- #1

- 128

- 8

## Summary:

- question on Galois correspondence

Question: Galois group of ##x^4 - 2 ## over Q.

The correspondence is:

$$

\begin{aligned}

(N_1) \leftarrow \rightarrow Q(\sqrt{2})

\end{aligned}

$$

$$

\begin{aligned}

(N_2) \leftarrow \rightarrow Q(i)

\end{aligned}

$$

$$

\begin{aligned}

(N_3) \leftarrow \rightarrow Q(i\sqrt{2})

\end{aligned}

$$

$$

\begin{aligned}

(N_4) \leftarrow \rightarrow Q(\sqrt{2}, i)

\end{aligned}

$$

$$

\begin{aligned}

(H_3) \leftarrow \rightarrow Q(\sqrt[4]{2} (1+i))

\end{aligned}

$$

$$

\begin{aligned}

(H_4) \leftarrow \rightarrow Q(\sqrt[4]{2} (1-i))

\end{aligned}

$$

My question is : How are these intermediate subfields (on the right side of the arrows) obtained?

(The eight subgroups are:)

$$

\begin{aligned}

(N_1) & = \left\{ 1,\xi, \eta^2 , \xi\eta^2 \right\} \\

(N_2) & = \left\{ 1, \eta , \eta^2 , \eta^3 \right\} \\

(N_3) & = \left\{ 1, \eta^2 , \xi\eta, \xi\eta^3 \right\} \\

(N_4) & = \left\{ 1, \eta^2 \right\} \\

(H_1) & = \left\{ 1,\xi \right\} \\

(H_2) & = \left\{ 1,\xi\eta^2 \right\} \\

(H_3) & = \left\{ 1,\xi\eta \right\} \\

(H_4) & = \left\{ 1,\xi\eta^3 \right\}

\end{aligned}

$$

in which:

$$

\begin{aligned}

\xi : \sqrt[4]{2} \rightarrow -\sqrt[4]{2} \\

i\sqrt[4]{2} \rightarrow i\sqrt[4]{2} \\

-\sqrt[4]{2} \rightarrow \sqrt[4]{2} \\

-i\sqrt[4]{2} \rightarrow -i\sqrt[4]{2}

\end{aligned}

$$

$$

\begin{aligned}

\eta : \sqrt[4]{2} \rightarrow -i\sqrt[4]{2} \\

i\sqrt[4]{2} \rightarrow -\sqrt[4]{2} \\

-\sqrt[4]{2} \rightarrow -i\sqrt[4]{2} \\

-i\sqrt[4]{2} \rightarrow \sqrt[4]{2}

\end{aligned}

$$

The correspondence is:

$$

\begin{aligned}

(N_1) \leftarrow \rightarrow Q(\sqrt{2})

\end{aligned}

$$

$$

\begin{aligned}

(N_2) \leftarrow \rightarrow Q(i)

\end{aligned}

$$

$$

\begin{aligned}

(N_3) \leftarrow \rightarrow Q(i\sqrt{2})

\end{aligned}

$$

$$

\begin{aligned}

(N_4) \leftarrow \rightarrow Q(\sqrt{2}, i)

\end{aligned}

$$

$$

\begin{aligned}

(H_3) \leftarrow \rightarrow Q(\sqrt[4]{2} (1+i))

\end{aligned}

$$

$$

\begin{aligned}

(H_4) \leftarrow \rightarrow Q(\sqrt[4]{2} (1-i))

\end{aligned}

$$

My question is : How are these intermediate subfields (on the right side of the arrows) obtained?

(The eight subgroups are:)

$$

\begin{aligned}

(N_1) & = \left\{ 1,\xi, \eta^2 , \xi\eta^2 \right\} \\

(N_2) & = \left\{ 1, \eta , \eta^2 , \eta^3 \right\} \\

(N_3) & = \left\{ 1, \eta^2 , \xi\eta, \xi\eta^3 \right\} \\

(N_4) & = \left\{ 1, \eta^2 \right\} \\

(H_1) & = \left\{ 1,\xi \right\} \\

(H_2) & = \left\{ 1,\xi\eta^2 \right\} \\

(H_3) & = \left\{ 1,\xi\eta \right\} \\

(H_4) & = \left\{ 1,\xi\eta^3 \right\}

\end{aligned}

$$

in which:

$$

\begin{aligned}

\xi : \sqrt[4]{2} \rightarrow -\sqrt[4]{2} \\

i\sqrt[4]{2} \rightarrow i\sqrt[4]{2} \\

-\sqrt[4]{2} \rightarrow \sqrt[4]{2} \\

-i\sqrt[4]{2} \rightarrow -i\sqrt[4]{2}

\end{aligned}

$$

$$

\begin{aligned}

\eta : \sqrt[4]{2} \rightarrow -i\sqrt[4]{2} \\

i\sqrt[4]{2} \rightarrow -\sqrt[4]{2} \\

-\sqrt[4]{2} \rightarrow -i\sqrt[4]{2} \\

-i\sqrt[4]{2} \rightarrow \sqrt[4]{2}

\end{aligned}

$$

Last edited: