Let $k$ be a field of characteristic zero. Let $A \subseteq B$ be two commutative $k$-algebras (which are integral domains) such that $B=A[b]$, for some $b \in B$, and $u,v \in B$.
Assume that: (i) $\frac{B}{\langle u,v \rangle}$ is a flat $A$-module. (ii) $\frac{B}{\langle u \rangle}$ is a flat $A$-module and $\frac{B}{\langle v \rangle}$ is a flat $A$-module.
Is there something interesting that can be said about $A,B,u,v$?
Notice that $\frac{B}{\langle v \rangle} \to \frac{B}{\langle u,v \rangle}$ is surjective, but I do not see how this helps.
Special case: The theorem presented here implies the following (if I am not wrong): If $p,q \in k[x,y]$ with $\operatorname{Jac}(p,q) \in k^{\times}$, then $\frac{k[x,y]}{\langle p,q \rangle}$ is flat as a $k[x]$-module and as a $k[y]$-module (take there $R=k[x]$ and $R=k[y]$).
Moreover, if we assume (this is possible by a change of variables) that $p,q$ are monic in $y$ (namely, $p=y^n+p_{n-1}y^{n-1}+\cdots+p_1y+p_0$, $q=y^n+q_{m-1}y^{m-1}+\cdots+q_1y+q_0$, where $p_i,q_j \in k[x]$), then $\frac{k[x,y]}{\langle p \rangle}$ is flat over $k[x]$ and $\frac{k[x,y]}{\langle q \rangle}$ is flat over $k[x]$. Similarly, $p,q$ are monic in $x$, so $\frac{k[x,y]}{\langle p \rangle}$ is flat over $k[y]$ and $\frac{k[x,y]}{\langle q \rangle}$ is flat over $k[y]$. This claim follows from a result presented here (and is mentioned in wikipedia. Originally, it seems to be a result of Nagata).
Therefore, this is indeed a special case with $A=k[x],B=k[x][y], u=p, v=q$ (or $A=k[y],B=k[y][x], u=p, v=q$).
See also the following questions: 1, 2 and 3.
Important remark: According to the first comment to this question, which says: "... if $R$ is a UFD and $R[T]/I$ is flat over $R$, $I$ a proper ideal, then $I$ must be principal", it follows that $\langle p,q \rangle =\langle r \rangle$ for some $r \in k[x][y]$ (unless $\langle p,q \rangle =k[x][y]$). Then $p=Ur, q=Vr$ for some $U,V \in k[x,y]$. It follows that $r=\lambda \in k$ (since $\operatorname{Jac}(p,q) \in k^{\times}$), hence $\langle p,q \rangle =\langle r \rangle= \langle \lambda \rangle= k[x][y]$ after all. Therefore, the Jacobian Conjecture is true, unless I have an error in my above arguments.
Any hints and comments are welcome; thank you!
