Field cut out by a character

Question

Let $p$ be an odd prime and let $\Delta:=\textrm{Gal}(\mathbb Q(\zeta_p)/\mathbb Q)$, where $\zeta_p$ is a primitive $p$-th root of unity. Consider the character: $ \chi: \textrm{Gal}(\overline{\mathbb Q}/\mathbb Q) \rightarrow \Delta \rightarrow \mathbb F_p^{\times}, $ where the first map is the projection and the second map is the canonical isomorphism. I am interested in the character $\chi^{1-k}$, where $k \geq 2$ even. The kernel of $\chi^{1-k}$ is of the form $\textrm{Gal}(\overline{\mathbb Q}/E)$, where $E$ is a finite extension of $\mathbb Q$. Why is $E$ contained in $\mathbb Q(\zeta_p)$ and moreover why is the degree of $E$ over $\mathbb Q$ equal to $\frac{p-1}{ \textrm{gcd}(p-1,k-1)}$?

Answer 1

At the end of the day, you have a character $\chi$ from a group $G$ to $\mathbb F_p^\times$ of order $p-1$. If $g\in \ker\chi$, then $\chi(g) = \chi(g)^{1-k} = 1$, so $g\in \ker\chi$. We see immediately that $$\mathrm{Gal}(\overline{ \mathbb Q}/E) = \ker\chi^{1-k} \supset \ker\chi= \mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q(\zeta_p))$$ from which it follows that $E\subset \mathbb Q(\zeta_p)$.

The degree of $E$ over $\mathbb Q$ is equal to the size of the image of $\chi^{1-k}$ which is exactly the order of $\chi^{1-k}$, i.e. $\frac{p-1}{\mathrm{gcd}(p-1, k-1)}$.