# Show that $E/F$ is Galois extension [duplicate]

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?

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

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

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.