Suppose $F(x,y,z)$ is a complex-coefficient homogeneous polynomial of degree $d$ and contains the term $z^d$. Then for each $\lambda \in \Bbb C$, $f_\lambda(z):=F(\lambda,1,z)$ is a polynomial in $z$ of degree $d$, so it has $d$ zeors (counted with multiplicity). How can we show that there is an open subset $U$ of $\Bbb C$ such that the $d$ zeros of $f_\lambda$ are all distinct, for each $\lambda \in U$?
This is asserted in p.142 of Miranda's book Algebraic Curves and Riemann Surfaces, showing that a plane curve defined by a degree $d$ polynomial has degree $d$. If I could show that the subset $\{\lambda\in \Bbb C:f_\lambda$ has a multiple root$\}$ is a proper closed subset then the result follows, but I can't see why.
Best Answer
A polynomial with multiple roots has zero discriminant, and the discriminant of a polynomial $P$ is itself a polynomial expression in the coefficients of $P$. So if you have an algebraic family of polynomials, the set of all polynomials in that family with multiple roots is Zariski closed, and in your situation it is enough to find one polynomial without multiple roots.
I am not quite sure I understand your definition of $f_\lambda$ though, because if you take $F(x,y,z)=z^d$ then the claim is obviously false.