specific examples of algebraic closure on finite field

Question

I want to confirm my understanding of algebraic closure for finite fields.

What sorts of elements do the algebraic closures $\overline {\mathbb{F}_2}$, $\overline {\mathbb{F}_3}$, $\overline {\mathbb{F}_7}$, and $\overline {\mathbb{F}_{25}}$ contain?

This answer seems most illuminating, specifically:

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 $E_1=\Bbb{F}_p$

Taking $p=3$, for example, such that $E_1 = \mathbb{F}_3 = \lbrace0, 1, 2\rbrace$

Then $E_2 =\Bbb{F}_{3^{2}} =\Bbb{F}_9 =\lbrace0, 1, 2, 3, 4, 5, 6, 7, 8\rbrace$

Indeed, $E_1\subset E_2$, but following this train of thought implies that the algebraic closure of $\Bbb{F}_p$ is just the positive integers, but if that were the case, surely one of the references I've read would have said so by now!

The only other way I've been able to interpret this is that $\overline {\Bbb{F}_3} = {0, 1, 2, 3, 9, 27, 81, ...}$ or similar.

Answer 1

This is a possible way to have a picture of the algebraic closure of $\Bbb F_p$. Consider all natural numbers $1,2,3,4,5,6,7,8,9,10,11,12,13,\dots$ and arrange them in a tree with levels $0,1,2,3,\dots$ as follows.

• In level $0$ we have the one and only $1$ .
• In level $1$ we have all the prime numbers, so $2,3,5,7,11,13,\dots$ .
• In level $2$ we have all natural numbers which are a product of two primes, so $4,6,9,10,\dots$ .
• In level $3$ we have all natural numbers which are a product of three primes, so $8,12,\dots$ .

And so on. These are the vertices of the tree. Now draw an edge / a connection (as arrow) $a\to b$ from one level to the next one if $a$ divides $b$.

It is hard to draw, but i will try.

:::::::::::::::::::::::::::::::::::::::
         |/ |/
level 3: 8 12 ...
         |/|
level 2: 4 6 9 10 ...
         |/|/ /
level 1: 2 3 5 7 11 13 ...
         |/ / / / /
level 0: 1 

Now imagine at each place $a$ placed the field $$ \Bbb F_{p^a} $$ with $p^a$ elements. Then for each edge $a\to b$ in the tree we have a field inclusion of the corresponding fields $\Bbb F_{p^a}\to \Bbb F_{p^b}$.

Consider now the union of all fields in all levels. This is $$ \bar {\Bbb F}_p\ . $$ The way to construct it shows, that all statements "in it", that depend only on elements, and only on finitely many elements can be checked in a (sufficiently big) finite field.