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November 17th, 2017, 12:44 PM   #1
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Non trivial subgroup

Let $\alpha=\sqrt{2+\sqrt{3}}$.

Let $L=\mathbb{Q}(\alpha)$.

>For each non trivial $H< \text{Gal}(L/\mathbb{Q})$ determine $\text{Fix}(H)$. Express the answer in form of $\mathbb{Q}(\beta)$ whith $\beta$ given explicitly in term of $\alpha$

This is what i wrote: but i can not conclude please help me to improve my answer and to continue:


$L$ is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z_2}$
(since its generators are $\alpha \to -\alpha,\alpha\to \frac{1}{\alpha}$).

For computing the fix field of $<\alpha \to -\alpha>$ we need to solve the equation $m+n\alpha+n'\alpha^2+n''\alpha^3=m-n\alpha+n'\alpha^2-n''\alpha^3$. Using the fact that $\{1,\alpha,\alpha^2,\alpha^3\}$ is a base for $\mathbb{Z}(\alpha)$ we have $n=n''=0$

For computing the fix field of $<\alpha \to \frac{1}{\alpha}>$ we need to solve the equation $m+n\alpha+n'\alpha^2+n'' \alpha^3=m+\frac{n}{\alpha}+\frac{n'}{\alpha^2}+ \frac{n''}{\alpha^3}$ samely since $\{1,\alpha,\alpha^2,\alpha^3\}$ is a base for $\mathbb{Z}(\alpha)$, we have $m\alpha^3+n+n'\alpha+n''\alpha^2=m\alpha^3+n \alpha^2+n ' \alpha + n''$
so $n=n''$

Last edited by greg1313; November 18th, 2017 at 01:51 PM.
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November 26th, 2017, 02:20 AM   #2
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The minimal polynomial of $\displaystyle \alpha$ over $\displaystyle \mathbb{Q} $ is
$\displaystyle m(x)=(x^2-2)^2-3. $

The root of $\displaystyle m(x) $ are $\displaystyle \alpha=\sqrt{2+\sqrt{3}},\ -\sqrt{2+\sqrt{3}},\
\sqrt{2-\sqrt{3}},\ -\sqrt{2-\sqrt{3}}. $


$\displaystyle \sqrt{2+\sqrt{3}}\cdot \sqrt{2-\sqrt{3}} = 1. $ It implies $\displaystyle \sqrt{2-\sqrt{3}}=\frac{1}{\alpha}. $

We get expressions of all the roots in terms of $\displaystyle \alpha $:

$\displaystyle \alpha, -\alpha,\ \frac{1}{\alpha},\ -\frac{1}{\alpha}. $

Galois group $\displaystyle \text{Gal}(L/\mathbb{Q}) $ of $\displaystyle L=\mathbb{Q}(\alpha) $ can be expressed as the group of following functions:
$\displaystyle
\varphi_1(x)=x,\ \varphi_2(x)=-x,\ \varphi_3(x)=\frac{1}{x},\ \varphi_4(x)=-\frac{1}{x}.
$


$\displaystyle \text{Gal}(L/\mathbb{Q}) $ is isomorphic to $\displaystyle \mathbb{Z}_2 \times \mathbb{Z_2} $.


Let the fix field of the subgroup generated by $\displaystyle <\alpha \to -\alpha> $ be $\displaystyle \mathbb{Q}(\beta) $ where $\displaystyle \beta\in\mathbb{Q}(\alpha) $:

$\displaystyle \beta = m+n\alpha+n'\alpha^2+n''\alpha^3$.

$\displaystyle m+n\alpha+n'\alpha^2+n''\alpha^3=m-n\alpha+n'\alpha^2-n''\alpha^3. $

$\displaystyle 2n\alpha+2n''\alpha^3=0. $

Using the fact that $\displaystyle \{1,\alpha,\alpha^2,\alpha^3\} $ is a base for $\displaystyle \mathbb{Q}(\alpha) $ (a basis of the linear space $\displaystyle \mathbb{Q}(\alpha) $) we have
$\displaystyle n=n''=0 $.

And $\displaystyle \beta = m+n'\alpha^2 $:
$\displaystyle \beta = \alpha^2$
Let $\displaystyle m=0, \ n'=1$.
$\displaystyle \mathbb{Q}(\alpha^2) $ is the fix field of$\displaystyle <\alpha \to -\alpha> $.
Thanks from Country Boy

Last edited by ABVictor; November 26th, 2017 at 02:34 AM.
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