I am studying subfields of cyclotomic extension, especially the following situation:
Let $\xi$ be a primitive $m$-th root of unity, and $\chi$ be a multiplicative character on $(\mathbb{Z}/m\mathbb{Z})^\times$ such that $Ker\chi={<}x{>}$ where $x$ is an order $d$ element of $(\mathbb{Z}/m\mathbb{Z})^\times$.
I found the following:
- The automorphism $\sigma_{x}\in Gal(\mathbb{Q}(\xi)/\mathbb{Q})$ which sends $\xi$ to $\xi^{x}$ fixes $\xi^{jx}+\xi^{jx^2}+\cdots+\xi^{jx^d}$ for every integer $j$.
- There exists $y\in(\mathbb{Z}/m\mathbb{Z})^\times$ such that $(\mathbb{Z}/m\mathbb{Z})^\times=y{<}x{>}+y^2{<}x{>}+\cdots+y^{\phi(m)/d}{<}x{>}$.
- Repeating to apply the automorphism $\sigma_{y}\in Gal(\mathbb{Q}(\xi)/\mathbb{Q})$ to $\xi^{x}+\xi^{x^2}+\cdots+\xi^{x^d}$ gives all $\xi^{jx}+\xi^{jx^2}+\cdots+\xi^{jx^d}$.
- In some cases, $\xi^{x}+\xi^{x^2}+\cdots+\xi^{x^d}=0$.
Now, I'm trying to figure out a sufficient condition for that $\xi^{x}+\xi^{x^2}+\cdots+\xi^{x^d}$ is non-zero and that $\xi^{jx}+\xi^{jx^2}+\cdots+\xi^{jx^d}$ are distinct.
If the sufficient condition follows, the minimal polynomial of $\xi^{x}+\xi^{x^2}+\cdots+\xi^{x^d}$ has distinct $\phi(m)/d$ roots, and therefore, the field fixed by $<\sigma_{x}>$ is $\mathbb{Q}(\xi^{x}+\xi^{x^2}+\cdots+\xi^{x^d})$.
I want $m$ to cover 8 while for each $m$, there may exist $\chi$ that does not satisfy that condition. I found the following proposition from Basis of primitive nth Roots in a Cyclotomic Extension? :
In general, the primitive $m$th roots of unity in the $m$th cyclotomic field form a normal basis over $\mathbf{Q}$ if and only if $m$ is squarefree.
This gives a powerful, sufficient condition. But $m$ cannot be 8. Thus, we can take $m$ as a multiple of square.
