Existence of algebraic closures

Let $K$ be a field.

Theorem (Existence of algebraic closure). There exists a field extension $\overline{K}/K$ such that:

1. $\overline{K}$ is algebraically closed (every nonconstant polynomial in $\overline{K}[x]$ splits into linear factors), and
2. $\overline{K}/K$ is an algebraic extension (equivalently, every element of $\overline{K}$ is an algebraic element over $K$).

Such an extension $\overline{K}$ is called an algebraic closure of $K$. A standard proof uses Zorn’s lemma (hence the axiom of choice) to build a maximal algebraic extension and then shows it must be algebraically closed.

Moreover, any two algebraic closures of $K$ are $K$-isomorphic; see uniqueness of algebraic closures.

Examples

1. $\mathbb{C}$ is an algebraic closure of $\mathbb{R}$: it is algebraically closed, and $[\mathbb{C}:\mathbb{R}]=2$, so every complex number is algebraic over $\mathbb{R}$.

1. The field $\overline{\mathbb{Q}}$ of algebraic numbers (elements algebraic over $\mathbb{Q}$) is an algebraic closure of $\mathbb{Q}$.

1. An algebraic closure of $\mathbb{F}_p$ can be realized as the union
   $$\overline{\mathbb{F}_p}=\bigcup_{n\ge1}\mathbb{F}_{p^n}$$
   inside a fixed ambient algebraic closure. This viewpoint is compatible with the existence and uniqueness of finite fields.