A Topology such that the continuous functions are exactly the polynomials (I originally asked this question on Math.SE, where it received a lot of attention, but no solution.)
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?
I don't expect a full answer to this question, but I would even be satisfied if someone could show the existence or nonexistence of such a topology even for a single field such as $\Bbb Q, \Bbb R,\Bbb C,\Bbb Q_p$ or $  \Bbb F_q^\text{alg}$. 
(Note that $(K,\tau)$ is not necessarily a topological field!)
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A short summary of the comments on Math.SE: Assume that you are given such a field $K$ with a topology $\tau$. Then $\tau$ is necessarily a $T_0$-space and connected. Also, the linear transformations $x \mapsto ax+b$ with $a \in K^\times$, $b \in K$ are homeomorphisms of $\tau$ (and there can be no other homeomorphisms). In the special case $K=\Bbb R$, for every $U \in \tau$ and $a \in \Bbb R$, we have $|(-\infty,a)\cap U |=|(a,\infty)\cap U |=2^{\aleph_0}$. More over, if there is an open interval $(a,b)$ and a $U \in \tau$ with $|(a,b)\cap U |<2^{\aleph_0}$, then for all $c,d \in \Bbb R$, $(-\infty,c)\cup (d,\infty) \in \tau$. As a proposal for a topology on $\Bbb R$, the coarsest topology such that all sets $p^{-1}(\Bbb Z)$ with $p \in \Bbb R[X]$ are closed, has been considered. Obviously every polynomial is continuous with respect to this topology. However, no one has given any reason why every continuous function in this topology should be a polynomial.
 A: In the case of a topological field (edit: containing  real numbers) the  answer is not. Let $K$ a infinite field, with a such topology $\tau$ coherent with the algebraic field structure then $(K, +, 0, \tau)$ is a topological groups. As you said $G$ must be $T_1$ (i.e. $\{0\}$ is closed, this is equivalent to $T_0$ in a topological group), then it is completely regular, i.e. continuous functions separate closed set from points, but if $K$ is infinite the topology $\tau$ cannot be discrete or indiscrete (in these cases any function become continuous, but a polynomial can have only a finite set of zeros , then exists a lot of non polynomial functions), neither $\tau$ could be cofinite (i.e. opens are exactly complements of finite sets) because cofinite  topology in a infinite set isn't $T_2$. Then exist a infinite closed $A$ and a $x\not\in A$ and a continuous function that separate these, but just because $A$ is infinite this cannot be  polynomial.
I wanted to write this answer as a comment, but it was too long
