Extension of degree 2 is normal

[mathmari]

Hey!

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 $K$. So, 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? (Wondering)

 

[mathmari]

• $[K:F(a)]=1$ 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 $K$. So, the extension $K/F$ is normal.

We have 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 ? (Wondering)