System of polynomial equations: P(x)=P(y) rather than P(x)=0

2013-01-11T17:25:59Z

$P$ is a system of polynomials in $n$ variables over $\mathbb{Q}$. $Q$ is a singe such polynomial. Let $V$ be the zeros of $Q$. I know from some symmetry argument that for every $y \in [0,1]^n \setminus V$ there are at least $k$ elements in $\{ x:\frac{P(x)}{Q(x)}=\frac{P(y)}{Q(y)} \}$. I want to prove that, for $y$ in a Lebesgue measure 1 subset of $[0,1]^n$, there are in fact exactly $k$ elements.

In order to do this, I am trying to show that for any $y \in [0,1]^n \setminus (V \cup V_1)$ (where $V_1$ is some other variety to be determined/guessed from the example at hand), there are exactly $k$ elements in $\{ x:Q(y) P(x) =Q(x)P(y) \} \setminus V$.

I thought of casting $Q(y) P(x) =Q(x)P(y)$ as a system of polynomials in $2n$ variables, and taking the ideal quotient with respect to the solutions I already know (from the $k$ solutions and from $V$ and $V1$), but this proved too computationally intensive (I called Singular from Mathematica). In a simple example $n=8$, $P$ has 60+ polynomials, and the Groebner basis for $Q(y) P(x) =Q(x)P(y)$ has 500+ elements. Of course this first approach is very naive and in particular does not exploit the two levels of symmetry (the $k$ solutions and the $Q(y)P(x) =Q(x)P(y)$ rather than $H(x,y)=0$ bit).

While doing this I found myself always starting by solving $Q(x_0) P(x) =Q(x)P(x_0)$ for a particular "generic" $x_0$, to check that I had exactly $k$ solutions outside of $V$. This is of course computationally much much faster.

Hence my question:

Can I formalize this intuition of genericity?

Of course, any comment, reference or alternative suggestion is welcome. Thanks!

Answer by Will Sawin for System of polynomial equations: P(x)=P(y) rather than P(x)=0

2013-01-11T20:14:45Z

Yes, you can formalize this intuition. You can look for solutions over the algebraic closure of the field $\mathbb Q(x_0)$, setting $y=x_0$. If there are just $k$ solutions, then at every point except for a positive codimension closed set there will just be $k$ solutions of the original problem.

The reason this works is that, if you consider the projection from the variety of points satisfing $Q(y)P(x)=Q(x)P(y)$, $x,y\not\in V$ to the variety of points $x$ satisfying $x\not \in V$, the fiber over a typical point (meaning, a point not on a specific closed subset of positive codimension) will be geometrically the same as the fiber over the generic point. One geometric property of a variety in this sense is the number of points it has over the algebraic closure of the base field.

If the fiber over the generic point has extra points, but for some reason these usually fail to be real or fail to lie in $[0,1]^n$, proving that is more subtle, and ideas of genericity may not be as helpful.

This is all assuming you are looking for real solutions. Things are different with rational solutions.

System of polynomial equations: P(x)=P(y) rather than P(x)=0 - MathOverflow

System of polynomial equations: P(x)=P(y) rather than P(x)=0

Answer by Will Sawin for System of polynomial equations: P(x)=P(y) rather than P(x)=0