In general, an algebraic closure of a field $K$ is denoted by $\overline{K}$. Typical examples arising in number theory are $K=\mathbb{Q}$, $K=\mathbb{F}_p(t)$, $K=\mathbb{Q}_p$. Usually one needs the axiom of choice in order to prove the existence of algebraic closures. There are (at least) two exceptions: For $K=\mathbb{R}$ we have $\overline{K}=\mathbb{C}$. Then, for every subfield $K \subseteq \mathbb{R}$, we may realize $\overline{K}$ as the subfield of $\mathbb{C}$ which consists of complex numbers which are algebraic over $K$. It also exists without AC. We may also replace $\mathbb{R}$ by a real closed field, one only has to adjoin $\sqrt{-1}$. For $K=\mathbb{F}_q$, a finite field, we have for every $n$ an extension $\mathbb{F}_{q^n}$ of degree $n$ and every divisibility relation $n|m$ induces a canonical $\mathbb{F}_q$-homomorphism $\mathbb{F}_{q^n} \to \mathbb{F}_{q^m}$. It follows that we may consider the colimit $\mathbb{F}_{q^{\infty}} := \varinjlim_{n} \mathbb{F}_{q^n}$ (often this is written as a union, which is not quite correct). This turns out to be an algebraic closure of $\mathbb{F}_q$.
Let me also share a quite nice construction of an algebraic closure: Consider the (infinite) tensor product $A$ of all the $K$-algebras $K[x]/(f)$, where $f \in K[x] \setminus \{0\}$. By linear algebra it is non-zero, hence has a maximal ideal $\mathfrak{m}$ (Zorn's Lemma!). Then $K' := A/\mathfrak{m}$ is a field extension of $K$, and by construction every $f \in K[x] \setminus \{0\}$ has a root in $K'$. It is a nontrivial result that this already is the algebraic closure; but even if we don't use this, we can just repeat this process $K \hookrightarrow K' \hookrightarrow K'' \hookrightarrow K''' \hookrightarrow \dotsc$ and observe that the colimit $\overline{K}$ is an algebraic closure of $K$. A similar reasoning can be obtained to show that every two algebraic closures are isomorphic (but not in a canonical way).
