A polynomial over $A[z]/(z^2-a)$ can be "completed" to a polynomial over $A$

Question

Let $A$ be a unique factorization domain and $a\in A$ be an element without repeated irreducible factors (I think some weaker conditions may suffice). Let $B:=A[z]/(z^2-a)$ and consider a polynomial $f\in B[x]$. I wonder if we can find a nonzero polynomial $g\in B[x]$ such that $fg\in A[x]$. If so, how to find $g$ (or how to prove the existence of such a $g$).

Answer 1

I assume that $2 \neq 0$ and that $2$ is not a zero-divisor in $A$.

The map $z \mapsto -z$ induces an involution of $B$, and the elements of $A$ are precisely those elements of $B$ fixed by this involution.

By applying this involution on coefficients, we furthermore get an involution of $B[x]$, and $A[x]$ is the subring fixed by this involution.

Since $f \overline{f}$ is fixed by this involution (where $\overline{f}$ means the result you get by applying the involution to $f$), it must be in $A[x]$, so you can take $g = \overline{f}$.