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 Fn/∼ where F is the field and ∼ 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 henceR); 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?)

**Answer**

Here is a version of continuity of the roots.

Consider the monic complex polynomial f(z)=zn+c1zn−1+...+cn∈C[z] and factor it as f(z)=(z−a1)...(z−an)(ak∈C)

where the roots ak are arranged in some order, and of course needn’t be distinct.

Then for every ϵ>0, there exists δ>0 such that every polynomial g(z)=zn+d1zn−1+...+dn∈C[z] satisfying |dk−ck|<δ(k=1,...,n) can be written

g(z)=(z−b1)...(z−bn)(bk∈C)

with |bk−ak|<ϵ(k=1,...,n).

A more geometric version is to consider the Viète map v:Cn→Cn sending, in the notation above, (a1,...,an) to (c1,...,cn) (identified with zn+c1zn−1+...+cn=(z−a1)...(z−an) ).

It is a polynomial map (and so certainly continuous!) since ck=(−1)ksk(a1,...,an), where sk is the k-th symmetric polynomial in n variables.

There is an obvious action of the symmetric group Sn on Cn and the theorem of continuity of the roots states that the Viète map descends to a homeomorphism w:Cn/Sn→Cn. It is trivial (by the definition of quotient topology) that w is a bijective continuous mapping, but continuity of the inverse is the difficult part.

The difficulty is concentrated at those points (c1,...,cn) corresponding to polynomials zn+c1zn−1+...+cn having multiple roots.

This, and much more, is proved in Whitney’s *Complex Analytic Varieties* (see App. V.4, pp. 363 ff).

**Algebraic geometry point of view** Since you are interested in general algebraically closed fields k, here is an interpretation for that case.

The symmetric group Sn acts on Ank and the problem is whether the quotient set Ank/Sn has a reasonable algebraic structure. The answer is yes and the Viète map again descends to an isomorphism *of algebraic varieties* Ank/Sn∼→Ank.

This is the geometric interpretation of the fundamental theorem on symmetric polynomials.

The crucial point is that the symmetric polynomials are a finitely generated k-algebra.

Hilbert’s 14th problem was whether more generally the invariants of a polynomial ring under the action of a linear group form a finitely generated algebra. Emmy Noether proved in 1926 that the answer is yes for a finite group (in any characteristic), as illustrated by Sn.

However Nagata anounced counterexamples (in all characteristics) to Hilbert’s 14th problem at the International Congress of Mathematicians in 1958 and published them in 1959.

**Attribution***Source : Link , Question Author : Harry Altman , Answer Author : Communicative Algebra*