$P$ 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!