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This question is closely related to the question A generic degree $d$ polynomial has $d$ distinct roots, but that question is missing the important hypothesis of nonsingularity, so the answer falls short of proving the existence of a polynomial in the family with $d$ distinct roots.

Suppose that $F(\lambda,z)$ is a nonsingular polynomial, and for each fixed $\lambda\in \mathbb{C}$ the polynomial $F$ has degree $d$ in $z$. (Nonsingular means that if $F(\lambda,z)=0$ then one of the partials in nonzero at $(\lambda,z)$.) Prove that for some fixed value of $\lambda$, the polynomial $F(\lambda,z)$ has $d$ distinct roots.

This should be true for "generic" values of $\lambda$ but I really only need it to hold for one value. Notice that if $F(\lambda,z)$ has a multiple root in $z$ then at the root, $F=0$ and$\frac{\partial F}{\partial z}=0,$ so $\frac{\partial F}{\partial \lambda} \neq 0.$

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Im coming from differential topology so Im not sure this is the way you want to think about this. But anyway lets denote the space of normalized polynomials of degree $d$ by $$ V_d := \{ z^d + a_{d-1} z^{d-1} + ... + a_0 \; | \; a_i \in \mathbb{C} \} \overset{\sim}{=} \mathbb{C}^d $$ as an affine space of (real) dimension $2d$. Since we want to think about roots, consider the smooth submersion (elementary symmetric polynomials appear in coordinates) \begin{align*} q:\mathbb{C}^d &\to V_d \\ (b_1,\ldots,b_d) &\mapsto (z-b_1) \ldots (z-b_d) \end{align*} and the smooth map $$ p:\mathbb{C}^d \to \mathbb{C} \\ (b_1, \ldots , b_d) \mapsto \prod\limits_{i,j} (b_i - b_j)$$ You can proove that $0$ is a regular value so $p^{-1}(0) \subset \mathbb{C}^d $ is submanifold of codimension $2$. Now $q:p^{-1}(0) \to V_d$ is an embedding and its image $V_d^0$ is the space of polynomials with multiple roots. Now you can consider your $F$ as a (hopefully smooth (better holomorphic)) map $$F: \mathbb{C} \to V_d \\ \lambda \mapsto F_\lambda$$

For $\lambda \in \mathbb{C}$ we have that $$ F_\lambda \in V^0_d \Leftrightarrow \exists z_0 \in \mathbb{C}: F_\lambda(z_0) = 0 \text{ and } \frac{\partial F_\lambda}{\partial z}(z_0) = 0$$ Your condition that $F$ is non-singular shows that for any $\lambda$ with $F_\lambda \in V^0_d$ we must have $\frac{\partial F}{\partial \lambda}(\lambda) =F'(\lambda) = D_\lambda F \neq 0$ (as a polynomial, since we know there exists $z_0$ with $\frac{\partial F}{\partial \lambda}(\lambda,z_0 ) \neq 0$). That is, the map $F: \mathbb{C} \to V_d$ is transversal to $V_d^0$. So $$F^{-1}(V^0_d) = \{\lambda \in \mathbb{C} \;| \; F(\lambda,z) \text{ has less than } d \text{ distinct roots} \}$$ is smooth submanifold of $\mathbb{C}$ with (real) codimension $2$, i.e. its a zero dimensional submanifold, i.e. its a discrete set of points.

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  • $\begingroup$ Very nice! I'm not sure if its right to say your map is an identification since it's not injective (invariant under permutations of roots). But generally, great! $\endgroup$
    – subrosar
    Commented Oct 28, 2021 at 5:02
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    $\begingroup$ Ah you are right, its not an identification. I guess its a covering. I will edit the answer later. Thanks! $\endgroup$
    – Jonas
    Commented Oct 28, 2021 at 7:39

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