TITLE: Paradox: Roots of a polynomial require less information to express than coefficients?
QUESTION [43 upvotes]: A somewhat information theoretical paradox occurred to me, and I was wondering if anyone could resolve it.
Let $p(x) = x^n + c_{n-1} x^{n-1} + \cdots + c_0 = (x - r_0) \cdots (x - r_{n-1})$ be a degree $n$ polynomial with leading coefficient $1$. Clearly, the polynomial can be specified exactly by its $n$ coefficients $c=\{c_{n-1}, \ldots, c_0\}$ OR by its $n$ roots $r=\{r_{n-1}, \ldots, r_0\}$.
So the roots and the coefficients contain the same information. However, it takes less information to specify the roots, because their order doesn't matter. (i.e. the roots of the polynomial require $\lg(n!)$ bits less information to specify than the coefficients).
Isn't this a paradox? Or is my logic off somewhere?
Edit: To clarify, all values belong to any algebraically closed field (such as the complex numbers). And note that the leading coefficient is specified to be 1, meaning that there is absolutely a one-to-one correspondence between the $n$ remaining coefficients $c$ and the $n$ roots $r$.

REPLY [3 votes]: The core of your paradox seems to be that you claim that in some sense there are "more" (ordered) tuples $(c_{n-1},\dots,c_0)$ than (unordered) sets $\{r_{n-1},\dots,r_0\}$. This is not true for infinite sets like $\mathbb C$, only for finite sets, and even then it would not be significant for your problem as will be seen below.
In the sequel we only consider polynomials having leading coefficient $1$ to avoid a "trival non-uniqueness".

It is in general not true that the set $R(p)$ of roots of a polynomial $p = p(x)$ allows to reconstruct $p$. Note that $R(p)$ is nothing else than a set and does not contain information about the multiplicity of the roots. The number $m(p)$ of elements of $R(p)$ may be anything between $1$ and $n = \deg(p)$, i.e. we have $1 \le m(p) \le n$. [Note that for $\deg(p) = 0$ this says that the set $R(p)$ is empty.] The notation $R(p) = \{r_{n-1},\dots,r_0\}$ is misleading since it suggests that the set $R(p)$ has $n$-elements. Instead we have $R(p) = \{ \rho_1, \dots, \rho_{m(p)}\}$ with $\rho_i \ne \rho_j$ for $i \ne j$.
Therefore the problem is this: To each ordered tuple $\tau = (c_{n-1},\dots,c_0)$ of finite length we can associate a unique polynomial $p_\tau$ having the $c_i$ as coefficients, but  to a finite set $R$ we can associate infinitely many polynomials having $R$ as a root set. There is exactly one such polynomial with minimal degree (in which each $\rho \in R$ occurs with multiplicity $1$, i.e.the degree equals the number of elements of $R$), but if we associate it to $R$, we do not get all polynomials having $R(p) = R$. The only exception is $R =\emptyset$; in that case $p(x) = 1$.

What we really need to consider is the set of pairs $\{ (\rho_1,k_1) \dots, (\rho_m,k_m) \}$ where $k_i \ge 1$ is the multiplicity of the root $\rho_i$. Then we get a unique polynomial $p$ of degree $n = \sum k_i$ such that $R(p) =  \{ \rho_1, \dots, \rho_{m(p)}\}$.
Alternatively we can consider $\mathbb C^n/\mathfrak S_n$, the quotient of $\mathbb C^n$ by the operation of the symmetric group of $n$ elements. That is, tupels are identified if they are related by permuting their components. One might be tempted to think of the elements of $\mathbb C^n/\mathfrak S_n$ as simple (unordered) sets, but it is not true. In fact, $\mathbb C^n/\mathfrak S_n$ can be identified with the collection of pairs $\{ (\rho_1,k_1) \dots, (\rho_m,k_m) \}$ with $k_i \ge 1$ and $n = \sum k_i$.


See also How to show symmetric product of complex space is homeomorphic to the complex space .