Q:

If $F$ is a field, then any two algebraic closures are isomorphic by an isomorphism that is the identity on $F$.

To start, suppose $K_1$ and $K_2$ are two algebraic closures of $F$.

(a) Let $P$ be the set of partial functions $f$ from $K_1$ to $K_2$ with the following properties:

$F$ is contained in the domain of $f$ and $f$ restricted to $F$ is the identity on $F$.
$f$ is a field ismorphism between its domain and range.

Order $P$ as follows: $f < g$ if the domain of $f$ is contained in the domain of $g$ and for $a \in \mathrm{dom}(f)$, $f(a) = g(a)$. Prove that Zorn's Lemma applies to this partial order.

This is easy, and I have done it.

(b) By Zorn's Lemma, choose a maximal $f$ in $P$. You need to prove two things:

The domain of $f$ is $K_1$.
The range of $f$ is $K_2$.

Hint: the two cases are similar. In the first case, imagine $a \in K_1$ which is not in the domain of $f$. Use the fact that $K_2$ is algebraically closed to extend $f$ to include $a$ in its domain.

So, I try to follow the hint. I can't see any way to do this directly so I try to find a contradiction. I suppose that $dom(f) \neq K_1$.
I can choose $a \in K_1 \setminus dom(f)$ and $a$ is algebraic over $dom(f)$, so it has a minimal polynomial $p$ over $dom(f)$. $p$ is also irreducible.
Nothing else seems to stand out to me. What am I missing?

A:

Maybe I'm missing something here, but what about this?
Suppose that $a \in K_1$ is not in the domain of $f$ and let $p(X) \in F[X]$ be the minimal polynomial of $a$. Then define $g(b) = f(b)$ for every $b$ in the domain of $f$, while $g(a) = \alpha$ where $\alpha$ is any fixed root of $p(X)$ in $K_2$ not in the range of $f$. After extending (pointwise) $g$ to a field morphism $\text{dom}(f)(a) \to \text{range}(f)(\alpha)$ it is clear that $f < g$, contradicting the maximality of $f$.
The other case is essentially the same.

