Proof of the existence of an algebraic closure

Proof

First we show that every field $F$ has an extension $F_1$ in which each nonconstant polynomial over $F$ has a root, with $F_1/F$ algebraic.

1. Let $S$ be the set of monic polynomials $f \in F[x]$ with $\deg f \ge 1$. Take one variable $X_f$ for each $f \in S$.
2. Let $R = F[X_f : f \in S]$ be the union of the polynomial rings in finitely many of these variables. Two elements of $R$ lie in a common such ring, where we add and multiply them. Let $I$ be the ideal generated by the elements $f(X_f)$ for $f \in S$.
3. We show $I \ne R$, by contradiction. Suppose $1 \in I$. Then $1 = \sum_{i=1}^n g_i \, f_i(X_{f_i})$ for some distinct $f_1, \dots, f_n \in S$ and some $g_i \in R$.
4. The $g_i$ involve finitely many variables. So there are distinct $f_{n+1}, \dots, f_N \in S$ with all $g_i \in R_0 = F[X_{f_1}, \dots, X_{f_N}]$.
5. Set $E_0 = F$. For $1 \le i \le n$, some extension $E_i$ of $E_{i-1}$ contains a root $a_i$ of $f_i$. By induction this gives a field $E = E_n$ containing $F$ and $a_1, \dots, a_n$.
6. Applying the universal property of the polynomial ring one variable at a time gives a ring homomorphism $\varphi \colon R_0 \to E$. It is the identity on $F$, sends $X_{f_i}$ to $a_i$ for $i \le n$, and sends $X_{f_i}$ to $0$ for $i > n$.
7. Applying $\varphi$ to Step 3 gives $1 = \sum_{i=1}^n \varphi(g_i) \, f_i(a_i) = 0$ in $E$. This is impossible in a field, so $I \ne R$.
8. By the existence of maximal ideals, some maximal ideal $M$ of $R$ contains $I$. Then $F_1 = R/M$ is a field, because R/M is a field iff M is maximal.
9. The map $F \to F_1$, $c \mapsto c + M$, is a ring homomorphism into the quotient ring. It is injective, so we identify $F$ with its image, a subfield of $F_1$.
10. For $f \in S$ let $x_f = X_f + M$. Then $f(x_f) = f(X_f) + M = 0$, since $f(X_f) \in I \subseteq M$. So each $f \in S$ has a root in $F_1$.
11. Let $g \in F[x]$ with $\deg g \ge 1$ and leading coefficient $c$. Then $c^{-1} g \in S$ has a root in $F_1$, which is also a root of $g$.
12. Each $x_f$ is algebraic over $F$, since $f \ne 0$ and $f(x_f) = 0$.
13. Let $u \in F_1$. Then $u = h + M$ with $h \in R$, and $h$ involves finitely many variables $X_{f_1}, \dots, X_{f_k}$. So $u = h(x_{f_1}, \dots, x_{f_k})$ lies in the generated field $G_k = F(x_{f_1}, \dots, x_{f_k})$.
14. Let $G_0 = F$ and $G_j = G_{j-1}(x_{f_j})$. The element $x_{f_j}$ is algebraic over $G_{j-1}$, since its polynomial $f_j$ has coefficients in $F \subseteq G_{j-1}$.
15. So $[G_j : G_{j-1}]$ is finite by the degree of a simple extension. By the tower law, $G_k/F$ is a finite extension.
16. Finite extensions are algebraic, so $u$ is algebraic over $F$. Hence $F_1/F$ is algebraic.

Now we iterate this construction for the field $K$ of the statement.

17. Let $K_0 = K$ and $K_{i+1} = (K_i)_1$, the field built from $F = K_i$ in Steps 1 to 16. Replacing each $K_{i+1}$ by an isomorphic copy, we may assume that $K_i$ is a subfield of $K_{i+1}$.
18. Let $\Omega = \bigcup_{i \ge 0} K_i$. Any two elements of $\Omega$ lie in a common $K_i$, since the $K_i$ increase. Define their sum and product in that $K_i$.
19. This does not depend on the choice of $i$, because each $K_i$ is a subfield of the later ones. Each field axiom involves finitely many elements, which lie in a common $K_i$. So $\Omega$ is a field containing $K$.
20. By induction on $i$, $K_i/K$ is algebraic. For $i = 0$, each $c \in K$ is a root of $x - c$. If $K_i/K$ is algebraic, then $K_{i+1}/K_i$ is algebraic by Step 16, so $K_{i+1}/K$ is algebraic by transitivity.
21. Every element of $\Omega$ lies in some $K_i$, so it is algebraic over $K$. Hence $\Omega/K$ is algebraic.
22. Let $g \in \Omega[x]$ with $\deg g \ge 1$. Its finitely many coefficients lie in some $K_N$. By Step 11 for $F = K_N$, $g$ has a root in $K_{N+1} \subseteq \Omega$.
23. So $\Omega$ is algebraically closed. With Step 21, $\Omega$ is an algebraic closure of $K$.