Let $K_{1}$ and $K_{2}$ be two isomorphic fields. Prove that an isomorphism $f \colon K_{1} \xrightarrow{\sim} K_{2}$ can be extended to an isomorphism $\sigma \colon \overline{K_{1}} \xrightarrow{\sim} \overline{K_{2}}$ between their respective closures.
I intended to do this using Zorn's Lemma (akin to the proof of uniqueness of algebraic closures up to isomorphism), this is my attempt:
Attempt:
Let $f \colon K_{1} \rightarrow K_{2}$ be a field isomorphism, and let $i_{1} \colon K_{1} \rightarrow \overline{K_{1}}$, $i_{2} \colon K_{2} \rightarrow \overline{K_{2}}$ be field homomorphisms.
Composing $f$ with $i_{2}$ gives us: $\phi = i_{2} \circ f \colon K_{1} \rightarrow \overline{K_{2}}$. We aim to prove the existence of $\sigma \colon \overline{K_{1}} \rightarrow \overline{K_{2}}$, such that $\phi = \sigma \circ i_{1}$. This can be done using Zorn's Lemma.
Let $\mathcal{A}$ be a collection of pairs $(M, \psi)$, where $M$ is a subfield of $\overline{K_{1}}$ containing $K_{1}$, and $\psi$ such that $\psi \circ i_{1} = \phi$. We define a partial ordering on $\mathcal{A}$ as follows: $(M, \psi) \leq (\tilde{M}, \tilde{\psi}) \iff M \subset \tilde{M}, \tilde{\psi}\vert_{M} = \psi$.
For the empty chains in $\mathcal{A}$, $(M, id)$ is an upper bound. Hence $\mathcal{A}$ is not empty. For the non-empty chains in $\mathcal{A}$ we create an upper bound by taking the unions. Then by Zorn's Lemma $\mathcal{A}$ has a maximal element, say $(F, \sigma)$ where $\sigma \colon F \rightarrow \overline{K_{2}}$.
Suppose, for the sake of contradiction, that $F \neq \overline{K_{1}}$. Then there exists some $\alpha \in \overline{K_{1}} \setminus F$. Then $\alpha$ is algebraic over $K_{1}$. So there exists a monic polynomial $f \in K_{1}[X]$ such that $f(\alpha) = 0$.
This is where I am very unsure of the proof, I'm having trouble finding a contradiction for the maximality of $(F, \sigma)$. I have tried doing so using a lemma regarding the existence of isomorphisms between splitting fields but it doesn't seem like the right approach.
Does anybody have an idea how to find a contradiction for the maximality of $(F, \sigma)$, and whether this is the best approach? Thanks in advance.
$\sigma:F\to\overline{K_2}$ can be extended to a homomorphism of the polynomial rings $F[x]\to\overline{K_2}[x]$ by $\sigma(\sum\limits_{i=0}^n a_ix^i)=\sum\limits_{i=0}^n\sigma(a_i)x^i$, i.e we simply apply $\sigma$ to the coefficients of the polynomial.
Now let $m\in F[x]$ be the minimal polynomial of $\alpha$ over $F$. Pick any root $\beta\in\overline{K_2}$ of the polynomial $\sigma(m)\in\overline{K_2}[x]$. Such a root exists because $\overline{K_2}$ is algebraically closed. Note that $\sigma(m)$ must be irreducible over the field $\sigma(F)$ (why?), and so $\sigma(m)$ is also the minimal polynomial of $\beta$ over $\sigma(F)$. And now we can define a map $\rho:F(\alpha)\to\overline{K_2}$ as follows:
$\rho(f(\alpha))=\sigma(f)(\beta)$
where $f\in F[x]$ is a polynomial. It's not hard to check that $\rho$ is a well defined homomorphism that extends $\sigma$, contradicting the maximality of $F$.
So indeed $F=\overline{K_1}$, we got our required extension. It remains to prove that $\sigma$ is an isomorphism. It is clearly injective because any field homomorphism is. Can you prove that it is surjective?