A ‘strong’ form of the Fundamental Theorem of Algebra Let $ n \in \mathbb{N} $ and $ a_{0},\ldots,a_{n-1} \in \mathbb{C} $ be constants. By the Fundamental Theorem of Algebra, the polynomial
$$
p(z) := z^{n} + \sum_{k=0}^{n-1} a_{k} z^{k} \in \mathbb{C}[z]
$$
has $ n $ roots, including multiplicity. If we vary the values of $ a_{0},\ldots,a_{n-1} $, the roots will obviously change, so it seems natural to ask the following question.

Do the $ n $ roots of $ p(z) $ depend on the coefficients in an analytic sort of way? More precisely, can we find holomorphic functions $ r_{1},\ldots,r_{n}: \mathbb{C}^{n} \to \mathbb{C} $ such that
  $$
z^{n} + \sum_{k=0}^{n-1} a_{k} z^{k} = \prod_{j=1}^{n} [z - {r_{j}}(a_{0},\ldots,a_{n-1})]?
$$

The definition of a holomorphic function of several complex variables is given as follows:

Definition Let $ n \in \mathbb{N} $ and $ \Omega \subseteq \mathbb{C}^{n} $ be a domain (i.e., a connected open subset). A function $ f: \Omega \to \mathbb{C} $ is said to be holomorphic if and only if it is holomorphic in the usual sense in each of its $ n $ variables.

The existence of $ r_{1},\ldots,r_{n}: \mathbb{C}^{n} \to \mathbb{C} $ that are continuous seems to be a well-known result (due to Ostrowski, perhaps?), but I am unable to find anything in the literature that is concerned with the holomorphicity of these functions.
Any help would be greatly appreciated. Thank you very much! 

Consider the map
$$E\colon = (x_1, x_2, \ldots, x_n) \mapsto (\sum x_i, \sum_{j< j} x_i x_j, \ldots, x_1 x_2 \cdots x_n)= (e_1, \ldots, e_n)$$
Assume that there exists a solution map $S$   defined locally around a point $(e_1, \ldots, e_n)$  such that
$$E \circ S = \textrm{Id}$$
Then taking the derivative maps at the point $(e_1, \ldots, e_n)$ we get
$$D E \circ D S = \textrm{Id}$$
Now, note that the Jacobian of $E$ at a point $(x_1, \ldots, x_n)$ has determinant $\prod_{i< j} (x_i - x_j)$  (interesting, but not difficult calculation). We conclude that for every $(e_1, \ldots, e_n)$ where the discriminant $\prod_{i<j} (x_i - x_j)^2$ (expressed in terms of $(e_1, \ldots, e_n)$) is $0$, every point $(x_1, \ldots, x_n)$ in the fibre of $(e_1, \ldots, e_n)$ is singular for the map $E$.  Therefore,

*

*we cannot invert smoothly the map $e$ around points in the image with a $0$ discriminant ( chain rule)


*we can invert smoothly around points in the image with a non-zero discriminant
(by inverse function theorem).
Note: this is just what Prof. Lubin is saying.
