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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?

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marked as duplicate by Thomas Shelby, Misha Lavrov, Paul Frost, Lord Shark the Unknown, Leucippus Apr 27 at 4:40

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  • $\begingroup$ The degree of an inseparable extension is divisible by the characteristic. $\endgroup$ – Lord Shark the Unknown Apr 17 at 17:25
  • $\begingroup$ What is your definition of separable? $\endgroup$ – Brahadeesh Apr 17 at 21:23
  • $\begingroup$ Minimal polynomial over F is separable, i.e. it has no repeated roots $\endgroup$ – You_know_who Apr 18 at 1:08
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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.

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