Exercise 5.1.3

Proof. Suppose that $[L:F]=2$. Then $F⊊L$, so there exists $\alpha \in L$ such that $\alpha \notin F$.

As $\alpha \notin F$, $F⊊F(\alpha )$, thus $[F(\alpha ):F]>1$. Since $F(\alpha )\subset L$, $[F(\alpha ):F]\le 2$, hence $[F(\alpha ):F]=2$, so $F(\alpha )=L$. Let $f$ be the minimal polynomial of $\alpha$ over $F$. Then $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=[F(\alpha ):F]=2$, so $f=x^2+\mathit{ax}+b,a,b\in F$.

Since $\alpha \in L$ is a root of $x^2+\mathit{ax}+b\in F[x]\subset L[x]$, there exists a polynomial $q(x)\in L[x]$ such that $x^2+\mathit{ax}+b=(x-\alpha )q(x)$, where $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(q)=1$ and $q$ is monic. Therefore there exists $\beta \in L$ such that $q(x)=x-\beta$. So $f=(x-\alpha )(x-\beta )$ splits completely over $L$, and since $\beta \in L$, $L=F(\alpha )=F(\alpha ,\beta )$. $L$ is a splitting field of $f$.

Conclusion: Every quadratic extension $L$ of a field $F$ is a splitting field (so is a normal extension). □