System of polynomial equations: P(x)=P(y) rather than P(x)=0 - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T21:14:24Z http://mathoverflow.net/feeds/question/118649 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118649/system-of-polynomial-equations-pxpy-rather-than-px0 System of polynomial equations: P(x)=P(y) rather than P(x)=0 bjncn 2013-01-11T17:25:59Z 2013-01-12T15:45:35Z <p>$P$ 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.</p> <p>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 $\{ x:Q(y) P(x) =Q(x)P(y) \} \setminus V$.</p> <p>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).</p> <p>While doing this I found myself always starting by solving $Q(x_0) P(x) =Q(x)P(x_0)$ for a particular "<strong>generic</strong>" $x_0$, to check that I had exactly $k$ solutions outside of $V$. This is of course computationally much much faster.</p> <p>Hence my question: </p> <ul> <li>Can I formalize this intuition of genericity?</li> </ul> <p>Of course, any comment, reference or alternative suggestion is welcome. Thanks!</p> http://mathoverflow.net/questions/118649/system-of-polynomial-equations-pxpy-rather-than-px0/118666#118666 Answer by Will Sawin for System of polynomial equations: P(x)=P(y) rather than P(x)=0 Will Sawin 2013-01-11T20:14:45Z 2013-01-12T15:45:35Z <p>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.</p> <p>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.</p> <p>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.</p> <p>This is all assuming you are looking for real solutions. Things are different with rational solutions.</p>