where we note all fields in are algebraic extensions of .
We show that every chain in has an upper bound.
Let which is a field.
If , then we define for some ,
which is clearly independent of the choice of .
Then is an upper bound of .
By Zorn's lemma, has a maximal element .
Since is algebraic,
We claim that , whence is the desired morphism.
Suppose towards contradiction there exists and consider the simple extension.
Since is algebraic over , it is algebraic over ,
thus it is the root of an irreducible .
Abusing notation to invoke the induced homomorphism
let , which is irreducible over ,
and has a root in — here we use that is algebraically closed.
Now by ^P2, the isomorphism lifts to an isomorphism
sending to , contradicting the maximality of .
Therefore , and we're done.