Difference in extension of finite fields and extension of infinite fields Let $F\subset K$ be finite field extension. It is clear $[K:F]$ is finite saying $K$ is a finite dimensional vector space. Consider the root of $x^2-a=0,a\in F$. 
Here is a statement saying if $F,K$ finite fields, $[K:F]=2l,l\in N-\{0\}$, $x\not\in F$ and $x^2-a=0$ with $a\in F$, then $x\in K$. I cannot prove this directly using degree formula as this does not give me any quite useful information. I want to show $[K(x):K]=1$. So $[K(x):F]=[K(x):K][K:F]=[K(x):F(x)][F(x):F]\leq 4l$. So $[K(x):K]\leq 2$
However consider $Q\to Q[2^{\frac{1}{4}}]$ degree 4 extension by $x^4-2$ being irreducible through eisenstein criterion where $Q$ is rational number and the map is embedding $Q$ as the subfield of $Q[2^{\frac{1}{4}}]$. Say $x^2+1=0,1\in Q$ for sure. I do not have $i\in Q[2^{\frac{1}{4}}]$. 
What is the reconcillation here? It seems finite field and infinite field behaves quite differently here.
 A: The difference between finite and infinite fields here is that for any given $k$ an infinite field may have many extensions of degree $k$, but a finite field has only one. 
A: The structure theorem of finite fields and finite extensions of finite fields is quite strong.


*

*There is one and only one finite field of order $p^n$ for every prime power $p^n$.


*Every extension of finite fields is a Galois extension and the Galois group is cyclic.


*If $m \mid n$ then $\operatorname{GF}(p^m) \subseteq \operatorname{GF}(p^n)$.

The first one is key here. Let's say $F = \operatorname{GF}(p^m)$.

*

*If $p = 2$ then the map $x \mapsto x^2$ is an automorphism. Thus every element of $F$ has a square root.


*If $p > 2$ then half the elements of $F^\times$ are squares.
(See The number of elements which are squares in a finite field. for details.)
So for $p = 2$ the equation $x^2 - a$ has all of its roots already in $F$. Next, suppose $p \ne 2$ and $a, b \in F^\times$ are two non-squares. Then
$$ F[x]/(x^2 - a) \text{ and }  F[x]/(x^2 - b) $$
are fields of order $p^{2m}$ and hence by point (1.) above they are isomorphic.
