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 $\displaystyle F$ 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.