Proposition. smallest algebraic closure [1606]

    Let $\Omega$ be an algebraically closed field containing $F$ and $$ F' = \{ \alpha \in \Omega \mid \alpha \text{ is algebraic over } F \}. $$ Then

    1. $F'$ is an algebraic closure of $F$.
    2. If $K \subseteq \Omega$ is an algebraic field extension of $F$, then $K' = F'$.

    Proof. sketch [1606A]

      1. Clearly $F' / F$ is algebraic. Take any $f(x) \in F[x]$, $f(x)$ splits completely over $\Omega$. By Proposition 1503, $F(\alpha_{1}, \dots, \alpha_{n})$ is a splitting field of $f(x)$, where $f(x) = a(x - \alpha_{1})\cdots(x - \alpha_{n}), a \in F, \alpha_{i} \in \Omega$. Hence each $\alpha_{i}$ is algebraic over $F$ and belongs to $F'$, which implies that $f(x)$ splits completely over $F'$. Therefore, $F'$ is an algebraic closure of $F$.
      2. $\alpha \in \Omega$ is algebraic over $F$ implies it is algebraic over $K$, so $\alpha \in K'$, which means $F' \subseteq K'$. By Proposition 14012, $K' / F$ is algebraic. So every $\beta \in K'$ is algebraic over $F$, hence $\beta \in F'$, it follows that $K' \subseteq F'$. So $K' = F'$.