Case 1: $K / F$ is finite.

We prove by induction on $[K / F] = n$. The initial case $n = 1$ implies $K = F$ and hence $\varphi = \psi$. Assume the theorem holds for $n-1$. Pick any $\alpha \in K - F$, we consider $m_{\alpha, F}(x) \in F[x]$. Our strategy is to use Theorem 1506 to deduce it from $n$ to $n-1$ and then use the assumption.

Clearly $\varphi(m_{\alpha, F}(x)) \in \Omega[x]$. Since $\Omega$ is algebraically closed, $\varphi(m_{\alpha, F}(x))$ has a root $\beta \in \Omega$. Moreover, $\varphi(m_{\alpha, F(x)})$ is monic and irreducible because $m_{\alpha, F}(x)$ is monic and irreducible. So $m_{\beta, \Omega}(x) = \varphi(m_{\alpha, F}(x))$. Now we can apply Theorem 1506 and obtain an isomorphism $\varphi' : F(\alpha) \to \Omega$such that $\varphi'|_{F} = \varphi$. Then, by our induction hypothesis, we can extend $\varphi'$ to $\psi : K \to \Omega$ since $$ [K : F(\alpha)] = \frac{[K : F]}{[F(\alpha) : F]} $$ and we have $[F(\alpha) : F] > 1$ (because $\alpha \not\in F$).

Case 2: $K / F$ is infinite.

In this case, we use Zorn’s Lemma and the proved case that $K / F$ is finite for assistance.

Let $S$ be the set of pairs $(L, \phi)$ where $L$ is a subfield of $K$ containing $F$ with $\phi : L \to \Omega$ a field embedding with $\phi|_{F} = \varphi$. First, $S$ is not empty since $(F, \operatorname{id}_{F}) \in S$. Second, we define an order on $S$ by $$ (L, \phi) \leq (L', \phi') \iff L \subset L' \text{ and } \phi'|_{L} = \phi. $$ Suppose $C \subseteq S$ is a chain. We show that $C$ has an upper bound in $S$. Let $$ E = \bigcup_{(L, \phi) \in C}L. $$ $E$ is indeed a subfield of $K$. Given $\alpha \in E$, there exists $(L, \phi) \in C$ such that $\alpha \in L$. Set $\sigma(\alpha) = \phi(\alpha)$. We show that $\sigma(\alpha)$ does not depend on the choice of $(L, \phi)$. If $(L', \phi')$ is another element in $C$ satisfying $\alpha \in L'$, then $(L', \phi') \leq (L, \phi)$ or $(L, \phi) \leq (L', \phi')$ because $C$ is a chain. If it is the former case, then $L' \subseteq L$ and $\phi |_{L'} = \phi'$, which implies $\phi'(\alpha) = \phi(\alpha)$. The latter case is almost the same. Therefore, for every $\alpha \in E$, we have defined a unique $\sigma(\alpha) \in \Omega$. So $\sigma : E \to \Omega, \alpha \mapsto \phi(\alpha)$ is a well-defined map. Moreover, $\sigma |_{F} = \varphi$ and $(E, \sigma) \in S$ and is an upper bound of $C$. By Zorn’s Lemma, $S$ has a maximal element $(L, \phi)$.

We then show that $L = K$ by contradiction. If $L \subsetneq K$, pick any $\alpha \in K - L$, then $L(\alpha) / L$ is finite. By the case of finite extension proved above, $\phi : L \to \Omega$ extends to $\phi': L(\alpha) \to \Omega$. Then $(L, \phi) \leq (L(\alpha), \varphi')$ and $(L, \phi) \neq (L(\alpha), \phi')$. This contradicts the maximality of $(L, \phi)$. Therefore, $L = K$ and $\phi$ is the desired extension of $\varphi$.

$\impliedby$: Trivial.

$\implies$: Take any non-constant $f(x) \in \Omega[x]$. By Theorem 1504, $f(x)$ has a splitting field $\Omega'$ and $\Omega' / \Omega$ is finite thus algebraic by Proposition 1404. Then by Proposition 14012, since $\Omega' / \Omega$ and $\Omega / F$ are algebraic, $\Omega' / F$ is algebraic too. Hence for any $\alpha \in \Omega'$, there exists $a(x) \in F[x]$ such that $a(\alpha) = 0$. Now, since $\Omega$ is an algebraic closure of $F$, $a(x)$ splits completely in $\Omega$, which implies $\alpha \in \Omega$. Therefore, $\Omega' = \Omega$. So every non-constant $f(x) \in \Omega[x]$ splits completely over $\Omega$, $\Omega$ is algebraically closed.

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 $K' = F'$.

1. TODO
2. Apply Proposition 1603 to $L / F$ and the natural field embedding $F \to L'$, then there exists a field embedding $\varphi : L \to L'$ such that $\varphi |_{F} = \operatorname{id}_{F}$. By Proposition 1605, $L$ and $L'$ are both algebraically closed. Then $L$ is algebraically closed $\implies$ $\varphi(L)$ is algebraically closed since $L \cong \varphi(L)$ (recall that a field homomorphism is either $0$ or its kernel is $0$). By Proposition 14012, $L' / F$ is algebraic $\implies$ $L' / \varphi(L)$ is algebraic. By part 4 of Proposition 1601, $L' = \varphi(L)$ since $L'$ is algebraically closed.

We often denote an algebraic closure by $\overline{F}$.