TITLE: Is the statement that every field has an algebraic closure known to be equivalent to the ultrafilter lemma?
QUESTION [31 upvotes]: The existence and uniqueness of algebraic closures is generally proven using Zorn's lemma.  A quick Google search leads to a 1992 paper of Banaschewski, which I don't have access to, asserting that the proof only requires the ultrafilter lemma.  Questions:

Is it known whether the two are equivalent in ZF?
Would anyone like to give a quick sketch of the construction assuming the ultrafilter lemma?  I dislike the usual construction and am looking for others.

REPLY [32 votes]: As I mentioned in a comment to Eivind Dahl's answer, it
seems that there is also an easy argument directly from the
Compactness theorem of first order logic. Since you said you are looking for alternative constructions, let me expand on the idea here.
The Compactness theorem asserts that if every finite subset
of a first order theory has a model, then the whole theory
has a model. Andres mentioned that this is equivalent to
the Ultrafilter lemma.
Existence. Let $F$ be a field. Let $T$ be the theory
consisting of the field axioms, the atomic diagram of $F$
(which asserts all the equations and negated equations true
in $F$, using constants for elements of $F$), plus the
assertions that every polynomial over $F$ has a root. This
last assertion is made separately as an assertion about
each particular polynomial over $F$, using the constants in
the language added for the elements of $F$. This theory is
easily seen to be finitely satisfiable, since any finite
subtheory mentions only finitely many polynomials of $F$,
and we can satisfy it in a finite extension of $F$. Thus,
by the Compactness theorem, the whole theory has a model
$K$. If we take the collection of elements of $K$ that are
algebraic over $F$, this will be algebraically closed.
Uniqueness. Let $F$ be a field and let $E$ and $K$
be algebraic closures of it. Let $T$ be the theory
consisting of the union of the atomic diagrams of $E$ and
$K$, plus the field axioms. (Note that we haven't added any
axioms saying that elements of $E$ and $K$ are distinct, and in the end these constants will be in effect melded together, providing the isomorphism.)
If $T_0$ is a finite subtheory, then only finitely many
elements of $E$ and $K$ appear. Those elements of $E$ and
$K$ appear in some $F[\vec a]\subset E$ and $F[\vec
b]\subset K$. We can embed both of these extensions of $F$
into a single finite extension $F[\vec u]$, which will
satisfy all assertions in $T_0$. (Note, this embedding
effectively decides a little piece of the isomorphism, by
mapping some of the $\vec b$'s to some of the $\vec a$'s.)
Thus, by Compactness, the whole theory is satisfiable. If
$G$ is a model of $T$, then since $G$ interprets the atomic
diagrams of $E$ and $K$, we get isomorphisms of $E$ and $K$
into subfields of $G$. These maps agree on $F$ and have a
common range, which is the set of elements of $G$ that are
algebraic over $F$. Thus, the composition is an isomorphism
of $E$ and $K$.