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I want to know about an algebraically closed field that is not of characteristic $0$.

I really don't know about infinite fields with characteristic $p$ so I will appreciate your comments.

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Any field has an algebraic closure—so the short answer to this is: Just take the algebraic closure of any field of characteristic $p$.

For finite fields, it is possible to describe the algebraic closure fairly explicitly. Let $K=\mathbb{F}_q$, the finite field of order $q$. Then, for any $n\geq 1$, there is a unique (up to isomorphism) field extension $K\to K_n \cong \mathbb{F}_{q^n}$ of degree $n$, with $K_n \subset K_m$ if and only if $n | m$. $K_n$ can be described as the splitting field of any irreducible polynomial of degree $n$ with coefficients in $K$.

Clearly, the algebraic closure of $K$ is just $\overline{K} = \bigcup_n K_n $, which we could also write as $\varinjlim K_n$.

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    $\begingroup$ Yes. Still, to be fair, for the fastidious, there can be some concern about the uniqueness of inclusions, blah-blah-blah, but everything turns out ok. Ascending union with uniquely determined inclusion maps... bingo. $\endgroup$ – paul garrett Oct 15 '13 at 1:02
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    $\begingroup$ And if you want the union to be a nested one, then you can use $$\overline{K}=\bigcup_nK_{n!}.$$ $\endgroup$ – Jyrki Lahtonen Oct 15 '13 at 7:24

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