Question about finite field extensions I have questions about the proof of this result: 

Theorem: Let $K:F$ be a extension and $A_F^K=\{\alpha\in K:\alpha$ is algebraic over $F\}$. Prove that if $K$ is algebraically closed, then $A_F^K$ is algebraically closed.

Proof (Arturo Magidin):
Let $A_F^K = \{ a\in K\mid a\text{ is algebraic over }F\}$, and let $f(x)\in A_F^K[x]$ be a non-constant polynomial. We may assume $f(x)$ is monic,
$$f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_0,\quad a_i\in F,\quad n\gt 0.$$
Since $a_i\in A_F^K$, then $a_i$ is algebraic over $F$ for all $i$, so the field $E=F(a_0,\ldots,a_{n-1})$ is a finite extension of $F$. 
Now $f(x)\in A_F^K[x]\subseteq K[x]$, so we know that $K$ contains a root $r$ of $f(x)$, since $K$ is algebraically closed. Therefore, $[E(r):E] \leq n$ is finite, so $[E(r):F] = [E(r):E][E:F]$ is also finite. Therefore, $[F(r):F] \leq [E(r):F]$ is finite, so $r$ is algebraic over $F$. Since $r$ is algebraic over $F$, then $r\in A_F^K$. Thus, $f(x)$ has a root in $A_F^K$.
This shows that every non-constant polynomial in $A_F^K(x)$ has a root in $A_F^K.$

Why $E$ is a finite extension of $F?$, why $[E(r):E] \leq n$, why $[F(r):F] \leq [E(r):F]$, and why did he concluded that $r$ is algebraic over $F?$

Thank you in advance :) 
 A: First, let’s gather some general facts. Let $E / F$ be any field extension.

  
*
  
*(A) For any $α ∈ E$, if there is some nonzero $f ∈ F[X]$ with $f(α) = 0$, then $F(α)$ is a finite extension of $F$ with $[F(α):F] ≤ \deg f$.
  
*(B) For any $α ∈ E$, the degree $[F(α):F]$ is finite if and only if $α$ is algebraic over $F$.
  
*(C) Whenever $F'$ is an intermediate field of $E / F$, then $[E : F] = [E : F'][F' : F]$ and in particular, $[F' : F] ≤ [E : F]$.
  

These are general facts you should get to know by studying Galois theory. Consult your lecture notes or your favorite algebra book.
Now, let’s turn to your special situation:


*

*Why is $E$ a finite extension of $F$? This follows inductively from (B) and (C). Consider the tower of field extensions, each obtained by adjoining an algebraic element $a_{n-1}, …, a_0$,
$$E = F(a_0,…,a_{n-1}) / F(a_0,…,a_{n-2}) / … / F(a_0) / F.$$

*Why is $[E(r) : E] ≤ n$? This follows immediately from (A) since $f(r) = 0$ and $f ∈ E[X]$.

*Why is $[F(r) : F] ≤ [E(r) : F]$? As $F ⊆ E$, $F(r)$ is an intermediate extension of $E(r) / F$, so the result follows by (C).

*Why did he conclude that $r$ is algebraic over $F$? By (3), $[F(r) : F]$ is finite, as $[E(r) : F] = [E(r) : E][E : F]$ itself is finite by (1) and (2), using (C). In particular, $F(r) / F$ is a finite extension. By (B), $r$ is algebraic.
A: Generally, if $\alpha$ is algebraic over $F$, then $F(\alpha)$ is a finite extension of $F$; because $1,\alpha,...,\alpha^n$ generates $F(\alpha)$ as an $F$-vector space when $n$ is sufficiently large (pick the degree of a nonzero polynomial vanishing at $\alpha$). By induction, it follows that $F(\alpha_1,...,\alpha_n)$ is a finite extension of $F$ if all the $\alpha_i$'s are algebraic over $F$, because $\alpha_{i+1}$ is algebraic over $F$, thus over $F(\alpha_1,...,\alpha_i)$; and because if $E:F, F:K$ are finite, then so is $E:K$.
$[E(r):E]\leq n$ because $r$ is cancelled by a polynomial of degree $n$, and so as earlier, $1,r,...,r^{n-1}$ spans $E(r)$ as an $E$-vector space.
$[F(r):F]\leq [E(r): F]$ because $[E(r):F]=[E(r): F(r)][F(r):F]$
From this the author concludes that $F(r):F$ is finite, therefore the family $(r^n)_{n\in \Bbb{N}}$ cannot be $F$-linearly independant: there is a nonzero polynomial $g\in F[X]$ such that $g(r)=0$ 
