Why is the algebraic closure of a finite field countable? An algebraic closure of a field K is an algebraic extension of K that is algebraically closed. It is one of many closures in mathematics.
But why is it a countable set?
 A: HINT: If $F$ is a field then there is a surjection from $F[x]\times\Bbb N$ onto the algebraic closure of $F$ (essentially count the roots of each polynomial, and map all the excessive points to $0$).
So it suffices to see that if $F$ is countable then $F[x]$ is countable.
[It should be noted that the axiom of choice is used here, and it is consistent with its failure that a countable field has an uncountable algebraic closure.]
A: It suffices to show that the prime field $\Bbb{F}_p$ has a countable algebraic closure, as $K$ shares its algebraic closure with its prime field.
We can form a nested chain of extensions
$$
E_1\subset E_2\subset\cdots \subset E_i\subset E_{i+1}\subset\cdots
$$
of finite fields $E_i$ for all positive integers $i$ such that $E_1=\Bbb{F}_p$ and then
declare $E_i$ to be the (up to isomorphism) unique degree $i$ extension of $E_{i-1}$. By induction we see that $E_i=\Bbb{F}_{p^{i!}}$ is the unique extension of $E_1$ of degree $i!$.
Note that during this process we also gave, for all $j$, a specific way of including $E_j$ as a subfield of $E_{j+1}$. Such inclusion maps allow us to form the direct limit (it would be a union, if we had a set that contains all the fields $E_i$ as subsets)
$$
E=\lim_{\to}E_i.
$$
Also observe that $E_k$ contains a unique copy of all the finite fields $\Bbb{F}_{p^\ell}$ such that $\ell\mid k!$. In particular a copy of $K$ is in there.
Now


*

*$E$ is a countable set, because it is a nested union of countably many finite subsets.

*All the elements of $E$ belong to one of the fields $E_i$, and hence they are all algebraic over $\Bbb{F}_p$.

*If $f(x)=\sum_{i=0}^n a_ix^i\in E[x]$ is a polynomial of degree $n$, then the coefficient $a_i\in E_{\ell_i}$ for some $\ell_i\in\Bbb{N}$. Thus $f(x)\in E_M$, where $M=\max\{\ell_i\mid i=0,1,\ldots,n\}$. Therefore $f(x)$ splits into linear factors in $E_{M+n}$, and thus also in $E$. Therefore $E$ is algebraically closed.

A: The reason is that the algebraic closure is the union of all its finite subextensions of $K$. If $K$ is finite (or even countable, like the rationals), each one is finite (respectively, countable), and there are countably many. The union of countably many countable sets is countable.
