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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
  • $\begingroup$ I find the concept of almost surjective polynomials a bit confusing. Can you give an example of an almost surjective polynomial over $\mathbb{C}$ that is not surjective? $\endgroup$ – Vincent Oct 2 at 10:00
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    $\begingroup$ Or wait, can your question be rephrased as 'If $K$ is an infinite field and polynomials are almost surjective over $K$, must polynomials be fully surjective over $K$?' ? $\endgroup$ – Vincent Oct 2 at 10:05
  • $\begingroup$ Here it is my 1 cent also. Suppose there exist a $T_0$, compact and first countable topology on K such that polynomials are continuous. There are no isolated points x, otherwise $x+k$ would be also isolated because polynomials are continuous, thus the topology is discrete. But being compact it should be finite. Now take a polynomial p with non taken values $s_1,\ldots, s_k$. Take $y_n$ in the $U_n $ (given by first countability) of $s_1$ , WLOG $x_n$ different by $s_i$. Take $x_n$ st $p(x_n)=y_n$. The space is sequentially compact, so WLOG $x_n\to x$. Then $p(x)=s$ and p is surjective. $\endgroup$ – Andrea Marino Oct 2 at 12:12
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If all non constant polynomials are surjective then K is algebraically closed. If there exist some non constant polynomial which is not surjective then there exists some element which is not in the image of this polynomial. If this element is zero then K is not algebraically closed. If this element is not zero then you can construct a new polynomial which has no roots in K and so K is not algebraically closed.

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    $\begingroup$ The question is about what happens if all nonconstant polynomials are almost surjective, not surjective. Of course, if some nonconstant polynomial is actually not surjective then the field is not algebraically closed, but to answer the question you'd need to find an example of such a field in which every nonconstant polynomial is still almost surjective. $\endgroup$ – Eric Wofsey Oct 2 at 15:25
  • $\begingroup$ In your definition of 'almost surjective' we could have all but zero points in the image of f, in which case K is algebraically closed. Your question was 'If 𝐾 is an infinite field and polynomials are almost surjective over 𝐾, must 𝐾 be algebraically closed?' and my answer is no. The question says nothing about whether such an infinite field could exist. $\endgroup$ – sandy Oct 2 at 19:26
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    $\begingroup$ This is like answering the question of the Riemann hypothesis ("must every nontrivial zero of the Riemann zeta function have real part $1/2$?") by saying "no, since a zero with real part $1/3$ would be a nontrivial zero whose real part is not $1/2$". $\endgroup$ – Eric Wofsey Oct 2 at 19:50
  • $\begingroup$ I see what you are saying now. It might be simpler to ask 'Does there exist an infinite field K which is not algebraically closed and where all polynomials over K are almost surjective?' $\endgroup$ – sandy Oct 6 at 21:31

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