MathOverflow is a question and answer site for professional mathematicians. 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

For a given a bivariate polynomial $P(x,y)$ with rational coefficients:

Q1. How compute such (invertible) substitutions of its variables that would transform the polynomial into a homogeneous one? In particular, how to represent $P(x,x)$ as in the form $H(S(x,y),T(x,y))$, if such representation exists, where $H,S,T$ are polynomials and $H$ is homogeneous.

Q2. Can the degree of $H$ be minimized? Can maximum degree of $H,S,T$ be minimized?

Example. The polynomial $$x^6 + x^5 + (y - 6)x^4 - 3x^3 + (-y + 12)x^2 + (3y^2 + 4)x + y^3 + 4y - 8$$ can be represented as $$u^3 + uv^2 + v^3,$$ where $u=x+y$ and $v=x^2-2$.

P.S. Trivial representation with $H(u,v)=u$ and $u=P(x,y)$, $v=0$ is unacceptable as non-invertible.

The above questions and restrictions may sound too vague, so I'd be thankful if someone can help me to formalize the problem I'm trying to pose.

share|cite|improve this question

Set $u=S(x,y)$, $v=T(x,y)$, and let $x=K(u,v)$, $y=L(u,v)$ be the inverse map with $H(u,v)=P(x,y)$ homogeneous in $u,v$. Over an algebraic closure of the (unspecified) base field, $H(u,v)$ splits into linear factors. Suppose that $v-\alpha u$ is such a factor. Then $0=P(K(u,\alpha u),L(u,\alpha u))$, so the algebraic curve given by $P(X,Y)=0$ has a component with a polynomial parameterization.

If the substitutions are allowed to be rational maps, then we analogously get that the curve has a component which admits a rational parameterization. More explicitly, let $Q(X,Y)$ be an irreducible divisor of $P(X,Y)$ such that $Q(K(u,\alpha u),L(u,\alpha u))=0$. This is equivalent to the following: The genus of the function field of the curve $Q(X,Y)=0$ is $0$. This can be checked by hand in small cases, or some computer algebra system.

However, even if the genus is $0$, it might still be difficult to find the parametrization.

share|cite|improve this answer

Edit corrected major mistake

One approach is to work symbolically and solve a system over the rationals.

Choose bounds for the degrees of $S,T,H$ and write them as $\sum a_m x^i y^j$ where each $a_m$ is a fresh variable and $H$ is homogeneous. $H(S(x,y),T(x,y))$ is a polynomial in $x,y$ with coefficients polynomials in $a_i$. Make a system by equating the coefficients of $P(x,y)=H(S(x,y),T(x,y))$ Solve the system over the rationals.

While this will work in theory, solving the system might be quite hard. Experimenting with your example and degrees $(2,2,3)$, maple found 4 solutions in about 2 minutes.

Partially optimistic might be the fact that the system is overdetermined.

share|cite|improve this answer

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.