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.

  • $\begingroup$ The fields you are looking for must be finite (because for infinite fields, polynomial maps are "the same as" polynomials), and of order $q=2^n$ (because your method discards more generally $\Bbb F_q$ for any odd prime power $q$). There remains to find out how to discard also (or not?) the candidates $\Bbb F_{2^n}$ for $n>1.$ $\endgroup$ Jan 12 at 23:36
  • $\begingroup$ For fields of infinite order, the polynomial maps and the (formal) polynomials coincide. This cannot hold for finite fields, since there are only finitely many polynomial maps, but infinitely many (formal) polynomials. $\endgroup$
    – Rob Arthan
    Jan 12 at 23:39
  • $\begingroup$ @RobArthan We know, but how does it help? $\endgroup$ Jan 12 at 23:46
  • 3
    $\begingroup$ Let $k$ be a finite field with more than two elements. The polynomial function $x \longmapsto x(x-1)$ is nonconstant and non-injective, so there is some $b \in k$ that it doesn’t reach, then $x^2-x-b$ is a nonconstant polynomial function without a root. $\endgroup$
    – Aphelli
    Jan 12 at 23:48
  • $\begingroup$ Wow @Aphelli I think you should edit this as an answer! (just add the easy explanations about infinite fields and about your non-injectivity). $\endgroup$ Jan 12 at 23:51

1 Answer 1


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.


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