Show that an extension $K/F$ with $|K:F| = 2$ is normal.

Question

My attempt:

It is well known that finite field extensions are algebraic.

If $a \in F$, then $\min_F(a) = X-a$ trivially splits.

If $a \in K \setminus F$, then $\{1,a\}$ is $F$-linearly independent and thus is an $F$-basis for $K$ because $|K:F| = 2$. Hence, $K = F[a]$ and thus

$$\deg \min_F(a)= |F[a]:F| = 2$$

In $K[X]$, the factor $X-a$ divides $\min_F(a)$ and thus the other irreducible factor in its decomposition must be linear as well, i.e. $\min_F(a)$ splits completely over $K$ and the extension is normal.

Is this correct?

Answer 1

Since $[K:F]=2$, we can have a basis $\{1,\alpha\}$ for $K$ over $F$ where $\alpha\in K\setminus F$. Since $K$ is a field, $\alpha^2\in K$, so $\alpha$ is a root of a degree two polynomial $x^2+ax+b$ for some $a,b\in F$. The roots of this polynomial are $\alpha$ and $-a-\alpha$. Hence $K=F(\alpha)$ is the splitting field of $x^2+ax+b$ hence a normal extension of $F$.

Answer 2

You could use the Galois correspondence, and the fact that any subgroup of index $2$ is normal.