TITLE: Can a non-surjective polynomial map from an infinite field to itself miss only finitely many points?
QUESTION [93 upvotes]: Is there an infinite field $k$ together with a polynomial $f \in k[x]$ such that the associated map $f \colon k \to k$ is not surjective but misses only finitely many elements in $k$ (i.e. only finitely many points $y \in k$ do not lie in the image of $f$)?
For finite fields $k$, there are such polynomials $f$. If such a poynomial $f$ exists, then $k$ cannot be algebraically closed; the field $\mathbb{R}$ doesn't work either.

REPLY [8 votes]: Recently, I have shown the following theorem with Hendrik Lenstra. 
Definition: A field $k$ is called large if every irreducible $k$-curve with a $k$-rational smooth point has infinitely many $k$-points. 
Some examples of large fields are $\mathbb{R}$, $\mathbb{Q}_p$ ($p$ prime), $l((t))$ (where $l$ is a field), infinite algebraic extensions of finite fields. Furthermore, finite extensions of large fields are large.
Theorem: Let $k$ be a perfect large field and let $f \in k[x]$. Consider the evaluation map $f_k: k \to k$. Assume that $f_k$ is not surjective. Then the set $k \setminus f_k(k)$ is infinite (in fact, it has cardinality $|k|$).
See my arXiv article for a proof.