## Check this proof.

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.
-------------------
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 $f(x)\in
F[x] \mbox{ and }f(x) \not \in F$
there is an $\alpha \in E
$
such as $f(\alpha)=0$.

Theorem 1: If $K$ is a finite extension of $E
$
and $E$ is finite extension of $F$. Then
$K$ if finite extension over $F$.

Theorem 2: If $E$ is a finite extension over $F
$
, then $E$ is an algebraic extension over $F
$
.

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 $\alpha
\in K$
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 $f(x)
\in E[x]$
such as, $f(\beta)=0$. Thus, there is a
polynomial $\alpha_0+\alpha_1x+...+\alpha_nx^n,\alpha_i\in
E$
. Since $\alpha_i$ are algebraic over $
F$
, $F(\alpha_0,...,\alpha_n)$ is finite extension over
$F$. Since $\beta$ is algebraic over $
F(\alpha_0,...,\alpha_n)$
,thus $F(\alpha_0,...,\alpha_n,\beta)
$
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)
$
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 $A \leq
B$
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
$
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 $
\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.