$A$-flatness of $\frac{B}{\langle b_i \rangle}$, $A \subseteq B=A[b]$, implies $A$-flatness of $\frac{B}{\langle b_1,b_2 \rangle}$?

Question

Let $A \subseteq B=A[b]$ be two commutative $k$-algebras, where $k$ is a field of characteristic zero and $b \in B-A$.

Let $b_1,b_2 \in B-A$ be two elements of $B$ such that $\frac{B}{\langle b_i \rangle}=\frac{A[b]}{\langle b_i \rangle}$ is flat over $A$, $1 \leq i \leq 2$.

In that case, is it true that $\frac{B}{\langle b_1,b_2 \rangle}=\frac{A[b]}{\langle b_1,b_2 \rangle}$ is flat over $A$? If not, it would be nice to see a counterexample.

I know that there are surjections $\frac{B}{\langle b_i \rangle} \to \frac{B}{\langle b_1,b_2 \rangle}$, $1 \leq i \leq 2$, but do not see the connection to flatness (if such connection exists). Am I missing something trivial here (probably yes)?

Any hints and comments are welcome!

Answer 1

Take $A=k[X]$, $B=k[X,Y]$, $b_1=X^2+Y$, $b_2=Y$. Then $B/\langle b_i\rangle$ is free over $A$ of rank $1$ (hence flat). However, $B/\langle b_1,b_2\rangle=A/(X^2)$ is certainly not flat over $A$.