I was wondering which fields $K$ can be equipped with a topology such that a function $f:K \to K$ is continuous if and only if it is a polynomial function $f(x)=a_nx^n+\cdots+a_0$. Obviously, the finite fields with the discrete topology have this property, since every function $f:\Bbb F_q \to \Bbb F_q$ can be written as a polynomial.
So what is with infinite fields. Does anyone see any field $K$ where such a topology can be found? If there is no such field, can anyone supply a proof that finding such a topology is impossible. I would even be satisfied if one could prove this nonexistence for only one special field (say $\Bbb Q, \Bbb R,\Bbb C$ or $ \Bbb F_q^\text{alg} $). I suspect that there is no such topology, but I have no idea how to prove that.
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(My humble ideas on the problem: Assume that you are given such a field $K$ with a topology $\tau$. Then for $a,b \in K$ , $a \ne 0$, $x \mapsto ax+b$ is a continuous function with continuous inverse, hence a homeomorphism. Thus $K$ is a homogenous space with doubly transitive homeomorphism group. Since $\tau$ cannot be indiscrete, there is an open set $U$, and $x,y \in K$ with $x \in U,y \not\in U$. Now for every $a \in K$, $a*(U-y)/x$ includes $a$ but not $0$, and thus $K\setminus\{0\}=\bigcup_{a \in K^\times}a*(U-y)/x$, is an open subset. Thus $K$ is a T1 space, i. e. every singleton set $\{x\}$ is closed. Also $K$ is connected: Otherwise, there would be a surjective continuous function $f:K \to \{0,1\} \subset K$, which is definitely not a polynomial.)
EDIT: This question asks the analogous question with polynomials replaced by holomorphic functions. Feel free to post anything which strikes you as a remarkable property of such a hypothetical topology.
Best Answer
Currently, this is more an elongated comment than an answer ...
Consider the case $K=\mathbb R$. Such a topology $\mathcal T$ has to be invariant under translations, scaling, and reflection because $x\mapsto ax+b$ with $a\ne 0$ is a homeomorphism.
(cf. JimBelks comments above) Assume there is a nonempty open set $U$ bounded from below, then wlog. (by translation invariance) $U\subseteq (0,\infty)$ and hence $(0,\infty)=\bigcup_{r>0}rU$ is open. By reflection, also $(-\infty,0)$ is open and by the pasting lemma, $x\mapsto|x|=\begin{cases}x&\text{if }x\ge0\\-x&\text{if }x\le 0\end{cases}$ is continuous, contradiction. Therefore all nonempty open sets are unbounded from below, and by symmetry also unbounded from above.
Especially, all nonempty open sets are infinite. Let $U$ be a nonempty open neighbourhood of $0$. Let $I$ be a standard-open interval, i.e. of the form$I=(-\infty,a)$, $I=(a,\infty)$, or $I=(a,b)$. Assume $|U\cap I|<|\mathbb R|$ and $0\notin I$. Then the set $\{\frac xy\mid x,y\in U\cap I\}$ does not cover all of $(0,1)$, hence for suitable $c\in(0,1)$, the open set $U\cap cU$ is disjoint from $I$ and nonempty (contains $0$). If $I$ itself is unbounded this contradicts the result above. Therefore (by symmetry) $$|U\cap(a,\infty)|=|U\cap(-\infty,a)|=|\mathbb R|$$ for all nonempty open $U$ and $a\in\mathbb R$. However, if $I=(a,b)$ is bounded and $U\cap I=\emptyset$, then $\bigcup_{ca+d<a\atop cb+d>b} (cU+d) =(-\infty,a)\cup (b,\infty)$ is open. Taking inverse images under a suitable cubic, one sees that all sets of the form $(-\infty,a)\cup c,d)\cup (b,\infty)$ are open and ultimately all open neighbourhoods (in the standard topology!) of the point at infinity in the one-point compactification of $\mathbb R$ are open. A topology containing only these sets would describe "continuity at infinity" and make polynomials continuous - but also many other functions. Anyway we have:
Since $x\mapsto |x|$ is continuous under the indiscrete topology, there exists an open set $\emptyset\ne U\ne \mathbb R$. Wlog. $0\notin U$. Then $\bigcup_{c>0}cU=K\setminus\{0\}$ is open, hence
So $\mathcal T$ is coarser than the cofinite topology. Since $x\mapsto |x|$ is continuous under the cofinite topology, it is strictly coarser, i.e. there exists an open set $U\ne \emptyset $ such that $\mathbb R\setminus U$ is infinite.