Show that $E/F$ is Galois extension If $F$ has characteristic $\neq$2 and $E/F$ is a field extension with $[E:F]=2$, then $E/F$ is Galois.
Normal and separable extension is Galois extension. Can we say that since the degree of extension is $2$ it is normal. But how to prove that it is separable?
 A: The field extension $E/F$ is of course Galois precisely when it is both separable and normal.
We recall that a field extension is separable if and only if the minimal polynomial $p(x) \in F[x]$ of any element $\mu \in E$ is itself a separable polynomial over $F$; this means, by definition, that $p(x)$ has no square factors in any extension of $F$ or, equivalently, that $p(x)$ and its derivative $p'(x)$ share no roots in any field extension of $F$; likewise, if $p(x)$ is inseparable, $p(x)$ and $p'(x)$ share a common zero; if we write 
$p(x) = x^2 + ax + b,  \tag 1$
we have
$p' = 2x + a = 0; \tag 2$
thus the unique root $r$ of $p'(x)$ satisfies
$2r + a = 0 \Longrightarrow r = -\dfrac{a}{2}; \tag 3$
if $r$ is also a zero of $p(x)$,
$\dfrac{a^2}{4} - \dfrac{a^2}{2} + b = \left (-\dfrac{a}{2} \right )^2 - a \dfrac{a}{2} + b = p(r) = 0, \tag 4$
whence
$ b - \dfrac{a^2}{4} =\dfrac{a^2}{4} - \dfrac{a^2}{2} + b = 0 \Longrightarrow b = \dfrac{a^2}{4}; \tag 5$
we next find that
$b = \dfrac{a^2}{4} \Longrightarrow p(x) = x^2 + ax + \dfrac{a^2}{4} = \left ( x + \dfrac{a}{2} \right )^2; \tag 6$
and, therefore,
$p(\mu) = 0 \Longrightarrow \left ( \mu + \dfrac{a}{4} \right )^2 = 0 \Longrightarrow \mu + \dfrac{a}{4} = 0 \Longrightarrow \mu \in F; \tag 7$
since $p(x)$ inseparable thus forces $\mu \in F$, we have shown that the minimal polynomial $p(x)$ of any $\mu \in E \setminus F$ is separable; and clearly, if $\mu \in F$, its minimal polynomial, being linear, is also separable, having precisely one zero; thus $E/F$ is separable extension.
It is furthermore easy to see that $E/F$ is normal, for if $r \in E \setminus F$ is a root of $p(x) \in F[x]$, then 
$r^2 + ar + b = 0, \tag 8$
whence
$r^2p(b/r) = r^2(b^2/r^2 + ab/r + b) = b^2 + abr + br^2 = b(r^2 + ar + b) = 0; \tag 9$
since $r \ne 0$ this forces
$p(b/r) = 0; \tag{10}$
we thus see that the minimal polynomial of any $\mu \in E \setminus F$ splits in $E[x]$; $E/F$ is thus, by definition, a normal field extension.
$E/F$ being both separable and normal is thus Galois.
