How do you prove that algebraically closed fields of characteristic $p$ exist?
I have also read:
For a finite field of prime power order $q$, the algebraic closure is a countably infinite field that contains a copy of the field of order $q^n$ for each positive integer $n$ (and is in fact the union of these copies).
Why would the algebraic closure have to be characteristic $p$?
Best Answer
This only answers the main question.
Just take a sequence of inclusions:
$$\mathbb F_p\to\mathbb F_{p^2}\to\mathbb F_{p^6}\to\cdots\to \mathbb F_{p^{n!}}\to\cdots$$
Then the direct limit (essentially the union) is algebraically closed and a field.
That's because any polynomial in elements of this field has coefficients contained in one $\mathbb F_{p^{k!}}$. Thus, it splits in $\mathbb F_{p^{d\cdot k!}}$ for some $d$, and then note that $d\cdot k!\mid(dk)!$. So the polynomial splits in $\mathbb F_{p^{(dk)!}}$.