# Continuity of the roots of a polynomial in terms of its coefficients

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?)

• It's not actually true for $\mathbb R$: the roots of $x^2+a$ disappear in a dramatically discontinuous manner when $a$ passes through $0$ from below. (It should be true if you restrict attention to polynomials without multiple roots, though). Commented Sep 9, 2011 at 21:08
• Well, I'm assuming that if the field isn't algebraically closed, but the algebraic closure has a natural value/topology extending it (as is the case with R or a complete non-Archimedean field), then we look at the roots there; of course then if it's true for the larger field, it's true for the smaller one! So in that sense I'm only asking about the algebraically closed case. Commented Sep 9, 2011 at 21:22
• Wilkinson's polynomial suggests that continuity may not prevent instability/ill-conditioning Commented Sep 10, 2011 at 17:23
• A proof for $\mathbb{C}$ can be found here ams.org/journals/proc/1987-100-02/S0002-9939-1987-0884486-8/… Commented Jan 11, 2014 at 16:08

Here is a version of continuity of the roots.
Consider the monic complex polynomial $f(z)=z^n+c_1z^{n-1}+...+c_n\in \mathbb C[z]$ and factor it as $$f(z)=(z-a_1)...(z-a_n) \quad (a_k\in \mathbb C)$$ where the roots $a_k$ are arranged in some order, and of course needn't be distinct.
Then for every $\epsilon \gt 0$, there exists $\delta \gt 0$ such that every polynomial $g(z) =z^n+d_1z^{n-1}+...+d_n\in \mathbb C[z]$ satisfying $|d_k-c_k|\lt \delta \quad (k=1,...,n)$ can be written $$g(z)=(z-b_1)...(z-b_n) \quad (b_k\in \mathbb C)$$
with $|b_k-a_k|\lt \epsilon \quad (k=1,...,n)$.

A more geometric version is to consider the Viète map $v:\mathbb C^n \to \mathbb C^n$ sending, in the notation above, $(a_1,...,a_n)$ to $(c_1,...,c_n)$ (identified with $z^n+c_1z^{n-1}+...+c_n=(z-a_1)...(z-a_n)$ ).
It is a polynomial map (and so certainly continuous!) since $c_k=(-1)^{k} s_k( a_1,...,a_n)$, where $s_k$ is the $k$-th symmetric polynomial in $n$ variables.
There is an obvious action of the symmetric group $S_n$ on $\mathbb C^n$ and the theorem of continuity of the roots states that the Viète map descends to a homeomorphism $w: \mathbb C^n / S_n \to \mathbb C^n$. 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 $(c_1,...,c_n)$ corresponding to polynomials $z^n+c_1z^{n-1}+...+c_n$ 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 $S_n$ acts on $\mathbb A_k^n$ and the problem is whether the quotient set $\mathbb A_k^n /S_n$ has a reasonable algebraic structure. The answer is yes and the Viète map again descends to an isomorphism of algebraic varieties $\mathbb A_k^n /S_n \stackrel {\sim }{\to} \mathbb A_k^n$.
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 $S_n$.
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.

• A nice reference on this subject is Morris Marden's book The Geometry of the Zeros of a Polynomial in a Complex Variable. Commented Sep 9, 2011 at 22:12
• I didn't know that book: thanks for the reference, Mariano. Commented Sep 9, 2011 at 22:15
• OK, upvoted, the proof the book gives is quite simple. But it assumes local compactness, which means (assuming algebraic closure) that it'll only work for C. (OK the general statement it gives will be true for other locally compact fields, but that statement only deals with polynomials that split over the given field.) I suppose I'll accept this, and if I really want to know about other cases I'll ask a new question... Commented Sep 14, 2011 at 1:02

I think there might be a proof of your statement using the following complex analysis trick (I don't know if a similar idea could work in $\mathbb{C}_p)$: if $U$ is an open subset with smooth boundary $\partial U$ consider,

$$N_{U}(p) = \frac{1}{2i \pi} \oint_{\partial U} \frac{p'(z)}{p(z)}dz$$

When it's defined, $N_U(p)$ is the number of zeros of $p$ in $U$ counted with multiplicity. Then fix a polynomial $p_0$ of degree $n$, and pick $U$ a neighborhood of its zeros. Then the map $p \mapsto N_U(p)$ is well defined and continuous in a neighborhood of $p_0$, but since it can only take integer values, it's constant and equal to $n$. So if $p$ in that neighborhood has degree $n$, all its roots are in $U$.

