Let $\zeta_{2023}=e^{2\pi i /2023}$ be a primitive $2023$th root of unity. Describe :
- Quadratic subfields of $\mathbb{Q}(\zeta_{2023})$.
- Quartic subfields $E$ of $\mathbb{Q}(\zeta_{2023})$ such that $\mathcal{Gal}(E/\mathbb{Q})\cong \mathbb{Z}/4\mathbb{Z}$.
Since $2023=7\cdot 17^2$, we have $\mathbb{Q}(\zeta_{2023})=\mathbb{Q}(\zeta_{7})\cdot\mathbb{Q}(\zeta_{17^2})$, and then
$$\begin{align}
\mathcal{Gal}(\mathbb{Q}(\zeta_{2023})/\mathbb{Q})&\cong\mathcal{Gal}(\mathbb{Q}(\zeta_{7})/\mathbb{Q})\times\mathcal{Gal}(\mathbb{Q}(\zeta_{17^2})/\mathbb{Q})\cong(\mathbb{Z}/7\mathbb{Z})^*\times(\mathbb{Z}/17^2\mathbb{Z})^* \\
&\cong\mathbb{Z}/6\mathbb{Z}\times\mathbb{Z}/17\cdot16\mathbb{Z}
\end{align}
$$
Furthermore $(\mathbb{Z}/7\mathbb{Z})^*=\langle [3]_7 \rangle$ and $(\mathbb{Z}/17^2\mathbb{Z})^*=\langle [3]_{17^2} \rangle$, thus $\mathcal{Gal}(\mathbb{Q}(\zeta_{7})/\mathbb{Q})=\langle\sigma\rangle$ with $\sigma(\zeta_7)=\zeta_{7}^3$, and $\mathcal{Gal}(\mathbb{Q}(\zeta_{17^2})/\mathbb{Q})=\langle\tau\rangle$ with $\tau(\zeta_{17^2})=\zeta_{17^2}^3$. Finally, by extending $\sigma$ and $\tau$ to automorphisms of $\mathbb{Q}(\zeta_{2023})$, we get
$$
\mathcal{Gal}(\mathbb{Q}(\zeta_{2023})/\mathbb{Q})=\langle\sigma\rangle\times\langle\tau\rangle
$$
-
The only subgroups of index $2$ in $\mathcal{Gal}(\mathbb{Q}(\zeta_{2023})/\mathbb{Q})$ are
$$
\begin{align}
H&:=\langle\sigma^2\rangle\times\langle\tau\rangle\cong\mathbb{Z/3Z}\times\mathbb{Z/17\cdot 16Z} \\
L&:=\langle\sigma^4\rangle\times\langle\tau\rangle\cong\mathbb{Z/3Z}\times\mathbb{Z/17\cdot 16Z} \\
K&:=\langle\sigma\rangle\times\langle\tau^2\rangle\cong\mathbb{Z/6Z}\times\mathbb{Z/17\cdot 8Z}
\end{align}
$$
Thus, the only quadratic subfield are $\mathbb{Q}(\zeta_{2023})^H$, $\mathbb{Q}(\zeta_{2023})^L$ and $\mathbb{Q}(\zeta_{2023})^K$. Unfortunately, I couldn't find an explicit description of these subfields through generators. -
Again, with some groups theory arguments I could find the desired subgroups of $\mathcal{Gal}(\mathbb{Q}(\zeta_{2023})/\mathbb{Q})$, but I couldn't get an explicit description of the corresponding subfields.
Best Answer
As in my comments, not all subgroups of a direct product is a direct product of subgroups of each factor (although the image of the projection onto a factor is a subgroup of that factor). The correct way to do this is to use Goursat's lemma:
For details, I suggest reading a nice answer by @Tobias Kildetoft.
