TITLE: Continuity of the roots of a polynomial in terms of its coefficients
QUESTION [111 upvotes]: It's commonly stated that the roots of a polynomial are a continuous function of the coefficients.  How is this statement formalized?  I would assume it's by restricting to polynomials of a fixed degree n (maybe monic? seems like that shouldn't matter), and considering the collection of roots as a a point in $F^n/\sim$ where F is the field and $\sim$ is permutation of coordinates, but is there something I'm missing?  More to the point, where would I find a proof?
At least, I've seen this stated for C (and hence R); is this even true in general -- say, for an algebraically closed valued field (and hence complete non-Archimedean field because those extend uniquely)?  I've seen it implied that it's not always true in the non-Archimedean case; is this correct?  What's a counterexample?  (If this is wrong and it is true in this generality, is it true in any greater generality?)

REPLY [8 votes]: Here is my favorite proof.  Let $f:\mathbb{C}^n\to\mathbb{C}^n$ be the map taking $(a_1,\dots,a_n)\in\mathbb{C}^n$ to the coefficients of the monic polynomial $\prod_{i=1}^n(x-a_i)$.  This map is clearly continuous and, since it is invariant under permuting the $a_i$, it descends to a continuous map $g:\mathbb{C}^n/S_n\to\mathbb{C}^n$ where $S_n$ acts on $\mathbb{C}^n$ by permuting the coordinates.  The claim is then that this map $g$ is a homeomorphism.
It is clear that $g$ is a bijection, since every monic polynomial of degree $n$ factors as a product $\prod_{i=1}^n(x-a_i)$, uniquely up to permuting the factors.  So, the hard part is to prove $g^{-1}$ is continuous.
The trick for this is to homogenize the polynomials to extend the maps to projective space so that compactness gives you continuity of the inverse for free.  Let us consider $\mathbb{C}$ as a subspace of $\mathbb{CP}^1$ and $\mathbb{C}^n$ as a subspace of $\mathbb{CP}^n$ in the usual way.  Then $f$ extends to a map $F:(\mathbb{CP}^1)^n\to\mathbb{CP}^n$ as follows.  Identify $\mathbb{CP}^1$ with the projectivization of the space of homogeneous linear polynomials in two variables, and identify $\mathbb{CP}^n$ with the projectivization of the space of homogeneous degree $n$ polynomials in two variables.  Then $F$ is just the map which takes $n$ linear homogeneous polynomials and multiplies them together to get a degree $n$ homogeneous polynomial.  (To see that this extends $f$, identify $a_i\in\mathbb{C}$ with the homogeneous polynomial $x-a_iy$, so then $F$ maps $(a_1,\dots,a_n)$ to $\prod_{i=1}^n(x-a_iy)$ whose coefficients are the same as those of $\prod_{i=1}^n(x-a_i)$.)
Just like $f$, this extension $F$ in invariant under permuting the inputs, so it descends to a continuous map $G:(\mathbb{CP}^1)^n/S_n\to\mathbb{CP}^n$ which extends $g$.  Just like $g$, this map $G$ is easily seen to be a bijection.  But now for the magic: since $(\mathbb{CP}^1)^n/S_n$ is compact and $\mathbb{CP}^n$ is Hausdorff, any continuous bijection between them is automatically a homeomorphism!  Thus $G$ is a homeomorphism.
There's one detail left to check: we now know that $g$ is a homeomorphism when you consider its domain as a subspace of $(\mathbb{CP}^1)^n/S_n$, but is that subspace topology the same as the quotient topology on $\mathbb{C}^n/S_n$?  The answer is yes, because $\mathbb{C}^n$ is open in $(\mathbb{CP}^1)^n$ and invariant under the action of $S_n$, so an $S_n$-invariant open subset of $\mathbb{C}^n$ is the same thing as an $S_n$-invariant open subset of $(\mathbb{CP}^1)^n$ that happens to be contained in $\mathbb{C}^n$.