Solving a polynomial equation with a parametrization Let $f(x,y)$ be an irreducible polynomial, in the two variables $x$ and $y$. 
It sometimes happens that a “lucky” change of variables $x=g(t)$, where $g$ is a non constant polynomial, transforms our irreducible equation in a nice completely factored equation : in other words, if $d$ is the degree of $f$ in $y$, there are univariate polynomials $c,h_1, \ldots ,h_d$ in $t$ such that the identity
$$
f(g(t),y)=c(t)\prod_{k=1}^d (y-h_k(t))
$$
holds. Is an algorithm known to decide if such a $g$ exists, or even better, to compute it explicitly ?
 Update 20 :00  As noted in a comment below, the answer probably depends on the field. But I believe that this dependence is not very strong and I’m basically interested in an answer over any (zero characteristic) field.
 A: If you have such polynomials, it means that you have an inclusion $K(x,y_1,\ldots,y_d)/(\prod (Y-y_k) = f(x,Y)) \subset K(t)$. By the primitive element theorem, you can even find a $t$ such that this is an equality (and so you can find a $t$ as a rational fraction of $x,y_1,\ldots,y_k$ !)
So this happens if and only if the field $K(x,y_1 \ldots, y_k)$ is isomorphic to $K(t)$, which also means that it is of genus $0$.
You can compute the genus rather easily using the Riemann - Hurwitz formula on the map $K(x) \to K(x,y_1,\ldots,y_n)$.
I am not an expert at finding the expressions of $t$ and vice-versa. It is doable though : a good look at the Riemann surface tells you, up to an automorphism of $K(t)$, how many zeros and poles $x$, and the $y_i$ should have. Then you put an indeterminate for each zero and pole, and try to find a solution of $\prod (Y-y_k) = f(x,Y)$ (assuming $K$ is algebraically closed).
And for bonus points you can represent the Galois group of $f$ with $K(x)-$automorphisms of $K(t)$
