Rest Stop Where Divisibility Becomes Composition

Before we rejoin the highway of broader themes, pause at a rest area for a crisp algebra puzzle. Let $k$ be a field and let $f,g\in k[x]$ with $x$ an indeterminate. Introduce $y$ as a second indeterminate independent of $x$ . Assume that in $k[x,y]$ the polynomial $g(y)-g(x)$ divides $f(y)-f(x)$ . Show that there exists a single $h\in k[t]$ with $f(x)=h\bigl(g(x)\bigr)$ . I will post a full solution later, together with a few ways to read this divisibility as a test for composition.

The same question extends to several variables. Let $x=(x_1,\dots,x_n)$ and $y=(y_1,\dots,y_n)$ be independent tuples, and let $f,g\in k[x_1,\dots,x_n]$ . If $g(y)-g(x)$ divides $f(y)-f(x)$ in $k[x_1,\dots,x_n,y_1,\dots,y_n]$ , prove that there is $h\in k[t]$ with $f=h\circ g$ . A follow up post will give the proof and sketch applications to functional equations, invariance along the fibers, and degree bounds.

As a teaser, here is a concrete application. Suppose $f(x)\in k[x]$ , where $k$ has characteristic not equal to $2$ , and assume that
$f(x)=f\left(\frac{-x+\sqrt{-3x^{2}-4}}{2}\right)=f\left(\frac{-x-\sqrt{-3x^{2}-4}}{2}\right)$
where the square root is interpreted in a suitable quadratic extension of $k(x)$ . Then $f(x)=G(x^{3}+x)$ for some $G\in k[x]$ . Indeed, over an algebraic closure of $k(x)$ one has the factorization
$y^{3}+y-(x^{3}+x)=(y-x)\left(y-\frac{-x+\sqrt{-3x^{2}-4}}{2}\right)\left(y-\frac{-x-\sqrt{-3x^{2}-4}}{2}\right).$
Under the stated equalities the difference $f(y)-f(x)$ vanishes at each of these three values of $y$ , hence $y^{3}+y-(x^{3}+x)$ divides $f(y)-f(x)$ in $k[x,y]$ . Invoking the main proposition with the inner map $H(x)=x^{3}+x$ yields the composition $f(x)=G\bigl(x^{3}+x\bigr)$ . Note also that $H'(x)=3x^{2}+1\neq 0$ , so the cubic in $y$ is separable over $k(x)$ , which makes the divisibility test especially transparent. A full proof and further examples will appear in the next post.