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A polynomial in the complex variable $z$, whose coefficients are themselves complex polynomials in another complex variable $a$, looks like $$ f\in\Bbb C[A][Z],\;\;f(z,a)=c_0(a)+c_1(a)z+\cdots+c_n(a)z^n $$ with $c_j\in\Bbb C[A]$ of degree less or equal than $1$. Assume not all $c_j$ are constant and the leading coefficient $c_n\equiv1$.

I need to prove that the roots of $z\mapsto f(z,a)$ are simple for all $a\in\Bbb C$ except for a finite number of $a$.

A root of $f(\cdot,a)$ is simple if and only if it is not a root of $f'(\cdot,a)$ and this happens if and only if the resultant $P(a)$ of $f(\cdot,a),f'(\cdot,a)$ doesn't vanish.

Now $P\in\Bbb C[A]$, hence if it is not constantly $0$, then the claim is true (if $P$ is non constant then it vanishes for a finite number of values of $a$, while if it is constant and non zero, $f(\cdot,a)$ and $f'(\cdot,a)$ don't share a zero for any $a$).

Is it always possible to exclude the case $P\equiv0$?

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    $\begingroup$ What about $f(z,a):=(z-a)^2$ ? $\endgroup$ Commented Oct 28, 2021 at 9:55
  • $\begingroup$ Sorry I forgot the assumption that $c_j$ have degree less or equal than $1$. I am going to edit $\endgroup$
    – Joe
    Commented Oct 28, 2021 at 11:29
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    $\begingroup$ $P=0$ if and only if $f$ has a multiple factor (a factor which is a power of some polynomial). $\endgroup$ Commented Oct 28, 2021 at 12:56
  • $\begingroup$ @AlexandreEremenko: sure, I wrote when I talked about the resultant. I was wondering how to prove that $P\equiv0$ is never the case in the setting above described $\endgroup$
    – Joe
    Commented Oct 28, 2021 at 13:56

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No, it's not true, as is shown by the polynomial $f(z,a):=(z-a)(z-1)^2$.

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