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.


2 Answers 2


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.


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.


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