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.