Field has Algebraic Closure

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Theorem

Every field has an algebraic closure.


Proof

Let $F$ be a field.

Let $\FF$ be the collection of all extensions of $F$.



Define an ordering on $\FF$ thus:

$\forall K, L \in \FF: K \preceq L \iff L$ is an extension of $K$.

Let $C$ be a chain in $\FF$.

Let $\ds E = \bigcup_{K \mathop \in C} K$.

$E$ satisfies all field axioms, so $E \in \FF$.



By Set is Subset of Union, $E$ is an upper bound for $C$.


By Zorn's Lemma, $C$ has a maximal element $m$.

Then $m$ is an algebraic closure of $F$.



$\blacksquare$


Boolean Prime Ideal Theorem

This theorem depends on the Boolean Prime Ideal Theorem (BPI).

Although not as strong as the Axiom of Choice, the BPI is similarly independent of the Zermelo-Fraenkel axioms.

As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.