• I know this is an old answer, and I like the general idea, but I have a concern. Doesn't the statement that $p\mapsto N_U(p)$ is continuous somehow uses what we want to prove? I mean if the roots are not continuous, it could be that suddenly one jumps onto $\partial U$ or out of $U$. How can we be sure that this does not happen? How have your established continuity of $N_U(p)$ in the first place? Commented Mar 19, 2018 at 11:06
• @M.Winter : The continuity of $N_U(p)$ with regards to $p$ comes from the integral formula. The only potential problem would be if $p$ had a zero on $\partial U$. To prove that it can't be the case, let's denote $\|p\|_{\infty, \partial U} = \max_{\partial U} |p|$. Since $p_0$ is continuous and non zero on the compact set $\partial U$, we have $\|p_0\|_{\infty, \partial U} > 0$. And because $p \mapsto \|p\|_{\infty, \partial U}$ is continuous, there's a neighborhood of $p_0$ where this quantity is still positive. Commented Mar 25, 2018 at 16:11
• @JoelCohen Maybe I am understanding something wrong, but why should $\|p\|_{\infty,\partial U}>0$ imply that $p$ has no zero on $\partial U$? Commented Mar 25, 2018 at 23:27
• @M.Winter : You're right ! My mistake, it should have been the minimum instead of the maximum (so the notation does not fit), let's say $m(p) = \min_{\partial U} |p|$. We have $m(p_0) > 0$ and still have $m(p) >0$ in some neighborhood of $p_0$. Commented Mar 26, 2018 at 1:02
• Which argument is that $N_U(p)$ is the number of zeros of the zeros of $p$ in $U$ counted with multiplicity. Commented May 25, 2018 at 7:50

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 $$\mathbb{C}^n$$.

• I really like this solution for its economy of means. Commented Nov 12, 2020 at 13:20

I posed this as a problem in a course on local fields I taught a little while ago. One of my students, David Krumm, solved it and wrote it up here. The context of David's solution is that $$K$$ is an arbitrary normed field, with some chosen extension of the norm to the algebraic closure of $$K$$. (If $$K$$ is complete or even Henselian, the norm extends uniquely; in general it does not.) Then he shows that for every $$\epsilon > 0$$ there exists some $$\delta > 0$$ so that if you perturb each of the coefficients of your polynomial $$f$$ by at most $$\delta$$, every root of $$f_{\delta}$$ is wtihin $$\epsilon$$ of some root of $$f$$ and vice versa. (I didn't think of this until just now, but I guess this is equivalent to saying that the sets of roots are within $$\epsilon$$ of each other for the Hausdorff metric.) He also shows that if $$f$$ itself has distinct roots, then for sufficiently small $$\delta$$ so does $$f_{\delta}$$ and then you can match up the roots in a canonical way.

After he solved this problem I looked into the literature and found a dozen papers or more on various refinements of it, including some very recent ones. At the moment these papers seem to be hiding from me, but if/when I find them I'll give some references.

• On the Hausdorff metric: Huh, so it is. However it seems weaker than saying that some permutation of the roots of $f_\delta$ is within $\varepsilon$ of some permutation of the roots of f. This still may well be enough for what I originally needed, though. :) Commented Sep 12, 2011 at 3:06

Suppose $P_a(z)=\sum\limits_{k=0}^na_kz^k$. Taking the partial of $P_a(z)=0$ with respect to $a_k$, we get $$0=P_a^{\;\prime}(z)\frac{\partial z}{\partial a_k} + z^k$$ Thus, we get that $$\frac{\partial z}{\partial a_k}=-\frac{z^k}{P_a^{\;\prime}(z)}$$ The existence of these partial derivatives are guaranteed by the Inverse Function Theorem.

Thus, as long as $P_a^{\;\prime}(z)\ne0$ when $P_a(z)=0$, $\frac{\partial z}{\partial a_k}$ will exist and be finite. Therefore, if $P_a$ has no repeated roots, $\frac{\partial z}{\partial a_k}$ is finite.

This shows that unless $P_a$ has repeated roots, each root is a smooth function of the coefficients.

