Question about the Galois correspondence

[thaiqi]

TL;DR

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:
$$
\begin{aligned}
\xi : \sqrt[4]{2} \rightarrow -\sqrt[4]{2} \\
i\sqrt[4]{2} \rightarrow i\sqrt[4]{2} \\
-\sqrt[4]{2} \rightarrow \sqrt[4]{2} \\
-i\sqrt[4]{2} \rightarrow -i\sqrt[4]{2}
\end{aligned}
$$
$$
\begin{aligned}
\eta : \sqrt[4]{2} \rightarrow -i\sqrt[4]{2} \\
i\sqrt[4]{2} \rightarrow -\sqrt[4]{2} \\
-\sqrt[4]{2} \rightarrow -i\sqrt[4]{2} \\
-i\sqrt[4]{2} \rightarrow \sqrt[4]{2}
\end{aligned}
$$

 

[Infrared]

First of all, notice that the splitting field of $x^4-2$ is $E=\mathbb{Q}(i,\sqrt[4]{2}).$
This is a degree $8$ extension of $\mathbb{Q}$ with basis $i^p\left(\sqrt[4]{2}\right)^q$ where $p=0,1$ and $q=0,1,2,3.$

In each case, it should be straightforward to check that the given field is indeed fixed by the Galois subgroups. You can also check that the index of the given field in $E$ equals the index of the Galois subgroup, which gives you what you want by the Galois correspondence theorem.

If you'd prefer to work it out directly (not using any theorems), let's take a look at $N_1$.

The subgroup $N_1$ is generated by $\xi$ and $\eta^2$. The element $\xi$ takes $\sqrt[4]{2}$ to $\sqrt[4]{2}$ and $i$ to $-i$, so an element of $E$ is fixed by $\xi$ if and only if it is real.

A real element of $E$ is of the form $a+b\sqrt[4]{2}+c\sqrt{2}+d\left(\sqrt[4]{2}\right)^3.$

The element $\eta$ takes $\sqrt[4]{2}$ to $-i\sqrt[4]{2}$ and fixes $i$, so $\eta^2$ takes $\sqrt[4]{2}$ to $-\sqrt[4]{2}$ and $\left(\sqrt[4]{2}\right)^3$ to $-\left(\sqrt[4]{2}\right)^3$ and fixes $\sqrt{2}$. It follows that the above general element of $E\cap\mathbb{R}$ is fixed by $\eta^2$ if and only if $b=d=0$. So, the subfield of $E$ fixed by $\xi$ and $\eta^2$ is generated by $\sqrt{2}$. The other correspondences are similar.

 

[thaiqi]

First of all, notice that the splitting field of $x^4-2$ is $E=\mathbb{Q}(i,\sqrt[4]{2}).$
This is a degree $8$ extension of $\mathbb{Q}$ with basis $i^p\left(\sqrt[4]{2}\right)^q$ where $p=0,1$ and $q=0,1,2,3.$

……

Thanks very much.
Can $H_{4}$ and $N_4$ be deduced in detail ? I felt them more difficult.

 

[Infrared]

I made a typo in my previous post: $\eta$ takes $i$ to $-i$ (but $\eta^2$ still fixes $i$ so it's okay).

The subgroup $N_4$ is generated by $\eta^2$, which I worked out above takes $\sqrt[4]{2}$ to $-\sqrt[4]{2}$ and fixes $i$. So the basis element $i^p(\sqrt[4]{2})^q$ is fixed if $q$ is even and taken to its negative if $q$ is odd. So, the fixed field of $N_4$ is generated by $i$ and $\sqrt{2}$.

Is it possible you have a typo in $H_4$? I think that the element $\xi\eta^3$ has order $4$, not $2$.

 

[thaiqi]

I made a typo in my previous post: $\eta$ takes $i$ to $-i$ (but $\eta^2$ still fixes $i$ so it's okay).

The subgroup $N_4$ is generated by $\eta^2$, which I worked out above takes $\sqrt[4]{2}$ to $-\sqrt[4]{2}$ and fixes $i$. So the basis element $i^p(\sqrt[4]{2})^q$ is fixed if $q$ is even and taken to its negative if $q$ is odd. So, the fixed field of $N_4$ is generated by $i$ and $\sqrt{2}$.

How does it come out that $N_4$ is generated by $i$ and $\sqrt{2}$？

Is it possible you have a typo in $H_4$? I think that the element $\xi\eta^3$ has order $4$, not $2$.

