The field $C$ is an algebraic closure for $F$

Question

Let $E/F$ be an algebraic extension and $C$ be an algebraic closure of $E$.
I want to show that the field $C$ is an algebraic closure also for $F$.

I have done the following:

$C$ is an algebraic closure of $E$, i.e., each polynomial of $E[x]$ has all the roots in $C$. Let $f\in F[x]$ be a non-zero polynomial. Since $E$ is an extension of $F$, we have that $f\in E[x]$. So, each polymnomial of $F[x]$ has all the roots in $C$.

To conclude that $C$ is an algebraic closure also for $F$, it is left to show that $C/F$ is algebraic, or not?

So, we have to show that each element of $C$ is algebraic over $F$, i.e., for each element $c\in C$ there is a non-zero polynomial $g(x)\in F[x]$, such that $g(c)=0$.

To show that we have to use the fact that $C/E$ is algebraic, or not? But how exactly?

Answer 1

A field extension $C$ of $F$ is an algebraic closure if

1. $C$ is algebraically closed;
2. $C$ is algebraic over $F$.

Property 1 is clearly satisfied.

Now, let's prove that if $K$ is algebraic over $L$ and $M$ is algebraic over $K$, then $M$ is algebraic over $L$. This will show property 2.

Let $c\in M$. Then $c$ satisfies a nonzero polynomial $f(x)\in K[x]$. If $f(x)=a_0+a_1x+\dots+a_nx^n$, then $c$ is algebraic over $L'=L(a_0,a_1,\dots,a_n)$, which is finite dimensional over $L$ (see “key fact” below).

By the “key fact”, $L'(c)$ is finite dimensional over $L'$. Thus $$ [L'(c):L]=[L'(c):L'][L':L] $$ so $c$ belongs to a finite dimensional extension of $L$ and therefore it is algebraic over $L$.

Key fact If $E$ is an extension field of $F$ and $c\in E$, then

1. $c$ is algebraic over $F$ if and only if $F(c)$ is finite dimensional over $F$;

2. if $E$ is finite dimensional over $F$, then $E$ is algebraic over $F$.

Answer 2

Suppose $F \subseteq E \subseteq L$ are fields. If $L$ is algebraic over $E$, and $E$ is algebraic over $F$, then $L$ is algebraic over $F$. This result should give you what you want.

Here is a sketch of the proof. In general, if $A \subseteq B$ are fields, and $b \in B$ is algebraic over $A$, then $A(b)$ (by definition, the intersection of all subfields of $B$ which contain $A$ and the element $b$) is a finite extension of $A$, i.e. it is finite dimensional as a vector space over $A$. It follows that if $b_1, ... , b_n \in B$ are algebraic over $A$, then $A(b_1, ... , b_n)$ is a finite extension of $A$. This follows because $A(b_1, ... , b_i) = A(b_1, ... , b_{i-1})(b_i)$ is a finite extension of $A(b_1, ... , b_{i-1})$.

If $x \in L$, then $x$ is the root of a polynomial $f(X) = a_0 + a_1X + \cdots + a_nX^n \in E[X]$. Now all the $a_i$ are algebraic over $F$, so $F(a_0, ... , a_n)$ is a finite extension of $F$. Clearly, $x$ is algebraic over $F(a_0, ... , a_n)$, so $F(a_0, ... , a_n)(x) = F(a_0,..., a_n,x)$ is a finite extension of $F(a_0, ... , a_n)$, hence of $F$.

In particular, $x$ is a member of a field which is a finite extension of $F$. Now, you just need to show that finite extensions are algebraic.