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

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

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

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

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

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

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