（I checked the book and cannot discern if it has error.）

 

[Infrared]

Again, you can avoid any computation if you allow yourself to use the correspondence theorem. That said,

How does it come out that $N_4$ is generated by $i$ and $\sqrt{2}$？

We see that $\eta^2$ takes $\sqrt[4]{2}$ to its negative and fixes $i$. Each basis element is either fixed or taken to its negative according to whether the exponent of $\sqrt[4]{2}$ is even or odd. So the fixed field of $\langle\eta^2\rangle$ is generated by the elements $\sqrt{2},i$.

（I checked the book and cannot discern if it has error.）

Well can you check how $\xi\eta^3$ acts and see if you agree with me?

 

[thaiqi]

Well can you check how $\xi\eta^3$ acts and see if you agree with me?

I computed the case of $\xi\eta^3$ which gives out the result of $Q(\sqrt[4]{2}(1-i))$ and did not find problem. (it takes $\sqrt[4]{2}$ into $-i\sqrt[4]{2}$ and $i$ into $-i$)I have worked $N_4$ out. But I met problem in the same way for $N_3$(for it is fixed for three mappings).
Can you illustrate how $N_3$ 's $i\sqrt 2$ is worked out? (I could check that $i\sqrt 2$ is invariant under those three mappings, but how can I compute it out? )

 

[Infrared]

I think you have a typo in your formula for $\eta$.

$$
\begin{aligned}
\eta : \sqrt[4]{2} \rightarrow -i\sqrt[4]{2} \\
i\sqrt[4]{2} \rightarrow -\sqrt[4]{2} \\
-\sqrt[4]{2} \rightarrow -i\sqrt[4]{2} \\
-i\sqrt[4]{2} \rightarrow \sqrt[4]{2}
\end{aligned}
$$

$\eta$ can't take both $\sqrt[4]{2}$ and $-\sqrt[4]{2}$ to $-i\sqrt[4]{2}$. Presumably, one of them should be taken to $i\sqrt[4]{2}$ instead. Could you double check this?

 

[thaiqi]

I think you have a typo in your formula for $\eta$.
$\eta$ can't take both $\sqrt[4]{2}$ and $-\sqrt[4]{2}$ to $-i\sqrt[4]{2}$. Presumably, one of them should be taken to $i\sqrt[4]{2}$ instead. Could you double check this?

You are correct. It should be taking $\sqrt[4]{2}$ to $i\sqrt[4]{2}$, (not $-i\sqrt[4]{2}$)Then how about my question above?

I computed the case of $\xi\eta^3$ which gives out the result of $Q(\sqrt[4]{2}(1-i))$ and did not find problem. (it takes $\sqrt[4]{2}$ into $-i\sqrt[4]{2}$ and $i$ into $-i$)I have worked $N_4$ out. But I met problem in the same way for $N_3$(for it is fixed for three mappings).
Can you illustrate how $N_3$ 's $i\sqrt 2$ is worked out? (I could check that $i\sqrt 2$ is invariant under those three mappings, but how can I compute it out? )

 

[Infrared]

The group $N_3$ is generated by $\eta^2$ and $\xi\eta$. The element $\eta^2$ takes $\sqrt[4]{2}$ to $-\sqrt[4]{2}$ and fixes $i$. So, the basis elements $i^p\left(\sqrt[4]{2}\right)^q$ are either fixed or taken to their negatives according to whether $q$ is even or odd.

So, a basis for the fixed field of $\eta^2$ is $\{1,\sqrt{2},i,i\sqrt{2}\}$. But here $\xi\eta$ changes the sign of $\sqrt{2}$ and $i$ and fixes the other two, so the field of elements fixed by both $\eta^2$ and $\xi\eta$ is generated by $i\sqrt{2}$.

 

[thaiqi]

I think you have explained it ($N_3$) very clearly.
Thank you very much!

 

[thaiqi]

In my book it writes: (as to $N_1$)
"
\eta^2(\sqrt 2) = \eta \eta (\sqrt[4]{2}) ^2 = \eta(i \sqrt[4]{2}) = (- \sqrt[4]{2})^2 = \sqrt{2}
"

I doubt there is a typo in it, it should be:
$$\eta(i \sqrt[4]{2})^2 $$
shouldn't it?

 

[zinq]

thaiqi: Did you mean to write "doubt" or "believe"?

 

[thaiqi]

thaiqi: Did you mean to write "doubt" or "believe"?

I think the book got an error here. But I am not self-confident.