Fix a tuple $(G_1,G_2,H_1,H_2,\varphi)$, and let $m=|G_1/G_2|=|H_1/H_2|$. It is not hard to see that $|S|=m|G_2||H_2|$. Note that $m\in\{1,2\}$ because $m$ divides $\gcd(2\cdot 3,2^4\cdot 17)$. If $m=1$, Goursat's lemma gives subgroups of the form $G_1\times H_1$, which you have already obtained: $$ \begin{array}{c|cc} \text{Index} & \text{Subgroups}\ (m=1)\\ \hline 2 & \langle \sigma^2 \rangle \times \langle \tau \rangle,\ \langle \sigma \rangle \times \langle \tau^2 \rangle \\ 4 & \langle\sigma^2\rangle \times \langle\tau^2\rangle,\ \langle\sigma\rangle \times \langle\tau^4\rangle \end{array} $$ Now suppose $m=2$. To find index $2$ subgroups, we have no choice but choosing $(G_1,G_2,H_1,H_2)=(\langle\sigma\rangle,\langle\sigma^2\rangle,\langle\tau\rangle,\langle\tau^2\rangle)$. We have only one choice for $\varphi$ because the only isomorphism $G_1/G_2\cong\mathbb{Z}/2\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}\cong H_1/H_2$ is the identity. Applying Goursat's lemma, we get $S=\langle\sigma^2\rangle\times\langle\tau^2\rangle\ \cup\ \sigma\langle\sigma^2\rangle\times\tau\langle\tau^2\rangle=\langle(\sigma,\tau)\rangle$. Finding index $4$ subgroups is similar. The only subgroup of index $4$ is $\langle (\sigma, \tau^2) \rangle$.
We have the following subgroup lattice of the Galois group, including only subgroups of index $1,2,4$:
Now we try to determine some generators for the corresponding fixed fields.$\newcommand{\rat}{\mathbb{Q}}\newcommand{\zt}{\zeta_{17}}\newcommand{\zs}{\zeta_{7}}$ As given by @Jyrki Lahtonen, $\sqrt{17}\in\rat(\zt)$ and $i\sqrt{7}\in\rat(\zs)$. More explicitly, $$ \begin{align*} \sqrt{17}&=\sum_{k=1}^{16} \left( \frac{k}{17} \right) \zt^k \\ i\sqrt{7}&=\sum_{k=1}^{6} \left( \frac{k}{7} \right) \zs^k \\ \end{align*} $$ where $(\frac{k}{p})$ is the Legendre symbol.$\newcommand{\l}{\langle}\newcommand{\r}{\rangle}\newcommand{\s}{\sigma}\newcommand{\t}{\tau}$ The nonzero quadratic residues modulo a prime $p$ form a subgroup of $(\mathbb{Z}/p\mathbb{Z})^*$. Using this fact, we see that $\sqrt{17}$ is fixed by $\s$ and $\t^2$ but not $\t$, because $3$ is not a quadratic residue modulo $17$ but $9$ is. Similarly, $i\sqrt{7}$ is fixed by $\t$ and $\s^2$ but not $\s$. Finally, $i\sqrt{7}\sqrt{17}$ has degree $2$ over $\rat$ and is fixed by $\sigma\tau$ (using the fact that $\s(i\sqrt{7})=-i\sqrt{7}$ and $\t(\sqrt{17})=-\sqrt{17}$). We have found generators for all quadratic subfields.
From the subgroup lattice, we see that only $\frac{\l\s\r\times\l\t\r}{\l\s\r\times\l\t^4\r}$ and $\frac{\l\s\r\times\l\t\r}{\l(\s,\t^2)\r}$ are isomorphic to $\mathbb{Z}/4\mathbb{Z}$. The number $\gamma=\zt+\zt^{-1}+\zt^4+\zt^{-4}$ is fixed by $\t^4$ but not $\t^2$. Let $$ \begin{align*} \alpha &= \zs-\zs^3+\zs^2-\zs^{-1}+\zs^{-3}-\zs^{-2} = i\sqrt{7} \\ \beta &= \zt-\zt^{-8}+\zt^{-4}-\zt^{-2}+\zt^{-1}-\zt^{8}+\zt^{4}-\zt^{2} \end{align*} $$ We have $\alpha\neq 0$, $\beta\neq 0$, and $$ \s(\alpha)=-\alpha,\ \t^2(\beta)=-\beta $$ so $\alpha\beta$ is fixed by $\s\t^2$ (but not $\s$, for example). We have thus found generators for all quartic subfields whose Galois group is isomorphic to $\mathbb{Z}/4\mathbb{Z}$.
Subfield lattice of $\mathbb{Q}(\zeta_{2023})$, including only subfields with degree $1,2,4$:
As a sidenote, apparently $\beta=\sqrt{\frac{17-\sqrt{17}}{2}}$.