Fall 2016 - Problem 7

Solution.

Let $n$ be the smallest integer such that $\xi^n = 1$ so that $\xi$ is a primitive $n$-th root of unity. Suppose $f\p{\xi} = 2^{1/4}$ so that $\Q \subseteq \Q\p{2^{1/4}} \subseteq \Q\p{\xi}$. Consider $\Gal\p{\Q\p{\xi}/\Q} \simeq C_n$, and let $H \leq \Gal\p{\Q\p{\xi}/\Q}$ be the subgroup whose fixed field is $\Q\p{2^{1/4}}$. Since $C_n$ is abelian, $H$ is a normal subgroup, but this implies that $\Q\p{2^{1/4}}/\Q$ is Galois, a contradiction since the extension is not normal.