To determine whether an infinite field ( K ) where polynomials are almost surjective must be algebraically closed, we proceed as follows:

Key Definitions and Setup

A field ( K ) is algebraically closed if every nonconstant polynomial in ( K[x] ) has a root in ( K ). Polynomials are almost surjective over ( K ) if for every nonconstant ( f(x) \in K[x] ), the image ( f(K) ...

Yes. If K is infinite and every nonconstant polynomial f ∈ K[x] has cofinite image f(K) ⊂ K, then K is algebraically closed.

Outline of the argument

1. Power maps are surjective

• For any n ≥ 1, consider f(x) = x^n. Its image is {0} ∪ (K*)^n.
• If (K*)^n ≠ K*, then K*(K*)^n is a nonempty union of cosets of (K*)^n in K*. Each coset is infinite, so the complement of f(K) in K is infinite, contrad...

Yes.
If an infinite field K has the property that every non-constant polynomial
f ∈ K[x] takes all but finitely many values of K, then K is necessarily algebraically closed. (Finite fields obviously have the property as well.)

Proof.

0. Notation. Throughout, “almost surjective’’ means that f(K) omits only finitely many elements of K.

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Yes – in fact one shows that as soon as an infinite field (K) admits a single irreducible polynomial of degree (>1), one can manufacture another polynomial (f\in K[x]) whose image (f(K)) misses infinitely many points of (K). Hence the “almost‐surjectivity’’ hypothesis forces every irreducible in (K[x]) to have degree 1, i.e. (K) is algebraically closed.

Here is one clean way to see...

I'll address this interesting question about the relationship between "almost surjectivity" of polynomials and algebraic closure.

The answer is no - there exist infinite fields where polynomials are almost surjective but which are not algebraically closed.

Counterexample: The Algebraic Closure of a Finite Field

Let be the algebraic closure of the finite fiel...

An excellent and deep question in field theory. The answer is yes: an infinite field over which polynomials are almost surjective must be algebraically closed.

This result is a known theorem in field theory, and its proof is not elementary. It relies on some deeper results about the value sets of polynomials and norms in field extensions.

Here is a sketch of the proof. The main strategy is proof...

Analyzing "Almost Surjective" Polynomials and Field Properties

This question asks an interesting question about the relationship between "almost surjective" polynomials and algebraically closed fields.

Understanding the Property

First, let's clarify what it means for polynomials to be "almost surjective" over a field K:

• For any nonconstant polynomial f(x) ∈ K[x], the image f(K) contains ...

<think> I need to figure out if an infinite field where polynomials are almost surjective must be algebraically closed. "Almost surjective" means that for any nonconstant polynomial, its image misses only finitely many points in the field.

First, I know that finite fields and algebraically closed fields satisfy this. For finite fields, since the field is finite, any function can miss only finitel...

Yes.

Let K be an infinite field in which

(*) for every non–constant f∈K[x] the set K\f(K) is finite.

We show that K must be algebraically closed.
Assume the contrary and divide the proof into characteristic-cases.

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1. Characteristic different from 2

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Because K is not algebraically c...