mathbb{Q}[a,b] \\
B & =A[x,y]/((y-b)^2-((x-a)^3+(x-a)^2))
\end{align}
$$
and $C$ again the integral closure of $B$ in its quotient field.
If one computes $\Omega_{C|A}$ one gets $C$, so it is free of rank $1$ as necessary for smoothness over $A$.
The computation of $\Omega_{C|A}$ goes over the sequence
$$(*) \quad I/I^2 \to \Omega_{S|A} \otimes_S C \to \Omega_{C|A} \to 0$$
where $C = S/I$ and $S=A[w,x,y]$, $I=(f_1,f_2,f_3) \subseteq S$ a certain ideal of three elements of $S$ that is found by Macaulay 2 on computing the integral closure $C$ of $B$.
It remains to prove flatness of $C$ and $B$ over $A$. Flatness of $B$ over $A$ follows from the following statement:
Let $R[X_1,\ldots,X_n] = S$ and $S_1=S/(f)$, $f \in S$. Then $S_1$ is flat over $R$ if for every maximal ideal $\mathfrak{m}$ of $R$ the ring $S_1 \otimes_R R/\mathfrak{m}$ is not equal to $S \otimes_R R/\mathfrak{m}$.
This can be found in Milne, Etale Cohomology, Chapter 1, Remark 2.6(a).
To prove the flatness of $C = S/I$ over $A$ seems to be more difficult and I currently can not prove it. The case $A=k[a]$ is considerably simpler in this regard, as one needs only check that every irreducible component of the domain dominates $\operatorname{spec}(A)$.