• Are you making the assumption that $z$ is a function of the coefficients ? Commented Sep 9, 2011 at 23:45
• I am making the assumption that $z$ is a root of $P_a$, i.e. $P_a(z)=0$. As the vector of coefficients, $a$, varies, the root $z$ is also going to vary. So in that sense, yes, $z$ is a function of the coefficients.
– robjohn
Commented Sep 10, 2011 at 0:11
• I think you proved the following: if $\frac{\partial z}{\partial a_k}$ does exist then it is equal to ... Commented Sep 25, 2012 at 20:11
• @vesszabo: since $P_a^\prime(z)$ exists and $z^k$ exists, if $P_a^\prime(z)\not=0$, the first equation guarantees the existence of $\frac{\partial z}{\partial a_k}$. So I qualify my conclusion by saying the $P_a$ has no repeated roots, which guarantees that if $P_a(z)=0$, then $P_a^\prime(z)\not=0$.
– robjohn
Commented Sep 25, 2012 at 21:00
• Maybe I misunderstand something. I give an example where the situation is similar. Let $f:(a,b)\to\mathbf{R}$ be invertible differentiable function, its inverse is $f^{(-1)}$. Then $f\circ f^{(-1)}=id$. Taking the differentiation and applying the chain rule we obtain that $f'\circ f^{(-1)}\cdot \left(f^{(-1)}\right)'=1$ from it we have $\left(f^{(-1)}\right)'=...$. However this proof is wrong, because we don't know $\left(f^{(-1)}\right)'$ does exist at all. So my problem is not the repeated roots. Commented Sep 26, 2012 at 10:05

I think this problem can be dealt with Rouche's theorem. Recall that Rouche's theorem in complex analysis says

(Rouche's theorem) If $$f(z)$$ and $$g(z)$$ are analytic interior to a simple closed Jordan curve $$C$$ and if they are continuous on $$C$$ and $$|g(z)|\leq |f(z)|,\quad z\in C,$$ then the function $$f(z)+g(z)$$ has the same number of zeros (counted with multiplicity) interior to $$C$$ as does $$f(z)$$.

Consider the polynomial $$f(z)=a_0+a_1z+\cdots+a_nz^n,\quad a_n\neq 0.$$ Let $$\zeta$$ be a root of $$f(z)$$ and $$\varepsilon>0$$. To prove the continuity, we need to show that there exists a $$\delta$$ such that the perturbed polynomial $$g(z)=a_0+\delta_0+(a_1+\delta_1)z+\cdots+(a_n+\delta_n)z^n$$ where $$|\delta_i|\leq \delta$$, has the same number of zeros as $$f(z)$$ inside the circle $$C(\zeta,\varepsilon)$$ with center $$\zeta$$ and radius $$\varepsilon$$,

We may suppose that $$\varepsilon$$ is smaller than the distances from $$\zeta$$ other zeroes of $$f(z)$$, so that $$f(z)$$ is nonzero on $$C(\zeta,\varepsilon)$$. Since the circle is compact, $$|f(z)|$$ attains its minimum $$m>0$$ on it.

Let $$h(z)=f(z)-g(z)$$. Then on $$C(\zeta,\varepsilon)$$, we have $$|h(z)|=|\delta_0+\delta_1z+\cdots+\delta_n z^n|\leq \delta \sum_{j=0}^{n}(|\zeta|+\varepsilon)^j.$$ Thus if we choose $$\delta< \frac{m}{\sum_{j=0}^n(|\zeta|+\varepsilon|)^j}$$, then $$|h(z)|<|f(z)|$$ on $$C(\zeta,r)$$. Hence by Rouche's theorem, we can conlude that $$f(z)$$ and $$g(z)$$ has the same number of zeros inside $$C(\zeta,r)$$.

In the complex case, if we ignore or forbid multiple roots and fix the degree: We can assume without loss of generality that the leading coefficient is always 1 -- normalizing the coefficients is a continuous transformation of coefficient space.

Now, then, the coefficients are a continuous injective function of the roots -- we can find them by multiplying linear polynomials with the given roots together. On the other hand, with the leading coefficient fixed to 1, both the space of possible coefficients and $\mathbb C^n/\sim$ minus multiple-root points are locally just copies of $\mathbb C^n$, so the inverse mapping from coefficients back to roots also has to be continuous.

This argument ought to work in any algebraically closed topological field (or would it? I'm not actually sure how wild a topological field is allowed to be). I'm not quite sure about how well it generalizes to situations involving multiple roots, though. The best arguments for that I can imagine right away are somewhat specific to $\mathbb C$.

