A question about algebraically closed fields

Question

A field $\mathbb{K}$ is said to be algebraically closed in practice if every polynomial over $\mathbb{K}$ of positive degree less than or equal to $10^{10}$ has zero belonging $\mathbb{K}$. The question arises: is it possible that an algebraically closed in practice field is not algebraically closed?

PS. The question still remains open in characteristic 0.

Answer 1

Let me expand on Dustan Levenstein's answer in the comments: inside the algebraic closure $\overline{\mathbb{F}_p}$ of $\mathbb{F}_p$, consider the union $K = \bigcup_i \mathbb{F}_{p^{(N!)^i}}$, for $N = 10^{10}$. Since this is a nested union, $K$ is a field. If $f(x) \in K[x]$ is a polynomial of degree $\le 10^{10}$, then $f(x) \in \mathbb{F}_{p^{(N!)^i}}[x]$ for some $i$ (since $f$ has finitely many coefficients). Setting $q = p^{(N!)^i}$, we have that $f(x) \in \mathbb{F}_q[x]$ has an irreducible factor of degree $\le 10^{10}$, and thus splits in $\mathbb{F}_{q^(N!)} = \mathbb{F}_{p^{(N!)^{i+1}}} \subseteq K$.

However, $K$ is not algebraically closed - if $q > N!$ is prime, then $\mathbb{F}_{p^q} \not \subseteq K$. If it were, then $\mathbb{F}_{p^q} \subseteq \mathbb{F}_{p^{(N!)^i}}$ for some $i$ (since $\mathbb{F}_{p^q}$ is a finite field), but $\mathbb{F}_{p^d} \subseteq \mathbb{F}_{p^e}$ iff $d \mid e$.

Answer 2

Phenomena like this can occur.

To see this, we can take a baby step and try for $2$ before we try $10^{10}$. There are fields called quadratically closed fields for which every element has a square root. Using that, you can rewrite any monic polynomial of degree $2$ in the form $(x+a)^2-b$ by completing the square. Then it factors into $((x+a)-b')((x+a)+b')$, after which you have a root in the field. So for this type of field, all polynomials of degree no greater than $2$ have a root.

As suggested in the comments, it's probably possible that you can take a field, look at the tower of extensions over the field which split polynomials of low degrees, and argue that the union is algebraically closed in practice.

Answer 3

Here's a construction that works in characteristic $0$. I'm really just expanding on rschwieb's suggested strategy, so I hope it qualifies as a serious answer.

Given $N$ (in this case $10^{10}$), call a subfield $K\subseteq\mathbb{C}$ a "small" field if there is a chain of fields $$\mathbb{Q}=K_0\subseteq K_1\subseteq\dots\subseteq K_n=K$$ where the degree of the extension $K_{i+1}/K_i$ is less than or equal to $N$ for all $i$.

If $$\mathbb{Q}=L_0\subseteq L_1\subseteq\dots\subseteq L_m=L$$ is another such chain, where $L_{i+1}=L_i(\gamma_i)$ is a simple extension (which we can assume, either using the fact that every finite extension of fields of characteristic $0$ is simple, or in a more elementary way by refining the chain of fields to adjoin one element at a time) then so is $$\mathbb{Q}=K_0\subseteq K_1\subseteq\dots\subseteq K_n\subseteq\langle K, L_1\rangle\subseteq\dots\subseteq\langle K,L\rangle,$$ since $$[\langle K,L_{i+1}\rangle:\langle K,L_i\rangle]=[\langle K,L_i\rangle(\gamma_i):\langle K,L_i\rangle]\leq [L_i(\gamma_i):L_i]=[L_{i+1}:L_i]\leq N,$$ (because the minimal polynomial of $\gamma_i$ over $\langle K,L_i\rangle$ divides its minimal polynomial over $L_i$), and by induction the join of finitely many small fields is small, and the union of all small fields is a field $\mathbb{K}$.

If $\alpha\in\mathbb{C}$ is a root of a polynomial $p(t)$ of degree less than or equal to $N$ over $\mathbb{K}$, then all of the coefficients of $p(t)$ are contained in some small field $K$, and the degree of $K(\alpha)$ over $K$ is less than or equal to $N$, so $K(\alpha)$ is small, and so $\alpha\in\mathbb{K}$. Therefore $\mathbb{K}$ is "algebraically closed in practice".

Let $p$ be a prime greater than $N$, and let $\beta\in\mathbb{C}$ be an algebraic number whose minimal polynomial over $\mathbb{Q}$ has degree $p$. Then $\beta$ is contained in no small field, and so $\mathbb{K}$ is not algebraically closed.