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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. $$ $$

(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

e 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

ot\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.