• I think you are first assuming the polynomial to have $n$ roots. Am I right? (The question does not explicitly say $F$ is algebraically closed.) Commented Sep 9, 2011 at 21:50
• Yes, you're right. I started to write the answer down for $\mathbb C$ and then generalized a bit too quickly. Will edit. Commented Sep 9, 2011 at 21:56
• Well, as stated in the comments, I essentially am assuming F is algebraically closed, so that's not problematic. Commented Sep 9, 2011 at 22:26
• The coefficients of a monic polynomial are simply symmetric polynomials of the roots, and therefore, continuous. The function from the roots to the coefficients as a map $\mathbb{C}^n\mapsto\mathbb{C}^n$ is definitely injective since the roots are a function of the coefficients. (In other words, I agree with you, but this seems simpler to me.)
– robjohn
Commented Sep 10, 2011 at 1:34

Here is a link to a short note establishing that the roots of a polynomial are $C^\infty$ functions of the coefficients using the implicit function theorem.

• I think that that note only deals with simple zeros (degenerate ones prevent the use of the implicit function theorem). (Following your link I discovered this monography which is quite interesting and useful for me, thank you). Commented Jan 8, 2016 at 23:14

Here's a proof without using complex analysis, differential geometry or the projective plane, and only using results from general topology.

Consider the equivalence relation on $$\mathbb{C}^n$$ given by $$x\sim y$$ iff $$y = sx := (x_{s(1)}, ..., x_{s(n)})$$ for some $$s\in S_n$$. Give $$\mathbb{C}^n/\sim$$ the quotient topology and let $$q:\mathbb{C}^n\to\mathbb{C}^n/\sim$$ be the quotient map. Note that $$q^{-1}q[x+B_r] = \bigcup_{s\in S_n} (sx+B_r)$$ is open, where $$B_r = \{y : |y| < r\}$$, so that $$q[x+B_r]$$ is open. If $$U\subseteq \mathbb{C}^n/\sim$$ is open and $$q(x)\in U$$ then by taking $$r > 0$$ such that $$x+B_r\subseteq q^{-1}[U]$$, it follows that $$\{q(x+B_r) : x\in \mathbb{C}^n, r > 0\}$$ forms a basis for $$\mathbb{C}^n/\sim$$. Let $$d(q(x), q(y)) = \min\{|x'-y'| : x'\sim x, y'\sim y\}.$$ Clearly $$d(q(x), q(y)) = 0$$ iff $$q(x) = q(y)$$ and $$d(q(x), q(y)) = d(q(y), q(x))$$. Given $$x, y, z\in\mathbb{C}^n$$, let $$x'\sim x$$ be such that $$d(q(x), q(y)) = |x'-y|$$. Then $$d(q(x), q(z))\leq |x'-z'| \leq |x'-y|+|y-z'| = d(q(x), q(y))+|y-z'|$$ for $$z'\sim z$$. Taking minimum over all $$z'\sim z$$, we obtain triangle inequality for $$d$$. So $$d$$ is a metric. Notice that $$q[x+B_r] = \{q(y) : d(q(x), q(y)) < r\}$$ so that $$\mathbb{C}^n/\sim$$ is metrizable with metric $$d$$.

Let $$\sigma:\mathbb{C}^n\to\mathbb{C}^n$$ map $$(\lambda_1, ..., \lambda_n)$$ to coefficients of $$(x-\lambda_1)...(x-\lambda_n) = x^n+a_1x^{n-1}+...+a_n$$. From Viete's formulas, $$\sigma$$ is continuous. Note the following theorem:

Theorem 1. Let $$f:X\to Y$$ be a continuous map between metrizable spaces. Then $$f$$ is perfect iff for all $$x_n\in X$$ such that $$f(x_n)$$ is convergent, it follows that $$x_n$$ has a convergent subsequence.

For reference of this see chapter 2 of Geometric Aspects of General Topology by K. Sakai.

Suppose that $$\lambda^{(m)} = (\lambda_1^{(m)}, ..., \lambda_n^{(m)})$$ is such that $$\sigma(\lambda^{(m)}) = (a_1^{(m)}, ..., a_n^{(m)})$$ is convergent. Note that $$|\lambda_i^{(m)}|^n = \left|\sum_{j=1}^n a_j^{(m)}(\lambda_i^{(m)})^{n-j}\right|\leq \sum_{j=1}^n |a_j^{(m)}||\lambda_i^{(m)}|^{n-j}.$$ If $$|\lambda_i^{(m)}|\geq 1$$ then by dividing above inequality by $$|\lambda_i^{(m)}|^{n-1}$$, it follows that $$|\lambda_i^{(m)}|\leq \sum_{j=1}^n |a_j^{(m)}|$$, thus for arbitrary $$\lambda_i^{(m)}$$ we have $$|\lambda_i^{(m)}|\leq \max(1, \sum_{j=1}^n |a_j^{(m)}|)$$ so that $$\lambda_i^{(m)}$$ are bounded. Thus we can take convergent subsequence $$\lambda^{(m_k)}$$ of $$\lambda^{(m)}$$. It follows that $$\sigma$$ is a perfect map.

