# A generic degree $d$ polynomial has $d$ distinct roots: part 2

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 $$\frac{\partial F}{\partial \lambda} \neq 0.$$

## 1 Answer

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.

• 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! Commented Oct 28, 2021 at 5:02
• Ah you are right, its not an identification. I guess its a covering. I will edit the answer later. Thanks! Commented Oct 28, 2021 at 7:39