## Existence of Algebraic Closures
**Theorem.** Every field $K$ has an algebraic closure.
[step: The set of algebraic extensions is nonempty and partially ordered]
Let $\mathcal{A}$ denote the collection of all algebraic extensions of $K$ (up to $K$-isomorphism) that can be realized inside a fixed "universe" set of sufficiently large cardinality. We partially order $\mathcal{A}$ by inclusion: $(L_1, i_1) \leq (L_2, i_2)$ when $L_1 \subseteq L_2$ and the embedding $i_2$ restricts to $i_1$ on $L_1$.
This poset is nonempty because $K$ itself is an algebraic extension of $K$.
[guided]
The technical device of fixing a universe set avoids the set-theoretic issue that "the class of all algebraic extensions" is not a set. Any set containing at least $\lvert K[x] \rvert$ elements suffices, since every algebraic extension of $K$ has cardinality at most $\lvert K[x] \rvert$.
[/guided]
[/step]
[step: Every chain has an upper bound]
Let $\{L_\alpha\}_{\alpha \in I}$ be a chain in $\mathcal{A}$. Define
\begin{align*}
L = \bigcup_{\alpha \in I} L_\alpha.
\end{align*}
We verify that $L$ is a field and an algebraic extension of $K$.
**$L$ is a field.** Given any $a, b \in L$, there exist indices $\alpha, \beta$ with $a \in L_\alpha$ and $b \in L_\beta$. Since the $L_\alpha$ form a chain, one contains the other — say $L_\alpha \subseteq L_\beta$. Then $a, b \in L_\beta$, so $a + b$, $a - b$, $ab$, and (when $b \neq 0$) $a/b$ all lie in $L_\beta \subseteq L$. The field axioms are inherited from the $L_\alpha$.
**$L$ is algebraic over $K$.** Every element $a \in L$ belongs to some $L_\alpha$, and $L_\alpha / K$ is algebraic, so $a$ is a root of a nonzero polynomial in $K[x]$.
Therefore $L \in \mathcal{A}$ and $L_\alpha \leq L$ for every $\alpha$, so $L$ is an upper bound for the chain.
[guided]
The key point is that the union of a chain of fields is again a field. This works precisely because the chain condition guarantees that any two elements share a common ambient field. A union of fields that merely form a directed system (without being linearly ordered) need not be a field, but for chains this closure under field operations is automatic.
[/guided]
[/step]
[step: The maximal element is algebraically closed]
By Zorn's lemma, $\mathcal{A}$ has a maximal element $\overline{K}$. We claim $\overline{K}$ is algebraically closed.
Suppose for contradiction that $\overline{K}$ is not algebraically closed. Then there exists an irreducible polynomial $f(x) \in \overline{K}[x]$ of degree $\geq 2$. Form the extension
\begin{align*}
\overline{K}' = \overline{K}[x] / (f(x)).
\end{align*}
This is a proper algebraic extension of $\overline{K}$. We now show $\overline{K}'$ is algebraic over $K$, contradicting maximality.
Take any $\beta \in \overline{K}'$. Then $\beta$ is algebraic over $\overline{K}$, so $\beta$ satisfies some polynomial $g(t) = t^n + c_{n-1}t^{n-1} + \cdots + c_0$ with $c_i \in \overline{K}$. Each $c_i$ is algebraic over $K$, so the extension
\begin{align*}
K \subseteq K(c_0, c_1, \ldots, c_{n-1}) \subseteq K(c_0, \ldots, c_{n-1}, \beta)
\end{align*}
is a tower of algebraic extensions. By the tower law, $\beta$ is algebraic over $K$, meaning $\overline{K}' / K$ is algebraic.
But then $\overline{K} \subsetneq \overline{K}'$ with $\overline{K}' \in \mathcal{A}$, contradicting the maximality of $\overline{K}$.
Therefore $\overline{K}$ is algebraically closed, and since $\overline{K} / K$ is algebraic, $\overline{K}$ is an algebraic closure of $K$. $\blacksquare$
[guided]
This is the heart of the proof. The contradiction argument has two layers: first, that adjoining a root of an irreducible polynomial over $\overline{K}$ produces a strictly larger extension; second, that this larger extension is still algebraic over $K$ (not merely over $\overline{K}$). The second layer uses the fact that an extension algebraic over an algebraic extension is itself algebraic — the "tower law for algebraic extensions," which follows from the multiplicativity of degrees in finite extensions.
[/guided]
[/step]