Theorem 2. Let $$f:X\to Y$$, $$q:X\to Z$$ a quotient map and $$g = f\circ q^{-1}:Z\to Y$$ be well-defined. If $$f$$ is continuous/open/closed, then so is $$g$$.

See e.g. chapter 6 of Dugundji's Topology.

Thus since $$\sigma$$ is a perfect map, it follows that $$\sigma\circ q^{-1}:\mathbb{C}^n/\sim\to \mathbb{C}^n$$ is a closed continuous bijection, thus a homeomorphism. As a corollary we can state:

Corollary 1. Let $$a_1, ..., a_n\in\mathbb{C}$$ and $$\lambda_1, ..., \lambda_n$$ be roots of $$x^n+a_1x^{n-1}+...+a_n$$, $$\varepsilon > 0$$, then there exists $$\delta > 0$$ such that if $$|a_i-b_i| < \delta$$ then there exists an arrangement $$\mu_1, ..., \mu_n$$ of roots of $$x^n+b_1x^{n-1}+...+b_n$$ such that $$|\lambda_i-\mu_i| < \varepsilon$$ for $$i = 1, ..., n$$.

Corollary 2. If $$a_1, ..., a_n\in\mathbb{R}$$, let $$\lambda_1, ..., \lambda_n$$ be roots of $$x^n+a_1x^{n-1}+...+a_n$$ ordered so that $$\text{Re}(\lambda_1) \leq ... \leq \text{Re}(\lambda_n)$$. Define $$f:\mathbb{R}^n\to\mathbb{R}^n$$ as $$f(a_1, ..., a_n) = (\text{Re}(\lambda_1), ..., \text{Re}(\lambda_n))$$. Then $$f$$ is continuous.

Proof: The map $$g:\mathbb{R}^n\to\mathbb{R}^n$$ ordering elements of $$\mathbb{R}^n$$ increasingly is continuous, since if $$a\in \mathbb{R}^n$$ and $$r_1 < ... < r_m$$ are distinct among elements in $$g(a)$$, then for any $$\varepsilon > 0$$, if $$0 < \delta < \min(\varepsilon, \frac{r_2-r_1}{2}, ..., \frac{r_m-r_{m-1}}{2})$$ and $$|a-b| < \delta$$, then $$|g(a)-g(b)| = |a-b| < \varepsilon$$. The map $$f$$ is then composition of $$(\sigma\circ q^{-1})^{-1}\restriction_{\mathbb{R}^n}$$, the map $$\mathbb{C}^n/\sim\to \mathbb{R}^n/\sim$$ induced by the map $$(z_1, ..., z_n)\to (\text{Re}(z_1), ..., \text{Re}(z_n))$$, and the map $$\mathbb{R}^n/\sim\to\mathbb{R}^n$$ induced by $$g$$, and as such is continuous. $$\square$$

• Hm, so this isn't using complex analysis or anything, but as best I can tell it is still using particularities of $\mathbb{C}$ like local compactness? This still doesn't transfer to other settings? Commented Aug 16, 2023 at 21:56
• Yeah, it uses Bolzano-Weierstrass theorem to deduce that $\sigma$ is a perfect map. I think the proof should generalize to local fields akin to what Pete Clark was saying in his answer, possibly with some tweaks in places since $\mathbb{C}$ is both a local field and an algebraically closed field. I didn't consider those since I was writing this answer mostly for myself, I study general topology and I was looking for a proof that wouldn't use Rouche's theorem or something similar. Commented Aug 16, 2023 at 22:11
• In principle I'm only using Bolzano-Weierstrass and that $\mathbb{C}$ has absolute value, so I think the proof should translate perfectly fine to algebraically closed local fields, but that's not really all, since if we consider local fields that aren't necessarily algebraically closed then we'll have to make tweaks because of the fact that we are considering two fields at once, which might have a little different topological properties (I'm not sure on this). Commented Aug 16, 2023 at 22:25
• I mean, the thing is that $\mathbb{C}$ is the only algebraically closed local field, which is why I just said $\mathbb{C}$. But yeah, obviously the other answers here are also similarly limited... Commented Aug 18, 2023 at 4:57