What are some algebraically closed fields? What are some examples of algebraically closed fields?  Wikipedia lists exactly two: $\mathbb{C}$ and the (complex) algebraic numbers.
EDIT: scrolling to the bottom of the Wikipedia article, they mention that every field has an essentially unique "algebraic closure", which is algebraically closed, and that proving this fact in full generality requires the Axiom of Choice.  
But that leaves open some questions: are there any algebraically closed fields that are used often enough that they have names?    When do you need AC to show $\mathbb{F}$ has an algebraic closures, and when don't you need AC? 
I could not find a similar question here, and Google was not helpful.
 A: 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).
