The Stacks project

Proof. Let $F$ be a field. By Lemma 9.8.9 the cardinality of an algebraic extension of $F$ is bounded by $\max (\aleph _0, |F|)$. Choose a set $S$ containing $F$ with $|S| > \max (\aleph _0, |F|)$. Let's consider triples $(E, \sigma _ E, \mu _ E)$ where

1. $E$ is a set with $F \subset E \subset S$, and

2. $\sigma _ E : E \times E \to E$ and $\mu _ E : E \times E \to E$ are maps of sets such that $(E, \sigma _ E, \mu _ E)$ defines the structure of a field extension of $F$ (in particular $\sigma _ E(a, b) = a +_ F b$ for $a, b \in F$ and similarly for $\mu _ E$), and

3. $E/F$ is an algebraic field extension.

The collection of all triples $(E, \sigma _ E, \mu _ E)$ forms a set $I$. For $i \in I$ we will denote $E_ i = (E_ i, \sigma _ i, \mu _ i)$ the corresponding field extension to $F$. We define a partial ordering on $I$ by declaring $i \leq i'$ if and only if $E_ i \subset E_{i'}$ (this makes sense as $E_ i$ and $E_{i'}$ are subsets of the same set $S$) and we have $\sigma _ i = \sigma _{i'}|_{E_ i \times E_ i}$ and $\mu _ i = \mu _{i'}|_{E_ i \times E_ i}$, in other words, $E_{i'}$ is a field extension of $E_ i$.

Let $T \subset I$ be a totally ordered subset. Then it is clear that $E_ T = \bigcup _{i \in T} E_ i$ with induced maps $\sigma _ T = \bigcup \sigma _ i$ and $\mu _ T = \bigcup \mu _ i$ is another element of $I$. In other words every totally order subset of $I$ has a upper bound in $I$. By Zorn's lemma there exists a maximal element $(E, \sigma _ E, \mu _ E)$ in $I$. We claim that $E$ is an algebraic closure. Since by definition of $I$ the extension $E/F$ is algebraic, it suffices to show that $E$ is algebraically closed.

To see this we argue by contradiction. Namely, suppose that $E$ is not algebraically closed. Then there exists an irreducible polynomial $P$ over $E$ of degree $> 1$, see Lemma 9.10.2. By Lemma 9.8.5 we obtain a nontrivial finite extension $E' = E[x]/(P)$. Observe that $E'/F$ is algebraic by Lemma 9.8.8. Thus the cardinality of $E'$ is $\leq \max (\aleph _0, |F|)$. By elementary set theory we can extend the given injection $E \subset S$ to an injection $E' \to S$. In other words, we may think of $E'$ as an element of our set $I$ contradicting the maximality of $E$. This contradiction completes the proof. $\square$