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Let $F$ be a field. For a uni-variate polynomial $f(x)$ over $F$,let $M_f(F)$ denote the number of values that $f$ misses, that is, the cardinality of the subset $F - f(F)$ in $F$. Assume that $f$ is not surjective, that is, $M_f(F) > 0$. Is the number $M_f(F)$ big?

If $F = {\mathbb F}_q$ is a finite field of $q$ elements and $f\in \mathbb{F}_q[x]$ is a polynomial of degree $d>0$, then a simple result shows that if $M_f(F_q) > 0$, then $$M_f(F_q) \geq \frac{q-1}{d}. $$ As $q$ goes to infinity, this lower bound goes to infinity. This suggests the following question for which I do not know the answer.

Question. Assume that $F$ is an infinite field. If $M_f(F)>0$. Is it true that $M_f(F) = \infty$? In other words, for an infinite field $F$, if $f \in F[x]$ misses one value, is it true that $f$ misses infinitely many values?

The answer is positive if $F$ is big in the sense that F is algebraically closed (${\mathbb C}$ etc), or topologically closed ($\mathbb{R}$ or $\mathbb{Q}_p$ etc) or if $F$ is a small field such as a Hilbertian field (number fields etc). Anyone has more examples or counter-examples?

Gjergji-Zaimi points out that the above question has been asked before, see Can a non-surjective polynomial map from an infinite field to itself miss only finitely many points?

To make my question somewhat newer, I will ask a multi-variable version of the question. It is unknown to me even in the case that the field is the complex numbers.

Question: Assume that $f=(f_1,..., f_n): F^n -> F^n$ is a FINITE polynomial map in n variables over an infinite field F. If f misses one value, is it true that f misses infinitely many values?

I do not know the answer even in the case when $F={\mathbb C}$ is the complex numbers if $n \geq 2$. Note that the finiteness assumption of the map $f$ cannot be dropped, as Kharlamov (2012) has given the example $$f=(u(uv-1), u^2-(uv-1)^2)$$ which misses precisely one point, the origin (0,0) of ${\mathbb C}^2$.

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This question has appeared before with no actual answer… – Gjergji Zaimi Jan 29 '13 at 4:29
Note that when I asked a related question, Bjorn gave a non-rigorous argument that there should be counterexamples. I think that "no actual answer" is too pessimistic; one can conjecture with evidence that the answer is no.… – Greg Kuperberg Jan 29 '13 at 5:27
This question is a special case of a problem which has been open for four decades: Is an infinite field $K$ for which every definable set $X \subseteq K$ is finite or cofinite [called a ``minimal field''] necessarily algebraically closed? See Podewski, Minimale Ringe, Math. Phys. Semesterber 22 (1975), no. 2, 193 – 197. It is known that a counterexample must have characteristic zero: Wagner, Minimal fields, J. Symbolic Logic 65 (2000), no. 4, 1833 – 1835. – Thomas Scanlon Jan 29 '13 at 6:18
Thomas Scanlon - it seems to me you could post your comment as an answer. – HJRW Jan 29 '13 at 10:53
Thomas Scanlon: I do not see how Podewski's conjecture implies an answer to my question. My confusion is that for my field F, I usually have a polynomial f such that its value set f(F) (a definable subset of F) is neither finite nor co-finite and thus the field F is not minimal. For example, as you mention that Podewski's conjecture is true in positive characteristic. What this implies for my value gap question when F is an infinite algebraic extension of a finite field, the case of most interest to me. Can you explain in more detail? – dwan Jan 29 '13 at 18:49

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