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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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Eric Wofsey
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    I believe this is open. At least, it was posted on MO with no definitive answer: http://mathoverflow.net/questions/6820/can-a-non-surjective-polynomial-map-from-an-infinite-field-to-itself-miss-only-fi – Qiaochu Yuan May 20 '16 at 17:27
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    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. – Matthé van der Lee Jun 08 '18 at 20:19
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    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? – Vincent Oct 02 '19 at 10:00
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    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 $K$?' ? – Vincent Oct 02 '19 at 10:05
  • 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. – Andrea Marino Oct 02 '19 at 12:12
  • How I would build such a topology: start from $\mathbb{Q}$ and $\mathbb{F}_p$, then show that the topology extends by transcendent, algebraic extensions and by infinite chains. By Zorn, this gives the topology on whatever field. Don't know if this route map can work!! – Andrea Marino Oct 02 '19 at 12:14
  • Sorry for double comments but I can't edit. It is enough for the topology to be locally compact, so that we recover the cases of $\mathbb{Q}, \mathbb{C}$. – Andrea Marino Oct 02 '19 at 12:19
  • @AndreaMarino: I think it is extremely unlikely that you can get such topology at the end of an infinite chain. Local compactness is also not enough (then you could have the discrete topology!), and you need $T_2$, not just $T_0$, since you need to know the limit of $y_n$ is unique to conclude $p(x)=s$. Note then that no countably infinite compact $T_2$ space is homogeneous, so you can never have such a topology on a countably infinite field. – Eric Wofsey Oct 02 '19 at 15:34
  • Ooh you are right. I thought it was strange to have such a regular topology for arbitrary fields without being known. – Andrea Marino Oct 03 '19 at 14:31
  • I am slightly confused. If $K$ is finite, then any polynomial is "almost surjective" in this sense, because there are only finitely many points that are not in the image of $f$ (even if the image of $f$ consists of only one point). Furthermore, finite fields are not algebraically closed.

    Do you assume your characteristic to be 0?

     – student91 Sep 13 '22 at 11:47
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    @student91: That's why I restrict $K$ to be infinite in the precise statement of the question. – Eric Wofsey Sep 13 '22 at 12:23
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    I do not know whether OP knew this but this was conjectured by Reinke in their 1975 paper "Minimale Gruppen", and as far as I know is still open. A similar conjecture in model theory is the so-called "minimal fields conjecture" of Podewski, asserting that infinite "minimal" fields are algebraically closed (where "minimal" means that every field formula with one free variable either defines a finite subset of the field or a cofinite one). Wagner solves the latter in his paper "Minimal fields" in positive characteristic, where he also solves a weaker variant of OP's question. – Uri George Peterzil Sep 10 '24 at 15:56
  • I guess the problem with the original reasoning is that it relies on the fact that the image of $\mathbb{C}$ has to be simply connected, but this is hard to use for abitrary fields, the only topology you could use would be zariski but knowing zariski would give you info about the algebraic closeness. Maybe studying homology would give some results? The only nice thing is that being infinite gives you irreducibility – julio_es_sui_glace Nov 16 '24 at 13:28

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