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 November 17th, 2017, 12:44 PM #1 Senior Member   Joined: Jan 2015 From: usa Posts: 104 Thanks: 1 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.
 November 26th, 2017, 02:20 AM #2 Member   Joined: May 2017 From: Russia Posts: 34 Thanks: 5 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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