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Bounty Ended with Pete L. Clark's answer chosen by Philipp Lampe
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Typo corrected.
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Philipp Lampe
Philipp Lampe
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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 neithereither.

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 neither.

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.

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Philipp Lampe
Philipp Lampe
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Can a non-surjective polynomial map from an infinite field to itself miss only finitely many points?

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 neither.