# 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.

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$f\in \mathbf{C}[x,y]$ or $f\in \mathbf{R}[x,y]$? I think the answer depends on the field. –  vesszabo Dec 4 '12 at 18:46

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 $K(t)$
I understand that the Riemann-Hurwitz formula allows one to compute the genus $g$ from the Euler characteristic $\chi$. But how does one compute $\chi$ from the initial data $f$ ? I don't know much about homology. –  Ewan Delanoy Dec 8 '12 at 8:55
@ Ewan : the relation between $\chi$ and $g$ isn't what the Riemann-Hurwitz formula is about. Rather, the formula tells you how to compute $g$ of the extension from $g$ of the base field $K(x)$ (which is $0$) and from the data of where the corresponding covering of Riemann surfaces ramify and how. Those points are precisely the values of $x$ where the discriminant of $f(x,Y)$ is zero, and the order depends on the multiplicity of that zero. So you have to start with computing the discriminant of $f$. –  mercio Dec 8 '12 at 15:18