I want to show that each extension of degree $2$ is normal.
I have done the following:
Let $K/F$ the field extension with $[F:K]=2$.
Let $a\in K\setminus F$. Then we have that $F\leq F(a)\leq K$.
We have that $[K:F]=2\Rightarrow [K:F(a)][F(a):F]=2$.
There are the following possibilities:
- $[K:F(a)]=1$ and $[F(a):F]=2$ In this case we have that $K=F(a)$ and $\deg m(a,F)=2$. In $K$, since $a\in K$, we have that $m(a,F)=(x-a)g(x)$, with $\deg g(x)=1$. Since $g$ is a linear polynomial of $K[x]$ , so it is of the form $x-c$, so its root $c$ must belong to $K$. That means that $m(a,K)$ splits in $K$, i.e., all the roots are in $K$. Does this mean then that all the irreducible polynomials that have one root in $K$, have all the roots in $K$ ? Or how do we conclude then that the extension $K/F$ is normal ?
- $[K:F(a)]=2$ and $[F(a):F]=1$ In this case we have that $F=F(a)$, and so $a\in F$, a contradiction.
Is this correct? Could I improve something?
The idea of your solution is ok but there are little mistakes. First of all, it's $[K:F]=2$ and not $[F:K]=2$. Now you have $\deg m(a,F)=2$, and since $a$ is a root of $m(a, F)$ we can write in $K$: $m(a, F)=(x-a)g(x)$, using that $\deg m(a, F)=2$ we get $\deg g(x)=1$, so $g(x)$ has a root $c\in K$. This means that $m(a, F)\in F[x]$ splits in $K$, so by theorem that I wrote in the commentaries we have that $K/F$ is normal.
Alternatively, you can use the well-known fact that if $G$ is a group and $H$ is a subgroup of $G$ such that $|G:H|=2$, then $H$ is a normal subgroup of $G$, and then apply the fundamental theorem of Galois theory.