Let $F$ be a field that can be well-ordered. This post proved that $F$ has a well-orderable algebraic closure, and I tried to prove that any two of them are isomorphic:
Proof. Let $F_1$ and $F_2$ be two well-ordered algebraic closures of $F$. Let $\alpha$ be the $\alpha$-th element in $F_1\setminus F$. For each $\alpha\in F_1\setminus F$, we construct a field $E_\alpha\leq F_1$ and an isomorphism $\sigma_\alpha$ from $E_\alpha$ to a subfield of $F_2$ such that for all $\beta<\alpha$,
(i) $E_\beta$ is a subfield of $E_\alpha$, and
(ii) $\sigma_\beta$ extends to $\sigma_\alpha$, and
(iii) $E_\alpha$ contains the splitting field of the minimal polynomial of $\alpha$ in $F[X]$.
Let $E_0$ be the splitting field of the minimal polynomial $m_0(X)\in F[X]$. Let $\sigma_0$ be an isomorphism from $E_0$ onto a subfield of $F_2$, and that $\sigma_0\restriction F=\textrm{id}$.
Assume $\alpha>0$ and that we have constructed $E_\beta$ and $\sigma_\beta$ for all $\beta<\alpha$. Let $B=\bigcup_{\beta<\alpha}E_\beta$ and $\sigma_B=\bigcup_{\beta<\alpha}\sigma_\beta$ be an isomorphism. Let $m_\alpha(X)\in F[X]$ be the minimal polynomial of $\alpha$. Let $E_\alpha\supseteq B$ be the splitting field of $m_\alpha(X)\in B[X]$. Now since $E_\alpha$ is a finite extension of $B$, $\sigma_B$ can extend to an isomorphism $\sigma_\alpha$ from $E_\alpha$ onto a subfield of $F_2$. $E_\alpha$ and $\sigma_\alpha$ satisfy (i), (ii) and (iii).
Now since $F_1=\bigcup_{\alpha}E_\alpha$, we have an isomorphism $\sigma=\bigcup_\alpha\sigma_\alpha$ from $F_1$ onto a subfield of $F_2$, and that $\sigma\restriction F=\textrm{id}$. We need to show that $\sigma(F_1)=F_2$. Let $f(X)\in F[X]$ be irreducible. Let $a\in F_1$ be any root of $f(X)$. Then $f(\sigma(a))=\sigma(f(a))=\sigma(0)=0$ and so $\sigma(a)\in F_2$ is a root of $f(X)$. It follows that $\sigma(F_1)$ is an algebraic closure of $F$ and so $\sigma(F_1)=F_2$.
Now the problem is that I didn't use the fact that $F_2$ is well-orderable. What's wrong?
