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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. Tags subgroup, trivial Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post zzzhhh Abstract Algebra 3 July 21st, 2017 05:46 AM Bhuvaneshnick Algebra 8 December 23rd, 2014 06:38 AM Bucephalus Linear Algebra 10 January 28th, 2012 09:08 PM envision Abstract Algebra 3 October 4th, 2009 10:37 PM envision Abstract Algebra 1 October 4th, 2009 03:24 AM

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