TITLE: A question about algebraically closed fields
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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.