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Let $K$ be a field. Say that polynomials are almost surjective over $K$ if for any nonconstant polynomial $f(x)\in K[x]$, the image of the map $f:K\to K$ contains all but finitely many points of $K$. That is, for all but finitely many $a\in K$, $f(x)-a$ has a root.

Clearly polynomials are almost surjective over any finite field, or over any algebraically closed field. My question is whether the converse holds. That is:

If $K$ is an infinite field and polynomials are almost surjective over $K$, must $K$ be algebraically closed?

(This answer to a similar question gave a simple proof that $\mathbb{C}$ is algebraically closed from the fact that polynomials are almost surjective over $\mathbb{C}$. However, this proof made heavy use of special properties of $\mathbb{C}$ such as its topology, so it does not generalize to arbitrary fields.)

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    $\begingroup$ I believe this is open. At least, it was posted on MO with no definitive answer: mathoverflow.net/questions/6820/… $\endgroup$ – Qiaochu Yuan May 20 '16 at 17:27
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    $\begingroup$ Here's my 5 cents. $K$ must contain an $n^{th}$ root for each of its elements (take $f:=X^{n}$; for $\alpha\in K^{*}$, $f$ cannot miss the infinitely many elements $\alpha\cdot\beta^{n}$ with $\beta\in K^{*}$). At least when $char(K)=0$, the roots of unity are expressible as radicals. This implies that $K$ does not admit any finite Galois extensions with solvable Galois group. I.e., any minimal non-trivial Galois extension of $K$ must have a simple non-abelian Galois group. $\endgroup$ – Matthé van der Lee Jun 8 '18 at 20:19

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