Any two bivariate algebraically dependent polynomials are always in the same ring generated by some bivariate polynomial?

Question

If $f(x,y)$ and $g(x,y)$ are two algebraically dependent polynomials over some field $k$, is it true that there exists a bivariate polynomial $p(x,y)$ such that both $f(x,y)$ and $g(x,y)$ are in the ring $k[p(x,y)]$?

Answer 1

Yes. Let $k[T]$ denote the $k$-subalgebra of $k(x,y)$ generated by a subset $T$, and let $k(T)$ denote its field of quotients. Luroth's Theorem says that every field $L$, $k \subset L \subset k(x,y)$, of transcendence degree one over $k$ is equal to $k(p)$ for some $p \in k(x,y)$. A sharpening of E. Noether (char $k$ = 0) and A. Schinzel (arbritrary $k$) says that if $L$ contains a nonconstant polynomial, then some $p \in k[x,y]$ suffices (Theorem 4, page 10 in "Selected Topics on Polynomials" , by A. Schinzel). The hypothesis that $f$ and $g$ are algebraically dependent implies that $k(f,g)$ has transcendence degree $\leq 1$ over $k$. Thus $f,g \in k(p)$ for some $p \in k[x,y]$. Unique factorization in $k[x,y]$ and $k[p]$ then implies that $f,g \in k[p]$.