Do simple roots on most slices imply a polynomial is square free?

Question

Consider a complex polynomial $p(x,y)$ in two variables. I am interested in what it implies for $p$ if for all $y\in\mathbb C$ the one-variable polynomial $p(\cdot,y)$ has or does not have multiple roots.

To ask this more carefully, let me define the "good set" $G\subset\mathbb C$ be the set of those $y$ for which every root of $p(\cdot,y)$ is simple.

A couple of observations:

1. The set $G$ appears to be Zariski-open: If $d$ is the highest degree of $x$ in $p(x,y)$, we can regard the $p(\cdot,y)$s as a one-parameter family of polynomials of degree $d$. There is a multiple zero if and only if the discriminant is zero, and the discriminant $D(y)$ of $p(\cdot,y)$ is a polynomial so its vanishing set is Zariski-closed. (Something can go wrong when the degree of $p(\cdot,y)$ drops below $d$, but that is so rare that it will not be important. This issue will not have an effect on the question.)

2. It is easy to see that if $p$ contains a square, e.g. $p(x,y)=q(x,y)^2r(x,y)$ for some non-constant polynomials $q$ and $r$, then every slice $p(\cdot,y)$ will have multiple roots — the roots of the slice of $q$. Therefore in this case $G=\emptyset$.

3. When $p(x,y)=x^2+y^2-1$, we have $G={\mathbb C}\setminus \{-1,1\}$. The set $G$ can therefore be nontrivial.

Question: Does $G=\emptyset$ imply that $p$ contains a square factor? In other words, is $G$ non-empty (and therefore large) whenever $p$ is square free?

Answer 1

Consider $p$ and the partial derivative $\partial_x p$. We can compute the GCD of $p$ and $\partial_x p$ as elements of $\mathbb{C}(y)[x]$ by the Euclidean algorithm, and if $G=\emptyset$ this GCD must be nonconstant (since the same steps of the Euclidean algorithm also compute the GCD of $p(\cdot,y)$ and $\partial_xp(\cdot,y)$ for all values of $y$ that don't end up making any denominators vanish). On the other hand, if $p$ is squarefree in $\mathbb{C}(y)[x]$, then the GCD of $p$ and $\partial_x p$ must be $1$. If $p$ is squarefree in $\mathbb{C}[x,y]$ then it is also squarefree in $\mathbb{C}(y)[x]$ by Gauss's lemma, so this implies the result you ask for.