The author from my book wrote a nasty proof of existence of algebraic
closure for any field. It is nasty because for some reason in the first
paragraph he speaks about something which seems to be unnecessary.
I redone his proof but I am thinking maybe his first paragraph which I
consider unnecessary is really a requirement in the proof. This would
be long and I am not writing out each detail for it will take a long
time. If someone can check this and say it is correct I will greatly
appreciate it.
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The theorems that will be used.
Kronecker's Theorem: For every field there exists
an extension field such as for some there is an such as .
Theorem 1: If is a finite extension of and is finite extension of . Then
if finite extension over .
Theorem 2: If is a finite extension over , then is an algebraic extension over .
Zorn's Lemma: If is a partially-ordered-set such as
every chain has an upper bound. Then has at least one
maximal element.
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Here is my version (check for errors please):
Lemma: If are fields with algebraic over and is
algebraic extension over . Then is
algebraic extension over F.
Proof: Since a simple extension (where the element is algebraic)
is a finite extension, and a finite extension is an algebraic extension
by Theorem 2, we have that
is an algebraic extension over .
Thus, for any there must be such as, . Thus, there is a
polynomial . Since are algebraic over , is finite extension over
. Since is algebraic over ,thus is finite extension over
. Thus, by Theorem 1, we have that,
is finite extension over
. Thus, by Theorem 2, is algebraic over .
Thus, is algebraic
over . Thus, any element of is
algebraic over . Thus, is algebraic
extension over .
Q.E.D.
Existence Theorem:Every field has an extension field that has
algebraic closure.
Proof:Let be any field. Define a set,
. This set is non-
empty because trivially. Define a partial ordering on
. This is ordering by "inclusion", we define if and only if is a subfield of .
Let be any chain. Then define,
. Thus, is a field with well-
defined binary operations as were defined for . Also,
the way it was constructed is algebriac over and also .
Thus, , thus every chain has an upper bound.
The necessary conditions of Zorn's Lemma are satisfied. Thus, there is a
at least one maximal element, . Now we prove that
is algebraically closed. Assume, that there is a
with .
Then by Kronecker's Theorem there is a field
with for . Thus, we have
that . Now if we can show that is algebraic over then we have a
contradiction because is a maximal element. Thus,
must be algebraically closed. To show, this we use the Lemma. Since is an extension field of with being algebraic over , and is algebraic over . Thus, is algebraic over . Thus, must be algebraically closed.
Q.E.D.
Notice in the proof the difference between,
and . They look similar but one is with a bar over it if you do not see it.
Thank you.