This question asks: If $f,g \in k[x,y]$ are two algebraically dependent polynomials over an arbitrary field $k$, is it true that there exists a polynomial $h \in k[x,y]$ such that $f,g \in k[h]$, namely, $f=u(h)$ and $g=v(h)$ for some $u(t),v(t) \in k[t]$; the answer is positive.

Is it possible to replace the field $k$ by an integral domain $D$? Namely: If $f,g \in D[x,y]$ are two algebraically dependent polynomials over an arbitrary integral domain $D$, is it true that there exists a polynomial $h \in D[x,y]$ such that $f,g \in D[h]$?

Denote the field of fractions of $D$ by $Q(D)$. It is clear that if $f,g \in D[x,y] \subset Q(D)[x,y]$ are two algebraically dependent polynomials over $D$, then from the above question there exists a polynomial $h \in Q(D)[x,y]$ such that $f,g \in Q(D)[h]$, namely, $f=u(h)$ and $g=v(h)$ for some $u(t),v(t) \in Q(D)[t]$.

I do not see why, for example, $D[x][y] \ni f=u(h)=u_mh^m+\cdots+u_1h+u_0$ should imply that $h \in D[x,y]$ and $u_j \in D$ (changing variables does not seem to help, namely if the leading term is $cy^l$, with $c \in Q(D)$).

Any comments are welcome.