Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
$f\in \mathbf{C}[x,y]$ or $f\in \mathbf{R}[x,y]$? I think the answer depends on the field. – vesszabo Dec 4 '12 at 18:46
up vote 3 down vote accepted

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 $K(t)$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.