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 F there exists
an extension field F\leq E such as for some f(x)\in<br />
 F[x] \mbox{ and }f(x) \not \in F there is an \alpha \in E<br />
such as f(\alpha)=0.

Theorem 1: If K is a finite extension of E<br />
and E is finite extension of F. Then
K if finite extension over F.

Theorem 2: If E is a finite extension over F<br />
, then E is an algebraic extension over F<br />
.

Zorn's Lemma: If S is a partially-ordered-set such as
every chain has an upper bound. Then S has at least one
maximal element.
-------------------
Here is my version (check for errors please):

Lemma: If K\leq E\leq F are fields with \alpha<br />
\in K algebraic over E and E is
algebraic extension over F. Then E(\alpha) 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
E(\alpha) is an algebraic extension over E.
Thus, for any \beta \in E(\alpha) there must be f(x)<br />
\in E[x] such as, f(\beta)=0. Thus, there is a
polynomial \alpha_0+\alpha_1x+...+\alpha_nx^n,\alpha_i\in <br />
E. Since \alpha_i are algebraic over <br />
F, F(\alpha_0,...,\alpha_n) is finite extension over
F. Since \beta is algebraic over <br />
F(\alpha_0,...,\alpha_n),thus F(\alpha_0,...,\alpha_n,\beta)<br />
is finite extension over
F(\alpha_0,...,\alpha_n). Thus, by Theorem 1, we have that,
F(\alpha_0,...,\alpha_n,\beta) is finite extension over
F. Thus, by Theorem 2, F(\alpha_0,...,\alpha_n,\beta)<br />
is algebraic over F.
Thus, \beta \in F(\alpha_0,...,\alpha_n,\beta) is algebraic
over F. Thus, any element of E(\alpha) is
algebraic over F. Thus, E(\alpha) is algebraic
extension over F.
Q.E.D.

Existence Theorem:Every field has an extension field that has
algebraic closure.
Proof:Let F be any field. Define a set,
S=\{F\leq E|E\mbox{ is algebraic over }F\}. This set is non-
empty because F\in S trivially. Define a partial ordering on
S. This is ordering by "inclusion", we define A \leq<br />
 B if and only if A is a subfield of B.
Let C=\{E_k\}\subseteq S be any chain. Then define,
W=\bigcup_kE_k. Thus, W is a field with well-
defined binary operations as were defined for E_k. Also,
W the way it was constructed is algebriac over F<br />
and also E_k\leq W,\forall E_k\in C.
Thus, W\in S, 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, \bar F. Now we prove that
\bar F is algebraically closed. Assume, that there is a
f(x)\in \bar F[x] with f(x)\not \in \bar F.
Then by Kronecker's Theorem there is a field K\geq \bar F
with f(\alpha)=0 for \alpha \in K. Thus, we have
that \bar F(\alpha)>\bar F. Now if we can show that <br />
\bar F(\alpha) is algebraic over F then we have a
contradiction because \bar F is a maximal element. Thus,
\bar F must be algebraically closed. To show, this we use the Lemma. Since K is an extension field of \bar F with \alpha \in K being algebraic over \bar F, and \bar F is algebraic over F. Thus, \bar F(\alpha) is algebraic over F. Thus, \bar F must be algebraically closed.
Q.E.D.

Notice in the proof the difference between,
F and \bar F. They look similar but one is with a bar over it if you do not see it.
Thank you.