classification of $2$-dimensional field extensions Let $F$ be a field and $K:F$ be a field extension such that $[K:F]=2$. Then
(i). If $Char(F)\neq 2$, then there exists $\alpha\in K^*$, $\alpha\notin F^*$, such that $K=F(\alpha)$ and $\alpha^2\in F^*$.
(ii). If $Char(F)=2$, then the above statement does not hold.
How to prove?
 A: Part (i) comes from the usual formula for finding a root of a quadratic equation. This breaks down of course in characteristic $2$, because of the $2$ at the denominator.
As to part (ii), consider the field $F$ with two elements, and the field $K$ with $4$ elements, so that $[K : F ] = 2$. Since the fields are finite, the Frobenius endomorphism $\sigma : x \mapsto x^{2}$ is an automorphism (you actually have $\sigma^{2} = I$, the identity on $K$ - that is $\alpha^{4} = \alpha$ for all $\alpha \in K$), so if $\alpha^{2} = \sigma(\alpha) \in F$, you get $\alpha = (\alpha^{2})^{2} \in F$.
A: Let $x$ be any element in $K$ not in $F$. Since $\{1,x\}$ is a basis of $K$ as a $F$-vector space, we must have
$$
x^2=ax+b
$$
for some $a,b\in F$. Now consider $f\in F$ and compute $(x+f)^2=(a+2f)x+b+f^2$. So, if we choose $f=-\frac12a$, the element $\alpha=x+f$ is as wanted.
Yet, this can be done only if the characteristic is not $2$.
A: In characteristic $2$, not every quadratic equation can be standardized to the form $X^2 = a$ .  There is a second type, $X^2 + X = b$. 
More invariantly, the problem is that $2=0$ collapses positive and negative square roots to be equal, and the minimal polynomial of any element in the degree $2$ extension (but not in $F$) has a double root for square-root extensions. It has  two distinct roots for the $X^2 + X = b$ extensions. Hence the two types of extension are not isomorphic in char.2.
