Before yesterday, I had heard of basically two examples of algebraically closed fields, the algebraic numbers $\overline{\mathbb{Q}}$ and $\mathbb{C}$.
I tried to come up with more examples and started thinking about $\mathbb{F}_2$, the field with two elements.
All the polynomial maps on $\mathbb{F}_2$ that are non-constant have a root because the domain consists of just $0$ and $1$ and thus any function without a root is constrained to be constantly 1.
However, there are indeed polynomials in $\mathbb{F}_2[x]$ that are nonconstant and lack roots, such as $x^2 + x + 1$.
This question shows the construction for the algebraic closure of a finite field of prime order.
Let's call the weaker property closure with respect to roots of polynomial maps.
I'm wondering which fields are closed with respect to roots of polynomial maps but not algebraically closed besides $\mathbb{F}_2$, if there are any.
$\mathbb{F}_3$ does not have this property because $x^2 + 1 \mapsto \{0+1, 1+1, 1+1\} = \{1, 2\}$ has no roots.
For the other finite fields of odd prime order, I think I can use the polynomial $x^{p – 1} + 1$ to witness the failure of the weaker property since the multiplicative group of $\mathbb{F}_p$ has order $p-1$ and this function is nonconstant since $p \ge 3$.
I think my question is distinct from this question about the distinction between polynomials and polynomial maps because I'm asking about consequences of using the "wrong" notion specifically when characterizing the algebraically closed fields, not what the difference is per se.
Best Answer
As suggested, I’m turning my answer into a comment.
Let $k$ be an infinite field. Then if $k$ isn’t algebraically closed, it has a polynomial $p(x)$ of degree at least $2$ without a root. As $k$ is infinite, $p(x)$ can’t be a constant polynomial function, so $k$ isn’t closed with respect to the roots of polynomial maps.
Let $k$ be a finite field with at least three elements. The polynomial map $x \in k \longmapsto x(x-1)$ is non-constant and non-injective, so there is a $b \in k$ which isn’t in its image. So $x^2-x-b$ is a non-constant polynomial function on $k$ without a root so $k$ isn’t closed with respect to the roots of polynomial maps.