Proposition 2

Let $\Omega$ be an algebraically closed field and K a subfield of $\Omega$. Then the relative algebraic closure $\bar{K}$ of K in $\Omega$ is an algebraically closed field.

Let f be a non-constant polynomial in $\bar{K}[X] \subset \Omega[X]$. Since $\Omega$ is algebraically closed, the polynomial f has at least one root in $\Omega$, and since this root is algebraic over $\bar{K}$, it belongs to $\bar{K}$ (V, p. 19, Prop. 4). Therefore $\bar{K}$ satisfies (AC').

\* Example 2. — By Prop. 2 the set of all complex numbers that are algebraic over Q (often called briefly algebraic numbers) is an algebraically closed field. \*